Integrand size = 23, antiderivative size = 444 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{7/2} \left (a^2+b^2\right )^3 d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{4 b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]
-1/4*a^(5/2)*(15*a^4+46*a^2*b^2+63*b^4)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^ (1/2))/b^(7/2)/(a^2+b^2)^3/d+1/2*(a-b)*(a^2+4*a*b+b^2)*arctan(-1+2^(1/2)*t an(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(a-b)*(a^2+4*a*b+b^2)*arctan(1+ 2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/4*(a+b)*(a^2-4*a*b+b^2)* ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a+b)* (a^2-4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)^3/d*2^ (1/2)+1/4*(15*a^4+31*a^2*b^2+8*b^4)*tan(d*x+c)^(1/2)/b^3/(a^2+b^2)^2/d-1/2 *a^2*tan(d*x+c)^(5/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-1/4*a^2*(5*a^2+13*b ^2)*tan(d*x+c)^(3/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))
Result contains complex when optimal does not.
Time = 5.72 (sec) , antiderivative size = 403, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 b^2 \tan ^{\frac {11}{2}}(c+d x)+\frac {2 a^2 \left (15 a^2+16 b^2\right ) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}{b^3}+\frac {2 a \left (5 a^2+4 b^2\right ) \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{b^2}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))}{b}+2 a \tan ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))-2 b \tan ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))-\frac {a (a+b \tan (c+d x)) \left (a \sqrt {b} \left (a^2+b^2\right ) \left (15 a^4+31 a^2 b^2+24 b^4\right ) \sqrt {\tan (c+d x)}+\left (-4 (-1)^{3/4} (a+i b)^3 b^{7/2} \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )-4 \sqrt [4]{-1} b^{7/2} (i a+b)^3 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right ) (a+b \tan (c+d x))\right )}{b^{7/2} \left (a^2+b^2\right )^2}}{4 a \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \]
(2*b^2*Tan[c + d*x]^(11/2) + (2*a^2*(15*a^2 + 16*b^2)*Sqrt[Tan[c + d*x]]*( a + b*Tan[c + d*x]))/b^3 + (2*a*(5*a^2 + 4*b^2)*Tan[c + d*x]^(3/2)*(a + b* Tan[c + d*x]))/b^2 - (2*a^2*Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x]))/b + 2 *a*Tan[c + d*x]^(7/2)*(a + b*Tan[c + d*x]) - 2*b*Tan[c + d*x]^(9/2)*(a + b *Tan[c + d*x]) - (a*(a + b*Tan[c + d*x])*(a*Sqrt[b]*(a^2 + b^2)*(15*a^4 + 31*a^2*b^2 + 24*b^4)*Sqrt[Tan[c + d*x]] + (-4*(-1)^(3/4)*(a + I*b)^3*b^(7/ 2)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + a^(5/2)*(15*a^4 + 46*a^2*b^2 + 63*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] - 4*(-1)^(1/4)*b^(7/2 )*(I*a + b)^3*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])*(a + b*Tan[c + d*x]) ))/(b^(7/2)*(a^2 + b^2)^2))/(4*a*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2)
Time = 2.12 (sec) , antiderivative size = 419, normalized size of antiderivative = 0.94, number of steps used = 27, number of rules used = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.130, Rules used = {3042, 4048, 27, 3042, 4128, 27, 3042, 4130, 27, 3042, 4136, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{9/2}}{(a+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (5 a^2-4 b \tan (c+d x) a+\left (5 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{2 (a+b \tan (c+d x))^2}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (5 a^2-4 b \tan (c+d x) a+\left (5 a^2+4 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^{3/2} \left (5 a^2-4 b \tan (c+d x) a+\left (5 a^2+4 b^2\right ) \tan (c+d x)^2\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-16 a \tan (c+d x) b^3+\left (15 a^4+31 b^2 a^2+8 b^4\right ) \tan ^2(c+d x)+3 a^2 \left (5 a^2+13 b^2\right )\right )}{2 (a+b \tan (c+d x))}dx}{b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-16 a \tan (c+d x) b^3+\left (15 a^4+31 b^2 a^2+8 b^4\right ) \tan ^2(c+d x)+3 a^2 \left (5 a^2+13 b^2\right )\right )}{a+b \tan (c+d x)}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-16 a \tan (c+d x) b^3+\left (15 a^4+31 b^2 a^2+8 b^4\right ) \tan (c+d x)^2+3 a^2 \left (5 a^2+13 b^2\right )\right )}{a+b \tan (c+d x)}dx}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {-8 \left (a^2-b^2\right ) \tan (c+d x) b^3+a \left (15 a^4+31 b^2 a^2+24 b^4\right ) \tan ^2(c+d x)+a \left (15 a^4+31 b^2 a^2+8 b^4\right )}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}+\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\int \frac {-8 \left (a^2-b^2\right ) \tan (c+d x) b^3+a \left (15 a^4+31 b^2 a^2+24 b^4\right ) \tan ^2(c+d x)+a \left (15 a^4+31 b^2 a^2+8 b^4\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\int \frac {-8 \left (a^2-b^2\right ) \tan (c+d x) b^3+a \left (15 a^4+31 b^2 a^2+24 b^4\right ) \tan (c+d x)^2+a \left (15 a^4+31 b^2 a^2+8 b^4\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4136 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {\int -\frac {8 \left (\left (3 a^2-b^2\right ) b^4+a \left (a^2-3 b^2\right ) \tan (c+d x) b^3\right )}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {8 \int \frac {\left (3 a^2-b^2\right ) b^4+a \left (a^2-3 b^2\right ) \tan (c+d x) b^3}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {8 \int \frac {\left (3 a^2-b^2\right ) b^4+a \left (a^2-3 b^2\right ) \tan (c+d x) b^3}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 \int \frac {b^3 \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \int \frac {b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 a^3 \left (15 a^4+46 a^2 b^2+63 b^4\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 \left (15 a^4+31 a^2 b^2+8 b^4\right ) \sqrt {\tan (c+d x)}}{b d}-\frac {\frac {2 a^{5/2} \left (15 a^4+46 a^2 b^2+63 b^4\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {16 b^3 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}}{2 b \left (a^2+b^2\right )}-\frac {a^2 \left (5 a^2+13 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}-\frac {a^2 \tan ^{\frac {5}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\) |
-1/2*(a^2*Tan[c + d*x]^(5/2))/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + ( (-(((2*a^(5/2)*(15*a^4 + 46*a^2*b^2 + 63*b^4)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)*d) - (16*b^3*(((a - b)*(a^2 + 4*a*b + b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sq rt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 - ((a + b)*(a^2 - 4*a*b + b^2)*(-1/2 *Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt [2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/ b) + (2*(15*a^4 + 31*a^2*b^2 + 8*b^4)*Sqrt[Tan[c + d*x]])/(b*d))/(2*b*(a^2 + b^2)) - (a^2*(5*a^2 + 13*b^2)*Tan[c + d*x]^(3/2))/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])))/(4*b*(a^2 + b^2))
3.6.100.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim p[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* (n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b *(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ [a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & & !GtQ[n, 0] && !LeQ[n, -1]
Time = 0.50 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{3}}+\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 a^{3} \left (\frac {\left (-\frac {9}{8} b \,a^{4}-\frac {13}{4} a^{2} b^{3}-\frac {17}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {a \left (7 a^{4}+22 a^{2} b^{2}+15 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}+46 a^{2} b^{2}+63 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(356\) |
| default | \(\frac {\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{b^{3}}+\frac {\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}-\frac {2 a^{3} \left (\frac {\left (-\frac {9}{8} b \,a^{4}-\frac {13}{4} a^{2} b^{3}-\frac {17}{8} b^{5}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {a \left (7 a^{4}+22 a^{2} b^{2}+15 b^{4}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{8}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (15 a^{4}+46 a^{2} b^{2}+63 b^{4}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3} \left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(356\) |
1/d*(2/b^3*tan(d*x+c)^(1/2)+2/(a^2+b^2)^3*(1/8*(3*a^2*b-b^3)*2^(1/2)*(ln(( 1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x +c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^ (1/2)))+1/8*(a^3-3*a*b^2)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c) ^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))-2*a^3/b^3/(a^2+b^2)^3*(((- 9/8*b*a^4-13/4*a^2*b^3-17/8*b^5)*tan(d*x+c)^(3/2)-1/8*a*(7*a^4+22*a^2*b^2+ 15*b^4)*tan(d*x+c)^(1/2))/(a+b*tan(d*x+c))^2+1/8*(15*a^4+46*a^2*b^2+63*b^4 )/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4179 vs. \(2 (392) = 784\).
Time = 1.91 (sec) , antiderivative size = 8384, normalized size of antiderivative = 18.88 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Timed out} \]
Time = 0.63 (sec) , antiderivative size = 434, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {{\left (15 \, a^{7} + 46 \, a^{5} b^{2} + 63 \, a^{3} b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (9 \, a^{5} b + 17 \, a^{3} b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (7 \, a^{6} + 15 \, a^{4} b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} b^{3} + 2 \, a^{4} b^{5} + a^{2} b^{7} + {\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} + 2 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )} - \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b^{3}}}{4 \, d} \]
-1/4*((15*a^7 + 46*a^5*b^2 + 63*a^3*b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt( a*b))/((a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9)*sqrt(a*b)) - (2*sqrt(2)*(a^ 3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(-1/2*sqrt(2)*(s qrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)* log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(a^3 - 3*a^2* b - 3*a*b^2 + b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a ^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - ((9*a^5*b + 17*a^3*b^3)*tan(d*x + c)^( 3/2) + (7*a^6 + 15*a^4*b^2)*sqrt(tan(d*x + c)))/(a^6*b^3 + 2*a^4*b^5 + a^2 *b^7 + (a^4*b^5 + 2*a^2*b^7 + b^9)*tan(d*x + c)^2 + 2*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*tan(d*x + c)) - 8*sqrt(tan(d*x + c))/b^3)/d
Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Timed out} \]
Time = 23.41 (sec) , antiderivative size = 13319, normalized size of antiderivative = 30.00 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \]
(log(- (((((((((64*a*b*(15*a^4 + 2*b^4 + 41*a^2*b^2))/d - 256*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^6))^(1/2))*(1i/(d ^2*(a*1i - b)^6))^(1/2))/2 - (8*a*tan(c + d*x)^(1/2)*(225*a^14 - 184*b^14 + 608*a^2*b^12 - 272*a^4*b^10 + 3937*a^6*b^8 + 5804*a^8*b^6 + 4006*a^10*b^ 4 + 1380*a^12*b^2))/(b^4*d^2*(a^2 + b^2)^4))*(1i/(d^2*(a*1i - b)^6))^(1/2) )/2 + (2*a^2*(1125*a^14 + 16*b^14 + 6112*a^2*b^12 - 17727*a^4*b^10 - 23239 *a^6*b^8 - 11174*a^8*b^6 + 2930*a^10*b^4 + 3525*a^12*b^2))/(b^4*d^3*(a^2 + b^2)^6))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 + (tan(c + d*x)^(1/2)*(32*b^18 - 225*a^18 + 128*a^2*b^16 + 192*a^4*b^14 - 3841*a^6*b^12 + 18050*a^8*b^10 + 26801*a^10*b^8 + 16860*a^12*b^6 + 4049*a^14*b^4 - 30*a^16*b^2))/(b^5*d^4 *(a^2 + b^2)^8))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 - (a^3*(225*a^12 + 504*b ^12 + 872*a^2*b^10 + 4457*a^4*b^8 + 5916*a^6*b^6 + 4006*a^8*b^4 + 1380*a^1 0*b^2))/(2*b^5*d^5*(a^2 + b^2)^8))*(-1/(b^6*d^2*1i - a^6*d^2*1i + 6*a*b^5* d^2 + 6*a^5*b*d^2 - a^2*b^4*d^2*15i - 20*a^3*b^3*d^2 + a^4*b^2*d^2*15i))^( 1/2))/2 - log(- (((((((((64*a*b*(15*a^4 + 2*b^4 + 41*a^2*b^2))/d + 256*b^3 *tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(1i/(d^2*(a*1i - b)^6))^(1/2 ))*(1i/(d^2*(a*1i - b)^6))^(1/2))/2 + (8*a*tan(c + d*x)^(1/2)*(225*a^14 - 184*b^14 + 608*a^2*b^12 - 272*a^4*b^10 + 3937*a^6*b^8 + 5804*a^8*b^6 + 400 6*a^10*b^4 + 1380*a^12*b^2))/(b^4*d^2*(a^2 + b^2)^4))*(1i/(d^2*(a*1i - b)^ 6))^(1/2))/2 + (2*a^2*(1125*a^14 + 16*b^14 + 6112*a^2*b^12 - 17727*a^4*...