Integrand size = 33, antiderivative size = 380 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2 (9 A b+5 a B)}{15 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (5 a^2 A-14 A b^2-15 a b B\right )}{5 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
1/2*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctan(-1+2^(1/2)*ta n(d*x+c)^(1/2))/d*2^(1/2)+1/2*(a^3*(A-B)-3*a*b^2*(A-B)-3*a^2*b*(A+B)+b^3*( A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*(3*a^2*b*(A-B)-b^3* (A-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d *2^(1/2)-1/4*(3*a^2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*ln(1+2^(1/2 )*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2/5*a*(5*A*a^2-14*A*b^2-15*B*a*b) /d/tan(d*x+c)^(1/2)-2/15*a^2*(9*A*b+5*B*a)/d/tan(d*x+c)^(3/2)-2/5*a*A*(a+b *tan(d*x+c))^2/d/tan(d*x+c)^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.44 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},1,-\frac {1}{4},-\tan ^2(c+d x)\right )+5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)+b \left (9 a A b+9 a^2 B-3 b^2 B+5 b (A b+3 a B) \tan (c+d x)+15 b^2 B \tan ^2(c+d x)\right )\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \]
(-2*(3*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] + 5*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Hyperge ometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(9*a*A*b + 9*a^ 2*B - 3*b^2*B + 5*b*(A*b + 3*a*B)*Tan[c + d*x] + 15*b^2*B*Tan[c + d*x]^2)) )/(15*d*Tan[c + d*x]^(5/2))
Time = 1.21 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.82, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4088, 27, 3042, 4118, 25, 3042, 4111, 27, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan (c+d x)^{7/2}}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle \frac {2}{5} \int \frac {(a+b \tan (c+d x)) \left (-b (a A-5 b B) \tan ^2(c+d x)-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (9 A b+5 a B)\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x)) \left (-b (a A-5 b B) \tan ^2(c+d x)-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (9 A b+5 a B)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x)) \left (-b (a A-5 b B) \tan (c+d x)^2-5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (9 A b+5 a B)\right )}{\tan (c+d x)^{5/2}}dx-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4118 |
\(\displaystyle \frac {1}{5} \left (\int -\frac {b^2 (a A-5 b B) \tan ^2(c+d x)+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 A a^2-15 b B a-14 A b^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (-\int \frac {b^2 (a A-5 b B) \tan ^2(c+d x)+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 A a^2-15 b B a-14 A b^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (-\int \frac {b^2 (a A-5 b B) \tan (c+d x)^2+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (5 A a^2-15 b B a-14 A b^2\right )}{\tan (c+d x)^{3/2}}dx-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4111 |
\(\displaystyle \frac {1}{5} \left (-\int \frac {5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (-5 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (-5 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4017 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \int \frac {B a^3+3 A b a^2-3 b^2 B a-A b^3-\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 1482 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{5} \left (\frac {2 a \left (5 a^2 A-15 a b B-14 A b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (5 a B+9 A b)}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\right )-\frac {2 a A (a+b \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
((-10*(-1/2*((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B ))*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2] *Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*(-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*Sqr t[2])))/2))/d - (2*a^2*(9*A*b + 5*a*B))/(3*d*Tan[c + d*x]^(3/2)) + (2*a*(5 *a^2*A - 14*A*b^2 - 15*a*b*B))/(d*Sqrt[Tan[c + d*x]]))/5 - (2*a*A*(a + b*T an[c + d*x])^2)/(5*d*Tan[c + d*x]^(5/2))
3.4.97.3.1 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_)^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_) + (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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((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 - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x ] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*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[c^2 + d^2, 0] && LtQ[n , -1]
Time = 0.04 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {-\frac {2 A \,a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(290\) |
default | \(\frac {-\frac {2 A \,a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 a^{2} \left (3 A b +B a \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(290\) |
parts | \(\frac {\left (A \,b^{3}+3 B 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 d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (-\frac {\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 {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\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}\right )}{d}+\frac {A \,a^{3} \left (\frac {\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 {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B \,b^{3} \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 d}\) | \(537\) |
1/d*(-2/5*A*a^3/tan(d*x+c)^(5/2)-2/3*a^2*(3*A*b+B*a)/tan(d*x+c)^(3/2)+2*a* (A*a^2-3*A*b^2-3*B*a*b)/tan(d*x+c)^(1/2)+1/4*(-3*A*a^2*b+A*b^3-B*a^3+3*B*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)))+1/4*(A*a^3-3*A*a*b^2-3*B*a^2*b+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))))
Leaf count of result is larger than twice the leaf count of optimal. 6162 vs. \(2 (340) = 680\).
Time = 1.45 (sec) , antiderivative size = 6162, normalized size of antiderivative = 16.22 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (3 \, A a^{3} - 15 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{60 \, d} \]
1/60*(30*sqrt(2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B )*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 30*sqrt(2)*( (A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*arctan(-1/2 *sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 15*sqrt(2)*((A + B)*a^3 + 3*( A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(sqrt(2)*sqrt(tan(d*x + c )) + tan(d*x + c) + 1) + 15*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 8*(3*A*a^3 - 15*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 + 5*( B*a^3 + 3*A*a^2*b)*tan(d*x + c))/tan(d*x + c)^(5/2))/d
Timed out. \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 14.60 (sec) , antiderivative size = 7591, normalized size of antiderivative = 19.98 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
2*atanh((32*B^2*a^6*d^3*tan(c + d*x)^(1/2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4* a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4 *a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2))/(16*B^3*b^9*d^2 + 16 *B*a^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^ 8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^( 1/2) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 736*B^3*a^6*b^3*d^2 + 4 8*B^3*a^8*b*d^2 - 48*B*a*b^2*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^1 2*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) - (32*B^2*b^6*d^3*tan(c + d*x)^(1/2)*((5*B^2*a ^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4 *a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2 *d^4)^(1/2)/(4*d^4) - (3*B^2*a*b^5)/(2*d^2) - (3*B^2*a^5*b)/(2*d^2))^(1/2) )/(16*B^3*b^9*d^2 + 16*B*a^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^1 2*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 288*B^3*a^2*b^7*d^2 + 960*B^3*a^4*b^5*d^2 - 7 36*B^3*a^6*b^3*d^2 + 48*B^3*a^8*b*d^2 - 48*B*a*b^2*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) + (480*B^2*a^2*b^4*d^3*t an(c + d*x)^(1/2)*((5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^1...