Integrand size = 23, antiderivative size = 317 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {b^2 \sqrt {\tan (c+d x)}}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]
b^(3/2)*(5*a^2+b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/a^(3/2)/(a^2+ b^2)^2/d+1/2*(a^2-2*a*b-b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2) ^2/d*2^(1/2)+1/2*(a^2-2*a*b-b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b ^2)^2/d*2^(1/2)-1/4*(a^2+2*a*b-b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ c))/(a^2+b^2)^2/d*2^(1/2)+1/4*(a^2+2*a*b-b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2 )+tan(d*x+c))/(a^2+b^2)^2/d*2^(1/2)+b^2*tan(d*x+c)^(1/2)/a/(a^2+b^2)/d/(a+ b*tan(d*x+c))
Result contains complex when optimal does not.
Time = 0.67 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\frac {\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right )}-\frac {\sqrt [4]{-1} a \left ((a+i b)^2 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+(a-i b)^2 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )}{a^2+b^2}+\frac {b^2 \sqrt {\tan (c+d x)}}{a+b \tan (c+d x)}}{a \left (a^2+b^2\right ) d} \]
((b^(3/2)*(5*a^2 + b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqr t[a]*(a^2 + b^2)) - ((-1)^(1/4)*a*((a + I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[ c + d*x]]] + (a - I*b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))/(a^2 + b ^2) + (b^2*Sqrt[Tan[c + d*x]])/(a + b*Tan[c + d*x]))/(a*(a^2 + b^2)*d)
Time = 1.14 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.91, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.870, Rules used = {3042, 4052, 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.
| \(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4052 |
| \(\displaystyle \frac {\int \frac {2 a^2-2 b \tan (c+d x) a+b^2+b^2 \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 a^2-2 b \tan (c+d x) a+b^2+b^2 \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a^2-2 b \tan (c+d x) a+b^2+b^2 \tan (c+d x)^2}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 4136 |
| \(\displaystyle \frac {\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int \frac {2 \left (a \left (a^2-b^2\right )-2 a^2 b \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 \int \frac {a \left (a^2-b^2\right )-2 a^2 b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 \int \frac {a \left (a^2-b^2\right )-2 a^2 b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {4 \int \frac {a \left (a^2-2 b \tan (c+d x) a-b^2\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 a \int \frac {a^2-2 b \tan (c+d x) a-b^2}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {4 a \left (\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {\frac {b^2 \left (5 a^2+b^2\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {4 a \left (\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {2 b^2 \left (5 a^2+b^2\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {4 a \left (\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {4 a \left (\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {2 b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}}{2 a \left (a^2+b^2\right )}+\frac {b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}\) |
((2*b^(3/2)*(5*a^2 + b^2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(S qrt[a]*(a^2 + b^2)*d) + (4*a*(((a^2 - 2*a*b - b^2)*(-(ArcTan[1 - Sqrt[2]*S qrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[ 2]))/2 + ((a^2 + 2*a*b - b^2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + T an[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))/(2*a*(a^2 + b^2)) + (b^2*Sqrt[Tan[c + d *x]])/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))
3.6.96.3.1 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^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
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[(((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 = 0.07 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 b^{2} \left (\frac {\left (a^{2}+b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 a \left (a +b \tan \left (d x +c \right )\right )}+\frac {\left (5 a^{2}+b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\) | \(281\) |
| default | \(\frac {\frac {\frac {\left (a^{2}-b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {2 b^{2} \left (\frac {\left (a^{2}+b^{2}\right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 a \left (a +b \tan \left (d x +c \right )\right )}+\frac {\left (5 a^{2}+b^{2}\right ) \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\) | \(281\) |
1/d*(2/(a^2+b^2)^2*(1/8*(a^2-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)*ta n(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))-1/4*a*b*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*b^2/(a^2+b^2)^2*(1/2*(a^2+b^2)/a*tan(d*x+c)^(1/2)/(a+b*tan(d* x+c))+1/2*(5*a^2+b^2)/a/(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. 2658 vs. \(2 (277) = 554\).
Time = 0.50 (sec) , antiderivative size = 5341, normalized size of antiderivative = 16.85 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{2} \sqrt {\tan {\left (c + d x \right )}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\frac {\frac {4 \, b^{2} \sqrt {\tan \left (d x + c\right )}}{a^{4} + a^{2} b^{2} + {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )} + \frac {4 \, {\left (5 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}}}{4 \, d} \]
1/4*(4*b^2*sqrt(tan(d*x + c))/(a^4 + a^2*b^2 + (a^3*b + a*b^3)*tan(d*x + c )) + 4*(5*a^2*b^2 + b^4)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^5 + 2* a^3*b^2 + a*b^4)*sqrt(a*b)) + (2*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sq rt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 - 2*a*b - b^2)*ar ctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4))/d
Timed out. \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\text {Timed out} \]
Time = 12.97 (sec) , antiderivative size = 8282, normalized size of antiderivative = 26.13 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]
(log((((((((((128*b^2*(2*b^6 - a^6 + 9*a^2*b^4 + 6*a^4*b^2))/(a*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)^4))^(1 /2))*(1i/(d^2*(a*1i - b)^4))^(1/2))/2 + (64*b^2*tan(c + d*x)^(1/2)*(2*b^8 - a^8 + 5*a^2*b^6 + 67*a^4*b^4 - a^6*b^2))/(a*d^2*(a^2 + b^2)^2))*(1i/(d^2 *(a*1i - b)^4))^(1/2))/2 - (32*b^5*(25*a^6 + b^6 - 13*a^2*b^4 - 85*a^4*b^2 ))/(a^2*d^3*(a^2 + b^2)^3))*(1i/(d^2*(a*1i - b)^4))^(1/2))/2 + (16*b^5*tan (c + d*x)^(1/2)*(b^6 - 27*a^6 + 7*a^2*b^4 + 11*a^4*b^2))/(a^2*d^4*(a^2 + b ^2)^4))*(1i/(d^2*(a*1i - b)^4))^(1/2))/2 + (16*b^6*(5*a^2 + b^2))/(a*d^5*( a^2 + b^2)^4))*(-1/(a^4*d^2*1i + b^4*d^2*1i + 4*a*b^3*d^2 - 4*a^3*b*d^2 - a^2*b^2*d^2*6i))^(1/2))/2 - atan(((-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4 i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4*d^2 + a* b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4*d^2 + b^4 *d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((-1i/(4*(a^4* d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^2*4i - 6*a^2*b^2*d^2)))^(1/2)*((16* (16*a*b^16*d^4 + 136*a^3*b^14*d^4 + 432*a^5*b^12*d^4 + 680*a^7*b^10*d^4 + 560*a^9*b^8*d^4 + 216*a^11*b^6*d^4 + 16*a^13*b^4*d^4 - 8*a^15*b^2*d^4))/(a ^10*d^5 + a^2*b^8*d^5 + 4*a^4*b^6*d^5 + 6*a^6*b^4*d^5 + 4*a^8*b^2*d^5) - ( 16*tan(c + d*x)^(1/2)*(-1i/(4*(a^4*d^2 + b^4*d^2 + a*b^3*d^2*4i - a^3*b*d^ 2*4i - 6*a^2*b^2*d^2)))^(1/2)*(32*a^2*b^17*d^4 + 160*a^4*b^15*d^4 + 288*a^ 6*b^13*d^4 + 160*a^8*b^11*d^4 - 160*a^10*b^9*d^4 - 288*a^12*b^7*d^4 - 1...