Integrand size = 17, antiderivative size = 70 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b} \]
[Out]
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2701, 294, 327, 213} \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b} \]
[In]
[Out]
Rule 213
Rule 294
Rule 327
Rule 2701
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^6}{\left (-1+x^2\right )^3} \, dx,x,\csc (a+b x)\right )}{b} \\ & = \frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {5 \text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{4 b} \\ & = \frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {15 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b} \\ & = -\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b}-\frac {15 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{8 b} \\ & = \frac {15 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {15 \csc (a+b x)}{8 b}+\frac {5 \csc (a+b x) \sec ^2(a+b x)}{8 b}+\frac {\csc (a+b x) \sec ^4(a+b x)}{4 b} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.39 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(a+b x)\right )}{b} \]
[In]
[Out]
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )}+\frac {5}{8 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}-\frac {15}{8 \sin \left (b x +a \right )}+\frac {15 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(68\) |
default | \(\frac {\frac {1}{4 \cos \left (b x +a \right )^{4} \sin \left (b x +a \right )}+\frac {5}{8 \cos \left (b x +a \right )^{2} \sin \left (b x +a \right )}-\frac {15}{8 \sin \left (b x +a \right )}+\frac {15 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(68\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}+40 \,{\mathrm e}^{7 i \left (b x +a \right )}+18 \,{\mathrm e}^{5 i \left (b x +a \right )}+40 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}-\frac {15 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}\) | \(126\) |
norman | \(\frac {-\frac {1}{2 b}+\frac {15 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {5 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}+\frac {15 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{4 b}-\frac {\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}-\frac {15 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{8 b}+\frac {15 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{8 b}\) | \(149\) |
parallelrisch | \(\frac {\left (-60 \cos \left (2 b x +2 a \right )-15 \cos \left (4 b x +4 a \right )-45\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (60 \cos \left (2 b x +2 a \right )+15 \cos \left (4 b x +4 a \right )+45\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\left (-170 \cos \left (b x +a \right )+60 \cos \left (2 b x +2 a \right )-30 \cos \left (3 b x +3 a \right )+140\right ) \cot \left (\frac {b x}{2}+\frac {a}{2}\right )-32 \sec \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )}{8 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\) | \(167\) |
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.36 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 15 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 30 \, \cos \left (b x + a\right )^{4} + 10 \, \cos \left (b x + a\right )^{2} + 4}{16 \, b \cos \left (b x + a\right )^{4} \sin \left (b x + a\right )} \]
[In]
[Out]
\[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\int \frac {\sec ^{5}{\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{4} - 25 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{5} - 2 \, \sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \]
[In]
[Out]
none
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.04 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (7 \, \sin \left (b x + a\right )^{3} - 9 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16}{\sin \left (b x + a\right )} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \]
[In]
[Out]
Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.96 \[ \int \csc ^2(a+b x) \sec ^5(a+b x) \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{8\,b}-\frac {\frac {15\,{\sin \left (a+b\,x\right )}^4}{8}-\frac {25\,{\sin \left (a+b\,x\right )}^2}{8}+1}{b\,\left ({\sin \left (a+b\,x\right )}^5-2\,{\sin \left (a+b\,x\right )}^3+\sin \left (a+b\,x\right )\right )} \]
[In]
[Out]