Integrand size = 27, antiderivative size = 101 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))} \]
[Out]
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 84} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))}-\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]
[In]
[Out]
Rule 12
Rule 84
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {a}{(a-x)^3 x (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \frac {1}{(a-x)^3 x (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \left (\frac {1}{2 a^2 (a-x)^3}+\frac {3}{4 a^3 (a-x)^2}+\frac {7}{8 a^4 (a-x)}+\frac {1}{a^4 x}-\frac {1}{8 a^4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {7 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {3 a^3}{4 d (a-a \sin (c+d x))} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \left (7 \log (1-\sin (c+d x))-8 \log (\sin (c+d x))+\log (1+\sin (c+d x))-\frac {2}{(-1+\sin (c+d x))^2}+\frac {6}{-1+\sin (c+d x)}\right )}{8 d} \]
[In]
[Out]
Time = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(100\) |
default | \(\frac {\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(100\) |
risch | \(-\frac {i \left (-3 a^{2} {\mathrm e}^{i \left (d x +c \right )}-8 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(127\) |
parallelrisch | \(-\frac {\left (\left (7 \cos \left (2 d x +2 c \right )+28 \sin \left (d x +c \right )-21\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-4 \cos \left (2 d x +2 c \right )-16 \sin \left (d x +c \right )+12\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \cos \left (2 d x +2 c \right )+10 \sin \left (d x +c \right )-4\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(149\) |
norman | \(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {4 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {7 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(298\) |
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.64 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \sin \left (d x + c\right ) - 8 \, a^{2} + 8 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
[In]
[Out]
Timed out. \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.83 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 7 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - 8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]
[In]
[Out]
none
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 14 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 16 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {21 \, a^{2} \sin \left (d x + c\right )^{2} - 54 \, a^{2} \sin \left (d x + c\right ) + 37 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {\frac {3\,a^2\,\sin \left (c+d\,x\right )}{4}-a^2}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}-\frac {7\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d} \]
[In]
[Out]