Integrand size = 27, antiderivative size = 64 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {a^3}{4 d (a-a \sin (c+d x))} \]
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Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2915, 12, 78, 212} \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d} \]
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Rule 12
Rule 78
Rule 212
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {x}{a (a-x)^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4 \text {Subst}\left (\int \frac {x}{(a-x)^3 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4 \text {Subst}\left (\int \left (\frac {1}{2 (a-x)^3}-\frac {1}{4 a (a-x)^2}-\frac {1}{4 a \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {a^3}{4 d (a-a \sin (c+d x))}-\frac {a^3 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 d} \\ & = -\frac {a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}-\frac {a^3}{4 d (a-a \sin (c+d x))} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^2 \left (\text {arctanh}(\sin (c+d x))-\frac {\sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{4 d} \]
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Result contains complex when optimal does not.
Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.44
method | result | size |
risch | \(\frac {i \left (a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\) | \(92\) |
parallelrisch | \(\frac {\left (\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \sin \left (d x +c \right )\right ) a^{2}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(105\) |
derivativedivides | \(\frac {\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(106\) |
default | \(\frac {\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}+2 a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {a^{2}}{4 \cos \left (d x +c \right )^{4}}}{d}\) | \(106\) |
norman | \(\frac {\frac {8 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {9 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {11 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {12 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}-\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(281\) |
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Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.88 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, a^{2} \sin \left (d x + c\right ) + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
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Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, a^{2} \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.48 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {a^{2} \log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right ) - a^{2} \log \left ({\left | \frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right ) + \frac {a^{2} {\left (\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right )\right )} - 6 \, a^{2}}{\frac {1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2}}{16 \, d} \]
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Time = 13.35 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.66 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d} \]
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