Integrand size = 27, antiderivative size = 116 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac {a \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {a^2 b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 \left (a^2-b^2\right ) d} \]
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Time = 0.16 (sec) , antiderivative size = 116, 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 = {2916, 12, 1661, 815} \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {a \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac {a \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]
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Rule 12
Rule 815
Rule 1661
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^3 \text {Subst}\left (\int \frac {x^2}{b^2 (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \frac {x^2}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {\sec ^2(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{2 d}+\frac {\text {Subst}\left (\int \frac {-\frac {a^2 b^2}{a^2-b^2}+\frac {a b^2 x}{a^2-b^2}}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{2 b d} \\ & = -\frac {\sec ^2(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{2 d}+\frac {\text {Subst}\left (\int \left (-\frac {a b}{2 (a+b)^2 (b-x)}+\frac {2 a^2 b^2}{(a-b)^2 (a+b)^2 (a+x)}-\frac {a b}{2 (a-b)^2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b d} \\ & = \frac {a \log (1-\sin (c+d x))}{4 (a+b)^2 d}-\frac {a \log (1+\sin (c+d x))}{4 (a-b)^2 d}+\frac {a^2 b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2 d}-\frac {\sec ^2(c+d x) \left (\frac {b}{a^2-b^2}-\frac {a \sin (c+d x)}{a^2-b^2}\right )}{2 d} \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-\frac {a \log (1-\sin (c+d x))}{(a+b)^2}+\frac {a \log (1+\sin (c+d x))}{(a-b)^2}-\frac {4 a^2 b \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{(a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \]
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Time = 0.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {\frac {a^{2} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}-\frac {a \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {a \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(112\) |
default | \(\frac {\frac {a^{2} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}-\frac {a \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {a \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}}{d}\) | \(112\) |
parallelrisch | \(\frac {2 b \,a^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+a \left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (a \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (a \sin \left (d x +c \right )+\frac {b \cos \left (2 d x +2 c \right )}{2}-\frac {b}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{2} \left (a +b \right )^{2} d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(173\) |
norman | \(\frac {\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) d}+\frac {a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {a^{2} b \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a^{2}+2 a b +b^{2}\right ) d}\) | \(214\) |
risch | \(-\frac {i a x}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i a c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}+\frac {i a x}{2 a^{2}-4 a b +2 b^{2}}+\frac {i a c}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {2 i a^{2} b x}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {2 i a^{2} b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {i a \,{\mathrm e}^{3 i \left (d x +c \right )}-i a \,{\mathrm e}^{i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left (-a^{2}+b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 \left (a^{2}-2 a b +b^{2}\right ) d}+\frac {a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\) | \(317\) |
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Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.33 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {4 \, a^{2} b \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]
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\[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Time = 0.23 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.14 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, a^{2} b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]
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Time = 0.45 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.45 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {4 \, a^{2} b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right )^{2} - a^{3} \sin \left (d x + c\right ) + a b^{2} \sin \left (d x + c\right ) - b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]
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Time = 12.62 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.78 \[ \int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}-\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}+\frac {a^2\,b\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )} \]
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