Integrand size = 27, antiderivative size = 133 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d} \]
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Time = 0.13 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2981, 2686, 8, 3554, 2814, 2739, 632, 210} \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {2 a^3 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}} \]
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Rule 8
Rule 210
Rule 632
Rule 2686
Rule 2739
Rule 2814
Rule 2981
Rule 3554
Rubi steps \begin{align*} \text {integral}& = \frac {a \int \sec (c+d x) \tan (c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \tan ^2(c+d x) \, dx}{a^2-b^2} \\ & = -\frac {a^2 x}{b \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^3 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac {b \int 1 \, dx}{a^2-b^2}+\frac {a \text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{\left (a^2-b^2\right ) d} \\ & = -\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right ) d} \\ & = -\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (4 a^3\right ) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right ) d} \\ & = -\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {a \sec (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \tan (c+d x)}{\left (a^2-b^2\right ) d} \\ \end{align*}
Time = 0.73 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 a^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {-\left (\left (a^2-b^2\right ) (c+d x) \cos (c+d x)\right )+b (a-b \sin (c+d x))}{(a-b) (a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{b d} \]
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Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {-\frac {16}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {16}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b \sqrt {a^{2}-b^{2}}}}{d}\) | \(130\) |
default | \(\frac {-\frac {16}{\left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {16}{\left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b \sqrt {a^{2}-b^{2}}}}{d}\) | \(130\) |
risch | \(-\frac {x}{b}+\frac {2 i \left (i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {i a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}\) | \(223\) |
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Time = 0.47 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.77 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{3} \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b - 2 \, a b^{3} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}, -\frac {\sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{3} b + a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}\right ] \]
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Timed out. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{{\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}}} - \frac {d x + c}{b} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}}{d} \]
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Time = 14.03 (sec) , antiderivative size = 1538, normalized size of antiderivative = 11.56 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
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