Integrand size = 29, antiderivative size = 183 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2} d}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d} \]
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Time = 0.19 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2981, 3554, 8, 2670, 14, 2825, 12, 2814, 2739, 632, 210} \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \cos (c+d x)}{b d \left (a^2-b^2\right )}-\frac {b \cos (c+d x)}{d \left (a^2-b^2\right )}+\frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {a x}{a^2-b^2}-\frac {2 a^4 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{3/2}}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )} \]
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Rule 8
Rule 12
Rule 14
Rule 210
Rule 632
Rule 2670
Rule 2739
Rule 2814
Rule 2825
Rule 2981
Rule 3554
Rubi steps \begin{align*} \text {integral}& = \frac {a \int \tan ^2(c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sin (c+d x) \tan ^2(c+d x) \, dx}{a^2-b^2} \\ & = \frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a \int 1 \, dx}{a^2-b^2}+\frac {a^2 \int \frac {a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac {b \text {Subst}\left (\int \frac {1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d} \\ & = -\frac {a x}{a^2-b^2}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a^3 \int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac {b \text {Subst}\left (\int \left (-1+\frac {1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{\left (a^2-b^2\right ) d} \\ & = -\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {a^4 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )} \\ & = -\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}-\frac {\left (2 a^4\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^2 \left (a^2-b^2\right ) d} \\ & = -\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {\left (4 a^4\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^2 \left (a^2-b^2\right ) d} \\ & = -\frac {a x}{a^2-b^2}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2} d}+\frac {a^2 \cos (c+d x)}{b \left (a^2-b^2\right ) d}-\frac {b \cos (c+d x)}{\left (a^2-b^2\right ) d}-\frac {b \sec (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {b^3-a^3 (c+d x)+a b^2 (c+d x)}{-a^2 b^2+b^4}-\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac {\cos (c+d x)}{b}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{d} \]
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Time = 0.62 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {-\frac {32}{\left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{2} \sqrt {a^{2}-b^{2}}}-\frac {32}{\left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(150\) |
default | \(\frac {-\frac {32}{\left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\frac {2 b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2}}-\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right ) \left (a +b \right ) b^{2} \sqrt {a^{2}-b^{2}}}-\frac {32}{\left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(150\) |
risch | \(\frac {a x}{b^{2}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {-2 i a +2 b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}\) | \(254\) |
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Time = 0.42 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.36 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {\sqrt {-a^{2} + b^{2}} a^{4} \cos \left (d x + c\right ) \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}\right ] \]
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\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.35 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (d x + c\right )} a}{b^{2}} + \frac {2 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )} {\left (a^{2} b - b^{3}\right )}}}{d} \]
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Time = 14.17 (sec) , antiderivative size = 1656, normalized size of antiderivative = 9.05 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]
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