Integrand size = 29, antiderivative size = 165 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a^2 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \]
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Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2981, 3852, 2686, 30, 2775, 12, 2739, 632, 210} \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 a^2 b^2 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {a \tan ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec ^3(c+d x)}{3 d \left (a^2-b^2\right )}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{d \left (a^2-b^2\right )^2} \]
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Rule 12
Rule 30
Rule 210
Rule 632
Rule 2686
Rule 2739
Rule 2775
Rule 2981
Rule 3852
Rubi steps \begin{align*} \text {integral}& = \frac {a \int \sec ^4(c+d x) \, dx}{a^2-b^2}-\frac {a^2 \int \frac {\sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2-b^2}-\frac {b \int \sec ^3(c+d x) \tan (c+d x) \, dx}{a^2-b^2} \\ & = \frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a^2 \int \frac {b^2}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{\left (a^2-b^2\right ) d}-\frac {b \text {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{\left (a^2-b^2\right ) d} \\ & = -\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (a^2 b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2} \\ & = -\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 b^2\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 )}{\left (a^2-b^2\right )^2 d} \\ & = -\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 b^2\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 )}{\left (a^2-b^2\right )^2 d} \\ & = \frac {2 a^2 b^2 \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 )^{5/2} d}-\frac {b \sec ^3(c+d x)}{3 \left (a^2-b^2\right ) d}+\frac {a^2 \sec (c+d x) (b-a \sin (c+d x))}{\left (a^2-b^2\right )^2 d}+\frac {a \tan (c+d x)}{\left (a^2-b^2\right ) d}+\frac {a \tan ^3(c+d x)}{3 \left (a^2-b^2\right ) d} \\ \end{align*}
Time = 1.15 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.21 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {48 a^2 b^2 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\sec ^3(c+d x) \left (-4 a^2 b-8 b^3+3 b \left (5 a^2+b^2\right ) \cos (c+d x)-12 a^2 b \cos (2 (c+d x))+5 a^2 b \cos (3 (c+d x))+b^3 \cos (3 (c+d x))-6 a^3 \sin (c+d x)+12 a b^2 \sin (c+d x)+2 a^3 \sin (3 (c+d x))+4 a b^2 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{24 d} \]
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Time = 0.56 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (8 a +8 b \right )}-\frac {4}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 a^{2} b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (8 a -8 b \right )}+\frac {4}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(207\) |
default | \(\frac {-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (8 a +8 b \right )}-\frac {4}{\left (8 a +8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2 a^{2} b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {8}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (8 a -8 b \right )}+\frac {4}{\left (8 a -8 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(207\) |
risch | \(\frac {-2 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-4 i b^{2} a \,{\mathrm e}^{2 i \left (d x +c \right )}+\frac {4 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}}{3}+\frac {8 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{3}-\frac {2 i a^{3}}{3}-\frac {4 i a \,b^{2}}{3}+2 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(302\) |
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Time = 0.41 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a^{2} b^{2} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, \sqrt {a^{2} - b^{2}} a^{2} b^{2} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{4} b - 2 \, a^{2} b^{3} + b^{5} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} - {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
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\[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.38 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2} b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{2} b - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]
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Time = 17.99 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2\,a^2\,b^2\,\mathrm {atan}\left (\frac {\frac {a^2\,b^2\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a^2\,b^2}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,\left (2\,a^2\,b+b^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^3+a\,b^2\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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