Integrand size = 27, antiderivative size = 138 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a b^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d} \]
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Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2945, 12, 2739, 632, 210} \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 a b^3 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 d \left (a^2-b^2\right )}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2} \]
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Rule 12
Rule 210
Rule 632
Rule 2739
Rule 2945
Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\int \frac {\sec ^2(c+d x) \left (-a b+2 b^2 \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )} \\ & = \frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\int -\frac {3 a b^3}{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2} \\ & = \frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\left (a b^3\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2} \\ & = \frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {\left (2 a b^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 )}{\left (a^2-b^2\right )^2 d} \\ & = \frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {\left (4 a b^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 )}{\left (a^2-b^2\right )^2 d} \\ & = -\frac {2 a b^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 )^{5/2} d}+\frac {\sec ^3(c+d x) (a-b \sin (c+d x))}{3 \left (a^2-b^2\right ) d}-\frac {\sec (c+d x) \left (3 a b^2-b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d} \\ \end{align*}
Time = 1.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {24 a b^3 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {\sec ^3(c+d x) \left (4 a^3-10 a b^2-\frac {3}{2} a \left (a^2-7 b^2\right ) \cos (c+d x)-6 a b^2 \cos (2 (c+d x))-\frac {1}{2} a^3 \cos (3 (c+d x))+\frac {7}{2} a b^2 \cos (3 (c+d x))-3 a^2 b \sin (c+d x)+6 b^3 \sin (c+d x)+a^2 b \sin (3 (c+d x))+2 b^3 \sin (3 (c+d x))\right )}{(a-b)^2 (a+b)^2}}{12 d} \]
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Time = 0.64 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.56
method | result | size |
derivativedivides | \(\frac {-\frac {2}{\left (4 a -4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (4 a -4 b \right )}-\frac {-a +2 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \,b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (4 a +4 b \right )}-\frac {2}{\left (4 a +4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a +2 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(215\) |
default | \(\frac {-\frac {2}{\left (4 a -4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (4 a -4 b \right )}-\frac {-a +2 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \,b^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}}-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (4 a +4 b \right )}-\frac {2}{\left (4 a +4 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a +2 b}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\) | \(215\) |
risch | \(\frac {2 i \left (3 i b^{2} a \,{\mathrm e}^{5 i \left (d x +c \right )}-4 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+10 i b^{2} a \,{\mathrm e}^{3 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+6 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{2} b +2 b^{3}\right )}{3 d \left (-a^{2}+b^{2}\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {i b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(301\) |
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Time = 0.37 (sec) , antiderivative size = 469, normalized size of antiderivative = 3.40 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} + b^{2}} a b^{3} \cos \left (d x + c\right )^{3} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{5} + 4 \, a^{3} b^{2} - 2 \, a b^{4} + 6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}, \frac {3 \, \sqrt {a^{2} - b^{2}} a b^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + a^{5} - 2 \, a^{3} b^{2} + a b^{4} - 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
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\[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Exception generated. \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.38 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 4 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} - 4 \, a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}\right )}}{3 \, d} \]
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Time = 17.38 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.74 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2\,\left (4\,a\,b^2-a^3\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {2\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,b^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^2+b^2\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a\,b^3\,\mathrm {atan}\left (\frac {\frac {a\,b^3\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^2\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{2\,a\,b^3}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
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