Integrand size = 27, antiderivative size = 212 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 a^4 \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}-\frac {2 a^2 \left (a^2-3 b^2\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \]
[Out]
Time = 0.19 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2976, 2727, 2743, 12, 2739, 632, 210} \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 a^2 \left (a^2-3 b^2\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}+\frac {2 a^4 \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}+\frac {a^3 \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]
[In]
[Out]
Rule 12
Rule 210
Rule 632
Rule 2727
Rule 2739
Rule 2743
Rule 2976
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2 (a+b)^2 (-1+\sin (c+d x))}-\frac {1}{2 (a-b)^2 (1+\sin (c+d x))}+\frac {a^3}{b \left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {a^2 \left (a^2-3 b^2\right )}{b \left (-a^2+b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx \\ & = -\frac {\int \frac {1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac {\int \frac {1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}-\frac {\left (a^2 \left (a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}+\frac {a^3 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{b \left (a^2-b^2\right )} \\ & = \frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {a^3 \int \frac {a}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}-\frac {\left (2 a^2 \left (a^2-3 b^2\right )\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 )^2 d} \\ & = \frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {a^4 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}+\frac {\left (4 a^2 \left (a^2-3 b^2\right )\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 )^2 d} \\ & = -\frac {2 a^2 \left (a^2-3 b^2\right ) \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 )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\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 \left (a^2-b^2\right )^2 d} \\ & = -\frac {2 a^2 \left (a^2-3 b^2\right ) \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 )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\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 \left (a^2-b^2\right )^2 d} \\ & = \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 \left (a^2-b^2\right )^{5/2} d}-\frac {2 a^2 \left (a^2-3 b^2\right ) \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 )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}+\frac {a^3 \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \\ \end{align*}
Time = 0.89 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.76 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {6 a^2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )+\frac {a^3 \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}}{d} \]
[In]
[Out]
Time = 0.73 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {-\frac {4 a^{2} \left (\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}-\frac {3 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(155\) |
default | \(\frac {-\frac {4 a^{2} \left (\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}-\frac {a}{2}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}-\frac {3 b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(155\) |
risch | \(-\frac {2 i \left (3 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+a^{4} {\mathrm e}^{3 i \left (d x +c \right )}+a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{3} b +2 i a \,b^{3}+a^{4} {\mathrm e}^{i \left (d x +c \right )}+3 a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}-b^{4} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left (a^{2}-b^{2}\right )^{2} d}+\frac {3 i b \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {3 i b \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(350\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.52 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\left [\frac {2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} + 2 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{3} b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )\right )}}, \frac {a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{3} b \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )}\right ] \]
[In]
[Out]
Timed out. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
[In]
[Out]
Exception generated. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
[In]
[Out]
none
Time = 0.44 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.05 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} - a b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )} {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}\right )}}{d} \]
[In]
[Out]
Time = 14.57 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.30 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2\,a\,\left (2\,a^2+b^2\right )}{{\left (a^2-b^2\right )}^2}-\frac {6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{{\left (a^2-b^2\right )}^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )}{{\left (a^2-b^2\right )}^2}-\frac {6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {6\,a^2\,b\,\mathrm {atan}\left (\frac {\frac {a^2\,b\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{a^2\,b}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
[In]
[Out]