Integrand size = 21, antiderivative size = 765 \[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {2^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac {2^{1+m} b \operatorname {AppellF1}\left (\frac {2+m}{2},1+m,1,\frac {4+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right ) d (2+m)}-\frac {2^{1+m} b \operatorname {AppellF1}\left (\frac {2+m}{2},1+m,1,\frac {4+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b+\sqrt {a^2+b^2}\right ) d (2+m)}+\frac {2^{1+m} a b \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )^2 d (3+m)}-\frac {2^{1+m} a b \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b+\sqrt {a^2+b^2}\right )^2 d (3+m)} \]
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Time = 6.24 (sec) , antiderivative size = 765, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {12, 6851, 6860, 371, 973, 524} \[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b 2^{m+1} \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \operatorname {AppellF1}\left (\frac {m+2}{2},m+1,1,\frac {m+4}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right )}{d (m+2) \sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )}-\frac {b 2^{m+1} \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \operatorname {AppellF1}\left (\frac {m+2}{2},m+1,1,\frac {m+4}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right )}{d (m+2) \sqrt {a^2+b^2} \left (\sqrt {a^2+b^2}+b\right )}+\frac {a b 2^{m+1} \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \operatorname {AppellF1}\left (\frac {m+3}{2},m+1,1,\frac {m+5}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right )}{d (m+3) \sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )^2}-\frac {a b 2^{m+1} \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \operatorname {AppellF1}\left (\frac {m+3}{2},m+1,1,\frac {m+5}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right )}{d (m+3) \sqrt {a^2+b^2} \left (\sqrt {a^2+b^2}+b\right )^2}+\frac {2^{m+1} \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}\right )^m \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^m \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},m+1,\frac {m+3}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a d (m+1)} \]
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Rule 12
Rule 371
Rule 524
Rule 973
Rule 6851
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {2^m \left (1-x^2\right ) \left (\frac {x}{1+x^2}\right )^{1+m}}{x \left (a+2 b x-a x^2\right )} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {2^{1+m} \text {Subst}\left (\int \frac {\left (1-x^2\right ) \left (\frac {x}{1+x^2}\right )^{1+m}}{x \left (a+2 b x-a x^2\right )} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {\left (2^{1+m} \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^m \left (1-x^2\right ) \left (1+x^2\right )^{-1-m}}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {\left (2^{1+m} \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \left (\frac {x^m \left (1+x^2\right )^{-1-m}}{a}-\frac {2 b x^{1+m} \left (1+x^2\right )^{-1-m}}{a \left (a+2 b x-a x^2\right )}\right ) \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{d} \\ & = \frac {\left (2^{1+m} \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int x^m \left (1+x^2\right )^{-1-m} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d}-\frac {\left (2^{2+m} b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{1+m} \left (1+x^2\right )^{-1-m}}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \\ & = \frac {2^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{a d (1+m)}-\frac {\left (2^{2+m} b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \left (-\frac {a x^{1+m} \left (1+x^2\right )^{-1-m}}{\sqrt {a^2+b^2} \left (2 b-2 \sqrt {a^2+b^2}-2 a x\right )}+\frac {a x^{1+m} \left (1+x^2\right )^{-1-m}}{\sqrt {a^2+b^2} \left (2 b+2 \sqrt {a^2+b^2}-2 a x\right )}\right ) \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \\ & = \frac {2^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac {\left (2^{2+m} b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{1+m} \left (1+x^2\right )^{-1-m}}{2 b-2 \sqrt {a^2+b^2}-2 a x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d}-\frac {\left (2^{2+m} b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{1+m} \left (1+x^2\right )^{-1-m}}{2 b+2 \sqrt {a^2+b^2}-2 a x} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d} \\ & = \frac {2^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac {\left (2^{3+m} a b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{2+m} \left (1+x^2\right )^{-1-m}}{\left (2 b-2 \sqrt {a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d}-\frac {\left (2^{3+m} a b \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{2+m} \left (1+x^2\right )^{-1-m}}{\left (2 b+2 \sqrt {a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d}+\frac {\left (2^{3+m} b \left (b-\sqrt {a^2+b^2}\right ) \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{1+m} \left (1+x^2\right )^{-1-m}}{\left (2 b-2 \sqrt {a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d}-\frac {\left (2^{3+m} b \left (b+\sqrt {a^2+b^2}\right ) \tan ^{-m}\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m\right ) \text {Subst}\left (\int \frac {x^{1+m} \left (1+x^2\right )^{-1-m}}{\left (2 b+2 \sqrt {a^2+b^2}\right )^2-4 a^2 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2+b^2} d} \\ & = \frac {2^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},1+m,\frac {3+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{a d (1+m)}+\frac {2^{1+m} b \operatorname {AppellF1}\left (\frac {2+m}{2},1+m,1,\frac {4+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right ) d (2+m)}-\frac {2^{1+m} b \operatorname {AppellF1}\left (\frac {2+m}{2},1+m,1,\frac {4+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b+\sqrt {a^2+b^2}\right ) d (2+m)}+\frac {2^{1+m} a b \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b-\sqrt {a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b-\sqrt {a^2+b^2}\right )^2 d (3+m)}-\frac {2^{1+m} a b \operatorname {AppellF1}\left (\frac {3+m}{2},1+m,1,\frac {5+m}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {a^2 \tan ^2\left (\frac {1}{2} (c+d x)\right )}{\left (b+\sqrt {a^2+b^2}\right )^2}\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}\right )^m \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{\sqrt {a^2+b^2} \left (b+\sqrt {a^2+b^2}\right )^2 d (3+m)} \\ \end{align*}
\[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx \]
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\[\int \frac {\sin ^{m}\left (d x +c \right )}{a +b \tan \left (d x +c \right )}d x\]
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\[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sin ^{m}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
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\[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int { \frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^m}{a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
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