Optimal. Leaf size=765 \[ \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 F_1\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 F_1\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 F_1\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 F_1\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 \, _2F_1\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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Rubi [A] time = 4.17, antiderivative size = 765, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {12, 6719, 6728, 364, 959, 510} \[ \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 F_1\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 F_1\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 {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 F_1\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 F_1\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 {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 \, _2F_1\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)} \]
Antiderivative was successfully verified.
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Rule 12
Rule 364
Rule 510
Rule 959
Rule 6719
Rule 6728
Rubi steps
\begin {align*} \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx &=\frac {2 \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \, _2F_1\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 ) \operatorname {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} \, _2F_1\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 ) \operatorname {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 ) \operatorname {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} \, _2F_1\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \, _2F_1\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 F_1\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 F_1\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 F_1\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 F_1\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*}
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Mathematica [F] time = 13.22, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^m(c+d x)}{a+b \tan (c+d x)} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.08, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{m}\left (d x +c \right )}{a +b \tan \left (d x +c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{m}}{b \tan \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (c+d\,x\right )}^m}{a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{m}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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