Optimal. Leaf size=229 \[ \frac{3 a^2 b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac{a^3 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)}+\frac{b^3 \sin ^{m+4}(c+d x) \, _2F_1\left (2,\frac{m+4}{2};\frac{m+6}{2};\sin ^2(c+d x)\right )}{d (m+4)} \]
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Rubi [A] time = 0.449863, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4401, 2643, 2564, 364, 2577} \[ \frac{3 a^2 b \sin ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{d (m+2)}+\frac{a^3 \cos (c+d x) \sin ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \sin ^{m+3}(c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )}{d (m+3)}+\frac{b^3 \sin ^{m+4}(c+d x) \, _2F_1\left (2,\frac{m+4}{2};\frac{m+6}{2};\sin ^2(c+d x)\right )}{d (m+4)} \]
Antiderivative was successfully verified.
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Rule 4401
Rule 2643
Rule 2564
Rule 364
Rule 2577
Rubi steps
\begin{align*} \int \sin ^m(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \sin ^m(c+d x)+3 a^2 b \sec (c+d x) \sin ^{1+m}(c+d x)+3 a b^2 \sec ^2(c+d x) \sin ^{2+m}(c+d x)+b^3 \sec ^3(c+d x) \sin ^{3+m}(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^m(c+d x) \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) \sin ^{1+m}(c+d x) \, dx+\left (3 a b^2\right ) \int \sec ^2(c+d x) \sin ^{2+m}(c+d x) \, dx+b^3 \int \sec ^3(c+d x) \sin ^{3+m}(c+d x) \, dx\\ &=\frac{a^3 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{x^{3+m}}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{a^3 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sin ^{1+m}(c+d x)}{d (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{3 a^2 b \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};\sin ^2(c+d x)\right ) \sin ^{2+m}(c+d x)}{d (2+m)}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{3+m}{2};\frac{5+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) \sin ^{3+m}(c+d x)}{d (3+m)}+\frac{b^3 \, _2F_1\left (2,\frac{4+m}{2};\frac{6+m}{2};\sin ^2(c+d x)\right ) \sin ^{4+m}(c+d x)}{d (4+m)}\\ \end{align*}
Mathematica [A] time = 2.51971, size = 205, normalized size = 0.9 \[ \frac{\sin ^{m+1}(c+d x) \left (b \sin (c+d x) \left (\frac{3 a^2 \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};\sin ^2(c+d x)\right )}{m+2}+b \left (\frac{3 a \sqrt{\cos ^2(c+d x)} \tan (c+d x) \, _2F_1\left (\frac{3}{2},\frac{m+3}{2};\frac{m+5}{2};\sin ^2(c+d x)\right )}{m+3}+\frac{b \sin ^2(c+d x) \, _2F_1\left (2,\frac{m+4}{2};\frac{m+6}{2};\sin ^2(c+d x)\right )}{m+4}\right )\right )+\frac{a^3 \sqrt{\cos ^2(c+d x)} \sec (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{m+1}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.392, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{m} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \tan \left (d x + c\right )^{3} + 3 \, a b^{2} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b \tan \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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