Optimal. Leaf size=132 \[ \frac{3 a^2 b \tan (c+d x)}{d}-\frac{3 a^2 b \cot (c+d x)}{d}+\frac{3 a^3 \sec (c+d x)}{2 d}-\frac{3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^3 \tan (c+d x)}{d} \]
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Rubi [A] time = 0.227833, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.31, Rules used = {2912, 3767, 8, 2622, 321, 207, 2620, 14, 288} \[ \frac{3 a^2 b \tan (c+d x)}{d}-\frac{3 a^2 b \cot (c+d x)}{d}+\frac{3 a^3 \sec (c+d x)}{2 d}-\frac{3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}+\frac{b^3 \tan (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 2912
Rule 3767
Rule 8
Rule 2622
Rule 321
Rule 207
Rule 2620
Rule 14
Rule 288
Rubi steps
\begin{align*} \int \csc ^3(c+d x) \sec ^2(c+d x) (a+b \sin (c+d x))^3 \, dx &=\int \left (b^3 \sec ^2(c+d x)+3 a b^2 \csc (c+d x) \sec ^2(c+d x)+3 a^2 b \csc ^2(c+d x) \sec ^2(c+d x)+a^3 \csc ^3(c+d x) \sec ^2(c+d x)\right ) \, dx\\ &=a^3 \int \csc ^3(c+d x) \sec ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \csc ^2(c+d x) \sec ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \csc (c+d x) \sec ^2(c+d x) \, dx+b^3 \int \sec ^2(c+d x) \, dx\\ &=\frac{a^3 \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{x^2} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac{3 a b^2 \sec (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{b^3 \tan (c+d x)}{d}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a^2 b \cot (c+d x)}{d}+\frac{3 a^3 \sec (c+d x)}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{3 a^2 b \tan (c+d x)}{d}+\frac{b^3 \tan (c+d x)}{d}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac{3 a^3 \tanh ^{-1}(\cos (c+d x))}{2 d}-\frac{3 a b^2 \tanh ^{-1}(\cos (c+d x))}{d}-\frac{3 a^2 b \cot (c+d x)}{d}+\frac{3 a^3 \sec (c+d x)}{2 d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{a^3 \csc ^2(c+d x) \sec (c+d x)}{2 d}+\frac{3 a^2 b \tan (c+d x)}{d}+\frac{b^3 \tan (c+d x)}{d}\\ \end{align*}
Mathematica [B] time = 0.564102, size = 267, normalized size = 2.02 \[ \frac{\csc ^4(c+d x) \left (-6 \left (a^3+2 a b^2\right ) \cos (2 (c+d x))-3 a \left (a^2+2 b^2\right ) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+12 a^2 b \sin (c+d x)-12 a^2 b \sin (3 (c+d x))+3 a^3 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-3 a^3 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 a^3+6 a b^2 \cos (3 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-6 a b^2 \cos (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a b^2+6 b^3 \sin (c+d x)-2 b^3 \sin (3 (c+d x))\right )}{2 d \left (\csc ^2\left (\frac{1}{2} (c+d x)\right )-\sec ^2\left (\frac{1}{2} (c+d x)\right )\right )} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.098, size = 161, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }}+{\frac{3\,{a}^{3}}{2\,d\cos \left ( dx+c \right ) }}+{\frac{3\,{a}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{{a}^{2}b}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-6\,{\frac{{a}^{2}b\cot \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{a{b}^{2}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.99946, size = 185, normalized size = 1.4 \begin{align*} \frac{a^{3}{\left (\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 6 \, a b^{2}{\left (\frac{2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} b{\left (\frac{1}{\tan \left (d x + c\right )} - \tan \left (d x + c\right )\right )} + 4 \, b^{3} \tan \left (d x + c\right )}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.73089, size = 479, normalized size = 3.63 \begin{align*} -\frac{4 \, a^{3} + 12 \, a b^{2} - 6 \,{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 4 \,{\left (3 \, a^{2} b + b^{3} -{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )\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 [A] time = 1.27563, size = 242, normalized size = 1.83 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \,{\left (a^{3} + 2 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{16 \,{\left (3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - \frac{18 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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