Optimal. Leaf size=205 \[ -\frac{a^2 b \sin ^3(c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{a^3 \cos (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{5 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
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Rubi [A] time = 0.188873, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3517, 2633, 2592, 302, 206, 2590, 270, 288} \[ -\frac{a^2 b \sin ^3(c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{a^3 \cos (c+d x)}{d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{5 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]
Antiderivative was successfully verified.
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Rule 3517
Rule 2633
Rule 2592
Rule 302
Rule 206
Rule 2590
Rule 270
Rule 288
Rubi steps
\begin{align*} \int \sin ^3(c+d x) (a+b \tan (c+d x))^3 \, dx &=\int \left (a^3 \sin ^3(c+d x)+3 a^2 b \sin ^3(c+d x) \tan (c+d x)+3 a b^2 \sin ^3(c+d x) \tan ^2(c+d x)+b^3 \sin ^3(c+d x) \tan ^3(c+d x)\right ) \, dx\\ &=a^3 \int \sin ^3(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin ^3(c+d x) \tan (c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^3(c+d x) \, dx\\ &=-\frac{a^3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{a^3 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \left (-2+\frac{1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=-\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}-\frac{a^2 b \sin ^3(c+d x)}{d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{5 b^3 \sin (c+d x)}{2 d}-\frac{a^2 b \sin ^3(c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{2 d}\\ &=\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{5 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^3 \cos (c+d x)}{d}+\frac{6 a b^2 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{a b^2 \cos ^3(c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}-\frac{3 a^2 b \sin (c+d x)}{d}+\frac{5 b^3 \sin (c+d x)}{2 d}-\frac{a^2 b \sin ^3(c+d x)}{d}+\frac{5 b^3 \sin ^3(c+d x)}{6 d}+\frac{b^3 \sin ^3(c+d x) \tan ^2(c+d x)}{2 d}\\ \end{align*}
Mathematica [B] time = 6.26557, size = 771, normalized size = 3.76 \[ -\frac{3 a \left (a^2-7 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{b \left (3 a^2-b^2\right ) \sin (3 (c+d x)) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{12 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{a \left (a^2-3 b^2\right ) \cos (3 (c+d x)) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{12 d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{3 b \left (5 a^2-3 b^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{\left (5 b^3-6 a^2 b\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{\left (6 a^2 b-5 b^3\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a b^2 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.053, size = 271, normalized size = 1.3 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d}}+{\frac{5\,{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d}}+{\frac{5\,{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{5\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+8\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) }{d}}+3\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+4\,{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}-{\frac{b{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}-3\,{\frac{b{a}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{b{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{3}}{3\,d}}-{\frac{2\,{a}^{3}\cos \left ( dx+c \right ) }{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.03738, size = 234, normalized size = 1.14 \begin{align*} \frac{4 \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{3} - 6 \,{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{2} b - 12 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a b^{2} +{\left (4 \, \sin \left (d x + c\right )^{3} - \frac{6 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, \sin \left (d x + c\right )\right )} b^{3}}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.39564, size = 455, normalized size = 2.22 \begin{align*} \frac{4 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 36 \, a b^{2} \cos \left (d x + c\right ) - 12 \,{\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (6 \, a^{2} b - 5 \, b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, b^{3} - 2 \,{\left (12 \, a^{2} b - 7 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \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(-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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