Optimal. Leaf size=146 \[ \frac{3 a^2 b \cos (c+d x)}{d}+\frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}+a^3 (-x)+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{9}{2} a b^2 x-\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{2 b^3 \cos (c+d x)}{d}+\frac{b^3 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.169452, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {2722, 3473, 8, 2590, 14, 2591, 288, 321, 203, 270} \[ \frac{3 a^2 b \cos (c+d x)}{d}+\frac{3 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}+a^3 (-x)+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{9}{2} a b^2 x-\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{2 b^3 \cos (c+d x)}{d}+\frac{b^3 \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 2722
Rule 3473
Rule 8
Rule 2590
Rule 14
Rule 2591
Rule 288
Rule 321
Rule 203
Rule 270
Rubi steps
\begin{align*} \int (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\int \left (a^3 \tan ^2(c+d x)+3 a^2 b \sin (c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^2(c+d x) \tan ^2(c+d x)+b^3 \sin ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^2(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin (c+d x) \tan ^2(c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{a^3 \tan (c+d x)}{d}-a^3 \int 1 \, dx-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-a^3 x+\frac{a^3 \tan (c+d x)}{d}-\frac{3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (9 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{b^3 \operatorname{Subst}\left (\int \left (-2+\frac{1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-a^3 x+\frac{3 a^2 b \cos (c+d x)}{d}+\frac{2 b^3 \cos (c+d x)}{d}-\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{3 a^2 b \sec (c+d x)}{d}+\frac{b^3 \sec (c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{\left (9 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^3 x-\frac{9}{2} a b^2 x+\frac{3 a^2 b \cos (c+d x)}{d}+\frac{2 b^3 \cos (c+d x)}{d}-\frac{b^3 \cos ^3(c+d x)}{3 d}+\frac{3 a^2 b \sec (c+d x)}{d}+\frac{b^3 \sec (c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}+\frac{9 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.729084, size = 113, normalized size = 0.77 \[ \frac{3 a \left (\left (8 a^2+27 b^2\right ) \tan (c+d x)-4 \left (2 a^2+9 b^2\right ) (c+d x)\right )+b \sec (c+d x) \left (4 \left (9 a^2+5 b^2\right ) \cos (2 (c+d x))+108 a^2+9 a b \sin (3 (c+d x))-b^2 \cos (4 (c+d x))+45 b^2\right )}{24 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 169, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +3\,{a}^{2}b \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cos \left ( dx+c \right ) \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 3.01256, size = 161, normalized size = 1.1 \begin{align*} -\frac{6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} + 9 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a b^{2} + 2 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{3} - 18 \, a^{2} b{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.48544, size = 271, normalized size = 1.86 \begin{align*} -\frac{2 \, b^{3} \cos \left (d x + c\right )^{4} + 3 \,{\left (2 \, a^{3} + 9 \, a b^{2}\right )} d x \cos \left (d x + c\right ) - 18 \, a^{2} b - 6 \, b^{3} - 6 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, a b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (c + d x \right )}\right )^{3} \tan ^{2}{\left (c + d x \right )}\, dx \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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