Optimal. Leaf size=220 \[ -\frac{3 a^2 b \cos (c+d x)}{d}+\frac{a^2 b \sec ^3(c+d x)}{d}-\frac{6 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}+a^3 x+\frac{5 a b^2 \tan ^3(c+d x)}{2 d}-\frac{15 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{15}{2} a b^2 x+\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{3 b^3 \cos (c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{3 b^3 \sec (c+d x)}{d} \]
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Rubi [A] time = 0.224299, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2722, 3473, 8, 2590, 270, 2591, 288, 302, 203} \[ -\frac{3 a^2 b \cos (c+d x)}{d}+\frac{a^2 b \sec ^3(c+d x)}{d}-\frac{6 a^2 b \sec (c+d x)}{d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}+a^3 x+\frac{5 a b^2 \tan ^3(c+d x)}{2 d}-\frac{15 a b^2 \tan (c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{15}{2} a b^2 x+\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{3 b^3 \cos (c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{3 b^3 \sec (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 2722
Rule 3473
Rule 8
Rule 2590
Rule 270
Rule 2591
Rule 288
Rule 302
Rule 203
Rubi steps
\begin{align*} \int (a+b \sin (c+d x))^3 \tan ^4(c+d x) \, dx &=\int \left (a^3 \tan ^4(c+d x)+3 a^2 b \sin (c+d x) \tan ^4(c+d x)+3 a b^2 \sin ^2(c+d x) \tan ^4(c+d x)+b^3 \sin ^3(c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^4(c+d x) \, dx+\left (3 a^2 b\right ) \int \sin (c+d x) \tan ^4(c+d x) \, dx+\left (3 a b^2\right ) \int \sin ^2(c+d x) \tan ^4(c+d x) \, dx+b^3 \int \sin ^3(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a^3 \tan ^3(c+d x)}{3 d}-a^3 \int \tan ^2(c+d x) \, dx-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^6}{\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 )^3}{x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a^3 \tan (c+d x)}{d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+a^3 \int 1 \, dx-\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}-\frac{2}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{b^3 \operatorname{Subst}\left (\int \left (3+\frac{1}{x^4}-\frac{3}{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{3 b^3 \cos (c+d x)}{d}+\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{6 a^2 b \sec (c+d x)}{d}-\frac{3 b^3 \sec (c+d x)}{d}+\frac{a^2 b \sec ^3(c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}+\frac{a^3 \tan ^3(c+d x)}{3 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{\left (15 a b^2\right ) \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=a^3 x-\frac{3 a^2 b \cos (c+d x)}{d}-\frac{3 b^3 \cos (c+d x)}{d}+\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{6 a^2 b \sec (c+d x)}{d}-\frac{3 b^3 \sec (c+d x)}{d}+\frac{a^2 b \sec ^3(c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}-\frac{15 a b^2 \tan (c+d x)}{2 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{5 a b^2 \tan ^3(c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{\left (15 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{15}{2} a b^2 x-\frac{3 a^2 b \cos (c+d x)}{d}-\frac{3 b^3 \cos (c+d x)}{d}+\frac{b^3 \cos ^3(c+d x)}{3 d}-\frac{6 a^2 b \sec (c+d x)}{d}-\frac{3 b^3 \sec (c+d x)}{d}+\frac{a^2 b \sec ^3(c+d x)}{d}+\frac{b^3 \sec ^3(c+d x)}{3 d}-\frac{a^3 \tan (c+d x)}{d}-\frac{15 a b^2 \tan (c+d x)}{2 d}+\frac{a^3 \tan ^3(c+d x)}{3 d}+\frac{5 a b^2 \tan ^3(c+d x)}{2 d}-\frac{3 a b^2 \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}\\ \end{align*}
Mathematica [A] time = 0.678412, size = 226, normalized size = 1.03 \[ \frac{\sec ^3(c+d x) \left (36 a \left (2 a^2+15 b^2\right ) (c+d x) \cos (c+d x)-3 \left (144 a^2 b+91 b^3\right ) \cos (2 (c+d x))-36 a^2 b \cos (4 (c+d x))-300 a^2 b-32 a^3 \sin (3 (c+d x))+24 a^3 c \cos (3 (c+d x))+24 a^3 d x \cos (3 (c+d x))-90 a b^2 \sin (c+d x)-195 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+180 a b^2 c \cos (3 (c+d x))+180 a b^2 d x \cos (3 (c+d x))-30 b^3 \cos (4 (c+d x))+b^3 \cos (6 (c+d x))-210 b^3\right )}{96 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 268, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) +3\,{a}^{2}b \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}- \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-4/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}-4/3\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) +5/2\,dx+5/2\,c \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{3\,\cos \left ( dx+c \right ) }}-{\frac{5\,\cos \left ( dx+c \right ) }{3} \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.82353, size = 225, normalized size = 1.02 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \,{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac{3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a b^{2} + 2 \,{\left (\cos \left (d x + c\right )^{3} - \frac{9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} b^{3} - 6 \, a^{2} b{\left (\frac{6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \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 = 1.59267, size = 369, normalized size = 1.68 \begin{align*} \frac{2 \, b^{3} \cos \left (d x + c\right )^{6} + 3 \,{\left (2 \, a^{3} + 15 \, a b^{2}\right )} d x \cos \left (d x + c\right )^{3} - 18 \,{\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} + 6 \, a^{2} b + 2 \, b^{3} - 18 \,{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} -{\left (9 \, a b^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{3} - 6 \, a b^{2} + 2 \,{\left (4 \, a^{3} + 21 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \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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