Optimal. Leaf size=144 \[ \frac{\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac{\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{\sec ^2(c+d x) \left (\left (a^2+4 b^2\right ) \sin (c+d x)+5 a b\right ) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.252769, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2837, 12, 1645, 819, 633, 31} \[ \frac{\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac{\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{\sec ^2(c+d x) \left (\left (a^2+4 b^2\right ) \sin (c+d x)+5 a b\right ) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 12
Rule 1645
Rule 819
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^2 (a+x)^3}{b^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \frac{x^2 (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^2 \left (-a b^2-4 b^2 x\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{a b^2 \left (a^2-9 b^2\right )-8 b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b d}\\ &=-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a^3-9 a b^2-8 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (a^3-9 a b^2+8 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac{\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac{\left (a^3-9 a b^2+8 b^3\right ) \log (1+\sin (c+d x))}{16 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x)) \left (5 a b+\left (a^2+4 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 0.393818, size = 140, normalized size = 0.97 \[ \frac{\left (a^3-9 a b^2-8 b^3\right ) \log (1-\sin (c+d x))-\left (a^3-9 a b^2+8 b^3\right ) \log (\sin (c+d x)+1)-\frac{(a-b)^3}{(\sin (c+d x)+1)^2}+\frac{(a-7 b) (a-b)^2}{\sin (c+d x)+1}+\frac{(a+b)^2 (a+7 b)}{\sin (c+d x)-1}+\frac{(a+b)^3}{(\sin (c+d x)-1)^2}}{16 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.074, size = 263, normalized size = 1.8 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{3}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{9\,a{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{9\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01877, size = 204, normalized size = 1.42 \begin{align*} -\frac{{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (a^{3} + 15 \, a b^{2}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{2} b - 6 \, b^{3} + 4 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} +{\left (a^{3} - 9 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.19204, size = 375, normalized size = 2.6 \begin{align*} -\frac{{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 12 \, a^{2} b - 4 \, b^{3} + 8 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (2 \, a^{3} + 6 \, a b^{2} -{\left (a^{3} + 15 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \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 = 1.23294, size = 227, normalized size = 1.58 \begin{align*} -\frac{{\left (a^{3} - 9 \, a b^{2} + 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (a^{3} - 9 \, a b^{2} - 8 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, b^{3} \sin \left (d x + c\right )^{4} + a^{3} \sin \left (d x + c\right )^{3} + 15 \, a b^{2} \sin \left (d x + c\right )^{3} + 12 \, a^{2} b \sin \left (d x + c\right )^{2} - 4 \, b^{3} \sin \left (d x + c\right )^{2} + a^{3} \sin \left (d x + c\right ) - 9 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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