Optimal. Leaf size=90 \[ -\frac{3 b \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (a b \sin (c+d x)+b^2\right )}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \]
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Rubi [A] time = 0.111033, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2837, 12, 805, 723, 206} \[ -\frac{3 b \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (a b \sin (c+d x)+b^2\right )}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 12
Rule 805
Rule 723
Rule 206
Rubi steps
\begin{align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x (a+x)^3}{b \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^4 \operatorname{Subst}\left (\int \frac{x (a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{(a+x)^2}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (b^2+a b \sin (c+d x)\right )}{8 d}-\frac{\left (3 b^2 \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac{3 b \left (a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{3 \sec ^2(c+d x) (a+b \sin (c+d x)) \left (b^2+a b \sin (c+d x)\right )}{8 d}\\ \end{align*}
Mathematica [B] time = 1.45126, size = 370, normalized size = 4.11 \[ \frac{\frac{1}{2} \left (10 a^2 b^3+5 a^4 b+b^5\right ) \left (6 b^2 \left (6 a^2+b^2\right ) \sin (c+d x)+12 a b^3 \sin ^2(c+d x)+3 \left ((a+b)^4 \log (1-\sin (c+d x))-(a-b)^4 \log (\sin (c+d x)+1)\right )+2 b^4 \sin ^3(c+d x)\right )-a b \left (a^2+b^2\right ) \left (6 b^3 \left (10 a^2+b^2\right ) \sin ^2(c+d x)+60 a b^2 \left (2 a^2+b^2\right ) \sin (c+d x)+20 a b^4 \sin ^3(c+d x)-6 (a-b)^5 \log (\sin (c+d x)+1)+6 (a+b)^5 \log (1-\sin (c+d x))+3 b^5 \sin ^4(c+d x)\right )+2 b \left (a^2-b^2\right ) \sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^5+b \sec ^2(c+d x) \left (-3 a \left (a^2+b^2\right ) \sin (c+d x)+5 a^2 b+b^3\right ) (a+b \sin (c+d x))^5+2 a \left (a^2-b^2\right )^2 \sec ^4(c+d x) (a+b \sin (c+d x))^4}{8 d \left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.066, size = 231, normalized size = 2.6 \begin{align*}{\frac{{a}^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{2}b\sin \left ( dx+c \right ) }{8\,d}}-{\frac{3\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{8\,d}}-{\frac{3\,{b}^{3}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.993212, size = 189, normalized size = 2.1 \begin{align*} -\frac{3 \,{\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (12 \, a b^{2} \sin \left (d x + c\right )^{2} +{\left (3 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} + 3 \,{\left (a^{2} b - b^{3}\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 = 2.01338, size = 340, normalized size = 3.78 \begin{align*} -\frac{3 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 24 \, a b^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{3} - 12 \, a b^{2} - 2 \,{\left (6 \, a^{2} b + 2 \, b^{3} -{\left (3 \, a^{2} b + 5 \, b^{3}\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 = 1.36876, size = 192, normalized size = 2.13 \begin{align*} -\frac{3 \,{\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \,{\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a^{2} b \sin \left (d x + c\right )^{3} + 5 \, b^{3} \sin \left (d x + c\right )^{3} + 12 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} b \sin \left (d x + c\right ) - 3 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} - 6 \, a b^{2}\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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