Optimal. Leaf size=142 \[ -\frac{3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac{3 b (3 a-5 b) (a-b) \log (\sin (c+d x)+1)}{16 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac{15 b^3 \sin (c+d x)}{8 d} \]
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Rubi [A] time = 0.303608, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2837, 12, 1645, 774, 633, 31} \[ -\frac{3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac{3 b (3 a-5 b) (a-b) \log (\sin (c+d x)+1)}{16 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac{15 b^3 \sin (c+d x)}{8 d} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 12
Rule 1645
Rule 774
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sec ^2(c+d x) (a+b \sin (c+d x))^3 \tan ^3(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^3 (a+x)^3}{b^3 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^2 \operatorname{Subst}\left (\int \frac{x^3 (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{\operatorname{Subst}\left (\int \frac{(a+x)^2 \left (-3 b^4-4 a b^2 x-4 b^2 x^2\right )}{\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{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+x) \left (9 a b^4+15 b^4 x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d}\\ &=-\frac{15 b^3 \sin (c+d x)}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac{\operatorname{Subst}\left (\int \frac{-9 a^2 b^4-15 b^6-24 a b^4 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^2 d}\\ &=-\frac{15 b^3 \sin (c+d x)}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}-\frac{(3 (3 a-5 b) (a-b) b) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{(3 b (a+b) (3 a+5 b)) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac{3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))}{16 d}+\frac{3 (3 a-5 b) (a-b) b \log (1+\sin (c+d x))}{16 d}-\frac{15 b^3 \sin (c+d x)}{8 d}+\frac{\sec ^4(c+d x) (a+b \sin (c+d x))^3}{4 d}-\frac{\sec ^2(c+d x) (a+b \sin (c+d x))^2 (4 a+7 b \sin (c+d x))}{8 d}\\ \end{align*}
Mathematica [A] time = 0.452349, size = 147, normalized size = 1.04 \[ \frac{\frac{(a-b)^3}{(\sin (c+d x)+1)^2}-\frac{3 (a-3 b) (a-b)^2}{\sin (c+d x)+1}+\frac{3 (a+b)^2 (a+3 b)}{\sin (c+d x)-1}+\frac{(a+b)^3}{(\sin (c+d x)-1)^2}+3 b (3 a-5 b) (a-b) \log (\sin (c+d x)+1)-3 b (a+b) (3 a+5 b) \log (1-\sin (c+d x))-16 b^3 \sin (c+d x)}{16 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.085, size = 297, normalized size = 2.1 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,{a}^{2}b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{9\,{a}^{2}b\sin \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-3\,{\frac{a{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}{b}^{3}}{8\,d}}-{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{8\,d}}-{\frac{15\,{b}^{3}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{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 = 0.990961, size = 234, normalized size = 1.65 \begin{align*} -\frac{16 \, b^{3} \sin \left (d x + c\right ) - 3 \,{\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (5 \, a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} - 2 \, a^{3} - 18 \, a b^{2} + 4 \,{\left (a^{3} + 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{2} -{\left (9 \, a^{2} b + 7 \, 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 = 2.14794, size = 423, normalized size = 2.98 \begin{align*} \frac{3 \,{\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{3} + 12 \, a b^{2} - 8 \,{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (8 \, b^{3} \cos \left (d x + c\right )^{4} - 6 \, a^{2} b - 2 \, b^{3} + 3 \,{\left (5 \, a^{2} b + 3 \, 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 = 1.26843, size = 254, normalized size = 1.79 \begin{align*} -\frac{16 \, b^{3} \sin \left (d x + c\right ) - 3 \,{\left (3 \, a^{2} b - 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (3 \, a^{2} b + 8 \, a b^{2} + 5 \, b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (18 \, a b^{2} \sin \left (d x + c\right )^{4} + 15 \, a^{2} b \sin \left (d x + c\right )^{3} + 9 \, b^{3} \sin \left (d x + c\right )^{3} + 4 \, a^{3} \sin \left (d x + c\right )^{2} - 12 \, a b^{2} \sin \left (d x + c\right )^{2} - 9 \, a^{2} b \sin \left (d x + c\right ) - 7 \, b^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\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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