Optimal. Leaf size=189 \[ -\frac{\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac{\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{a b \sin ^2(c+d x)}{d}-\frac{\sec ^2(c+d x) (9 a \sin (c+d x)+11 b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{b^2 \sin ^3(c+d x)}{3 d} \]
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Rubi [A] time = 0.358873, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2837, 12, 1645, 1810, 633, 31} \[ -\frac{\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac{\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)}{16 d}-\frac{a b \sin ^2(c+d x)}{d}-\frac{\sec ^2(c+d x) (9 a \sin (c+d x)+11 b) (a+b \sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^2}{4 d}-\frac{b^2 \sin ^3(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 2837
Rule 12
Rule 1645
Rule 1810
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{x^6 (a+x)^2}{b^6 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^6 (a+x)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{(a+x) \left (-a b^6-3 b^6 x-4 a b^4 x^2-4 b^4 x^3-4 a b^2 x^4-4 b^2 x^5\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 b^3 d}\\ &=-\frac{\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{b^6 \left (7 a^2+11 b^2\right )+32 a b^6 x+8 b^4 \left (a^2+2 b^2\right ) x^2+16 a b^4 x^3+8 b^4 x^4}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac{\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \left (-8 b^4 \left (a^2+3 b^2\right )-16 a b^4 x-8 b^4 x^2+\frac{5 b^6 \left (3 a^2+7 b^2\right )+48 a b^6 x}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac{\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac{a b \sin ^2(c+d x)}{d}-\frac{b^2 \sin ^3(c+d x)}{3 d}-\frac{\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{5 b^6 \left (3 a^2+7 b^2\right )+48 a b^6 x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 b^5 d}\\ &=-\frac{\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac{a b \sin ^2(c+d x)}{d}-\frac{b^2 \sin ^3(c+d x)}{3 d}-\frac{\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}-\frac{\left (15 a^2-48 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac{\left (15 a^2+48 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=-\frac{\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac{\left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac{\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac{a b \sin ^2(c+d x)}{d}-\frac{b^2 \sin ^3(c+d x)}{3 d}-\frac{\sec ^2(c+d x) (11 b+9 a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac{\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d}\\ \end{align*}
Mathematica [A] time = 1.58653, size = 186, normalized size = 0.98 \[ \frac{-48 \left (a^2+3 b^2\right ) \sin (c+d x)-3 \left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 \left (15 a^2-48 a b+35 b^2\right ) \log (\sin (c+d x)+1)-48 a b \sin ^2(c+d x)+\frac{3 (a+b) (9 a+13 b)}{\sin (c+d x)-1}+\frac{3 (9 a-13 b) (a-b)}{\sin (c+d x)+1}+\frac{3 (a+b)^2}{(\sin (c+d x)-1)^2}-\frac{3 (a-b)^2}{(\sin (c+d x)+1)^2}-16 b^2 \sin ^3(c+d x)}{48 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.077, size = 355, normalized size = 1.9 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,{a}^{2} \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}{a}^{2}}{8\,d}}-{\frac{5\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,{a}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d}}-{\frac{3\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-3\,{\frac{ab \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-6\,{\frac{ab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{5\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d}}-{\frac{7\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{35\,{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24\,d}}-{\frac{35\,{b}^{2}\sin \left ( dx+c \right ) }{8\,d}}+{\frac{35\,{b}^{2}\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 = 0.996051, size = 243, normalized size = 1.29 \begin{align*} -\frac{16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} - 3 \,{\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 48 \,{\left (a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac{6 \,{\left (24 \, a b \sin \left (d x + c\right )^{2} +{\left (9 \, a^{2} + 13 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 20 \, a b -{\left (7 \, a^{2} + 11 \, 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}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.52149, size = 502, normalized size = 2.66 \begin{align*} \frac{48 \, a b \cos \left (d x + c\right )^{6} - 24 \, a b \cos \left (d x + c\right )^{4} + 3 \,{\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 144 \, a b \cos \left (d x + c\right )^{2} + 24 \, a b + 2 \,{\left (8 \, b^{2} \cos \left (d x + c\right )^{6} - 8 \,{\left (3 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 3 \,{\left (9 \, a^{2} + 13 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2} + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, 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 = 1.27115, size = 267, normalized size = 1.41 \begin{align*} -\frac{16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} + 48 \, a^{2} \sin \left (d x + c\right ) + 144 \, b^{2} \sin \left (d x + c\right ) - 3 \,{\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (36 \, a b \sin \left (d x + c\right )^{4} + 9 \, a^{2} \sin \left (d x + c\right )^{3} + 13 \, b^{2} \sin \left (d x + c\right )^{3} - 48 \, a b \sin \left (d x + c\right )^{2} - 7 \, a^{2} \sin \left (d x + c\right ) - 11 \, b^{2} \sin \left (d x + c\right ) + 16 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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