Optimal. Leaf size=160 \[ \frac{b \left (15 a^2+4 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac{b \left (15 a^2+4 b^2\right ) \cos (e+f x)}{5 f}-\frac{a \left (4 a^2+9 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a x \left (4 a^2+9 b^2\right )-\frac{11 a b^2 \sin ^3(e+f x) \cos (e+f x)}{20 f}-\frac{b^2 \sin ^3(e+f x) \cos (e+f x) (a+b \sin (e+f x))}{5 f} \]
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Rubi [A] time = 0.215738, antiderivative size = 180, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2791, 2753, 2734} \[ \frac{\left (-52 a^2 b^2+3 a^4-16 b^4\right ) \cos (e+f x)}{30 b f}+\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \left (6 a^2-71 b^2\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} a x \left (4 a^2+9 b^2\right )-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f} \]
Antiderivative was successfully verified.
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Rule 2791
Rule 2753
Rule 2734
Rubi steps
\begin{align*} \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (4 b-a \sin (e+f x)) (a+b \sin (e+f x))^3 \, dx}{5 b}\\ &=\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sin (e+f x)\right ) \, dx}{20 b}\\ &=\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sin (e+f x)\right ) \, dx}{60 b}\\ &=\frac{1}{8} a \left (4 a^2+9 b^2\right ) x+\frac{\left (3 a^4-52 a^2 b^2-16 b^4\right ) \cos (e+f x)}{30 b f}+\frac{a \left (6 a^2-71 b^2\right ) \cos (e+f x) \sin (e+f x)}{120 f}+\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\\ \end{align*}
Mathematica [A] time = 0.635079, size = 117, normalized size = 0.73 \[ \frac{15 a \left (4 \left (4 a^2+9 b^2\right ) (e+f x)-8 \left (a^2+3 b^2\right ) \sin (2 (e+f x))+3 b^2 \sin (4 (e+f x))\right )-60 b \left (18 a^2+5 b^2\right ) \cos (e+f x)+10 \left (12 a^2 b+5 b^3\right ) \cos (3 (e+f x))-6 b^3 \cos (5 (e+f x))}{480 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.027, size = 124, normalized size = 0.8 \begin{align*}{\frac{1}{f} \left ( -{\frac{{b}^{3}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+3\,a{b}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{a}^{2}b \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +{a}^{3} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.73668, size = 163, normalized size = 1.02 \begin{align*} \frac{120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} + 480 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} b + 45 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2} - 32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{3}}{480 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.67325, size = 285, normalized size = 1.78 \begin{align*} -\frac{24 \, b^{3} \cos \left (f x + e\right )^{5} - 40 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} f x + 120 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right ) - 15 \,{\left (6 \, a b^{2} \cos \left (f x + e\right )^{3} -{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 3.52925, size = 284, normalized size = 1.78 \begin{align*} \begin{cases} \frac{a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{a^{3} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{3 a^{2} b \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a^{2} b \cos ^{3}{\left (e + f x \right )}}{f} + \frac{9 a b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{9 a b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{9 a b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{15 a b^{2} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{9 a b^{2} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{b^{3} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 b^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{8 b^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{3} \sin ^{2}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.71505, size = 174, normalized size = 1.09 \begin{align*} -\frac{b^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{3 \, a b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} x + \frac{{\left (12 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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