Optimal. Leaf size=112 \[ \frac{\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac{\left (a^2+b^2\right ) \cos (e+f x)}{f}-\frac{a b \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac{3 a b \sin (e+f x) \cos (e+f x)}{4 f}+\frac{3 a b x}{4}-\frac{b^2 \cos ^5(e+f x)}{5 f} \]
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Rubi [A] time = 0.104428, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 2635, 8, 3013, 373} \[ \frac{\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac{\left (a^2+b^2\right ) \cos (e+f x)}{f}-\frac{a b \sin ^3(e+f x) \cos (e+f x)}{2 f}-\frac{3 a b \sin (e+f x) \cos (e+f x)}{4 f}+\frac{3 a b x}{4}-\frac{b^2 \cos ^5(e+f x)}{5 f} \]
Antiderivative was successfully verified.
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Rule 2789
Rule 2635
Rule 8
Rule 3013
Rule 373
Rubi steps
\begin{align*} \int \sin ^3(e+f x) (a+b \sin (e+f x))^2 \, dx &=(2 a b) \int \sin ^4(e+f x) \, dx+\int \sin ^3(e+f x) \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{a b \cos (e+f x) \sin ^3(e+f x)}{2 f}+\frac{1}{2} (3 a b) \int \sin ^2(e+f x) \, dx-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a^2+b^2-b^2 x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{3 a b \cos (e+f x) \sin (e+f x)}{4 f}-\frac{a b \cos (e+f x) \sin ^3(e+f x)}{2 f}+\frac{1}{4} (3 a b) \int 1 \, dx-\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1+\frac{b^2}{a^2}\right )-\left (a^2+2 b^2\right ) x^2+b^2 x^4\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=\frac{3 a b x}{4}-\frac{\left (a^2+b^2\right ) \cos (e+f x)}{f}+\frac{\left (a^2+2 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac{b^2 \cos ^5(e+f x)}{5 f}-\frac{3 a b \cos (e+f x) \sin (e+f x)}{4 f}-\frac{a b \cos (e+f x) \sin ^3(e+f x)}{2 f}\\ \end{align*}
Mathematica [A] time = 0.34122, size = 91, normalized size = 0.81 \[ \frac{-30 \left (6 a^2+5 b^2\right ) \cos (e+f x)+5 \left (4 a^2+5 b^2\right ) \cos (3 (e+f x))-3 b (b \cos (5 (e+f x))-5 a (12 (e+f x)-8 \sin (2 (e+f x))+\sin (4 (e+f x))))}{240 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.023, size = 95, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{{b}^{2}\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 ) }+2\,ab \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 ) -{\frac{{a}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.85758, size = 127, normalized size = 1.13 \begin{align*} \frac{80 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} + 15 \,{\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 - 16 \,{\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^{2}}{240 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.912, size = 232, normalized size = 2.07 \begin{align*} -\frac{12 \, b^{2} \cos \left (f x + e\right )^{5} - 45 \, a b f x - 20 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 60 \,{\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right ) - 15 \,{\left (2 \, a b \cos \left (f x + e\right )^{3} - 5 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{60 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 3.37078, size = 221, normalized size = 1.97 \begin{align*} \begin{cases} - \frac{a^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 a b x \sin ^{4}{\left (e + f x \right )}}{4} + \frac{3 a b x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} + \frac{3 a b x \cos ^{4}{\left (e + f x \right )}}{4} - \frac{5 a b \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{4 f} - \frac{3 a b \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} - \frac{b^{2} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 b^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{8 b^{2} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{2} \sin ^{3}{\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 = 1.56763, size = 173, normalized size = 1.54 \begin{align*} \frac{3}{4} \, a b x - \frac{b^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{a b \sin \left (4 \, f x + 4 \, e\right )}{16 \, f} - \frac{a b \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} + \frac{{\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (2 \, a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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