Optimal. Leaf size=102 \[ \frac{a^2 x^4}{4}+\frac{a b \sin \left (c+d x^2\right )}{d^2}-\frac{a b x^2 \cos \left (c+d x^2\right )}{d}+\frac{b^2 \sin ^2\left (c+d x^2\right )}{8 d^2}-\frac{b^2 x^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}+\frac{b^2 x^4}{8} \]
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Rubi [A] time = 0.133626, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3379, 3317, 3296, 2637, 3310, 30} \[ \frac{a^2 x^4}{4}+\frac{a b \sin \left (c+d x^2\right )}{d^2}-\frac{a b x^2 \cos \left (c+d x^2\right )}{d}+\frac{b^2 \sin ^2\left (c+d x^2\right )}{8 d^2}-\frac{b^2 x^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}+\frac{b^2 x^4}{8} \]
Antiderivative was successfully verified.
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Rule 3379
Rule 3317
Rule 3296
Rule 2637
Rule 3310
Rule 30
Rubi steps
\begin{align*} \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b \sin (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x+2 a b x \sin (c+d x)+b^2 x \sin ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}+(a b) \operatorname{Subst}\left (\int x \sin (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x \sin ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}-\frac{a b x^2 \cos \left (c+d x^2\right )}{d}-\frac{b^2 x^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac{b^2 \sin ^2\left (c+d x^2\right )}{8 d^2}+\frac{1}{4} b^2 \operatorname{Subst}\left (\int x \, dx,x,x^2\right )+\frac{(a b) \operatorname{Subst}\left (\int \cos (c+d x) \, dx,x,x^2\right )}{d}\\ &=\frac{a^2 x^4}{4}+\frac{b^2 x^4}{8}-\frac{a b x^2 \cos \left (c+d x^2\right )}{d}+\frac{a b \sin \left (c+d x^2\right )}{d^2}-\frac{b^2 x^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac{b^2 \sin ^2\left (c+d x^2\right )}{8 d^2}\\ \end{align*}
Mathematica [A] time = 0.220874, size = 92, normalized size = 0.9 \[ \frac{4 a^2 d^2 x^4+16 a b \sin \left (c+d x^2\right )-16 a b d x^2 \cos \left (c+d x^2\right )-2 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )+2 b^2 d^2 x^4}{16 d^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.016, size = 93, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}{x}^{4}}{4}}+{\frac{{b}^{2}{x}^{4}}{8}}-{\frac{{b}^{2}}{2} \left ({\frac{{x}^{2}\sin \left ( 2\,d{x}^{2}+2\,c \right ) }{4\,d}}+{\frac{\cos \left ( 2\,d{x}^{2}+2\,c \right ) }{8\,{d}^{2}}} \right ) }+2\,ab \left ( -1/2\,{\frac{{x}^{2}\cos \left ( d{x}^{2}+c \right ) }{d}}+1/2\,{\frac{\sin \left ( d{x}^{2}+c \right ) }{{d}^{2}}} \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01859, size = 117, normalized size = 1.15 \begin{align*} \frac{1}{4} \, a^{2} x^{4} - \frac{{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} a b}{d^{2}} + \frac{{\left (2 \, d^{2} x^{4} - 2 \, d x^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) - \cos \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{16 \, d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.97841, size = 188, normalized size = 1.84 \begin{align*} \frac{{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{4} - 8 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \,{\left (b^{2} d x^{2} \cos \left (d x^{2} + c\right ) - 4 \, a b\right )} \sin \left (d x^{2} + c\right )}{8 \, d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 2.25549, size = 136, normalized size = 1.33 \begin{align*} \begin{cases} \frac{a^{2} x^{4}}{4} - \frac{a b x^{2} \cos{\left (c + d x^{2} \right )}}{d} + \frac{a b \sin{\left (c + d x^{2} \right )}}{d^{2}} + \frac{b^{2} x^{4} \sin ^{2}{\left (c + d x^{2} \right )}}{8} + \frac{b^{2} x^{4} \cos ^{2}{\left (c + d x^{2} \right )}}{8} - \frac{b^{2} x^{2} \sin{\left (c + d x^{2} \right )} \cos{\left (c + d x^{2} \right )}}{4 d} + \frac{b^{2} \sin ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} & \text{for}\: d \neq 0 \\\frac{x^{4} \left (a + b \sin{\left (c \right )}\right )^{2}}{4} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.16025, size = 166, normalized size = 1.63 \begin{align*} \frac{\frac{4 \,{\left ({\left (d x^{2} + c\right )}^{2} - 2 \,{\left (d x^{2} + c\right )} c\right )} a^{2}}{d} - \frac{16 \,{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} a b}{d} - \frac{{\left (2 \, d x^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) - 2 \,{\left (d x^{2} + c\right )}^{2} + 4 \,{\left (d x^{2} + c\right )} c + \cos \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{d}}{16 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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