Optimal. Leaf size=163 \[ \frac {a^2 x^6}{6}+\frac {2 a b \cos \left (c+d x^2\right )}{d^3}+\frac {2 a b x^2 \sin \left (c+d x^2\right )}{d^2}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {b^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{8 d^3}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}-\frac {b^2 x^4 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}-\frac {b^2 x^2}{8 d^2}+\frac {b^2 x^6}{12} \]
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Rubi [A] time = 0.25, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3379, 3317, 3296, 2638, 3311, 30, 2635, 8} \[ \frac {a^2 x^6}{6}+\frac {2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac {2 a b \cos \left (c+d x^2\right )}{d^3}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}+\frac {b^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{8 d^3}-\frac {b^2 x^4 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{4 d}-\frac {b^2 x^2}{8 d^2}+\frac {b^2 x^6}{12} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2635
Rule 2638
Rule 3296
Rule 3311
Rule 3317
Rule 3379
Rubi steps
\begin {align*} \int x^5 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 (a+b \sin (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (a^2 x^2+2 a b x^2 \sin (c+d x)+b^2 x^2 \sin ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac {a^2 x^6}{6}+(a b) \operatorname {Subst}\left (\int x^2 \sin (c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int x^2 \sin ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac {a^2 x^6}{6}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}-\frac {b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}+\frac {1}{4} b^2 \operatorname {Subst}\left (\int x^2 \, dx,x,x^2\right )-\frac {b^2 \operatorname {Subst}\left (\int \sin ^2(c+d x) \, dx,x,x^2\right )}{4 d^2}+\frac {(2 a b) \operatorname {Subst}\left (\int x \cos (c+d x) \, dx,x,x^2\right )}{d}\\ &=\frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac {b^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{8 d^3}-\frac {b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}-\frac {(2 a b) \operatorname {Subst}\left (\int \sin (c+d x) \, dx,x,x^2\right )}{d^2}-\frac {b^2 \operatorname {Subst}\left (\int 1 \, dx,x,x^2\right )}{8 d^2}\\ &=-\frac {b^2 x^2}{8 d^2}+\frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}+\frac {2 a b \cos \left (c+d x^2\right )}{d^3}-\frac {a b x^4 \cos \left (c+d x^2\right )}{d}+\frac {2 a b x^2 \sin \left (c+d x^2\right )}{d^2}+\frac {b^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{8 d^3}-\frac {b^2 x^4 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac {b^2 x^2 \sin ^2\left (c+d x^2\right )}{4 d^2}\\ \end {align*}
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Mathematica [A] time = 0.39, size = 122, normalized size = 0.75 \[ \frac {8 a^2 d^3 x^6-48 a b \left (d^2 x^4-2\right ) \cos \left (c+d x^2\right )+96 a b d x^2 \sin \left (c+d x^2\right )-6 b^2 d^2 x^4 \sin \left (2 \left (c+d x^2\right )\right )+3 b^2 \sin \left (2 \left (c+d x^2\right )\right )-6 b^2 d x^2 \cos \left (2 \left (c+d x^2\right )\right )+4 b^2 d^3 x^6}{48 d^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.70, size = 121, normalized size = 0.74 \[ \frac {2 \, {\left (2 \, a^{2} + b^{2}\right )} d^{3} x^{6} - 6 \, b^{2} d x^{2} \cos \left (d x^{2} + c\right )^{2} + 3 \, b^{2} d x^{2} - 24 \, {\left (a b d^{2} x^{4} - 2 \, a b\right )} \cos \left (d x^{2} + c\right ) + 3 \, {\left (16 \, a b d x^{2} - {\left (2 \, b^{2} d^{2} x^{4} - b^{2}\right )} \cos \left (d x^{2} + c\right )\right )} \sin \left (d x^{2} + c\right )}{24 \, d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.61, size = 181, normalized size = 1.11 \[ \frac {8 \, a^{2} d x^{6} + 48 \, {\left (\frac {2 \, x^{2} \sin \left (d x^{2} + c\right )}{d} - \frac {{\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c + c^{2} - 2\right )} \cos \left (d x^{2} + c\right )}{d^{2}}\right )} a b - {\left (\frac {6 \, x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right )}{d} + \frac {3 \, {\left (2 \, {\left (d x^{2} + c\right )}^{2} - 4 \, {\left (d x^{2} + c\right )} c + 2 \, c^{2} - 1\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d^{2}} - \frac {4 \, {\left ({\left (d x^{2} + c\right )}^{3} - 3 \, {\left (d x^{2} + c\right )}^{2} c + 3 \, {\left (d x^{2} + c\right )} c^{2}\right )}}{d^{2}}\right )} b^{2}}{48 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 140, normalized size = 0.86 \[ \frac {a^{2} x^{6}}{6}+\frac {b^{2} x^{6}}{12}-\frac {b^{2} \left (\frac {x^{4} \sin \left (2 d \,x^{2}+2 c \right )}{4 d}-\frac {-\frac {x^{2} \cos \left (2 d \,x^{2}+2 c \right )}{4 d}+\frac {\sin \left (2 d \,x^{2}+2 c \right )}{8 d^{2}}}{d}\right )}{2}+2 a b \left (-\frac {x^{4} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\frac {x^{2} \sin \left (d \,x^{2}+c \right )}{d}+\frac {\cos \left (d \,x^{2}+c \right )}{d^{2}}}{d}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.87, size = 106, normalized size = 0.65 \[ \frac {1}{6} \, a^{2} x^{6} + \frac {{\left (2 \, d x^{2} \sin \left (d x^{2} + c\right ) - {\left (d^{2} x^{4} - 2\right )} \cos \left (d x^{2} + c\right )\right )} a b}{d^{3}} + \frac {{\left (4 \, d^{3} x^{6} - 6 \, d x^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 3 \, {\left (2 \, d^{2} x^{4} - 1\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{48 \, d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.39, size = 149, normalized size = 0.91 \[ \frac {\frac {3\,b^2\,\sin \left (2\,d\,x^2+2\,c\right )}{2}-96\,a\,b\,{\sin \left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2+4\,a^2\,d^3\,x^6+2\,b^2\,d^3\,x^6+3\,b^2\,d\,x^2\,\left (2\,{\sin \left (d\,x^2+c\right )}^2-1\right )-3\,b^2\,d^2\,x^4\,\sin \left (2\,d\,x^2+2\,c\right )+24\,a\,b\,d^2\,x^4\,\left (2\,{\sin \left (\frac {d\,x^2}{2}+\frac {c}{2}\right )}^2-1\right )+48\,a\,b\,d\,x^2\,\sin \left (d\,x^2+c\right )}{24\,d^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 6.79, size = 209, normalized size = 1.28 \[ \begin {cases} \frac {a^{2} x^{6}}{6} - \frac {a b x^{4} \cos {\left (c + d x^{2} \right )}}{d} + \frac {2 a b x^{2} \sin {\left (c + d x^{2} \right )}}{d^{2}} + \frac {2 a b \cos {\left (c + d x^{2} \right )}}{d^{3}} + \frac {b^{2} x^{6} \sin ^{2}{\left (c + d x^{2} \right )}}{12} + \frac {b^{2} x^{6} \cos ^{2}{\left (c + d x^{2} \right )}}{12} - \frac {b^{2} x^{4} \sin {\left (c + d x^{2} \right )} \cos {\left (c + d x^{2} \right )}}{4 d} + \frac {b^{2} x^{2} \sin ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} - \frac {b^{2} x^{2} \cos ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} + \frac {b^{2} \sin {\left (c + d x^{2} \right )} \cos {\left (c + d x^{2} \right )}}{8 d^{3}} & \text {for}\: d \neq 0 \\\frac {x^{6} \left (a + b \sin {\relax (c )}\right )^{2}}{6} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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