Optimal. Leaf size=78 \[ \frac {32 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3}-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]
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Rubi [A] time = 0.07, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {446, 88, 63, 203} \[ \frac {32 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3}-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]
Antiderivative was successfully verified.
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Rule 63
Rule 88
Rule 203
Rule 446
Rubi steps
\begin {align*} \int \frac {x^8}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {5 c}{d^2 \sqrt {c+d x}}+\frac {\sqrt {c+d x}}{d^2}+\frac {16 c^2}{d^2 \sqrt {c+d x} (4 c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (16 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (32 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{3 c+x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d^3}\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {32 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 65, normalized size = 0.83 \[ \frac {32 \sqrt {3} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )+2 \left (d x^3-14 c\right ) \sqrt {c+d x^3}}{9 d^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 129, normalized size = 1.65 \[ \left [\frac {2 \, {\left (8 \, \sqrt {3} \sqrt {-c} c \log \left (\frac {d x^{3} + 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 14 \, c\right )}\right )}}{9 \, d^{3}}, \frac {2 \, {\left (16 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 14 \, c\right )}\right )}}{9 \, d^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 64, normalized size = 0.82 \[ \frac {32 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{9 \, d^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{6} - 15 \, \sqrt {d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.24, size = 467, normalized size = 5.99 \[ \frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, x^{3}}{9 d}-\frac {4 \sqrt {d \,x^{3}+c}\, c}{9 d^{2}}\right ) d -\frac {8 \sqrt {d \,x^{3}+c}\, c}{3 d}}{d^{2}}-\frac {16 i c \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )}{6 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{9 d^{5} \sqrt {d \,x^{3}+c}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.34, size = 53, normalized size = 0.68 \[ \frac {2 \, {\left (16 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} - 15 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.38, size = 88, normalized size = 1.13 \[ \frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^2}-\frac {28\,c\,\sqrt {d\,x^3+c}}{9\,d^3}+\frac {\sqrt {3}\,c^{3/2}\,\ln \left (\frac {\sqrt {3}\,d\,x^3-2\,\sqrt {3}\,c+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,16{}\mathrm {i}}{9\,d^3} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\sqrt {c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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