Optimal. Leaf size=163 \[ \frac {(2 b c-5 a d) \sqrt {c+d x^3}}{3 b^3}+\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}} \]
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Rubi [A]
time = 0.09, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 79, 52, 65,
214} \begin {gather*} -\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}}+\frac {\sqrt {c+d x^3} (2 b c-5 a d)}{3 b^3}+\frac {\left (c+d x^3\right )^{3/2} (2 b c-5 a d)}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \end {gather*}
Antiderivative was successfully verified.
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Rule 52
Rule 65
Rule 79
Rule 214
Rule 457
Rubi steps
\begin {align*} \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{\left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x (c+d x)^{3/2}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {(2 b c-5 a d) \text {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )}{6 b (b c-a d)}\\ &=\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {(2 b c-5 a d) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b^2}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^3}}{3 b^3}+\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {((2 b c-5 a d) (b c-a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{6 b^3}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^3}}{3 b^3}+\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}+\frac {((2 b c-5 a d) (b c-a d)) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 b^3 d}\\ &=\frac {(2 b c-5 a d) \sqrt {c+d x^3}}{3 b^3}+\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}}\\ \end {align*}
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Mathematica [A]
time = 0.25, size = 125, normalized size = 0.77 \begin {gather*} \frac {\sqrt {c+d x^3} \left (-15 a^2 d+a b \left (11 c-10 d x^3\right )+2 b^2 x^3 \left (4 c+d x^3\right )\right )}{9 b^3 \left (a+b x^3\right )}-\frac {(2 b c-5 a d) \sqrt {-b c+a d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.43, size = 983, normalized size = 6.03
method | result | size |
elliptic | \(-\frac {a \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{3} \left (b \,x^{3}+a \right )}+\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9 b^{2}}+\frac {2 \left (-\frac {2 d \left (a d -b c \right )}{b^{3}}-\frac {2 c d}{3 b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-5 a^{2} d^{2}+7 a b c d -2 b^{2} c^{2}\right ) \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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 {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{6 b^{3} d^{2}}\) | \(539\) |
risch | \(\text {Expression too large to display}\) | \(964\) |
default | \(\text {Expression too large to display}\) | \(983\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 2.11, size = 314, normalized size = 1.93 \begin {gather*} \left [-\frac {3 \, {\left ({\left (2 \, b^{2} c - 5 \, a b d\right )} x^{3} + 2 \, a b c - 5 \, a^{2} d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (2 \, b^{2} d x^{6} + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{3} + 11 \, a b c - 15 \, a^{2} d\right )} \sqrt {d x^{3} + c}}{18 \, {\left (b^{4} x^{3} + a b^{3}\right )}}, -\frac {3 \, {\left ({\left (2 \, b^{2} c - 5 \, a b d\right )} x^{3} + 2 \, a b c - 5 \, a^{2} d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (2 \, b^{2} d x^{6} + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{3} + 11 \, a b c - 15 \, a^{2} d\right )} \sqrt {d x^{3} + c}}{9 \, {\left (b^{4} x^{3} + a b^{3}\right )}}\right ] \end {gather*}
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 \begin {gather*} \int \frac {x^{5} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.20, size = 173, normalized size = 1.06 \begin {gather*} \frac {{\left (2 \, b^{2} c^{2} - 7 \, a b c d + 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{3}} + \frac {\sqrt {d x^{3} + c} a b c d - \sqrt {d x^{3} + c} a^{2} d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{4} + 3 \, \sqrt {d x^{3} + c} b^{4} c - 6 \, \sqrt {d x^{3} + c} a b^{3} d\right )}}{9 \, b^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 7.38, size = 229, normalized size = 1.40 \begin {gather*} \frac {2\,d\,x^3\,\sqrt {d\,x^3+c}}{9\,b^2}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {2\,a\,d^2}{b^3}+\frac {4\,c\,d}{3\,b^2}\right )}{3\,d}+\frac {a\,\left (\frac {2\,b\,c^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {a\,\left (\frac {2\,a\,d^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {4\,b\,c\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )}{b}\right )\,\sqrt {d\,x^3+c}}{b\,\left (b\,x^3+a\right )}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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