Optimal. Leaf size=152 \[ \frac {a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{4 \sqrt {b} e^{5/2}}+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+4 A b)}{6 a e^4}+\frac {(e x)^{3/2} \sqrt {a+b x^3} (a B+4 A b)}{4 e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}} \]
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Rubi [A] time = 0.10, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {453, 279, 329, 275, 217, 206} \begin {gather*} \frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2} (a B+4 A b)}{6 a e^4}+\frac {(e x)^{3/2} \sqrt {a+b x^3} (a B+4 A b)}{4 e^4}+\frac {a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{4 \sqrt {b} e^{5/2}}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 217
Rule 275
Rule 279
Rule 329
Rule 453
Rubi steps
\begin {align*} \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx &=-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(4 A b+a B) \int \sqrt {e x} \left (a+b x^3\right )^{3/2} \, dx}{a e^3}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(3 (4 A b+a B)) \int \sqrt {e x} \sqrt {a+b x^3} \, dx}{4 e^3}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(3 a (4 A b+a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^3}} \, dx}{8 e^3}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(3 a (4 A b+a B)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^6}{e^3}}} \, dx,x,\sqrt {e x}\right )}{4 e^4}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(a (4 A b+a B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^2}{e^3}}} \, dx,x,(e x)^{3/2}\right )}{4 e^4}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {(a (4 A b+a B)) \operatorname {Subst}\left (\int \frac {1}{1-\frac {b x^2}{e^3}} \, dx,x,\frac {(e x)^{3/2}}{\sqrt {a+b x^3}}\right )}{4 e^4}\\ &=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {a (4 A b+a B) \tanh ^{-1}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{4 \sqrt {b} e^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 126, normalized size = 0.83 \begin {gather*} \frac {x \sqrt {a+b x^3} \left (3 \sqrt {a} x^{3/2} (a B+4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )+\sqrt {b} \sqrt {\frac {b x^3}{a}+1} \left (-8 a A+5 a B x^3+4 A b x^3+2 b B x^6\right )\right )}{12 \sqrt {b} (e x)^{5/2} \sqrt {\frac {b x^3}{a}+1}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.76, size = 125, normalized size = 0.82 \begin {gather*} \frac {\sqrt {a+b x^3} \left (-8 a A e^6+5 a B e^6 x^3+4 A b e^6 x^3+2 b B e^6 x^6\right )}{12 e^7 (e x)^{3/2}}-\frac {\sqrt {\frac {b}{e^3}} \left (a^2 B+4 a A b\right ) \log \left (\sqrt {a+b x^3}-\sqrt {\frac {b}{e^3}} (e x)^{3/2}\right )}{4 b e} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.45, size = 255, normalized size = 1.68 \begin {gather*} \left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {b e} x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b e} \sqrt {e x}\right ) + 4 \, {\left (2 \, B b^{2} x^{6} + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b e^{3} x^{2}}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b e} x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b e} \sqrt {e x} x}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (2 \, B b^{2} x^{6} + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, b e^{3} x^{2}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 1.18, size = 7108, normalized size = 46.76 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}}}{\left (e x\right )^{\frac {5}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{3/2}}{{\left (e\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 24.32, size = 289, normalized size = 1.90 \begin {gather*} - \frac {2 A a^{\frac {3}{2}}}{3 e^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} b x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} - \frac {2 A \sqrt {a} b x^{\frac {3}{2}}}{3 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{e^{\frac {5}{2}}} + \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} + \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}}}{12 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B \sqrt {a} b x^{\frac {9}{2}}}{4 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{4 \sqrt {b} e^{\frac {5}{2}}} + \frac {B b^{2} x^{\frac {15}{2}}}{6 \sqrt {a} e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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