3.2017 \(\int \frac {x^5}{\sqrt {a+\frac {b}{x^3}}} \, dx\)

Optimal. Leaf size=74 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {b x^3 \sqrt {a+\frac {b}{x^3}}}{4 a^2}+\frac {x^6 \sqrt {a+\frac {b}{x^3}}}{6 a} \]

[Out]

1/4*b^2*arctanh((a+b/x^3)^(1/2)/a^(1/2))/a^(5/2)-1/4*b*x^3*(a+b/x^3)^(1/2)/a^2+1/6*x^6*(a+b/x^3)^(1/2)/a

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {b x^3 \sqrt {a+\frac {b}{x^3}}}{4 a^2}+\frac {x^6 \sqrt {a+\frac {b}{x^3}}}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b/x^3],x]

[Out]

-(b*Sqrt[a + b/x^3]*x^3)/(4*a^2) + (Sqrt[a + b/x^3]*x^6)/(6*a) + (b^2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(4*a^(
5/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {a+\frac {b}{x^3}}} \, dx &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )\right )\\ &=\frac {\sqrt {a+\frac {b}{x^3}} x^6}{6 a}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )}{4 a}\\ &=-\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{4 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^6}{6 a}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )}{8 a^2}\\ &=-\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{4 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^6}{6 a}-\frac {b \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right )}{4 a^2}\\ &=-\frac {b \sqrt {a+\frac {b}{x^3}} x^3}{4 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^6}{6 a}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 97, normalized size = 1.31 \[ \frac {\sqrt {a} x^{3/2} \left (2 a^2 x^6-a b x^3-3 b^2\right )+3 b^2 \sqrt {a x^3+b} \tanh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {a x^3+b}}\right )}{12 a^{5/2} x^{3/2} \sqrt {a+\frac {b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a + b/x^3],x]

[Out]

(Sqrt[a]*x^(3/2)*(-3*b^2 - a*b*x^3 + 2*a^2*x^6) + 3*b^2*Sqrt[b + a*x^3]*ArcTanh[(Sqrt[a]*x^(3/2))/Sqrt[b + a*x
^3]])/(12*a^(5/2)*Sqrt[a + b/x^3]*x^(3/2))

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fricas [A]  time = 1.14, size = 179, normalized size = 2.42 \[ \left [\frac {3 \, \sqrt {a} b^{2} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) + 4 \, {\left (2 \, a^{2} x^{6} - 3 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{48 \, a^{3}}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) - 2 \, {\left (2 \, a^{2} x^{6} - 3 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{24 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b/x^3)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(3*sqrt(a)*b^2*log(-8*a^2*x^6 - 8*a*b*x^3 - b^2 - 4*(2*a*x^6 + b*x^3)*sqrt(a)*sqrt((a*x^3 + b)/x^3)) + 4
*(2*a^2*x^6 - 3*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a^3, -1/24*(3*sqrt(-a)*b^2*arctan(2*sqrt(-a)*x^3*sqrt((a*x^3 +
b)/x^3)/(2*a*x^3 + b)) - 2*(2*a^2*x^6 - 3*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a^3]

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giac [A]  time = 0.23, size = 99, normalized size = 1.34 \[ -\frac {1}{12} \, b^{2} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {\frac {a x^{3} + b}{x^{3}}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {5 \, a \sqrt {\frac {a x^{3} + b}{x^{3}}} - \frac {3 \, {\left (a x^{3} + b\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{x^{3}}}{{\left (a - \frac {a x^{3} + b}{x^{3}}\right )}^{2} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b/x^3)^(1/2),x, algorithm="giac")

[Out]

-1/12*b^2*(3*arctan(sqrt((a*x^3 + b)/x^3)/sqrt(-a))/(sqrt(-a)*a^2) - (5*a*sqrt((a*x^3 + b)/x^3) - 3*(a*x^3 + b
)*sqrt((a*x^3 + b)/x^3)/x^3)/((a - (a*x^3 + b)/x^3)^2*a^2))

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maple [C]  time = 0.03, size = 3567, normalized size = 48.20 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a+b/x^3)^(1/2),x)

[Out]

1/12/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/a^4*(-18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/
2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1
/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(
1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1
/2)*x^2*a^2*b^2+2*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(
-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*3^(1/2)*x^4*a^3-18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1
)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*
b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*
EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(
1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*3^(1/2)*b^2-3*I*(a*x^4+b*x)^(1/2)*((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3
^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)*3^(1/2)*x*
a^2*b+36*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a
^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3
^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2
),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/2)*x*a*b^2-36*I*(-(I*3^
(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*
3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-
a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(
I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*3^(1/2)*x*a*b^2+1
8*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/
3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)
/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(
1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^2+18*I*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x
+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3
)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*Ellipti
cPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*
(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*3^(1/2)*x^2*a^2*b^2-18*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(
-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3))
)^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticP
i((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I
*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^2-6*x^4*(a*x^4+b*x)^(1/2)*a^3*((-a*x+(-a^2*b)^(1/3))
*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)
-36*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(
1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-
1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3
^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*x*a*b^2+36*(-(I*3^(1/2)-3)/(I*3^(1/
2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-
a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1
/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*
3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(1/3)*x*a*b^2+18*I*(-(I*3^(1/2)-3)/(I*3^
(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x
+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))
^(1/2)*EllipticPi((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),(
(I*3^(1/2)+3)*(I*3^(1/2)-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*3^(1/2)*b^2+18*(-(I*3^(1/2)-3)/
(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/
(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1
/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),((I*3^(1/2)+3)*(I*3^(1/2)
-1)/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*b^2-18*(-(I*3^(1/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/
3))*a*x)^(1/2)*((2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))/(1+I*3^(1/2))/(-a*x+(-a^2*b)^(1/3)))^(1/2)*((-
2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3)))^(1/2)*EllipticPi((-(I*3^(1
/2)-3)/(I*3^(1/2)-1)/(-a*x+(-a^2*b)^(1/3))*a*x)^(1/2),(I*3^(1/2)-1)/(I*3^(1/2)-3),((I*3^(1/2)+3)*(I*3^(1/2)-1)
/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-a^2*b)^(2/3)*b^2+9*b*x*(a*x^4+b*x)^(1/2)*a^2*((-a*x+(-a^2*b)^(1/3))*(2*
a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2))/((
a*x^3+b)*x)^(1/2)/(I*3^(1/2)-3)/((-a*x+(-a^2*b)^(1/3))*(2*a*x+I*3^(1/2)*(-a^2*b)^(1/3)+(-a^2*b)^(1/3))*(-2*a*x
+I*3^(1/2)*(-a^2*b)^(1/3)-(-a^2*b)^(1/3))/a^2*x)^(1/2)

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maxima [A]  time = 2.01, size = 104, normalized size = 1.41 \[ -\frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a + \frac {b}{x^{3}}} a b^{2}}{12 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{2} a^{2} - 2 \, {\left (a + \frac {b}{x^{3}}\right )} a^{3} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b/x^3)^(1/2),x, algorithm="maxima")

[Out]

-1/8*b^2*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/a^(5/2) - 1/12*(3*(a + b/x^3)^(3/2)*b^2
- 5*sqrt(a + b/x^3)*a*b^2)/((a + b/x^3)^2*a^2 - 2*(a + b/x^3)*a^3 + a^4)

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mupad [B]  time = 1.74, size = 79, normalized size = 1.07 \[ \frac {x^6\,\sqrt {a+\frac {b}{x^3}}}{6\,a}+\frac {b^2\,\ln \left (x^6\,\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )\,{\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )}^3\right )}{8\,a^{5/2}}-\frac {b\,x^3\,\sqrt {a+\frac {b}{x^3}}}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b/x^3)^(1/2),x)

[Out]

(x^6*(a + b/x^3)^(1/2))/(6*a) + (b^2*log(x^6*((a + b/x^3)^(1/2) - a^(1/2))*((a + b/x^3)^(1/2) + a^(1/2))^3))/(
8*a^(5/2)) - (b*x^3*(a + b/x^3)^(1/2))/(4*a^2)

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sympy [A]  time = 5.79, size = 102, normalized size = 1.38 \[ \frac {x^{\frac {15}{2}}}{6 \sqrt {b} \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {\sqrt {b} x^{\frac {9}{2}}}{12 a \sqrt {\frac {a x^{3}}{b} + 1}} - \frac {b^{\frac {3}{2}} x^{\frac {3}{2}}}{4 a^{2} \sqrt {\frac {a x^{3}}{b} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{4 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a+b/x**3)**(1/2),x)

[Out]

x**(15/2)/(6*sqrt(b)*sqrt(a*x**3/b + 1)) - sqrt(b)*x**(9/2)/(12*a*sqrt(a*x**3/b + 1)) - b**(3/2)*x**(3/2)/(4*a
**2*sqrt(a*x**3/b + 1)) + b**2*asinh(sqrt(a)*x**(3/2)/sqrt(b))/(4*a**(5/2))

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