3.2010 \(\int (a+\frac {b}{x^3})^{3/2} x^5 \, dx\)
Optimal. Leaf size=68 \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{4} b x^3 \sqrt {a+\frac {b}{x^3}}+\frac {1}{6} x^6 \left (a+\frac {b}{x^3}\right )^{3/2} \]
[Out]
1/6*(a+b/x^3)^(3/2)*x^6+1/4*b^2*arctanh((a+b/x^3)^(1/2)/a^(1/2))/a^(1/2)+1/4*b*x^3*(a+b/x^3)^(1/2)
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Rubi [A] time = 0.04, antiderivative size = 68, 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, 47, 63, 208} \[ \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 \sqrt {a}}+\frac {1}{6} x^6 \left (a+\frac {b}{x^3}\right )^{3/2}+\frac {1}{4} b x^3 \sqrt {a+\frac {b}{x^3}} \]
Antiderivative was successfully verified.
[In]
Int[(a + b/x^3)^(3/2)*x^5,x]
[Out]
(b*Sqrt[a + b/x^3]*x^3)/4 + ((a + b/x^3)^(3/2)*x^6)/6 + (b^2*ArcTanh[Sqrt[a + b/x^3]/Sqrt[a]])/(4*Sqrt[a])
Rule 47
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 \left (a+\frac {b}{x^3}\right )^{3/2} x^5 \, dx &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3} \, dx,x,\frac {1}{x^3}\right )\right )\\ &=\frac {1}{6} \left (a+\frac {b}{x^3}\right )^{3/2} x^6-\frac {1}{4} b \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,\frac {1}{x^3}\right )\\ &=\frac {1}{4} b \sqrt {a+\frac {b}{x^3}} x^3+\frac {1}{6} \left (a+\frac {b}{x^3}\right )^{3/2} x^6-\frac {1}{8} b^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x^3}\right )\\ &=\frac {1}{4} b \sqrt {a+\frac {b}{x^3}} x^3+\frac {1}{6} \left (a+\frac {b}{x^3}\right )^{3/2} x^6-\frac {1}{4} b \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x^3}}\right )\\ &=\frac {1}{4} b \sqrt {a+\frac {b}{x^3}} x^3+\frac {1}{6} \left (a+\frac {b}{x^3}\right )^{3/2} x^6+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x^3}}}{\sqrt {a}}\right )}{4 \sqrt {a}}\\ \end {align*}
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Mathematica [A] time = 0.19, size = 80, normalized size = 1.18 \[ \frac {1}{12} x^{3/2} \sqrt {a+\frac {b}{x^3}} \left (\frac {3 b^{3/2} \sinh ^{-1}\left (\frac {\sqrt {a} x^{3/2}}{\sqrt {b}}\right )}{\sqrt {a} \sqrt {\frac {a x^3}{b}+1}}+x^{3/2} \left (2 a x^3+5 b\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b/x^3)^(3/2)*x^5,x]
[Out]
(Sqrt[a + b/x^3]*x^(3/2)*(x^(3/2)*(5*b + 2*a*x^3) + (3*b^(3/2)*ArcSinh[(Sqrt[a]*x^(3/2))/Sqrt[b]])/(Sqrt[a]*Sq
rt[1 + (a*x^3)/b])))/12
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fricas [A] time = 1.18, size = 179, normalized size = 2.63 \[ \left [\frac {3 \, \sqrt {a} b^{2} \log \left (-8 \, a^{2} x^{6} - 8 \, a b x^{3} - b^{2} - 4 \, {\left (2 \, a x^{6} + b x^{3}\right )} \sqrt {a} \sqrt {\frac {a x^{3} + b}{x^{3}}}\right ) + 4 \, {\left (2 \, a^{2} x^{6} + 5 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{48 \, a}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} x^{3} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{2 \, a x^{3} + b}\right ) - 2 \, {\left (2 \, a^{2} x^{6} + 5 \, a b x^{3}\right )} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{24 \, a}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x^3)^(3/2)*x^5,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 + 5*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a, -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 + 5*a*b*x^3)*sqrt((a*x^3 + b)/x^3))/a]
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giac [A] time = 0.32, size = 51, normalized size = 0.75 \[ \frac {1}{12} \, \sqrt {a x^{4} + b x} {\left (2 \, a x^{3} + 5 \, b\right )} x - \frac {b^{2} \arctan \left (\frac {\sqrt {a + \frac {b}{x^{3}}}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x^3)^(3/2)*x^5,x, algorithm="giac")
[Out]
1/12*sqrt(a*x^4 + b*x)*(2*a*x^3 + 5*b)*x - 1/4*b^2*arctan(sqrt(a + b/x^3)/sqrt(-a))/sqrt(-a)
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maple [C] time = 0.01, size = 3569, normalized size = 52.49 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b/x^3)^(3/2)*x^5,x)
[Out]
1/12*((a*x^3+b)/x^3)^(3/2)*x^5/a^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)*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+5*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+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+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^2-36*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+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))*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)*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))*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)*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))*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)*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+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^2-15*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)/
((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 = 1.86, size = 98, normalized size = 1.44 \[ -\frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x^{3}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{3}}} + \sqrt {a}}\right )}{8 \, \sqrt {a}} + \frac {5 \, {\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {a + \frac {b}{x^{3}}} a b^{2}}{12 \, {\left ({\left (a + \frac {b}{x^{3}}\right )}^{2} - 2 \, {\left (a + \frac {b}{x^{3}}\right )} a + a^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x^3)^(3/2)*x^5,x, algorithm="maxima")
[Out]
-1/8*b^2*log((sqrt(a + b/x^3) - sqrt(a))/(sqrt(a + b/x^3) + sqrt(a)))/sqrt(a) + 1/12*(5*(a + b/x^3)^(3/2)*b^2
- 3*sqrt(a + b/x^3)*a*b^2)/((a + b/x^3)^2 - 2*(a + b/x^3)*a + a^2)
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mupad [B] time = 1.86, size = 74, normalized size = 1.09 \[ \frac {a\,x^6\,\sqrt {a+\frac {b}{x^3}}}{6}+\frac {5\,b\,x^3\,\sqrt {a+\frac {b}{x^3}}}{12}+\frac {b^2\,\ln \left (x^6\,\left (\sqrt {a+\frac {b}{x^3}}-\sqrt {a}\right )\,{\left (\sqrt {a+\frac {b}{x^3}}+\sqrt {a}\right )}^3\right )}{8\,\sqrt {a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^5*(a + b/x^3)^(3/2),x)
[Out]
(a*x^6*(a + b/x^3)^(1/2))/6 + (5*b*x^3*(a + b/x^3)^(1/2))/12 + (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^(1/2))
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sympy [A] time = 3.87, size = 76, normalized size = 1.12 \[ \frac {a \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a x^{3}}{b} + 1}}{6} + \frac {5 b^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {\frac {a x^{3}}{b} + 1}}{12} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} x^{\frac {3}{2}}}{\sqrt {b}} \right )}}{4 \sqrt {a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b/x**3)**(3/2)*x**5,x)
[Out]
a*sqrt(b)*x**(9/2)*sqrt(a*x**3/b + 1)/6 + 5*b**(3/2)*x**(3/2)*sqrt(a*x**3/b + 1)/12 + b**2*asinh(sqrt(a)*x**(3
/2)/sqrt(b))/(4*sqrt(a))
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