3.2045 \(\int \frac {x^4}{(a+\frac {b}{x^3})^{3/2}} \, dx\)
Optimal. Leaf size=291 \[ -\frac {91 b x^2 \sqrt {a+\frac {b}{x^3}}}{60 a^3}+\frac {13 x^5 \sqrt {a+\frac {b}{x^3}}}{15 a^2}-\frac {91 \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{60 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}} \]
[Out]
-2/3*x^5/a/(a+b/x^3)^(1/2)-91/60*b*x^2*(a+b/x^3)^(1/2)/a^3+13/15*x^5*(a+b/x^3)^(1/2)/a^2-91/180*b^(5/3)*(a^(1/
3)+b^(1/3)/x)*EllipticF((b^(1/3)/x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)/x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^
(1/2)+1/2*2^(1/2))*((a^(2/3)+b^(2/3)/x^2-a^(1/3)*b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/a
^3/(a+b/x^3)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)/x)/(b^(1/3)/x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)
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Rubi [A] time = 0.15, antiderivative size = 291, 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 =
{335, 290, 325, 218} \[ -\frac {91 \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{60 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {13 x^5 \sqrt {a+\frac {b}{x^3}}}{15 a^2}-\frac {91 b x^2 \sqrt {a+\frac {b}{x^3}}}{60 a^3}-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}} \]
Antiderivative was successfully verified.
[In]
Int[x^4/(a + b/x^3)^(3/2),x]
[Out]
(-91*b*Sqrt[a + b/x^3]*x^2)/(60*a^3) - (2*x^5)/(3*a*Sqrt[a + b/x^3]) + (13*Sqrt[a + b/x^3]*x^5)/(15*a^2) - (91
*Sqrt[2 + Sqrt[3]]*b^(5/3)*(a^(1/3) + b^(1/3)/x)*Sqrt[(a^(2/3) + b^(2/3)/x^2 - (a^(1/3)*b^(1/3))/x)/((1 + Sqrt
[3])*a^(1/3) + b^(1/3)/x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(
1/3)/x)], -7 - 4*Sqrt[3]])/(60*3^(1/4)*a^3*Sqrt[a + b/x^3]*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)/x))/((1 + Sqrt[3])
*a^(1/3) + b^(1/3)/x)^2])
Rule 218
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]
Rule 290
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 325
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 335
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]
Rubi steps
\begin {align*} \int \frac {x^4}{\left (a+\frac {b}{x^3}\right )^{3/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^6 \left (a+b x^3\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}}-\frac {13 \operatorname {Subst}\left (\int \frac {1}{x^6 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{3 a}\\ &=-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}}+\frac {13 \sqrt {a+\frac {b}{x^3}} x^5}{15 a^2}+\frac {(91 b) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{30 a^2}\\ &=-\frac {91 b \sqrt {a+\frac {b}{x^3}} x^2}{60 a^3}-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}}+\frac {13 \sqrt {a+\frac {b}{x^3}} x^5}{15 a^2}-\frac {\left (91 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{120 a^3}\\ &=-\frac {91 b \sqrt {a+\frac {b}{x^3}} x^2}{60 a^3}-\frac {2 x^5}{3 a \sqrt {a+\frac {b}{x^3}}}+\frac {13 \sqrt {a+\frac {b}{x^3}} x^5}{15 a^2}-\frac {91 \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}+\frac {b^{2/3}}{x^2}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{60 \sqrt [4]{3} a^3 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}\\ \end {align*}
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Mathematica [C] time = 0.04, size = 80, normalized size = 0.27 \[ \frac {12 a^2 x^6+91 b^2 \sqrt {\frac {a x^3}{b}+1} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {a x^3}{b}\right )-39 a b x^3-91 b^2}{60 a^3 x \sqrt {a+\frac {b}{x^3}}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^4/(a + b/x^3)^(3/2),x]
[Out]
(-91*b^2 - 39*a*b*x^3 + 12*a^2*x^6 + 91*b^2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b)]
)/(60*a^3*Sqrt[a + b/x^3]*x)
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fricas [F] time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{10} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{a^{2} x^{6} + 2 \, a b x^{3} + b^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4/(a+b/x^3)^(3/2),x, algorithm="fricas")
[Out]
integral(x^10*sqrt((a*x^3 + b)/x^3)/(a^2*x^6 + 2*a*b*x^3 + b^2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4/(a+b/x^3)^(3/2),x, algorithm="giac")
[Out]
integrate(x^4/(a + b/x^3)^(3/2), x)
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maple [B] time = 0.02, size = 2302, normalized size = 7.91 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4/(a+b/x^3)^(3/2),x)
[Out]
-1/60/((a*x^3+b)/x^3)^(3/2)/x^5*(a*x^3+b)/a^4/(-a^2*b)^(1/3)*(182*I*3^(1/2)*(-(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)*Ellipt
icF((-(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*x^3+b)*x)^(1/2)*x^2*a^2*b^2-364*I*(-a^2*b)^(1/3)*3^(1/2)*(-(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*x^3+b)*x)^(1/2)*x*a*b^2+182*I*(-a^2*b)^(2/3)*3^(1/2)*(-(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*x^3+b)*x)^(1/2)*b^2-12*I*(-a^2*b)^(1/3)*3^(1/2)*(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)*((a*x^3+b)*x)^(1/2)*x^3*a^2-182*(-(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*x^3
+b)*x)^(1/2)*x^2*a^2*b^2+364*(-(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)*((a*x^3+b)
*x)^(1/2)*x*a*b^2-182*(-(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)*((a*x^3+b)*x)^(1/
2)*b^2+36*(-a^2*b)^(1/3)*(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)*((a*x^3+b)*x)^(1/2)*x^3*a^2+51*I*(-a^2*b)^(1
/3)*3^(1/2)*(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)*((a*x^3+b)*x)^(1/2)*a*b+40*I*(-a^2*b)^(1/3)*3^(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)*x*a*b^2-153*(-a^2*b)^(1/3)*(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)*((a*x^3+b)*x)^(1/2)*a*b-
120*(-a^2*b)^(1/3)*((-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)*x*a*b^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 [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (a + \frac {b}{x^{3}}\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4/(a+b/x^3)^(3/2),x, algorithm="maxima")
[Out]
integrate(x^4/(a + b/x^3)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{{\left (a+\frac {b}{x^3}\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4/(a + b/x^3)^(3/2),x)
[Out]
int(x^4/(a + b/x^3)^(3/2), x)
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sympy [A] time = 1.38, size = 46, normalized size = 0.16 \[ - \frac {x^{5} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {3}{2} \\ - \frac {2}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (- \frac {2}{3}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**4/(a+b/x**3)**(3/2),x)
[Out]
-x**5*gamma(-5/3)*hyper((-5/3, 3/2), (-2/3,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3/2)*gamma(-2/3))
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