3.2025 \(\int \frac {x^4}{\sqrt {a+\frac {b}{x^3}}} \, dx\)

Optimal. Leaf size=270 \[ -\frac {7 b x^2 \sqrt {a+\frac {b}{x^3}}}{20 a^2}-\frac {7 \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{20 \sqrt [4]{3} a^2 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {x^5 \sqrt {a+\frac {b}{x^3}}}{5 a} \]

[Out]

-7/20*b*x^2*(a+b/x^3)^(1/2)/a^2+1/5*x^5*(a+b/x^3)^(1/2)/a-7/60*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^2/(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 = 0.12, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {335, 325, 218} \[ -\frac {7 \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{20 \sqrt [4]{3} a^2 \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}-\frac {7 b x^2 \sqrt {a+\frac {b}{x^3}}}{20 a^2}+\frac {x^5 \sqrt {a+\frac {b}{x^3}}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a + b/x^3],x]

[Out]

(-7*b*Sqrt[a + b/x^3]*x^2)/(20*a^2) + (Sqrt[a + b/x^3]*x^5)/(5*a) - (7*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[A
rcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)/x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)/x)], -7 - 4*Sqrt[3]])/(20*3^(1/4)*a
^2*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 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}{\sqrt {a+\frac {b}{x^3}}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^6 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {a+\frac {b}{x^3}} x^5}{5 a}+\frac {(7 b) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{10 a}\\ &=-\frac {7 b \sqrt {a+\frac {b}{x^3}} x^2}{20 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^5}{5 a}-\frac {\left (7 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{40 a^2}\\ &=-\frac {7 b \sqrt {a+\frac {b}{x^3}} x^2}{20 a^2}+\frac {\sqrt {a+\frac {b}{x^3}} x^5}{5 a}-\frac {7 \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 )}{20 \sqrt [4]{3} a^2 \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 = 0.03, size = 80, normalized size = 0.30 \[ \frac {4 a^2 x^6+7 b^2 \sqrt {\frac {a x^3}{b}+1} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\frac {a x^3}{b}\right )-3 a b x^3-7 b^2}{20 a^2 x \sqrt {a+\frac {b}{x^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a + b/x^3],x]

[Out]

(-7*b^2 - 3*a*b*x^3 + 4*a^2*x^6 + 7*b^2*Sqrt[1 + (a*x^3)/b]*Hypergeometric2F1[1/6, 1/2, 7/6, -((a*x^3)/b)])/(2
0*a^2*Sqrt[a + b/x^3]*x)

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{7} \sqrt {\frac {a x^{3} + b}{x^{3}}}}{a x^{3} + b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^(1/2),x, algorithm="fricas")

[Out]

integral(x^7*sqrt((a*x^3 + b)/x^3)/(a*x^3 + b), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(a + b/x^3), x)

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maple [B]  time = 0.01, size = 2017, normalized size = 7.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a+b/x^3)^(1/2),x)

[Out]

-1/20/((a*x^3+b)/x^3)^(1/2)/x*(a*x^3+b)/a^3/(-a^2*b)^(1/3)*(14*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)*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^2-28*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))*x*a*b^2+14*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))*b^2-4*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)*x^3*a^2-14*(-(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+28*(-(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-
14*(-(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+12*(a*x^4+b*x)^(1/2)*(-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)*a^2*x^3+7*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*b-21*
(a*x^4+b*x)^(1/2)*(-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)*a*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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\sqrt {a + \frac {b}{x^{3}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(a+b/x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(a + b/x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{\sqrt {a+\frac {b}{x^3}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b/x^3)^(1/2),x)

[Out]

int(x^4/(a + b/x^3)^(1/2), x)

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sympy [A]  time = 1.20, size = 46, normalized size = 0.17 \[ - \frac {x^{5} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, \frac {1}{2} \\ - \frac {2}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \sqrt {a} \Gamma \left (- \frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(a+b/x**3)**(1/2),x)

[Out]

-x**5*gamma(-5/3)*hyper((-5/3, 1/2), (-2/3,), b*exp_polar(I*pi)/(a*x**3))/(3*sqrt(a)*gamma(-2/3))

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