3.1999 \(\int \sqrt {a+\frac {b}{x^3}} x^4 \, dx\)

Optimal. Leaf size=267 \[ \frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{20 a \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {1}{5} x^5 \sqrt {a+\frac {b}{x^3}}+\frac {3 b x^2 \sqrt {a+\frac {b}{x^3}}}{20 a} \]

[Out]

3/20*b*x^2*(a+b/x^3)^(1/2)/a+1/5*x^5*(a+b/x^3)^(1/2)+1/20*3^(3/4)*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)/a/(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 = 267, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {335, 277, 325, 218} \[ \frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{5/3} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right ) \sqrt {\frac {a^{2/3}-\frac {\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac {b^{2/3}}{x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}}\right )|-7-4 \sqrt {3}\right )}{20 a \sqrt {a+\frac {b}{x^3}} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\frac {\sqrt [3]{b}}{x}\right )^2}}}+\frac {1}{5} x^5 \sqrt {a+\frac {b}{x^3}}+\frac {3 b x^2 \sqrt {a+\frac {b}{x^3}}}{20 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^3]*x^4,x]

[Out]

(3*b*Sqrt[a + b/x^3]*x^2)/(20*a) + (Sqrt[a + b/x^3]*x^5)/5 + (3^(3/4)*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[Ar
cSin[((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*a*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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && 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 \sqrt {a+\frac {b}{x^3}} x^4 \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^3}}{x^6} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{5} \sqrt {a+\frac {b}{x^3}} x^5-\frac {1}{10} (3 b) \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {3 b \sqrt {a+\frac {b}{x^3}} x^2}{20 a}+\frac {1}{5} \sqrt {a+\frac {b}{x^3}} x^5+\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^3}} \, dx,x,\frac {1}{x}\right )}{40 a}\\ &=\frac {3 b \sqrt {a+\frac {b}{x^3}} x^2}{20 a}+\frac {1}{5} \sqrt {a+\frac {b}{x^3}} x^5+\frac {3^{3/4} \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 a \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 = 0.06, size = 80, normalized size = 0.30 \[ \frac {x^2 \sqrt {a+\frac {b}{x^3}} \left (\left (a x^3+b\right ) \sqrt {\frac {a x^3}{b}+1}-b \, _2F_1\left (-\frac {1}{2},\frac {1}{6};\frac {7}{6};-\frac {a x^3}{b}\right )\right )}{5 a \sqrt {\frac {a x^3}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^3]*x^4,x]

[Out]

(Sqrt[a + b/x^3]*x^2*((b + a*x^3)*Sqrt[1 + (a*x^3)/b] - b*Hypergeometric2F1[-1/2, 1/6, 7/6, -((a*x^3)/b)]))/(5
*a*Sqrt[1 + (a*x^3)/b])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{4} \sqrt {\frac {a x^{3} + b}{x^{3}}}, x\right ) \]

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^4*sqrt((a*x^3 + b)/x^3), x)

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

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(sqrt(a + b/x^3)*x^4, x)

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maple [B]  time = 0.03, size = 2010, normalized size = 7.53 \[ \text {result too large to display} \]

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^2/a^2/(-a^2*b)^(1/3)*(6*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^2-12*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+4*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)*(-a^2*b)^(1/3)*3^(1/2)*x^3*a^2+6*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^2-6*(-(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+12*(-(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^2-12*x^3*(a*x^
4+b*x)^(1/2)*a^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)-6*(-(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+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)*(-a^2*b)^(1/3)*3^(1/2)*a*b-9*b*(a*x^4+b*x)^
(1/2)*a*(-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*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 \[ \int \sqrt {a + \frac {b}{x^{3}}} x^{4}\,{d x} \]

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(sqrt(a + b/x^3)*x^4, x)

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

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 = 1.29, size = 48, normalized size = 0.18 \[ - \frac {\sqrt {a} x^{5} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, - \frac {1}{2} \\ - \frac {2}{3} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 \Gamma \left (- \frac {2}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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