3.2025 \(\int \frac{x^4}{\sqrt{a+\frac{b}{x^3}}} \, dx\)
Optimal. Leaf size=270 \[ -\frac{7 \sqrt{2+\sqrt{3}} b^{5/3} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right ) \sqrt{\frac{a^{2/3}-\frac{\sqrt [3]{a} \sqrt [3]{b}}{x}+\frac{b^{2/3}}{x^2}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}}\right ),-7-4 \sqrt{3}\right )}{20 \sqrt [4]{3} a^2 \sqrt{a+\frac{b}{x^3}} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\frac{\sqrt [3]{b}}{x}\right )^2}}}-\frac{7 b x^2 \sqrt{a+\frac{b}{x^3}}}{20 a^2}+\frac{x^5 \sqrt{a+\frac{b}{x^3}}}{5 a} \]
[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])
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Rubi [A] time = 0.119266, antiderivative size = 270, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]
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 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]
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*}
Mathematica [C] time = 0.0240482, size = 80, normalized size = 0.3 \[ \frac{4 a^2 x^6+7 b^2 \sqrt{\frac{a x^3}{b}+1} \, _2F_1\left (\frac{1}{6},\frac{1}{2};\frac{7}{6};-\frac{a x^3}{b}\right )-3 a b x^3-7 b^2}{20 a^2 x \sqrt{a+\frac{b}{x^3}}} \]
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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Maple [B] time = 0.01, size = 2017, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}
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/(-b*a^2)^(1/3)*(-14*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*
a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)
*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3
^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/
2)-3))^(1/2))*3^(1/2)*x^2*a^2*b^2+28*I*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(
1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3
)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2
))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(1/
3)*3^(1/2)*x*a*b^2+4*I*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^
2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2)*(-b*a^2)^(1/3)*3^(1/2)*x^3*a^2-14*I*(-(I*3^(1
/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3
^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b
*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(
-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*(-b*a^2)^(2/3)*3^(1/2)*b^2+14*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1
/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2
)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*E
llipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^
(1/2))/(I*3^(1/2)-3))^(1/2))*x^2*a^2*b^2-28*(-b*a^2)^(1/3)*(-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(
1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3
^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)
-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))
^(1/2))*x*a*b^2-12*x^3*(a*x^4+b*x)^(1/2)*a^2*(-b*a^2)^(1/3)*(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)
^(1/3)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2)+14*(-b*a^2)^(2/3)*(-(I*3^(
1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))/(1+I*
3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2)*((I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3))/(-1+I*3^(1/2))/(-a*x+(-
b*a^2)^(1/3)))^(1/2)*EllipticF((-(I*3^(1/2)-3)*x*a/(-1+I*3^(1/2))/(-a*x+(-b*a^2)^(1/3)))^(1/2),((I*3^(1/2)+3)*
(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2))*b^2-7*I*(a*x^4+b*x)^(1/2)*(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I
*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2)*(-b*a^2)^
(1/3)*3^(1/2)*a*b+21*b*(a*x^4+b*x)^(1/2)*a*(-b*a^2)^(1/3)*(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(
1/3)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)-2*a*x-(-b*a^2)^(1/3)))^(1/2))/(x*(a*x^3+b))^(1/2)/(I*3^(1
/2)-3)/(1/a^2*x*(-a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3)+2*a*x+(-b*a^2)^(1/3))*(I*3^(1/2)*(-b*a^2)^(1/3
)-2*a*x-(-b*a^2)^(1/3)))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{a + \frac{b}{x^{3}}}}\,{d x} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{7} \sqrt{\frac{a x^{3} + b}{x^{3}}}}{a x^{3} + b}, x\right ) \end{align*}
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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Sympy [A] time = 1.4921, size = 46, normalized size = 0.17 \begin{align*} - \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 )} \end{align*}
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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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{a + \frac{b}{x^{3}}}}\,{d x} \end{align*}
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)