3.411 \(\int \frac{x^6}{\sqrt{a+b x^3}} \, dx\)

Optimal. Leaf size=254 \[ \frac{32 \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{16 a x \sqrt{a+b x^3}}{55 b^2}+\frac{2 x^4 \sqrt{a+b x^3}}{11 b} \]

[Out]

(-16*a*x*Sqrt[a + b*x^3])/(55*b^2) + (2*x^4*Sqrt[a + b*x^3])/(11*b) + (32*Sqrt[2 + Sqrt[3]]*a^2*(a^(1/3) + b^(
1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcS
in[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(55*3^(1/4)*b^(7
/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.0666739, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {321, 218} \[ \frac{32 \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{16 a x \sqrt{a+b x^3}}{55 b^2}+\frac{2 x^4 \sqrt{a+b x^3}}{11 b} \]

Antiderivative was successfully verified.

[In]

Int[x^6/Sqrt[a + b*x^3],x]

[Out]

(-16*a*x*Sqrt[a + b*x^3])/(55*b^2) + (2*x^4*Sqrt[a + b*x^3])/(11*b) + (32*Sqrt[2 + Sqrt[3]]*a^2*(a^(1/3) + b^(
1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcS
in[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(55*3^(1/4)*b^(7
/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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^6}{\sqrt{a+b x^3}} \, dx &=\frac{2 x^4 \sqrt{a+b x^3}}{11 b}-\frac{(8 a) \int \frac{x^3}{\sqrt{a+b x^3}} \, dx}{11 b}\\ &=-\frac{16 a x \sqrt{a+b x^3}}{55 b^2}+\frac{2 x^4 \sqrt{a+b x^3}}{11 b}+\frac{\left (16 a^2\right ) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{55 b^2}\\ &=-\frac{16 a x \sqrt{a+b x^3}}{55 b^2}+\frac{2 x^4 \sqrt{a+b x^3}}{11 b}+\frac{32 \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt{3}\right )}{55 \sqrt [4]{3} b^{7/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0234394, size = 79, normalized size = 0.31 \[ \frac{2 \left (8 a^2 x \sqrt{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )-8 a^2 x-3 a b x^4+5 b^2 x^7\right )}{55 b^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/Sqrt[a + b*x^3],x]

[Out]

(2*(-8*a^2*x - 3*a*b*x^4 + 5*b^2*x^7 + 8*a^2*x*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/
a)]))/(55*b^2*Sqrt[a + b*x^3])

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Maple [A]  time = 0.012, size = 320, normalized size = 1.3 \begin{align*}{\frac{2\,{x}^{4}}{11\,b}\sqrt{b{x}^{3}+a}}-{\frac{16\,ax}{55\,{b}^{2}}\sqrt{b{x}^{3}+a}}-{\frac{{\frac{32\,i}{165}}{a}^{2}\sqrt{3}}{{b}^{3}}\sqrt [3]{-{b}^{2}a}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}\sqrt{{ \left ( x-{\frac{1}{b}\sqrt [3]{-{b}^{2}a}} \right ) \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}}\sqrt{{-i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}{\it EllipticF} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}b \left ( x+{\frac{1}{2\,b}\sqrt [3]{-{b}^{2}a}}-{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ){\frac{1}{\sqrt [3]{-{b}^{2}a}}}}}},\sqrt{{\frac{i\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a} \left ( -{\frac{3}{2\,b}\sqrt [3]{-{b}^{2}a}}+{\frac{{\frac{i}{2}}\sqrt{3}}{b}\sqrt [3]{-{b}^{2}a}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{b{x}^{3}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^3+a)^(1/2),x)

[Out]

2/11*x^4*(b*x^3+a)^(1/2)/b-16/55*a*x*(b*x^3+a)^(1/2)/b^2-32/165*I*a^2/b^3*3^(1/2)*(-b^2*a)^(1/3)*(I*(x+1/2/b*(
-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)*((x-1/b*(-b^2*a)^(1/3))/(-3/2/b*
(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/
3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(
1/2)/b*(-b^2*a)^(1/3))*3^(1/2)*b/(-b^2*a)^(1/3))^(1/2),(I*3^(1/2)/b*(-b^2*a)^(1/3)/(-3/2/b*(-b^2*a)^(1/3)+1/2*
I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(b*x^3 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{6}}{\sqrt{b x^{3} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

integral(x^6/sqrt(b*x^3 + a), x)

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Sympy [A]  time = 1.01177, size = 37, normalized size = 0.15 \begin{align*} \frac{x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt{a} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**3+a)**(1/2),x)

[Out]

x**7*gamma(7/3)*hyper((1/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*gamma(10/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{\sqrt{b x^{3} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(b*x^3 + a), x)