3.550 \(\int \frac{x^7}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=38 \[ \frac{x^8 \left (a+b x^3\right )^{2/3} \, _2F_1\left (1,\frac{10}{3};\frac{11}{3};-\frac{b x^3}{a}\right )}{8 a} \]

[Out]

(x^8*(a + b*x^3)^(2/3)*Hypergeometric2F1[1, 10/3, 11/3, -((b*x^3)/a)])/(8*a)

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Rubi [A]  time = 0.0143734, antiderivative size = 51, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {365, 364} \[ \frac{x^8 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{8}{3};\frac{11}{3};-\frac{b x^3}{a}\right )}{8 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^3)^(1/3),x]

[Out]

(x^8*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 8/3, 11/3, -((b*x^3)/a)])/(8*(a + b*x^3)^(1/3))

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{x^7}{\sqrt [3]{a+b x^3}} \, dx &=\frac{\sqrt [3]{1+\frac{b x^3}{a}} \int \frac{x^7}{\sqrt [3]{1+\frac{b x^3}{a}}} \, dx}{\sqrt [3]{a+b x^3}}\\ &=\frac{x^8 \sqrt [3]{1+\frac{b x^3}{a}} \, _2F_1\left (\frac{1}{3},\frac{8}{3};\frac{11}{3};-\frac{b x^3}{a}\right )}{8 \sqrt [3]{a+b x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0096362, size = 51, normalized size = 1.34 \[ \frac{x^8 \sqrt [3]{\frac{b x^3}{a}+1} \, _2F_1\left (\frac{1}{3},\frac{8}{3};\frac{11}{3};-\frac{b x^3}{a}\right )}{8 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^3)^(1/3),x]

[Out]

(x^8*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 8/3, 11/3, -((b*x^3)/a)])/(8*(a + b*x^3)^(1/3))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{{x}^{7}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^3+a)^(1/3),x)

[Out]

int(x^7/(b*x^3+a)^(1/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

integrate(x^7/(b*x^3 + a)^(1/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{7}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

integral(x^7/(b*x^3 + a)^(1/3), x)

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Sympy [C]  time = 1.0914, size = 37, normalized size = 0.97 \begin{align*} \frac{x^{8} \Gamma \left (\frac{8}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{8}{3} \\ \frac{11}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{11}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**3+a)**(1/3),x)

[Out]

x**8*gamma(8/3)*hyper((1/3, 8/3), (11/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(11/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate(x^7/(b*x^3 + a)^(1/3), x)