3.589 \(\int \frac{x^m}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=39 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{a^2 (m+1)} \]

[Out]

(x^(1 + m)*Hypergeometric2F1[2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(a^2*(1 + m))

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Rubi [A]  time = 0.0078223, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {364} \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^3)^2,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(a^2*(1 + m))

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^m}{\left (a+b x^3\right )^2} \, dx &=\frac{x^{1+m} \, _2F_1\left (2,\frac{1+m}{3};\frac{4+m}{3};-\frac{b x^3}{a}\right )}{a^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0072708, size = 41, normalized size = 1.05 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{3};\frac{m+1}{3}+1;-\frac{b x^3}{a}\right )}{a^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^3)^2,x]

[Out]

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

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( b{x}^{3}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x^3 + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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Sympy [C]  time = 109.39, size = 515, normalized size = 13.21 \begin{align*} - \frac{a m^{2} x x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{a m x x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{3 a m x x^{m} \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{2 a x x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{3 a x x^{m} \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} - \frac{b m^{2} x^{4} x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{b m x^{4} x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} + \frac{2 b x^{4} x^{m} \Phi \left (\frac{b x^{3} e^{i \pi }}{a}, 1, \frac{m}{3} + \frac{1}{3}\right ) \Gamma \left (\frac{m}{3} + \frac{1}{3}\right )}{27 a^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right ) + 27 a^{2} b x^{3} \Gamma \left (\frac{m}{3} + \frac{4}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*m**2*x*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)/(27*a**3*gamma(m/3 + 4/3) + 2
7*a**2*b*x**3*gamma(m/3 + 4/3)) + a*m*x*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)
/(27*a**3*gamma(m/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3)) + 3*a*m*x*x**m*gamma(m/3 + 1/3)/(27*a**3*gamma(m
/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3)) + 2*a*x*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gam
ma(m/3 + 1/3)/(27*a**3*gamma(m/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3)) + 3*a*x*x**m*gamma(m/3 + 1/3)/(27*a
**3*gamma(m/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3)) - b*m**2*x**4*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a,
1, m/3 + 1/3)*gamma(m/3 + 1/3)/(27*a**3*gamma(m/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3)) + b*m*x**4*x**m*le
rchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)/(27*a**3*gamma(m/3 + 4/3) + 27*a**2*b*x**3*gam
ma(m/3 + 4/3)) + 2*b*x**4*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)/(27*a**3*gamm
a(m/3 + 4/3) + 27*a**2*b*x**3*gamma(m/3 + 4/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x^3 + a)^2, x)