3.6.89 \(\int \frac {x^m}{(a+b x^3)^2} \, dx\) [589]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 = {371} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], 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*}

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Mathematica [A]
time = 0.03, size = 41, normalized size = 1.05 \begin {gather*} \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{3};1+\frac {1+m}{3};-\frac {b x^3}{a}\right )}{a^2 (1+m)} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b \,x^{3}+a \right )^{2}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.35, size = 26, normalized size = 0.67 \begin {gather*} {\rm integral}\left (\frac {x^{m}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end {gather*}

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] Result contains complex when optimal does not.
time = 25.26, size = 515, normalized size = 13.21 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^m}{{\left (b\,x^3+a\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^3)^2,x)

[Out]

int(x^m/(a + b*x^3)^2, x)

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