3.6.67 \(\int \frac {1}{x^3 (a+b x^n+c x^{2 n})} \, dx\) [567]

Optimal. Leaf size=140 \[ \frac {c \, _2F_1\left (1,-\frac {2}{n};-\frac {2-n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) x^2}+\frac {c \, _2F_1\left (1,-\frac {2}{n};-\frac {2-n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) x^2} \]

[Out]

c*hypergeom([1, -2/n],[(-2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/x^2/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))+c*hyper
geom([1, -2/n],[(-2+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/x^2/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))

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Rubi [A]
time = 0.03, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1397, 371} \begin {gather*} \frac {c \, _2F_1\left (1,-\frac {2}{n};-\frac {2-n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{x^2 \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {c \, _2F_1\left (1,-\frac {2}{n};-\frac {2-n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{x^2 \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

(c*Hypergeometric2F1[1, -2/n, -((2 - n)/n), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c - b*Sqrt[b^2 -
4*a*c])*x^2) + (c*Hypergeometric2F1[1, -2/n, -((2 - n)/n), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c
+ b*Sqrt[b^2 - 4*a*c])*x^2)

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])

Rule 1397

Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]
}, Dist[2*(c/q), Int[(d*x)^m/(b - q + 2*c*x^n), x], x] - Dist[2*(c/q), Int[(d*x)^m/(b + q + 2*c*x^n), x], x]]
/; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]
time = 0.34, size = 258, normalized size = 1.84 \begin {gather*} \frac {2^{\frac {2+n}{n}} c \left (\frac {\left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{2/n} \, _2F_1\left (\frac {2+n}{n},\frac {2+n}{n};2+\frac {2}{n};\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{-b+\sqrt {b^2-4 a c}-2 c x^n}+\frac {x^{-n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {2+n}{n}} \, _2F_1\left (\frac {2+n}{n},\frac {2+n}{n};2+\frac {2}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{c}\right )}{\sqrt {b^2-4 a c} (2+n) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^n + c*x^(2*n))),x]

[Out]

(2^((2 + n)/n)*c*((((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(2/n)*Hypergeometric2F1[(2 + n)/n, (2 + n)/n, 2
+ 2/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(-b + Sqrt[b^2 - 4*a*c] - 2*c*x^n) + (((c*
x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((2 + n)/n)*Hypergeometric2F1[(2 + n)/n, (2 + n)/n, 2 + 2/n, (b + Sqrt
[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(c*x^n)))/(Sqrt[b^2 - 4*a*c]*(2 + n)*x^2)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{3} \left (a +b \,x^{n}+c \,x^{2 n}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/x^3/(a+b*x^n+c*x^(2*n)),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(1/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

integrate(1/((c*x^(2*n) + b*x^n + a)*x^3), x)

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Fricas [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(1/x^3/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

integral(1/(c*x^3*x^(2*n) + b*x^3*x^n + a*x^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate(1/((c*x^(2*n) + b*x^n + a)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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