3.6.63 \(\int \frac {x}{a+b x^n+c x^{2 n}} \, dx\) [563]

Optimal. Leaf size=136 \[ -\frac {c x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1397, 371} \begin {gather*} -\frac {c x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}-\frac {c x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b \sqrt {b^2-4 a c}-4 a c+b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^n + c*x^(2*n)),x]

[Out]

-((c*x^2*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])) - (c*x^2*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])

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

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Mathematica [A]
time = 0.42, size = 263, normalized size = 1.93 \begin {gather*} -c x^2 \left (\frac {1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-2/n} \, _2F_1\left (-\frac {2}{n},-\frac {2}{n};\frac {-2+n}{n};\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {1-4^{-1/n} \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},-\frac {2}{n};\frac {-2+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x^2*((1 - Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x
^n)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(2/n))/(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + (1 - Hypergeom
etric2F1[-2/n, -2/n, (-2 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(4^n^(-1)*((c*x^n)
/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(2/n)))/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c]))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{a + b x^{n} + c x^{2 n}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*x**n + c*x**(2*n)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(x/(a + b*x^n + c*x^(2*n)), x)

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