3.501 \(\int \frac{x^{-1+\frac{n}{3}}}{b x^n+c x^{2 n}} \, dx\)

Optimal. Leaf size=160 \[ -\frac{c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac{c^{2/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 b^{5/3} n}+\frac{\sqrt{3} c^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{\sqrt{3} \sqrt [3]{b}}\right )}{b^{5/3} n}-\frac{3 x^{-2 n/3}}{2 b n} \]

[Out]

-3/(2*b*n*x^((2*n)/3)) + (Sqrt[3]*c^(2/3)*ArcTan[(b^(1/3) - 2*c^(1/3)*x^(n/3))/(Sqrt[3]*b^(1/3))])/(b^(5/3)*n)
- (c^(2/3)*Log[b^(1/3) + c^(1/3)*x^(n/3)])/(b^(5/3)*n) + (c^(2/3)*Log[b^(2/3) - b^(1/3)*c^(1/3)*x^(n/3) + c^(
2/3)*x^((2*n)/3)])/(2*b^(5/3)*n)

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Rubi [A]  time = 0.132147, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {1584, 362, 345, 200, 31, 634, 617, 204, 628} \[ -\frac{c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac{c^{2/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 b^{5/3} n}+\frac{\sqrt{3} c^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{\sqrt{3} \sqrt [3]{b}}\right )}{b^{5/3} n}-\frac{3 x^{-2 n/3}}{2 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(2*b*n*x^((2*n)/3)) + (Sqrt[3]*c^(2/3)*ArcTan[(b^(1/3) - 2*c^(1/3)*x^(n/3))/(Sqrt[3]*b^(1/3))])/(b^(5/3)*n)
- (c^(2/3)*Log[b^(1/3) + c^(1/3)*x^(n/3)])/(b^(5/3)*n) + (c^(2/3)*Log[b^(2/3) - b^(1/3)*c^(1/3)*x^(n/3) + c^(
2/3)*x^((2*n)/3)])/(2*b^(5/3)*n)

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 362

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[x^(m + 1)/(a*(m + 1)), x] - Dist[b/a, Int[x^Simplify
[m + n]/(a + b*x^n), x], x] /; FreeQ[{a, b, m, n}, x] && FractionQ[Simplify[(m + 1)/n]] && SumSimplerQ[m, n]

Rule 345

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^{-1+\frac{n}{3}}}{b x^n+c x^{2 n}} \, dx &=\int \frac{x^{-1-\frac{2 n}{3}}}{b+c x^n} \, dx\\ &=-\frac{3 x^{-2 n/3}}{2 b n}-\frac{c \int \frac{x^{\frac{1}{3} (-3+n)}}{b+c x^n} \, dx}{b}\\ &=-\frac{3 x^{-2 n/3}}{2 b n}-\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{b+c x^3} \, dx,x,x^{1+\frac{1}{3} (-3+n)}\right )}{b n}\\ &=-\frac{3 x^{-2 n/3}}{2 b n}-\frac{c \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{b}+\sqrt [3]{c} x} \, dx,x,x^{1+\frac{1}{3} (-3+n)}\right )}{b^{5/3} n}-\frac{c \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{b}-\sqrt [3]{c} x}{b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,x^{1+\frac{1}{3} (-3+n)}\right )}{b^{5/3} n}\\ &=-\frac{3 x^{-2 n/3}}{2 b n}-\frac{c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac{c^{2/3} \operatorname{Subst}\left (\int \frac{-\sqrt [3]{b} \sqrt [3]{c}+2 c^{2/3} x}{b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,x^{1+\frac{1}{3} (-3+n)}\right )}{2 b^{5/3} n}-\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+c^{2/3} x^2} \, dx,x,x^{1+\frac{1}{3} (-3+n)}\right )}{2 b^{4/3} n}\\ &=-\frac{3 x^{-2 n/3}}{2 b n}-\frac{c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac{c^{2/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 b^{5/3} n}-\frac{\left (3 c^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{c} x^{1+\frac{1}{3} (-3+n)}}{\sqrt [3]{b}}\right )}{b^{5/3} n}\\ &=-\frac{3 x^{-2 n/3}}{2 b n}+\frac{\sqrt{3} c^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{\sqrt{3} \sqrt [3]{b}}\right )}{b^{5/3} n}-\frac{c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac{c^{2/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 b^{5/3} n}\\ \end{align*}

Mathematica [C]  time = 0.0096091, size = 34, normalized size = 0.21 \[ -\frac{3 x^{-2 n/3} \, _2F_1\left (-\frac{2}{3},1;\frac{1}{3};-\frac{c x^n}{b}\right )}{2 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Hypergeometric2F1[-2/3, 1, 1/3, -((c*x^n)/b)])/(2*b*n*x^((2*n)/3))

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Maple [C]  time = 0.058, size = 54, normalized size = 0.3 \begin{align*} -{\frac{3}{2\,bn} \left ({x}^{{\frac{n}{3}}} \right ) ^{-2}}+\sum _{{\it \_R}={\it RootOf} \left ({b}^{5}{n}^{3}{{\it \_Z}}^{3}+{c}^{2} \right ) }{\it \_R}\,\ln \left ({x}^{{\frac{n}{3}}}-{\frac{{b}^{2}n{\it \_R}}{c}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2/b/n/(x^(1/3*n))^2+sum(_R*ln(x^(1/3*n)-b^2*n/c*_R),_R=RootOf(_Z^3*b^5*n^3+c^2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -c \int \frac{x^{\frac{1}{3} \, n}}{b c x x^{n} + b^{2} x}\,{d x} - \frac{3}{2 \, b n x^{\frac{2}{3} \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70976, size = 502, normalized size = 3.14 \begin{align*} \frac{2 \, \sqrt{3} x^{2} x^{\frac{2}{3} \, n - 2} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} b x x^{\frac{1}{3} \, n - 1} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{2}{3}} - \sqrt{3} c}{3 \, c}\right ) + 2 \, x^{2} x^{\frac{2}{3} \, n - 2} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (\frac{c x x^{\frac{1}{3} \, n - 1} - b \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{1}{3}}}{x}\right ) - x^{2} x^{\frac{2}{3} \, n - 2} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (\frac{c^{2} x^{2} x^{\frac{2}{3} \, n - 2} + b c x x^{\frac{1}{3} \, n - 1} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{1}{3}} + b^{2} \left (-\frac{c^{2}}{b^{2}}\right )^{\frac{2}{3}}}{x^{2}}\right ) - 3}{2 \, b n x^{2} x^{\frac{2}{3} \, n - 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+1/3*n)/(b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(3)*x^2*x^(2/3*n - 2)*(-c^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*x^(1/3*n - 1)*(-c^2/b^2)^(2/3) - s
qrt(3)*c)/c) + 2*x^2*x^(2/3*n - 2)*(-c^2/b^2)^(1/3)*log((c*x*x^(1/3*n - 1) - b*(-c^2/b^2)^(1/3))/x) - x^2*x^(2
/3*n - 2)*(-c^2/b^2)^(1/3)*log((c^2*x^2*x^(2/3*n - 2) + b*c*x*x^(1/3*n - 1)*(-c^2/b^2)^(1/3) + b^2*(-c^2/b^2)^
(2/3))/x^2) - 3)/(b*n*x^2*x^(2/3*n - 2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{1}{3} \, n - 1}}{c x^{2 \, n} + b x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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