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

Optimal. Leaf size=176 \[ -\frac{c^{4/3} \log \left (\sqrt [3]{b} x^{-n/3}+\sqrt [3]{c}\right )}{b^{7/3} n}+\frac{c^{4/3} \log \left (b^{2/3} x^{-2 n/3}-\sqrt [3]{b} \sqrt [3]{c} x^{-n/3}+c^{2/3}\right )}{2 b^{7/3} n}+\frac{\sqrt{3} c^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{b} x^{-n/3}}{\sqrt{3} \sqrt [3]{c}}\right )}{b^{7/3} n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{3 x^{-4 n/3}}{4 b n} \]

[Out]

-3/(4*b*n*x^((4*n)/3)) + (3*c)/(b^2*n*x^(n/3)) + (Sqrt[3]*c^(4/3)*ArcTan[(c^(1/3) - (2*b^(1/3))/x^(n/3))/(Sqrt
[3]*c^(1/3))])/(b^(7/3)*n) - (c^(4/3)*Log[c^(1/3) + b^(1/3)/x^(n/3)])/(b^(7/3)*n) + (c^(4/3)*Log[c^(2/3) + b^(
2/3)/x^((2*n)/3) - (b^(1/3)*c^(1/3))/x^(n/3)])/(2*b^(7/3)*n)

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Rubi [A]  time = 0.142492, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {1584, 362, 345, 193, 321, 200, 31, 634, 617, 204, 628} \[ -\frac{c^{4/3} \log \left (\sqrt [3]{b} x^{-n/3}+\sqrt [3]{c}\right )}{b^{7/3} n}+\frac{c^{4/3} \log \left (b^{2/3} x^{-2 n/3}-\sqrt [3]{b} \sqrt [3]{c} x^{-n/3}+c^{2/3}\right )}{2 b^{7/3} n}+\frac{\sqrt{3} c^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{b} x^{-n/3}}{\sqrt{3} \sqrt [3]{c}}\right )}{b^{7/3} n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{3 x^{-4 n/3}}{4 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(4*b*n*x^((4*n)/3)) + (3*c)/(b^2*n*x^(n/3)) + (Sqrt[3]*c^(4/3)*ArcTan[(c^(1/3) - (2*b^(1/3))/x^(n/3))/(Sqrt
[3]*c^(1/3))])/(b^(7/3)*n) - (c^(4/3)*Log[c^(1/3) + b^(1/3)/x^(n/3)])/(b^(7/3)*n) + (c^(4/3)*Log[c^(2/3) + b^(
2/3)/x^((2*n)/3) - (b^(1/3)*c^(1/3))/x^(n/3)])/(2*b^(7/3)*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 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
 && IntegerQ[p]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

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{4 n}{3}}}{b+c x^n} \, dx\\ &=-\frac{3 x^{-4 n/3}}{4 b n}-\frac{c \int \frac{x^{-1-\frac{n}{3}}}{b+c x^n} \, dx}{b}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{(3 c) \operatorname{Subst}\left (\int \frac{1}{b+\frac{c}{x^3}} \, dx,x,x^{-n/3}\right )}{b n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{(3 c) \operatorname{Subst}\left (\int \frac{x^3}{c+b x^3} \, dx,x,x^{-n/3}\right )}{b n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{c+b x^3} \, dx,x,x^{-n/3}\right )}{b^2 n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{c^{4/3} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{c}+\sqrt [3]{b} x} \, dx,x,x^{-n/3}\right )}{b^2 n}-\frac{c^{4/3} \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{c}-\sqrt [3]{b} x}{c^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+b^{2/3} x^2} \, dx,x,x^{-n/3}\right )}{b^2 n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{b} x^{-n/3}\right )}{b^{7/3} n}+\frac{c^{4/3} \operatorname{Subst}\left (\int \frac{-\sqrt [3]{b} \sqrt [3]{c}+2 b^{2/3} x}{c^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+b^{2/3} x^2} \, dx,x,x^{-n/3}\right )}{2 b^{7/3} n}-\frac{\left (3 c^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{c^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x+b^{2/3} x^2} \, dx,x,x^{-n/3}\right )}{2 b^2 n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{3 c x^{-n/3}}{b^2 n}-\frac{c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{b} x^{-n/3}\right )}{b^{7/3} n}+\frac{c^{4/3} \log \left (c^{2/3}+b^{2/3} x^{-2 n/3}-\sqrt [3]{b} \sqrt [3]{c} x^{-n/3}\right )}{2 b^{7/3} n}-\frac{\left (3 c^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} x^{-n/3}}{\sqrt [3]{c}}\right )}{b^{7/3} n}\\ &=-\frac{3 x^{-4 n/3}}{4 b n}+\frac{3 c x^{-n/3}}{b^2 n}+\frac{\sqrt{3} c^{4/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x^{-n/3}}{\sqrt [3]{c}}}{\sqrt{3}}\right )}{b^{7/3} n}-\frac{c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{b} x^{-n/3}\right )}{b^{7/3} n}+\frac{c^{4/3} \log \left (c^{2/3}+b^{2/3} x^{-2 n/3}-\sqrt [3]{b} \sqrt [3]{c} x^{-n/3}\right )}{2 b^{7/3} n}\\ \end{align*}

Mathematica [C]  time = 0.0078412, size = 34, normalized size = 0.19 \[ -\frac{3 x^{-4 n/3} \, _2F_1\left (-\frac{4}{3},1;-\frac{1}{3};-\frac{c x^n}{b}\right )}{4 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Hypergeometric2F1[-4/3, 1, -1/3, -((c*x^n)/b)])/(4*b*n*x^((4*n)/3))

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Maple [C]  time = 0.065, size = 73, normalized size = 0.4 \begin{align*} 3\,{\frac{c}{{b}^{2}n{x}^{n/3}}}-{\frac{3}{4\,bn} \left ({x}^{{\frac{n}{3}}} \right ) ^{-4}}+\sum _{{\it \_R}={\it RootOf} \left ({b}^{7}{n}^{3}{{\it \_Z}}^{3}+{c}^{4} \right ) }{\it \_R}\,\ln \left ({x}^{{\frac{n}{3}}}+{\frac{{b}^{5}{n}^{2}{{\it \_R}}^{2}}{{c}^{3}}} \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*c/b^2/n/(x^(1/3*n))-3/4/b/n/(x^(1/3*n))^4+sum(_R*ln(x^(1/3*n)+b^5*n^2/c^3*_R^2),_R=RootOf(_Z^3*b^7*n^3+c^4))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} c^{2} \int \frac{x^{\frac{2}{3} \, n}}{b^{2} c x x^{n} + b^{3} x}\,{d x} + \frac{3 \,{\left (4 \, c x^{n} - b\right )}}{4 \, b^{2} n x^{\frac{4}{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^2*integrate(x^(2/3*n)/(b^2*c*x*x^n + b^3*x), x) + 3/4*(4*c*x^n - b)/(b^2*n*x^(4/3*n))

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Fricas [A]  time = 1.67486, size = 427, normalized size = 2.43 \begin{align*} -\frac{3 \, b x^{4} x^{-\frac{4}{3} \, n - 4} - 12 \, c x x^{-\frac{1}{3} \, n - 1} - 4 \, \sqrt{3} c \left (-\frac{c}{b}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} b x x^{-\frac{1}{3} \, n - 1} \left (-\frac{c}{b}\right )^{\frac{2}{3}} - \sqrt{3} c}{3 \, c}\right ) - 4 \, c \left (-\frac{c}{b}\right )^{\frac{1}{3}} \log \left (\frac{x x^{-\frac{1}{3} \, n - 1} - \left (-\frac{c}{b}\right )^{\frac{1}{3}}}{x}\right ) + 2 \, c \left (-\frac{c}{b}\right )^{\frac{1}{3}} \log \left (\frac{x^{2} x^{-\frac{2}{3} \, n - 2} + x x^{-\frac{1}{3} \, n - 1} \left (-\frac{c}{b}\right )^{\frac{1}{3}} + \left (-\frac{c}{b}\right )^{\frac{2}{3}}}{x^{2}}\right )}{4 \, b^{2} 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="fricas")

[Out]

-1/4*(3*b*x^4*x^(-4/3*n - 4) - 12*c*x*x^(-1/3*n - 1) - 4*sqrt(3)*c*(-c/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*x^(-
1/3*n - 1)*(-c/b)^(2/3) - sqrt(3)*c)/c) - 4*c*(-c/b)^(1/3)*log((x*x^(-1/3*n - 1) - (-c/b)^(1/3))/x) + 2*c*(-c/
b)^(1/3)*log((x^2*x^(-2/3*n - 2) + x*x^(-1/3*n - 1)*(-c/b)^(1/3) + (-c/b)^(2/3))/x^2))/(b^2*n)

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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)