3.2644 \(\int \frac {x^{-1-\frac {n}{3}}}{a+b x^n} \, dx\)

Optimal. Leaf size=158 \[ -\frac {\sqrt [3]{b} \log \left (a^{2/3} x^{-2 n/3}-\sqrt [3]{a} \sqrt [3]{b} x^{-n/3}+b^{2/3}\right )}{2 a^{4/3} n}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a} x^{-n/3}+\sqrt [3]{b}\right )}{a^{4/3} n}-\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x^{-n/3}}{\sqrt {3} \sqrt [3]{b}}\right )}{a^{4/3} n}-\frac {3 x^{-n/3}}{a n} \]

[Out]

-3/a/n/(x^(1/3*n))+b^(1/3)*ln(b^(1/3)+a^(1/3)/(x^(1/3*n)))/a^(4/3)/n-1/2*b^(1/3)*ln(b^(2/3)+a^(2/3)/(x^(2/3*n)
)-a^(1/3)*b^(1/3)/(x^(1/3*n)))/a^(4/3)/n-b^(1/3)*arctan(1/3*(b^(1/3)-2*a^(1/3)/(x^(1/3*n)))/b^(1/3)*3^(1/2))*3
^(1/2)/a^(4/3)/n

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Rubi [A]  time = 0.11, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {345, 193, 321, 200, 31, 634, 617, 204, 628} \[ -\frac {\sqrt [3]{b} \log \left (a^{2/3} x^{-2 n/3}-\sqrt [3]{a} \sqrt [3]{b} x^{-n/3}+b^{2/3}\right )}{2 a^{4/3} n}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a} x^{-n/3}+\sqrt [3]{b}\right )}{a^{4/3} n}-\frac {\sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x^{-n/3}}{\sqrt {3} \sqrt [3]{b}}\right )}{a^{4/3} n}-\frac {3 x^{-n/3}}{a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n/3)/(a + b*x^n),x]

[Out]

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

Rule 31

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

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 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 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 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 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 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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.20 \[ -\frac {3 x^{-n/3} \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};-\frac {b x^n}{a}\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n/3)/(a + b*x^n),x]

[Out]

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

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fricas [A]  time = 0.79, size = 145, normalized size = 0.92 \[ -\frac {6 \, x x^{-\frac {1}{3} \, n - 1} - 2 \, \sqrt {3} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x x^{-\frac {1}{3} \, n - 1} \left (\frac {b}{a}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 2 \, \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (\frac {x x^{-\frac {1}{3} \, n - 1} + \left (\frac {b}{a}\right )^{\frac {1}{3}}}{x}\right ) + \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (\frac {x^{2} x^{-\frac {2}{3} \, n - 2} - x x^{-\frac {1}{3} \, n - 1} \left (\frac {b}{a}\right )^{\frac {1}{3}} + \left (\frac {b}{a}\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 \, a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*x*x^(-1/3*n - 1) - 2*sqrt(3)*(b/a)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*x^(-1/3*n - 1)*(b/a)^(2/3) - sqrt(3
)*b)/b) - 2*(b/a)^(1/3)*log((x*x^(-1/3*n - 1) + (b/a)^(1/3))/x) + (b/a)^(1/3)*log((x^2*x^(-2/3*n - 2) - x*x^(-
1/3*n - 1)*(b/a)^(1/3) + (b/a)^(2/3))/x^2))/(a*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{-\frac {1}{3} \, n - 1}}{b x^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.07, size = 57, normalized size = 0.36 \[ \RootOf \left (a^{4} n^{3} \textit {\_Z}^{3}-b \right ) \ln \left (\frac {\RootOf \left (a^{4} n^{3} \textit {\_Z}^{3}-b \right )^{2} a^{3} n^{2}}{b}+x^{\frac {n}{3}}\right )-\frac {3 x^{-\frac {n}{3}}}{a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/a/n/(x^(1/3*n))+sum(_R*ln(x^(1/3*n)+a^3*n^2/b*_R^2),_R=RootOf(_Z^3*a^4*n^3-b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -b \int \frac {x^{\frac {2}{3} \, n}}{a b x x^{n} + a^{2} x}\,{d x} - \frac {3}{a n x^{\frac {1}{3} \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{\frac {n}{3}+1}\,\left (a+b\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(n/3 + 1)*(a + b*x^n)),x)

[Out]

int(1/(x^(n/3 + 1)*(a + b*x^n)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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