3.2640 \(\int \frac {x^{-1-\frac {2 n}{3}}}{a+b x^n} \, dx\)

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

[Out]

-3/2/a/n/(x^(2/3*n))-b^(2/3)*ln(a^(1/3)+b^(1/3)*x^(1/3*n))/a^(5/3)/n+1/2*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x^
(1/3*n)+b^(2/3)*x^(2/3*n))/a^(5/3)/n+b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x^(1/3*n))/a^(1/3)*3^(1/2))*3^(1/2)
/a^(5/3)/n

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Rubi [A]  time = 0.11, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {362, 345, 200, 31, 634, 617, 204, 628} \[ -\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x^{n/3}\right )}{a^{5/3} n}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x^{n/3}+b^{2/3} x^{2 n/3}\right )}{2 a^{5/3} n}+\frac {\sqrt {3} b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x^{n/3}}{\sqrt {3} \sqrt [3]{a}}\right )}{a^{5/3} n}-\frac {3 x^{-2 n/3}}{2 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - (2*n)/3)/(a + b*x^n),x]

[Out]

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

Rule 31

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

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

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Mathematica [C]  time = 0.02, size = 34, normalized size = 0.21 \[ -\frac {3 x^{-2 n/3} \, _2F_1\left (-\frac {2}{3},1;\frac {1}{3};-\frac {b x^n}{a}\right )}{2 a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 178, normalized size = 1.11 \[ -\frac {3 \, x x^{-\frac {2}{3} \, n - 1} - 2 \, \sqrt {3} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a \sqrt {x} x^{-\frac {1}{3} \, n - \frac {1}{2}} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + \sqrt {3} b}{3 \, b}\right ) + \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {a \sqrt {x} x^{-\frac {1}{3} \, n - \frac {1}{2}} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - b x x^{-\frac {2}{3} \, n - 1} + b \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}}{x}\right ) - 2 \, \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (\frac {b x x^{-\frac {1}{3} \, n - \frac {1}{2}} + a \sqrt {x} \left (-\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}}{x}\right )}{2 \, a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 54, normalized size = 0.34 \[ \RootOf \left (a^{5} n^{3} \textit {\_Z}^{3}+b^{2}\right ) \ln \left (-\frac {\RootOf \left (a^{5} n^{3} \textit {\_Z}^{3}+b^{2}\right ) a^{2} n}{b}+x^{\frac {n}{3}}\right )-\frac {3 x^{-\frac {2 n}{3}}}{2 a n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.47, size = 192, normalized size = 1.20 \[ \frac {x^{- \frac {2 n}{3}} \Gamma \left (- \frac {2}{3}\right )}{a n \Gamma \left (\frac {1}{3}\right )} - \frac {2 b^{\frac {2}{3}} e^{- \frac {i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{\frac {i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {2}{3}\right )}{3 a^{\frac {5}{3}} n \Gamma \left (\frac {1}{3}\right )} + \frac {2 b^{\frac {2}{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{i \pi }}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {2}{3}\right )}{3 a^{\frac {5}{3}} n \Gamma \left (\frac {1}{3}\right )} - \frac {2 b^{\frac {2}{3}} e^{\frac {i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{\frac {5 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {2}{3}\right )}{3 a^{\frac {5}{3}} n \Gamma \left (\frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**(-2*n/3)*gamma(-2/3)/(a*n*gamma(1/3)) - 2*b**(2/3)*exp(-I*pi/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(I*pi/3)
/a**(1/3))*gamma(-2/3)/(3*a**(5/3)*n*gamma(1/3)) + 2*b**(2/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(I*pi)/a**(1/
3))*gamma(-2/3)/(3*a**(5/3)*n*gamma(1/3)) - 2*b**(2/3)*exp(I*pi/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(5*I*pi/
3)/a**(1/3))*gamma(-2/3)/(3*a**(5/3)*n*gamma(1/3))

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