3.2647 \(\int \frac {x^{-1-\frac {4 n}{3}}}{a+b x^n} \, dx\)

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

[Out]

-3/4/a/n/(x^(4/3*n))+3*b/a^2/n/(x^(1/3*n))-b^(4/3)*ln(b^(1/3)+a^(1/3)/(x^(1/3*n)))/a^(7/3)/n+1/2*b^(4/3)*ln(b^
(2/3)+a^(2/3)/(x^(2/3*n))-a^(1/3)*b^(1/3)/(x^(1/3*n)))/a^(7/3)/n+b^(4/3)*arctan(1/3*(1-2*a^(1/3)/b^(1/3)/(x^(1
/3*n)))*3^(1/2))*3^(1/2)/a^(7/3)/n

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Rubi [A]  time = 0.11, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {362, 345, 193, 321, 200, 31, 634, 617, 204, 628} \[ -\frac {b^{4/3} \log \left (\sqrt [3]{a} x^{-n/3}+\sqrt [3]{b}\right )}{a^{7/3} n}+\frac {b^{4/3} \log \left (a^{2/3} x^{-2 n/3}-\sqrt [3]{a} \sqrt [3]{b} x^{-n/3}+b^{2/3}\right )}{2 a^{7/3} n}+\frac {\sqrt {3} b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x^{-n/3}}{\sqrt {3} \sqrt [3]{b}}\right )}{a^{7/3} n}+\frac {3 b x^{-n/3}}{a^2 n}-\frac {3 x^{-4 n/3}}{4 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - (4*n)/3)/(a + b*x^n),x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.19 \[ -\frac {3 x^{-4 n/3} \, _2F_1\left (-\frac {4}{3},1;-\frac {1}{3};-\frac {b x^n}{a}\right )}{4 a n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - (4*n)/3)/(a + b*x^n),x]

[Out]

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

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fricas [A]  time = 0.76, size = 180, normalized size = 1.02 \[ -\frac {3 \, a x x^{-\frac {4}{3} \, n - 1} - 4 \, \sqrt {3} b \left (-\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x^{\frac {1}{4}} x^{-\frac {1}{3} \, n - \frac {1}{4}} \left (-\frac {b}{a}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) + 2 \, b \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (\frac {x^{\frac {3}{4}} x^{-\frac {1}{3} \, n - \frac {1}{4}} \left (-\frac {b}{a}\right )^{\frac {1}{3}} + x x^{-\frac {2}{3} \, n - \frac {1}{2}} + \sqrt {x} \left (-\frac {b}{a}\right )^{\frac {2}{3}}}{x}\right ) - 4 \, b \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (\frac {x x^{-\frac {1}{3} \, n - \frac {1}{4}} - x^{\frac {3}{4}} \left (-\frac {b}{a}\right )^{\frac {1}{3}}}{x}\right ) - 12 \, b x^{\frac {1}{4}} x^{-\frac {1}{3} \, n - \frac {1}{4}}}{4 \, a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4/3*n)/(a+b*x^n),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{-\frac {4}{3} \, n - 1}}{b x^{n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4/3*n)/(a+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(-4/3*n - 1)/(b*x^n + a), x)

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maple [C]  time = 0.08, size = 73, normalized size = 0.41 \[ \RootOf \left (a^{7} n^{3} \textit {\_Z}^{3}+b^{4}\right ) \ln \left (\frac {\RootOf \left (a^{7} n^{3} \textit {\_Z}^{3}+b^{4}\right )^{2} a^{5} n^{2}}{b^{3}}+x^{\frac {n}{3}}\right )-\frac {3 x^{-\frac {4 n}{3}}}{4 a n}+\frac {3 b \,x^{-\frac {n}{3}}}{a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-4/3*n)/(b*x^n+a),x)

[Out]

3*b/a^2/n/(x^(1/3*n))-3/4/a/n/(x^(1/3*n))^4+sum(_R*ln(x^(1/3*n)+a^5*n^2/b^3*_R^2),_R=RootOf(_Z^3*a^7*n^3+b^4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{2} \int \frac {x^{\frac {2}{3} \, n}}{a^{2} b x x^{n} + a^{3} x}\,{d x} + \frac {3 \, {\left (4 \, b x^{n} - a\right )}}{4 \, a^{2} n x^{\frac {4}{3} \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-4/3*n)/(a+b*x^n),x, algorithm="maxima")

[Out]

b^2*integrate(x^(2/3*n)/(a^2*b*x*x^n + a^3*x), x) + 3/4*(4*b*x^n - a)/(a^2*n*x^(4/3*n))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{\frac {4\,n}{3}+1}\,\left (a+b\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^((4*n)/3 + 1)*(a + b*x^n)),x)

[Out]

int(1/(x^((4*n)/3 + 1)*(a + b*x^n)), x)

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sympy [C]  time = 4.34, size = 230, normalized size = 1.31 \[ \frac {x^{- \frac {4 n}{3}} \Gamma \left (- \frac {4}{3}\right )}{a n \Gamma \left (- \frac {1}{3}\right )} - \frac {4 b x^{- \frac {n}{3}} \Gamma \left (- \frac {4}{3}\right )}{a^{2} n \Gamma \left (- \frac {1}{3}\right )} + \frac {4 b^{\frac {4}{3}} e^{- \frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{\frac {i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {4}{3}\right )}{3 a^{\frac {7}{3}} n \Gamma \left (- \frac {1}{3}\right )} + \frac {4 b^{\frac {4}{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{i \pi }}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {4}{3}\right )}{3 a^{\frac {7}{3}} n \Gamma \left (- \frac {1}{3}\right )} + \frac {4 b^{\frac {4}{3}} e^{\frac {2 i \pi }{3}} \log {\left (1 - \frac {\sqrt [3]{b} x^{\frac {n}{3}} e^{\frac {5 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (- \frac {4}{3}\right )}{3 a^{\frac {7}{3}} n \Gamma \left (- \frac {1}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-4/3*n)/(a+b*x**n),x)

[Out]

x**(-4*n/3)*gamma(-4/3)/(a*n*gamma(-1/3)) - 4*b*x**(-n/3)*gamma(-4/3)/(a**2*n*gamma(-1/3)) + 4*b**(4/3)*exp(-2
*I*pi/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(I*pi/3)/a**(1/3))*gamma(-4/3)/(3*a**(7/3)*n*gamma(-1/3)) + 4*b**(
4/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(I*pi)/a**(1/3))*gamma(-4/3)/(3*a**(7/3)*n*gamma(-1/3)) + 4*b**(4/3)*e
xp(2*I*pi/3)*log(1 - b**(1/3)*x**(n/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(-4/3)/(3*a**(7/3)*n*gamma(-1/3))

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