3.6.1 \(\int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx\) [501]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {1598, 369, 352, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {\sqrt {3} c^{2/3} \text {ArcTan}\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}-\frac {3 x^{-2 n/3}}{2 b n} \end {gather*}

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 31

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

Rule 206

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 210

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

Rule 352

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 369

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 631

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 642

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 648

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 1598

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]

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) \text {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 \text {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 \text {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} \text {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) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 34, normalized size = 0.21 \begin {gather*} -\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} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.20, size = 54, normalized size = 0.34

method	result	size
risch	\(-\frac {3 x^{-\frac {2 n}{3}}}{2 b n}+\left (\munderset {\textit {\_R} =\RootOf \left (b^{5} n^{3} \textit {\_Z}^{3}+c^{2}\right )}{\sum }\textit {\_R} \ln \left (x^{\frac {n}{3}}-\frac {b^{2} n \textit {\_R}}{c}\right )\right )\)	\(54\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.37, size = 212, normalized size = 1.32 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 [A]
time = 6.08, size = 136, normalized size = 0.85 \begin {gather*} \frac {\frac {2 \, c \left (-\frac {b}{c}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3} \, n} - \left (-\frac {b}{c}\right )^{\frac {1}{3}} \right |}\right )}{b^{2}} - \frac {2 \, \sqrt {3} \left (-b c^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3} \, n} + \left (-\frac {b}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{c}\right )^{\frac {1}{3}}}\right )}{b^{2}} - \frac {\left (-b c^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {1}{3} \, n} \left (-\frac {b}{c}\right )^{\frac {1}{3}} + {\left (x^{n}\right )}^{\frac {2}{3}} + \left (-\frac {b}{c}\right )^{\frac {2}{3}}\right )}{b^{2}} - \frac {3}{b {\left (x^{n}\right )}^{\frac {2}{3}}}}{2 \, n} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{\frac {n}{3}-1}}{b\,x^n+c\,x^{2\,n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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