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

3.6.1.1 Optimal result

Integrand size = 25, antiderivative size = 160 \[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=-\frac {3 x^{-2 n/3}}{2 b n}+\frac {\sqrt {3} c^{2/3} \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{\sqrt {3} \sqrt [3]{b}}\right )}{b^{5/3} n}-\frac {c^{2/3} \log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{b^{5/3} n}+\frac {c^{2/3} \log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 b^{5/3} n} \]

output

-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
 

3.6.1.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.21 \[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=-\frac {3 x^{-2 n/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},1,\frac {1}{3},-\frac {c x^n}{b}\right )}{2 b n} \]

input

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

output

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

3.6.1.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {10, 886, 868, 750, 16, 1142, 25, 27, 1082, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 10

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

\(\Big \downarrow \) 886

\(\displaystyle -\frac {c \int \frac {x^{\frac {n-3}{3}}}{c x^n+b}dx}{b}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 868

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

\(\Big \downarrow \) 750

\(\displaystyle -\frac {3 c \left (\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{c} x^{n/3}}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}}{3 b^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{c} x^{n/3}+\sqrt [3]{b}}dx^{n/3}}{3 b^{2/3}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 16

\(\displaystyle -\frac {3 c \left (\frac {\int \frac {2 \sqrt [3]{b}-\sqrt [3]{c} x^{n/3}}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 1142

\(\displaystyle -\frac {3 c \left (\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}-\frac {\int -\frac {\sqrt [3]{c} \left (\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}\right )}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}}{2 \sqrt [3]{c}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {3 c \left (\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}+\frac {\int \frac {\sqrt [3]{c} \left (\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}\right )}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}}{2 \sqrt [3]{c}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3 c \left (\frac {\frac {3}{2} \sqrt [3]{b} \int \frac {1}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}+\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {3 c \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}+\frac {3 \int \frac {1}{-x^{2 n/3}-3}d\left (1-\frac {2 \sqrt [3]{c} x^{n/3}}{\sqrt [3]{b}}\right )}{\sqrt [3]{c}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {3 c \left (\frac {\frac {1}{2} \int \frac {\sqrt [3]{b}-2 \sqrt [3]{c} x^{n/3}}{-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}+b^{2/3}}dx^{n/3}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{c} x^{n/3}}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{c}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {3 c \left (\frac {-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{c} x^{n/3}}{\sqrt [3]{b}}}{\sqrt {3}}\right )}{\sqrt [3]{c}}-\frac {\log \left (b^{2/3}-\sqrt [3]{b} \sqrt [3]{c} x^{n/3}+c^{2/3} x^{2 n/3}\right )}{2 \sqrt [3]{c}}}{3 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}+\sqrt [3]{c} x^{n/3}\right )}{3 b^{2/3} \sqrt [3]{c}}\right )}{b n}-\frac {3 x^{-2 n/3}}{2 b n}\)

input

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

output

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

3.6.1.3.1 Defintions of rubi rules used

rule 10

Int[(u_.)*((e_.)*(x_))^(m_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x 
_Symbol] :> Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*(a + b*x^(s - r))^p, x], 
 x] /; FreeQ[{a, b, e, m, r, s}, x] && IntegerQ[p] && (IntegerQ[p*r] || GtQ 
[e, 0]) && PosQ[s - r]
 

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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 868

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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 886

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[x^(m + 1)/(a*(m + 
 1)), x] - Simp[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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

3.6.1.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.67 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.34

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

input

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

output

-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))
 

3.6.1.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.32 \[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=\frac {2 \, \sqrt {3} x^{2} x^{\frac {2}{3} \, n - 2} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x x^{\frac {1}{3} \, n - 1} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {2}{3}} - \sqrt {3} c}{3 \, c}\right ) + 2 \, x^{2} x^{\frac {2}{3} \, n - 2} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c x x^{\frac {1}{3} \, n - 1} - b \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {1}{3}}}{x}\right ) - x^{2} x^{\frac {2}{3} \, n - 2} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {c^{2} x^{2} x^{\frac {2}{3} \, n - 2} + b c x x^{\frac {1}{3} \, n - 1} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {1}{3}} + b^{2} \left (-\frac {c^{2}}{b^{2}}\right )^{\frac {2}{3}}}{x^{2}}\right ) - 3}{2 \, b n x^{2} x^{\frac {2}{3} \, n - 2}} \]

input

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

output

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) - sqrt(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))
 

3.6.1.6 Sympy [F]

\[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=\int \frac {x^{- n} x^{\frac {n}{3} - 1}}{b + c x^{n}}\, dx \]

input

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

output

Integral(x**(n/3 - 1)/(x**n*(b + c*x**n)), x)
 

3.6.1.7 Maxima [F]

\[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=\int { \frac {x^{\frac {1}{3} \, n - 1}}{c x^{2 \, n} + b x^{n}} \,d x } \]

input

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

output

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

3.6.1.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.85 \[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=\frac {\frac {2 \, c \left (-\frac {b}{c}\right )^{\frac {1}{3}} \log \left ({\left | x^{\frac {1}{3} \, n} - \left (-\frac {b}{c}\right )^{\frac {1}{3}} \right |}\right )}{b^{2}} - \frac {2 \, \sqrt {3} \left (-b c^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3} \, n} + \left (-\frac {b}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b}{c}\right )^{\frac {1}{3}}}\right )}{b^{2}} - \frac {\left (-b c^{2}\right )^{\frac {1}{3}} \log \left (x^{\frac {1}{3} \, n} \left (-\frac {b}{c}\right )^{\frac {1}{3}} + {\left (x^{n}\right )}^{\frac {2}{3}} + \left (-\frac {b}{c}\right )^{\frac {2}{3}}\right )}{b^{2}} - \frac {3}{b {\left (x^{n}\right )}^{\frac {2}{3}}}}{2 \, n} \]

input

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

output

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
 

3.6.1.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1+\frac {n}{3}}}{b x^n+c x^{2 n}} \, dx=\int \frac {x^{\frac {n}{3}-1}}{b\,x^n+c\,x^{2\,n}} \,d x \]

input

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

output

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