\(\int \frac {1}{x (a+b x^n+c x^{2 n})^{3/2}} \, dx\) [237]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \sqrt {a+b x^n+c x^{2 n}}}-\frac {\text {arctanh}\left (\frac {2 a+b x^n}{2 \sqrt {a} \sqrt {a+b x^n+c x^{2 n}}}\right )}{a^{3/2} n} \] Output:

2*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))^(1/2)-arctanh(1 
/2*(2*a+b*x^n)/a^(1/2)/(a+b*x^n+c*x^(2*n))^(1/2))/a^(3/2)/n
 

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\frac {2 \left (-\frac {\sqrt {a} \left (-b^2+2 a c-b c x^n\right )}{\left (b^2-4 a c\right ) \sqrt {a+x^n \left (b+c x^n\right )}}+\text {arctanh}\left (\frac {\sqrt {c} x^n-\sqrt {a+x^n \left (b+c x^n\right )}}{\sqrt {a}}\right )\right )}{a^{3/2} n} \] Input:

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

Output:

(2*(-((Sqrt[a]*(-b^2 + 2*a*c - b*c*x^n))/((b^2 - 4*a*c)*Sqrt[a + x^n*(b + 
c*x^n)])) + ArcTanh[(Sqrt[c]*x^n - Sqrt[a + x^n*(b + c*x^n)])/Sqrt[a]]))/( 
a^(3/2)*n)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1693, 1165, 27, 1154, 219}

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.

\(\displaystyle \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1693

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

\(\Big \downarrow \) 1165

\(\displaystyle \frac {\frac {2 \left (-2 a c+b^2+b c x^n\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^n+c x^{2 n}}}-\frac {2 \int -\frac {\left (b^2-4 a c\right ) x^{-n}}{2 \sqrt {b x^n+c x^{2 n}+a}}dx^n}{a \left (b^2-4 a c\right )}}{n}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {x^{-n}}{\sqrt {b x^n+c x^{2 n}+a}}dx^n}{a}+\frac {2 \left (-2 a c+b^2+b c x^n\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^n+c x^{2 n}}}}{n}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\frac {2 \left (-2 a c+b^2+b c x^n\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^n+c x^{2 n}}}-\frac {2 \int \frac {1}{4 a-x^{2 n}}d\frac {b x^n+2 a}{\sqrt {b x^n+c x^{2 n}+a}}}{a}}{n}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 \left (-2 a c+b^2+b c x^n\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x^n+c x^{2 n}}}-\frac {\text {arctanh}\left (\frac {2 a+b x^n}{2 \sqrt {a} \sqrt {a+b x^n+c x^{2 n}}}\right )}{a^{3/2}}}{n}\)

Input:

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

Output:

((2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*Sqrt[a + b*x^n + c*x^(2*n)]) 
 - ArcTanh[(2*a + b*x^n)/(2*Sqrt[a]*Sqrt[a + b*x^n + c*x^(2*n)])]/a^(3/2)) 
/n
 

Defintions of rubi rules used

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 219

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

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1165

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e) 
*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^ 
2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))   Int[(d 
+ e*x)^m*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p 
+ 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + 
 b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] 
 && IntQuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1693

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, 
x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ 
[Simplify[(m + 1)/n]]
 

Maple [F]

\[\int \frac {1}{x \left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (88) = 176\).

Time = 0.27 (sec) , antiderivative size = 449, normalized size of antiderivative = 4.58 \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {a} x^{2 \, n} + {\left (b^{3} - 4 \, a b c\right )} \sqrt {a} x^{n} + {\left (a b^{2} - 4 \, a^{2} c\right )} \sqrt {a}\right )} \log \left (-\frac {8 \, a b x^{n} + 8 \, a^{2} + {\left (b^{2} + 4 \, a c\right )} x^{2 \, n} - 4 \, {\left (\sqrt {a} b x^{n} + 2 \, a^{\frac {3}{2}}\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{x^{2 \, n}}\right ) + 4 \, {\left (a b c x^{n} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} n x^{2 \, n} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} n x^{n} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n\right )}}, \frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} \sqrt {-a} x^{2 \, n} + {\left (b^{3} - 4 \, a b c\right )} \sqrt {-a} x^{n} + {\left (a b^{2} - 4 \, a^{2} c\right )} \sqrt {-a}\right )} \arctan \left (\frac {{\left (\sqrt {-a} b x^{n} + 2 \, \sqrt {-a} a\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{2 \, {\left (a c x^{2 \, n} + a b x^{n} + a^{2}\right )}}\right ) + 2 \, {\left (a b c x^{n} + a b^{2} - 2 \, a^{2} c\right )} \sqrt {c x^{2 \, n} + b x^{n} + a}}{{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} n x^{2 \, n} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} n x^{n} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} n}\right ] \] Input:

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

Output:

[1/2*(((b^2*c - 4*a*c^2)*sqrt(a)*x^(2*n) + (b^3 - 4*a*b*c)*sqrt(a)*x^n + ( 
a*b^2 - 4*a^2*c)*sqrt(a))*log(-(8*a*b*x^n + 8*a^2 + (b^2 + 4*a*c)*x^(2*n) 
- 4*(sqrt(a)*b*x^n + 2*a^(3/2))*sqrt(c*x^(2*n) + b*x^n + a))/x^(2*n)) + 4* 
(a*b*c*x^n + a*b^2 - 2*a^2*c)*sqrt(c*x^(2*n) + b*x^n + a))/((a^2*b^2*c - 4 
*a^3*c^2)*n*x^(2*n) + (a^2*b^3 - 4*a^3*b*c)*n*x^n + (a^3*b^2 - 4*a^4*c)*n) 
, (((b^2*c - 4*a*c^2)*sqrt(-a)*x^(2*n) + (b^3 - 4*a*b*c)*sqrt(-a)*x^n + (a 
*b^2 - 4*a^2*c)*sqrt(-a))*arctan(1/2*(sqrt(-a)*b*x^n + 2*sqrt(-a)*a)*sqrt( 
c*x^(2*n) + b*x^n + a)/(a*c*x^(2*n) + a*b*x^n + a^2)) + 2*(a*b*c*x^n + a*b 
^2 - 2*a^2*c)*sqrt(c*x^(2*n) + b*x^n + a))/((a^2*b^2*c - 4*a^3*c^2)*n*x^(2 
*n) + (a^2*b^3 - 4*a^3*b*c)*n*x^n + (a^3*b^2 - 4*a^4*c)*n)]
 

Sympy [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {1}{x \left (a + b x^{n} + c x^{2 n}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral(1/(x*(a + b*x**n + c*x**(2*n))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}} x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {3}{2}} x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a+b\,x^n+c\,x^{2\,n}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x \left (a+b x^n+c x^{2 n}\right )^{3/2}} \, dx=\int \frac {\sqrt {x^{2 n} c +x^{n} b +a}}{x^{4 n} c^{2} x +2 x^{3 n} b c x +2 x^{2 n} a c x +x^{2 n} b^{2} x +2 x^{n} a b x +a^{2} x}d x \] Input:

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

Output:

int(sqrt(x**(2*n)*c + x**n*b + a)/(x**(4*n)*c**2*x + 2*x**(3*n)*b*c*x + 2* 
x**(2*n)*a*c*x + x**(2*n)*b**2*x + 2*x**n*a*b*x + a**2*x),x)