\(\int \frac {1}{x^3 (a+b (c x^n)^{\frac {1}{n}})^2} \, dx\) [117]

Optimal result

Integrand size = 19, antiderivative size = 125 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=-\frac {1}{2 a^2 x^2}+\frac {2 b \left (c x^n\right )^{\frac {1}{n}}}{a^3 x^2}+\frac {b^2 \left (c x^n\right )^{2/n}}{a^3 x^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 b^2 \left (c x^n\right )^{2/n} \log (x)}{a^4 x^2}-\frac {3 b^2 \left (c x^n\right )^{2/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^4 x^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {\left (c x^n\right )^{2/n} \left (a \left (-a \left (c x^n\right )^{-2/n}+4 b \left (c x^n\right )^{-1/n}+\frac {2 b^2}{a+b \left (c x^n\right )^{\frac {1}{n}}}\right )+6 b^2 \log (x)-6 b^2 \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )\right )}{2 a^4 x^2} \] Input:

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

Output:

((c*x^n)^(2/n)*(a*(-(a/(c*x^n)^(2/n)) + (4*b)/(c*x^n)^n^(-1) + (2*b^2)/(a 
+ b*(c*x^n)^n^(-1))) + 6*b^2*Log[x] - 6*b^2*Log[a + b*(c*x^n)^n^(-1)]))/(2 
*a^4*x^2)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {892, 54, 2009}

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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 54

\(\displaystyle \frac {\left (c x^n\right )^{2/n} \int \left (\frac {\left (c x^n\right )^{-3/n}}{a^2}-\frac {2 b \left (c x^n\right )^{-2/n}}{a^3}+\frac {3 b^2 \left (c x^n\right )^{-1/n}}{a^4}-\frac {3 b^3}{a^4 \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )}-\frac {b^3}{a^3 \left (b \left (c x^n\right )^{\frac {1}{n}}+a\right )^2}\right )d\left (c x^n\right )^{\frac {1}{n}}}{x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (c x^n\right )^{2/n} \left (\frac {3 b^2 \log \left (\left (c x^n\right )^{\frac {1}{n}}\right )}{a^4}-\frac {3 b^2 \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{a^4}+\frac {b^2}{a^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {2 b \left (c x^n\right )^{-1/n}}{a^3}-\frac {\left (c x^n\right )^{-2/n}}{2 a^2}\right )}{x^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 54

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && 
 ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
 

rule 892

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1))   Subst[Int[x^m*(a + b 
*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x 
] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (warning: unable to verify)

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

Time = 0.26 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.03

method	result	size
risch	\(\frac {1}{a \,x^{2} \left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right )}-\frac {3}{2 a^{2} x^{2}}+\frac {3 \left (x^{n}\right )^{\frac {2}{n}} c^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}} b^{2} \ln \left (x \right )}{a^{4} x^{2}}+\frac {3 b \left (x^{n}\right )^{\frac {1}{n}} c^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}}{a^{3} x^{2}}-\frac {3 \ln \left (b \,c^{\frac {1}{n}} \left (x^{n}\right )^{\frac {1}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{2 n}}+a \right ) b^{2} \left (x^{n}\right )^{\frac {2}{n}} c^{\frac {2}{n}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \,x^{n}\right ) \left (-\operatorname {csgn}\left (i x^{n}\right )+\operatorname {csgn}\left (i c \,x^{n}\right )\right ) \left (\operatorname {csgn}\left (i c \right )-\operatorname {csgn}\left (i c \,x^{n}\right )\right )}{n}}}{a^{4} x^{2}}\)	\(379\)

Input:

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

Output:

1/a/x^2/(b*c^(1/n)*(x^n)^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+cs 
gn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)-3/2/a^2/x^2+3/a^4/x^2*(x^n)^( 
2/n)*c^(2/n)*exp(I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c 
)-csgn(I*c*x^n))/n)*b^2*ln(x)+3/a^3*b/x^2*(x^n)^(1/n)*c^(1/n)*exp(1/2*I*Pi 
*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)-3 
/a^4*ln(b*c^(1/n)*(x^n)^(1/n)*exp(1/2*I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csg 
n(I*c*x^n))*(csgn(I*c)-csgn(I*c*x^n))/n)+a)*b^2/x^2*((x^n)^(1/n))^2*(c^(1/ 
n))^2*exp(I*Pi*csgn(I*c*x^n)*(-csgn(I*x^n)+csgn(I*c*x^n))*(csgn(I*c)-csgn( 
I*c*x^n))/n)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {6 \, b^{3} c^{\frac {3}{n}} x^{3} \log \left (x\right ) + 3 \, a^{2} b c^{\left (\frac {1}{n}\right )} x - a^{3} + 6 \, {\left (a b^{2} x^{2} \log \left (x\right ) + a b^{2} x^{2}\right )} c^{\frac {2}{n}} - 6 \, {\left (b^{3} c^{\frac {3}{n}} x^{3} + a b^{2} c^{\frac {2}{n}} x^{2}\right )} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right )}{2 \, {\left (a^{4} b c^{\left (\frac {1}{n}\right )} x^{3} + a^{5} x^{2}\right )}} \] Input:

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

Output:

1/2*(6*b^3*c^(3/n)*x^3*log(x) + 3*a^2*b*c^(1/n)*x - a^3 + 6*(a*b^2*x^2*log 
(x) + a*b^2*x^2)*c^(2/n) - 6*(b^3*c^(3/n)*x^3 + a*b^2*c^(2/n)*x^2)*log(b*c 
^(1/n)*x + a))/(a^4*b*c^(1/n)*x^3 + a^5*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {-6 c^{\frac {3}{n}} \mathrm {log}\left (c^{\frac {1}{n}} b x +a \right ) b^{3} x^{3}+6 c^{\frac {3}{n}} \mathrm {log}\left (x \right ) b^{3} x^{3}-6 c^{\frac {3}{n}} b^{3} x^{3}-6 c^{\frac {2}{n}} \mathrm {log}\left (c^{\frac {1}{n}} b x +a \right ) a \,b^{2} x^{2}+6 c^{\frac {2}{n}} \mathrm {log}\left (x \right ) a \,b^{2} x^{2}+3 c^{\frac {1}{n}} a^{2} b x -a^{3}}{2 a^{4} x^{2} \left (c^{\frac {1}{n}} b x +a \right )} \] Input:

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

Output:

( - 6*c**(3/n)*log(c**(1/n)*b*x + a)*b**3*x**3 + 6*c**(3/n)*log(x)*b**3*x* 
*3 - 6*c**(3/n)*b**3*x**3 - 6*c**(2/n)*log(c**(1/n)*b*x + a)*a*b**2*x**2 + 
 6*c**(2/n)*log(x)*a*b**2*x**2 + 3*c**(1/n)*a**2*b*x - a**3)/(2*a**4*x**2* 
(c**(1/n)*b*x + a))