\(\int \frac {x^3}{(a+b (c x^n)^{\frac {1}{n}})^2} \, dx\) [111]

Optimal result

Integrand size = 19, antiderivative size = 114 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=-\frac {2 a x^4 \left (c x^n\right )^{-3/n}}{b^3}+\frac {x^4 \left (c x^n\right )^{-2/n}}{2 b^2}+\frac {a^3 x^4 \left (c x^n\right )^{-4/n}}{b^4 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}+\frac {3 a^2 x^4 \left (c x^n\right )^{-4/n} \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{b^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {x^4 \left (c x^n\right )^{-4/n} \left (-4 a b \left (c x^n\right )^{\frac {1}{n}}+b^2 \left (c x^n\right )^{2/n}+\frac {2 a^3}{a+b \left (c x^n\right )^{\frac {1}{n}}}+6 a^2 \log \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )\right )}{2 b^4} \] Input:

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

Output:

(x^4*(-4*a*b*(c*x^n)^n^(-1) + b^2*(c*x^n)^(2/n) + (2*a^3)/(a + b*(c*x^n)^n 
^(-1)) + 6*a^2*Log[a + b*(c*x^n)^n^(-1)]))/(2*b^4*(c*x^n)^(4/n))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {892, 49, 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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[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 = 0.16 (sec) , antiderivative size = 463, normalized size of antiderivative = 4.06

method	result	size
risch	\(\frac {x^{4}}{a \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 {\left (x^{n}\right )^{-\frac {1}{n}} c^{-\frac {1}{n}} x^{4} {\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 b}+\frac {3 \left (x^{n}\right )^{-\frac {2}{n}} c^{-\frac {2}{n}} x^{4} {\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}}}{2 b^{2}}-\frac {3 a \left (x^{n}\right )^{-\frac {3}{n}} c^{-\frac {3}{n}} x^{4} {\mathrm e}^{-\frac {3 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}}}{b^{3}}+\frac {3 a^{2} \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 ) c^{-\frac {1}{n}} \left (x^{n}\right )^{-\frac {1}{n}} \left (x^{n}\right )^{-\frac {3}{n}} c^{-\frac {3}{n}} x^{4} {\mathrm e}^{-\frac {2 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^{4}}\)	\(463\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {b^{3} c^{\frac {3}{n}} x^{3} - 3 \, a b^{2} c^{\frac {2}{n}} x^{2} - 4 \, a^{2} b c^{\left (\frac {1}{n}\right )} x + 2 \, a^{3} + 6 \, {\left (a^{2} b c^{\left (\frac {1}{n}\right )} x + a^{3}\right )} \log \left (b c^{\left (\frac {1}{n}\right )} x + a\right )}{2 \, {\left (b^{5} c^{\frac {5}{n}} x + a b^{4} c^{\frac {4}{n}}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2} \, dx=\frac {c^{\frac {3}{n}} b^{3} x^{3}-3 c^{\frac {2}{n}} a \,b^{2} x^{2}+6 c^{\frac {1}{n}} \mathrm {log}\left (c^{\frac {1}{n}} b x +a \right ) a^{2} b x -6 c^{\frac {1}{n}} a^{2} b x +6 \,\mathrm {log}\left (c^{\frac {1}{n}} b x +a \right ) a^{3}}{2 c^{\frac {4}{n}} b^{4} \left (c^{\frac {1}{n}} b x +a \right )} \] Input:

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

Output:

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