\(\int x^3 (a+b (c x^n)^{\frac {1}{n}})^p \, dx\) [120]

Optimal result

Integrand size = 19, antiderivative size = 171 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=-\frac {a^3 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p}}{b^4 (1+p)}+\frac {3 a^2 x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{2+p}}{b^4 (2+p)}-\frac {3 a x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{3+p}}{b^4 (3+p)}+\frac {x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{4+p}}{b^4 (4+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {x^4 \left (c x^n\right )^{-4/n} \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^{1+p} \left (-\frac {a^3}{1+p}+\frac {3 a^2 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )}{2+p}-\frac {3 a \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^2}{3+p}+\frac {\left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^3}{4+p}\right )}{b^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {892, 53, 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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 53

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 53

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, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[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 = 0.73 (sec) , antiderivative size = 1127, normalized size of antiderivative = 6.59

method	result	size
risch	\(\text {Expression too large to display}\)	\(1127\)

Input:

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

Output:

(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)^(p+1)/(c^(1/n))*x^4/((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)/b/(p+1)-3/b/(p+1)*(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)^(p+1)/(4+p 
)/((x^n)^(1/n))*x^4/(c^(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)-3/b^2/(p+1)*(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)^(p+1)/(4+p)/(3+p)*a*x^4/((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)*p-3/b^2/(p+1)*(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)^(p+1)/(4+p)/(3+p)*a* 
x^4/((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)-6/b^4/(p+1)*(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)^(p+1)/(4+p)*a^3/(2+p)*x^4/((x^n)^(1/n))^4/(c^(1/n))^4/(3+p) 
*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)+6/b^3/(p+1)*(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)^(p+1)/(4+p)/(2 
+p)/(3+p)*a^2*x^4/((x^n)^(1/n))^3/(c^(1/n))^3*exp(-3/2*I*Pi*csgn(I*c*x^...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.07 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (6 \, a^{3} b c^{\left (\frac {1}{n}\right )} p x + {\left (b^{4} p^{3} + 6 \, b^{4} p^{2} + 11 \, b^{4} p + 6 \, b^{4}\right )} c^{\frac {4}{n}} x^{4} + {\left (a b^{3} p^{3} + 3 \, a b^{3} p^{2} + 2 \, a b^{3} p\right )} c^{\frac {3}{n}} x^{3} - 6 \, a^{4} - 3 \, {\left (a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} c^{\frac {2}{n}} x^{2}\right )} {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p}}{{\left (b^{4} p^{4} + 10 \, b^{4} p^{3} + 35 \, b^{4} p^{2} + 50 \, b^{4} p + 24 \, b^{4}\right )} c^{\frac {4}{n}}} \] Input:

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

Output:

(6*a^3*b*c^(1/n)*p*x + (b^4*p^3 + 6*b^4*p^2 + 11*b^4*p + 6*b^4)*c^(4/n)*x^ 
4 + (a*b^3*p^3 + 3*a*b^3*p^2 + 2*a*b^3*p)*c^(3/n)*x^3 - 6*a^4 - 3*(a^2*b^2 
*p^2 + a^2*b^2*p)*c^(2/n)*x^2)*(b*c^(1/n)*x + a)^p/((b^4*p^4 + 10*b^4*p^3 
+ 35*b^4*p^2 + 50*b^4*p + 24*b^4)*c^(4/n))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (179) = 358\).

Time = 0.18 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.25 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {{\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p^{3} x^{4} + {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p^{3} x^{3} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p^{2} x^{4} + 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p^{2} x^{3} + 11 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} p x^{4} - 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac {2}{n}} p^{2} x^{2} + 2 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a b^{3} c^{\frac {3}{n}} p x^{3} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} b^{4} c^{\frac {4}{n}} x^{4} - 3 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{2} b^{2} c^{\frac {2}{n}} p x^{2} + 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{3} b c^{\left (\frac {1}{n}\right )} p x - 6 \, {\left (b c^{\left (\frac {1}{n}\right )} x + a\right )}^{p} a^{4}}{b^{4} c^{\frac {4}{n}} p^{4} + 10 \, b^{4} c^{\frac {4}{n}} p^{3} + 35 \, b^{4} c^{\frac {4}{n}} p^{2} + 50 \, b^{4} c^{\frac {4}{n}} p + 24 \, b^{4} c^{\frac {4}{n}}} \] Input:

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

Output:

((b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p^3*x^4 + (b*c^(1/n)*x + a)^p*a*b^3*c^(3/ 
n)*p^3*x^3 + 6*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p^2*x^4 + 3*(b*c^(1/n)*x + 
a)^p*a*b^3*c^(3/n)*p^2*x^3 + 11*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*p*x^4 - 3* 
(b*c^(1/n)*x + a)^p*a^2*b^2*c^(2/n)*p^2*x^2 + 2*(b*c^(1/n)*x + a)^p*a*b^3* 
c^(3/n)*p*x^3 + 6*(b*c^(1/n)*x + a)^p*b^4*c^(4/n)*x^4 - 3*(b*c^(1/n)*x + a 
)^p*a^2*b^2*c^(2/n)*p*x^2 + 6*(b*c^(1/n)*x + a)^p*a^3*b*c^(1/n)*p*x - 6*(b 
*c^(1/n)*x + a)^p*a^4)/(b^4*c^(4/n)*p^4 + 10*b^4*c^(4/n)*p^3 + 35*b^4*c^(4 
/n)*p^2 + 50*b^4*c^(4/n)*p + 24*b^4*c^(4/n))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.31 \[ \int x^3 \left (a+b \left (c x^n\right )^{\frac {1}{n}}\right )^p \, dx=\frac {\left (c^{\frac {1}{n}} b x +a \right )^{p} \left (c^{\frac {4}{n}} b^{4} p^{3} x^{4}+6 c^{\frac {4}{n}} b^{4} p^{2} x^{4}+11 c^{\frac {4}{n}} b^{4} p \,x^{4}+6 c^{\frac {4}{n}} b^{4} x^{4}+c^{\frac {3}{n}} a \,b^{3} p^{3} x^{3}+3 c^{\frac {3}{n}} a \,b^{3} p^{2} x^{3}+2 c^{\frac {3}{n}} a \,b^{3} p \,x^{3}-3 c^{\frac {2}{n}} a^{2} b^{2} p^{2} x^{2}-3 c^{\frac {2}{n}} a^{2} b^{2} p \,x^{2}+6 c^{\frac {1}{n}} a^{3} b p x -6 a^{4}\right )}{c^{\frac {4}{n}} b^{4} \left (p^{4}+10 p^{3}+35 p^{2}+50 p +24\right )} \] Input:

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

Output:

((c**(1/n)*b*x + a)**p*(c**(4/n)*b**4*p**3*x**4 + 6*c**(4/n)*b**4*p**2*x** 
4 + 11*c**(4/n)*b**4*p*x**4 + 6*c**(4/n)*b**4*x**4 + c**(3/n)*a*b**3*p**3* 
x**3 + 3*c**(3/n)*a*b**3*p**2*x**3 + 2*c**(3/n)*a*b**3*p*x**3 - 3*c**(2/n) 
*a**2*b**2*p**2*x**2 - 3*c**(2/n)*a**2*b**2*p*x**2 + 6*c**(1/n)*a**3*b*p*x 
 - 6*a**4))/(c**(4/n)*b**4*(p**4 + 10*p**3 + 35*p**2 + 50*p + 24))