\(\int F^{c (a+b x)^n} x^3 \, dx\) [179]

Optimal result

Integrand size = 15, antiderivative size = 207 \[ \int F^{c (a+b x)^n} x^3 \, dx=-\frac {(a+b x)^4 \Gamma \left (\frac {4}{n},-c (a+b x)^n \log (F)\right ) \left (-c (a+b x)^n \log (F)\right )^{-4/n}}{b^4 n}+\frac {3 a (a+b x)^3 \Gamma \left (\frac {3}{n},-c (a+b x)^n \log (F)\right ) \left (-c (a+b x)^n \log (F)\right )^{-3/n}}{b^4 n}-\frac {3 a^2 (a+b x)^2 \Gamma \left (\frac {2}{n},-c (a+b x)^n \log (F)\right ) \left (-c (a+b x)^n \log (F)\right )^{-2/n}}{b^4 n}+\frac {a^3 (a+b x) \Gamma \left (\frac {1}{n},-c (a+b x)^n \log (F)\right ) \left (-c (a+b x)^n \log (F)\right )^{-1/n}}{b^4 n} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.88 \[ \int F^{c (a+b x)^n} x^3 \, dx=-\frac {(a+b x) \left (-c (a+b x)^n \log (F)\right )^{-4/n} \left ((a+b x)^3 \Gamma \left (\frac {4}{n},-c (a+b x)^n \log (F)\right )-a \left (-c (a+b x)^n \log (F)\right )^{\frac {1}{n}} \left (3 (a+b x)^2 \Gamma \left (\frac {3}{n},-c (a+b x)^n \log (F)\right )+a \left (-c (a+b x)^n \log (F)\right )^{\frac {1}{n}} \left (-3 (a+b x) \Gamma \left (\frac {2}{n},-c (a+b x)^n \log (F)\right )+a \Gamma \left (\frac {1}{n},-c (a+b x)^n \log (F)\right ) \left (-c (a+b x)^n \log (F)\right )^{\frac {1}{n}}\right )\right )\right )}{b^4 n} \] Input:

Integrate[F^(c*(a + b*x)^n)*x^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2656, 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.

\(\displaystyle \int x^3 F^{c (a+b x)^n} \, dx\)

\(\Big \downarrow \) 2656

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

\(\Big \downarrow \) 2009

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

Input:

Int[F^(c*(a + b*x)^n)*x^3,x]
 

Output:

-(((a + b*x)^4*Gamma[4/n, -(c*(a + b*x)^n*Log[F])])/(b^4*n*(-(c*(a + b*x)^ 
n*Log[F]))^(4/n))) + (3*a*(a + b*x)^3*Gamma[3/n, -(c*(a + b*x)^n*Log[F])]) 
/(b^4*n*(-(c*(a + b*x)^n*Log[F]))^(3/n)) - (3*a^2*(a + b*x)^2*Gamma[2/n, - 
(c*(a + b*x)^n*Log[F])])/(b^4*n*(-(c*(a + b*x)^n*Log[F]))^(2/n)) + (a^3*(a 
 + b*x)*Gamma[n^(-1), -(c*(a + b*x)^n*Log[F])])/(b^4*n*(-(c*(a + b*x)^n*Lo 
g[F]))^n^(-1))
 

Defintions of rubi rules used

rule 2009

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

rule 2656

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ 
ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, 
 b, c, d, n}, x] && PolynomialQ[Px, x]
 

Maple [F]

\[\int F^{c \left (b x +a \right )^{n}} x^{3}d x\]

Input:

int(F^(c*(b*x+a)^n)*x^3,x)
 

Output:

int(F^(c*(b*x+a)^n)*x^3,x)
 

Fricas [F]

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

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

Output:

integral(F^((b*x + a)^n*c)*x^3, x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a)**n)*x**3,x)
 

Output:

Integral(F**(c*(a + b*x)**n)*x**3, x)
 

Maxima [F]

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

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

Output:

integrate(F^((b*x + a)^n*c)*x^3, x)
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)^n*c)*x^3, x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x)^n)*x^3,x)
 

Output:

int(F^(c*(a + b*x)^n)*x^3, x)
 

Reduce [F]

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

int(F^(c*(b*x+a)^n)*x^3,x)
 

Output:

int(f**((a + b*x)**n*c)*x**3,x)