\(\int f^{\frac {c}{(a+b x)^3}} x^4 \, dx\) [165]

Optimal result

Integrand size = 15, antiderivative size = 239 \[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=\frac {2 a^2 f^{\frac {c}{(a+b x)^3}} (a+b x)^3}{b^5}-\frac {2 a^2 c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^3}\right ) \log (f)}{b^5}+\frac {a^4 (a+b x) \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}}}{3 b^5}-\frac {4 a^3 (a+b x)^2 \Gamma \left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3}}{3 b^5}-\frac {4 a (a+b x)^4 \Gamma \left (-\frac {4}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{4/3}}{3 b^5}+\frac {(a+b x)^5 \Gamma \left (-\frac {5}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{5/3}}{3 b^5} \] Output:

2*a^2*f^(c/(b*x+a)^3)*(b*x+a)^3/b^5-2*a^2*c*Ei(c*ln(f)/(b*x+a)^3)*ln(f)/b^ 
5+1/3*a^4*(b*x+a)*GAMMA(-1/3,-c*ln(f)/(b*x+a)^3)*(-c*ln(f)/(b*x+a)^3)^(1/3 
)/b^5-4/3*a^3*(b*x+a)^2*GAMMA(-2/3,-c*ln(f)/(b*x+a)^3)*(-c*ln(f)/(b*x+a)^3 
)^(2/3)/b^5-4/3*a*(b*x+a)^4*GAMMA(-4/3,-c*ln(f)/(b*x+a)^3)*(-c*ln(f)/(b*x+ 
a)^3)^(4/3)/b^5+1/3*(b*x+a)^5*GAMMA(-5/3,-c*ln(f)/(b*x+a)^3)*(-c*ln(f)/(b* 
x+a)^3)^(5/3)/b^5
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92 \[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=\frac {6 a^2 f^{\frac {c}{(a+b x)^3}} (a+b x)^3-6 a^2 c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^3}\right ) \log (f)+a^4 (a+b x) \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}}+4 a c (a+b x) \Gamma \left (-\frac {4}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \log (f) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}}-4 a^3 (a+b x)^2 \Gamma \left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3}+(a+b x)^5 \Gamma \left (-\frac {5}{3},-\frac {c \log (f)}{(a+b x)^3}\right ) \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{5/3}}{3 b^5} \] Input:

Integrate[f^(c/(a + b*x)^3)*x^4,x]
 

Output:

(6*a^2*f^(c/(a + b*x)^3)*(a + b*x)^3 - 6*a^2*c*ExpIntegralEi[(c*Log[f])/(a 
 + b*x)^3]*Log[f] + a^4*(a + b*x)*Gamma[-1/3, -((c*Log[f])/(a + b*x)^3)]*( 
-((c*Log[f])/(a + b*x)^3))^(1/3) + 4*a*c*(a + b*x)*Gamma[-4/3, -((c*Log[f] 
)/(a + b*x)^3)]*Log[f]*(-((c*Log[f])/(a + b*x)^3))^(1/3) - 4*a^3*(a + b*x) 
^2*Gamma[-2/3, -((c*Log[f])/(a + b*x)^3)]*(-((c*Log[f])/(a + b*x)^3))^(2/3 
) + (a + b*x)^5*Gamma[-5/3, -((c*Log[f])/(a + b*x)^3)]*(-((c*Log[f])/(a + 
b*x)^3))^(5/3))/(3*b^5)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 239, 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.

\(\displaystyle \int x^4 f^{\frac {c}{(a+b x)^3}} \, dx\)

\(\Big \downarrow \) 2656

\(\displaystyle \int \left (\frac {a^4 f^{\frac {c}{(a+b x)^3}}}{b^4}-\frac {4 a^3 (a+b x) f^{\frac {c}{(a+b x)^3}}}{b^4}+\frac {6 a^2 (a+b x)^2 f^{\frac {c}{(a+b x)^3}}}{b^4}+\frac {(a+b x)^4 f^{\frac {c}{(a+b x)^3}}}{b^4}-\frac {4 a (a+b x)^3 f^{\frac {c}{(a+b x)^3}}}{b^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^4 (a+b x) \sqrt [3]{-\frac {c \log (f)}{(a+b x)^3}} \Gamma \left (-\frac {1}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^5}-\frac {4 a^3 (a+b x)^2 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^5}-\frac {2 a^2 c \log (f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{(a+b x)^3}\right )}{b^5}+\frac {2 a^2 (a+b x)^3 f^{\frac {c}{(a+b x)^3}}}{b^5}+\frac {(a+b x)^5 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{5/3} \Gamma \left (-\frac {5}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^5}-\frac {4 a (a+b x)^4 \left (-\frac {c \log (f)}{(a+b x)^3}\right )^{4/3} \Gamma \left (-\frac {4}{3},-\frac {c \log (f)}{(a+b x)^3}\right )}{3 b^5}\)

Input:

Int[f^(c/(a + b*x)^3)*x^4,x]
 

Output:

(2*a^2*f^(c/(a + b*x)^3)*(a + b*x)^3)/b^5 - (2*a^2*c*ExpIntegralEi[(c*Log[ 
f])/(a + b*x)^3]*Log[f])/b^5 + (a^4*(a + b*x)*Gamma[-1/3, -((c*Log[f])/(a 
+ b*x)^3)]*(-((c*Log[f])/(a + b*x)^3))^(1/3))/(3*b^5) - (4*a^3*(a + b*x)^2 
*Gamma[-2/3, -((c*Log[f])/(a + b*x)^3)]*(-((c*Log[f])/(a + b*x)^3))^(2/3)) 
/(3*b^5) - (4*a*(a + b*x)^4*Gamma[-4/3, -((c*Log[f])/(a + b*x)^3)]*(-((c*L 
og[f])/(a + b*x)^3))^(4/3))/(3*b^5) + ((a + b*x)^5*Gamma[-5/3, -((c*Log[f] 
)/(a + b*x)^3)]*(-((c*Log[f])/(a + b*x)^3))^(5/3))/(3*b^5)
 

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]

\[\int f^{\frac {c}{\left (b x +a \right )^{3}}} x^{4}d x\]

Input:

int(f^(c/(b*x+a)^3)*x^4,x)
 

Output:

int(f^(c/(b*x+a)^3)*x^4,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.04 \[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=-\frac {20 \, a^{2} c {\rm Ei}\left (\frac {c \log \left (f\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) \log \left (f\right ) - {\left (20 \, a^{3} b^{2} - 3 \, b^{2} c \log \left (f\right )\right )} \left (-\frac {c \log \left (f\right )}{b^{3}}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -\frac {c \log \left (f\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) + 10 \, {\left (a^{4} b - 3 \, a b c \log \left (f\right )\right )} \left (-\frac {c \log \left (f\right )}{b^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {c \log \left (f\right )}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\right ) - {\left (2 \, b^{5} x^{5} + 2 \, a^{5} + 3 \, {\left (b^{2} c x^{2} - 8 \, a b c x - 9 \, a^{2} c\right )} \log \left (f\right )\right )} f^{\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{10 \, b^{5}} \] Input:

integrate(f^(c/(b*x+a)^3)*x^4,x, algorithm="fricas")
 

Output:

-1/10*(20*a^2*c*Ei(c*log(f)/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3))*log 
(f) - (20*a^3*b^2 - 3*b^2*c*log(f))*(-c*log(f)/b^3)^(2/3)*gamma(1/3, -c*lo 
g(f)/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)) + 10*(a^4*b - 3*a*b*c*log( 
f))*(-c*log(f)/b^3)^(1/3)*gamma(2/3, -c*log(f)/(b^3*x^3 + 3*a*b^2*x^2 + 3* 
a^2*b*x + a^3)) - (2*b^5*x^5 + 2*a^5 + 3*(b^2*c*x^2 - 8*a*b*c*x - 9*a^2*c) 
*log(f))*f^(c/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)))/b^5
 

Sympy [F]

\[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=\int f^{\frac {c}{\left (a + b x\right )^{3}}} x^{4}\, dx \] Input:

integrate(f**(c/(b*x+a)**3)*x**4,x)
 

Output:

Integral(f**(c/(a + b*x)**3)*x**4, x)
 

Maxima [F]

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

integrate(f^(c/(b*x+a)^3)*x^4,x, algorithm="maxima")
 

Output:

1/10*(2*b^4*x^5 + 3*b*c*x^2*log(f) - 24*a*c*x*log(f))*f^(c/(b^3*x^3 + 3*a* 
b^2*x^2 + 3*a^2*b*x + a^3))/b^4 + integrate(3/10*(20*a^2*b^3*c*x^3*log(f) 
+ 8*a^5*c*log(f) + (40*a^3*b^2*c*log(f) + 3*b^2*c^2*log(f)^2)*x^2 + 6*(5*a 
^4*b*c*log(f) - 4*a*b*c^2*log(f)^2)*x)*f^(c/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2 
*b*x + a^3))/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^ 
4), x)
 

Giac [F]

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

integrate(f^(c/(b*x+a)^3)*x^4,x, algorithm="giac")
 

Output:

integrate(f^(c/(b*x + a)^3)*x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=\int f^{\frac {c}{{\left (a+b\,x\right )}^3}}\,x^4 \,d x \] Input:

int(f^(c/(a + b*x)^3)*x^4,x)
 

Output:

int(f^(c/(a + b*x)^3)*x^4, x)
 

Reduce [F]

\[ \int f^{\frac {c}{(a+b x)^3}} x^4 \, dx=\text {too large to display} \] Input:

int(f^(c/(b*x+a)^3)*x^4,x)
 

Output:

( - 54*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**4*a* 
b*c**4*x + 27*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f 
)**4*b**2*c**4*x**2 - 18*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x 
**3))*log(f)**3*a**5*c**3 + 810*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + 
 b**3*x**3))*log(f)**3*a**4*b*c**3*x + 1134*f**(c/(a**3 + 3*a**2*b*x + 3*a 
*b**2*x**2 + b**3*x**3))*log(f)**3*a**3*b**2*c**3*x**2 + 198*f**(c/(a**3 + 
 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**3*a**2*b**3*c**3*x**3 - 
36*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**3*b**5*c 
**3*x**5 + 306*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log( 
f)**2*a**8*c**2 + 1836*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x** 
3))*log(f)**2*a**7*b*c**2*x - 1278*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x** 
2 + b**3*x**3))*log(f)**2*a**6*b**2*c**2*x**2 - 5616*f**(c/(a**3 + 3*a**2* 
b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**2*a**5*b**3*c**2*x**3 - 5190*f** 
(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**2*a**4*b**4*c* 
*2*x**4 - 2076*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log( 
f)**2*a**3*b**5*c**2*x**5 - 342*f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + 
 b**3*x**3))*log(f)**2*a**2*b**6*c**2*x**6 - 24*f**(c/(a**3 + 3*a**2*b*x + 
 3*a*b**2*x**2 + b**3*x**3))*log(f)**2*a*b**7*c**2*x**7 + 12*f**(c/(a**3 + 
 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)**2*b**8*c**2*x**8 - 3912* 
f**(c/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))*log(f)*a**10*b*c...