\(\int (f+g x)^2 (a+b \log (c (d+e x)^n))^n \, dx\) [171]

Optimal result

Integrand size = 24, antiderivative size = 348 \[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\frac {3^{-1-n} e^{-\frac {3 a}{b n}} g^2 (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \Gamma \left (1+n,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3}+\frac {2^{-n} e^{-\frac {2 a}{b n}} g (e f-d g) (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \Gamma \left (1+n,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3}+\frac {e^{-\frac {a}{b n}} (e f-d g)^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n} \Gamma \left (1+n,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3} \] Output:

3^(-1-n)*g^2*(e*x+d)^3*GAMMA(1+n,(-3*a-3*b*ln(c*(e*x+d)^n))/b/n)*(a+b*ln(c 
*(e*x+d)^n))^n/e^3/exp(3*a/b/n)/((c*(e*x+d)^n)^(3/n))/((-(a+b*ln(c*(e*x+d) 
^n))/b/n)^n)+g*(-d*g+e*f)*(e*x+d)^2*GAMMA(1+n,(-2*a-2*b*ln(c*(e*x+d)^n))/b 
/n)*(a+b*ln(c*(e*x+d)^n))^n/(2^n)/e^3/exp(2*a/b/n)/((c*(e*x+d)^n)^(2/n))/( 
(-(a+b*ln(c*(e*x+d)^n))/b/n)^n)+(-d*g+e*f)^2*(e*x+d)*GAMMA(1+n,-(a+b*ln(c* 
(e*x+d)^n))/b/n)*(a+b*ln(c*(e*x+d)^n))^n/e^3/exp(a/b/n)/((c*(e*x+d)^n)^(1/ 
n))/((-(a+b*ln(c*(e*x+d)^n))/b/n)^n)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.75 \[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\frac {2^{-n} 3^{-1-n} e^{-\frac {3 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-3/n} \left (2^n g^2 (d+e x)^2 \Gamma \left (1+n,-\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )+3^{1+n} e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \left (g (d+e x) \Gamma \left (1+n,-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )+2^n e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \Gamma \left (1+n,-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^n \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{-n}}{e^3} \] Input:

Integrate[(f + g*x)^2*(a + b*Log[c*(d + e*x)^n])^n,x]
 

Output:

(3^(-1 - n)*(d + e*x)*(2^n*g^2*(d + e*x)^2*Gamma[1 + n, (-3*(a + b*Log[c*( 
d + e*x)^n]))/(b*n)] + 3^(1 + n)*E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n 
^(-1)*(g*(d + e*x)*Gamma[1 + n, (-2*(a + b*Log[c*(d + e*x)^n]))/(b*n)] + 2 
^n*E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1)*Gamma[1 + n, -((a + b*Lo 
g[c*(d + e*x)^n])/(b*n))]))*(a + b*Log[c*(d + e*x)^n])^n)/(2^n*e^3*E^((3*a 
)/(b*n))*(c*(d + e*x)^n)^(3/n)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n)
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2848, 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.

\(\displaystyle \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx\)

\(\Big \downarrow \) 2848

\(\displaystyle \int \left (\frac {(e f-d g)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}+\frac {2 g (d+e x) (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}+\frac {g^2 (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n}{e^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(f + g*x)^2*(a + b*Log[c*(d + e*x)^n])^n,x]
 

Output:

(3^(-1 - n)*g^2*(d + e*x)^3*Gamma[1 + n, (-3*(a + b*Log[c*(d + e*x)^n]))/( 
b*n)]*(a + b*Log[c*(d + e*x)^n])^n)/(e^3*E^((3*a)/(b*n))*(c*(d + e*x)^n)^( 
3/n)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n) + (g*(e*f - d*g)*(d + e*x)^2 
*Gamma[1 + n, (-2*(a + b*Log[c*(d + e*x)^n]))/(b*n)]*(a + b*Log[c*(d + e*x 
)^n])^n)/(2^n*e^3*E^((2*a)/(b*n))*(c*(d + e*x)^n)^(2/n)*(-((a + b*Log[c*(d 
 + e*x)^n])/(b*n)))^n) + ((e*f - d*g)^2*(d + e*x)*Gamma[1 + n, -((a + b*Lo 
g[c*(d + e*x)^n])/(b*n))]*(a + b*Log[c*(d + e*x)^n])^n)/(e^3*E^(a/(b*n))*( 
c*(d + e*x)^n)^n^(-1)*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^n)
 

Defintions of rubi rules used

rule 2009

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

rule 2848

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. 
)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d 
 + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - 
 d*g, 0] && IGtQ[q, 0]
 

Maple [F]

\[\int \left (g x +f \right )^{2} {\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{n}d x\]

Input:

int((g*x+f)^2*(a+b*ln(c*(e*x+d)^n))^n,x)
 

Output:

int((g*x+f)^2*(a+b*ln(c*(e*x+d)^n))^n,x)
 

Fricas [F]

\[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\int { {\left (g x + f\right )}^{2} {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{n} \,d x } \] Input:

integrate((g*x+f)^2*(a+b*log(c*(e*x+d)^n))^n,x, algorithm="fricas")
 

Output:

integral((g^2*x^2 + 2*f*g*x + f^2)*(b*log((e*x + d)^n*c) + a)^n, x)
 

Sympy [F]

\[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\int \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{n} \left (f + g x\right )^{2}\, dx \] Input:

integrate((g*x+f)**2*(a+b*ln(c*(e*x+d)**n))**n,x)
 

Output:

Integral((a + b*log(c*(d + e*x)**n))**n*(f + g*x)**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((g*x+f)^2*(a+b*log(c*(e*x+d)^n))^n,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [F]

\[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\int { {\left (g x + f\right )}^{2} {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{n} \,d x } \] Input:

integrate((g*x+f)^2*(a+b*log(c*(e*x+d)^n))^n,x, algorithm="giac")
 

Output:

integrate((g*x + f)^2*(b*log((e*x + d)^n*c) + a)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\int {\left (f+g\,x\right )}^2\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^n \,d x \] Input:

int((f + g*x)^2*(a + b*log(c*(d + e*x)^n))^n,x)
 

Output:

int((f + g*x)^2*(a + b*log(c*(d + e*x)^n))^n, x)
 

Reduce [F]

\[ \int (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^n \, dx=\text {too large to display} \] Input:

int((g*x+f)^2*(a+b*log(c*(e*x+d)^n))^n,x)
 

Output:

(2*(log((d + e*x)**n*c)*b + a)**n*log((d + e*x)**n*c)*b*d**3*g**2*n + 2*(l 
og((d + e*x)**n*c)*b + a)**n*a*d**3*g**2*n + 6*(log((d + e*x)**n*c)*b + a) 
**n*a*e**3*f**2*n*x + 6*(log((d + e*x)**n*c)*b + a)**n*a*e**3*f**2*x + 6*( 
log((d + e*x)**n*c)*b + a)**n*a*e**3*f*g*n*x**2 + 6*(log((d + e*x)**n*c)*b 
 + a)**n*a*e**3*f*g*x**2 + 2*(log((d + e*x)**n*c)*b + a)**n*a*e**3*g**2*n* 
x**3 + 2*(log((d + e*x)**n*c)*b + a)**n*a*e**3*g**2*x**3 - 2*(log((d + e*x 
)**n*c)*b + a)**n*b*d**2*e*g**2*n**3*x - 2*(log((d + e*x)**n*c)*b + a)**n* 
b*d**2*e*g**2*n**2*x + (log((d + e*x)**n*c)*b + a)**n*b*d*e**2*g**2*n**3*x 
**2 + (log((d + e*x)**n*c)*b + a)**n*b*d*e**2*g**2*n**2*x**2 + 2*(log((d + 
 e*x)**n*c)*b + a)**n*b*e**3*f**2*n**3*x + 2*(log((d + e*x)**n*c)*b + a)** 
n*b*e**3*f**2*n**2*x + 2*(log((d + e*x)**n*c)*b + a)**n*b*e**3*f*g*n**3*x* 
*2 + 2*(log((d + e*x)**n*c)*b + a)**n*b*e**3*f*g*n**2*x**2 - 18*int(((log( 
(d + e*x)**n*c)*b + a)**n*x**2)/(3*log((d + e*x)**n*c)*a*b*d + 3*log((d + 
e*x)**n*c)*a*b*e*x + log((d + e*x)**n*c)*b**2*d*n**2 + log((d + e*x)**n*c) 
*b**2*e*n**2*x + 3*a**2*d + 3*a**2*e*x + a*b*d*n**2 + a*b*e*n**2*x),x)*a** 
2*b*e**4*f*g*n**3 - 18*int(((log((d + e*x)**n*c)*b + a)**n*x**2)/(3*log((d 
 + e*x)**n*c)*a*b*d + 3*log((d + e*x)**n*c)*a*b*e*x + log((d + e*x)**n*c)* 
b**2*d*n**2 + log((d + e*x)**n*c)*b**2*e*n**2*x + 3*a**2*d + 3*a**2*e*x + 
a*b*d*n**2 + a*b*e*n**2*x),x)*a**2*b*e**4*f*g*n**2 - 3*int(((log((d + e*x) 
**n*c)*b + a)**n*x**2)/(3*log((d + e*x)**n*c)*a*b*d + 3*log((d + e*x)**...