3.4.69 \(\int \log (f x^m) (a+b \log (c (d+e x)^n))^2 \, dx\) [369]

3.4.69.1 Optimal result

Integrand size = 23, antiderivative size = 309 \[ \int \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \, dx=2 a b m n x-4 b^2 m n^2 x+2 b m n (a-b n) x-2 a b n x \log \left (f x^m\right )+2 b^2 n^2 x \log \left (f x^m\right )+\frac {4 b^2 m n (d+e x) \log \left (c (d+e x)^n\right )}{e}+\frac {2 b^2 d m n \log \left (-\frac {e x}{d}\right ) \log \left (c (d+e x)^n\right )}{e}-\frac {2 b^2 n (d+e x) \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )}{e}-\frac {m (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}-\frac {d m \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}+\frac {(d+e x) \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}+\frac {2 b^2 d m n^2 \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e}-\frac {2 b d m n \left (a+b \log \left (c (d+e x)^n\right )\right ) \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e}+\frac {2 b^2 d m n^2 \operatorname {PolyLog}\left (3,1+\frac {e x}{d}\right )}{e} \]

output

2*a*b*m*n*x-4*b^2*m*n^2*x+2*b*m*n*(-b*n+a)*x-2*a*b*n*x*ln(f*x^m)+2*b^2*n^2 
*x*ln(f*x^m)+4*b^2*m*n*(e*x+d)*ln(c*(e*x+d)^n)/e+2*b^2*d*m*n*ln(-e*x/d)*ln 
(c*(e*x+d)^n)/e-2*b^2*n*(e*x+d)*ln(f*x^m)*ln(c*(e*x+d)^n)/e-m*(e*x+d)*(a+b 
*ln(c*(e*x+d)^n))^2/e-d*m*ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))^2/e+(e*x+d)*ln( 
f*x^m)*(a+b*ln(c*(e*x+d)^n))^2/e+2*b^2*d*m*n^2*polylog(2,1+e*x/d)/e-2*b*d* 
m*n*(a+b*ln(c*(e*x+d)^n))*polylog(2,1+e*x/d)/e+2*b^2*d*m*n^2*polylog(3,1+e 
*x/d)/e
 

3.4.69.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.48 \[ \int \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2 \, dx=\frac {-b^2 n^2 \left (m \log (x)-\log \left (f x^m\right )\right ) \left (2 e x-2 (d+e x) \log (d+e x)+(d+e x) \log ^2(d+e x)\right )+2 b n \left (m \log (x)-\log \left (f x^m\right )\right ) (e x-(d+e x) \log (d+e x)) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )+e m x \log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2-e x \left (m+m \log (x)-\log \left (f x^m\right )\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )^2+2 b m n \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \left (d+e x-e x (-1+\log (x))-(d+e x) \log (d+e x)+e x \log (x) \log (d+e x)+d \left (\log (x) \log \left (1+\frac {e x}{d}\right )+\operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )\right )\right )+b^2 m n^2 \left (-6 e x+2 e x \log (x)+4 d \log (d+e x)+4 e x \log (d+e x)-2 e x \log (x) \log (d+e x)-d \log ^2(d+e x)-e x \log ^2(d+e x)+d \log (x) \log ^2(d+e x)+e x \log (x) \log ^2(d+e x)-d \log \left (-\frac {e x}{d}\right ) \log ^2(d+e x)-2 d \log (x) \log \left (1+\frac {e x}{d}\right )-2 d \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-2 d \log (d+e x) \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )+2 d \operatorname {PolyLog}\left (3,1+\frac {e x}{d}\right )\right )}{e} \]

input

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

output

(-(b^2*n^2*(m*Log[x] - Log[f*x^m])*(2*e*x - 2*(d + e*x)*Log[d + e*x] + (d 
+ e*x)*Log[d + e*x]^2)) + 2*b*n*(m*Log[x] - Log[f*x^m])*(e*x - (d + e*x)*L 
og[d + e*x])*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]) + e*m*x*Log[x]* 
(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 - e*x*(m + m*Log[x] - Log[ 
f*x^m])*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])^2 + 2*b*m*n*(a - b*n 
*Log[d + e*x] + b*Log[c*(d + e*x)^n])*(d + e*x - e*x*(-1 + Log[x]) - (d + 
e*x)*Log[d + e*x] + e*x*Log[x]*Log[d + e*x] + d*(Log[x]*Log[1 + (e*x)/d] + 
 PolyLog[2, -((e*x)/d)])) + b^2*m*n^2*(-6*e*x + 2*e*x*Log[x] + 4*d*Log[d + 
 e*x] + 4*e*x*Log[d + e*x] - 2*e*x*Log[x]*Log[d + e*x] - d*Log[d + e*x]^2 
- e*x*Log[d + e*x]^2 + d*Log[x]*Log[d + e*x]^2 + e*x*Log[x]*Log[d + e*x]^2 
 - d*Log[-((e*x)/d)]*Log[d + e*x]^2 - 2*d*Log[x]*Log[1 + (e*x)/d] - 2*d*Po 
lyLog[2, -((e*x)/d)] - 2*d*Log[d + e*x]*PolyLog[2, 1 + (e*x)/d] + 2*d*Poly 
Log[3, 1 + (e*x)/d]))/e
 

3.4.69.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2870, 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.

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

\(\Big \downarrow \) 2870

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

\(\Big \downarrow \) 2009

\(\displaystyle -m \left (\frac {2 b d n \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}+\frac {d \log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}-2 a b n x-2 b n x (a-b n)-\frac {4 b^2 n (d+e x) \log \left (c (d+e x)^n\right )}{e}-\frac {2 b^2 d n \log \left (-\frac {e x}{d}\right ) \log \left (c (d+e x)^n\right )}{e}-\frac {2 b^2 d n^2 \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e}-\frac {2 b^2 d n^2 \operatorname {PolyLog}\left (3,\frac {e x}{d}+1\right )}{e}+4 b^2 n^2 x\right )+\frac {(d+e x) \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{e}-2 a b n x \log \left (f x^m\right )-\frac {2 b^2 n (d+e x) \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )}{e}+2 b^2 n^2 x \log \left (f x^m\right )\)

input

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

output

-2*a*b*n*x*Log[f*x^m] + 2*b^2*n^2*x*Log[f*x^m] - (2*b^2*n*(d + e*x)*Log[f* 
x^m]*Log[c*(d + e*x)^n])/e + ((d + e*x)*Log[f*x^m]*(a + b*Log[c*(d + e*x)^ 
n])^2)/e - m*(-2*a*b*n*x + 4*b^2*n^2*x - 2*b*n*(a - b*n)*x - (4*b^2*n*(d + 
 e*x)*Log[c*(d + e*x)^n])/e - (2*b^2*d*n*Log[-((e*x)/d)]*Log[c*(d + e*x)^n 
])/e + ((d + e*x)*(a + b*Log[c*(d + e*x)^n])^2)/e + (d*Log[-((e*x)/d)]*(a 
+ b*Log[c*(d + e*x)^n])^2)/e - (2*b^2*d*n^2*PolyLog[2, 1 + (e*x)/d])/e + ( 
2*b*d*n*(a + b*Log[c*(d + e*x)^n])*PolyLog[2, 1 + (e*x)/d])/e - (2*b^2*d*n 
^2*PolyLog[3, 1 + (e*x)/d])/e)
 

3.4.69.3.1 Defintions of rubi rules used

rule 2009

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

rule 2870

Int[Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_ 
.))^(p_), x_Symbol] :> With[{u = IntHide[(a + b*Log[c*(d + e*x)^n])^p, x]}, 
 Simp[Log[f*x^m]   u, x] - Simp[m   Int[1/x   u, x], x]] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] && IGtQ[p, 1]
 

3.4.69.4 Maple [F]

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

input

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

output

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

3.4.69.5 Fricas [F]

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

input

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

output

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

3.4.69.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.4.69.7 Maxima [F]

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

input

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

output

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

3.4.69.8 Giac [F]

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

input

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

output

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

3.4.69.9 Mupad [F(-1)]

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

input

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

output

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