\(\int \frac {(f+g x) (a+b \log (c x^n))^3}{(d+e x)^3} \, dx\) [454]

Optimal result

Integrand size = 25, antiderivative size = 295 \[ \int \frac {(f+g x) \left (a+b \log \left (c x^n\right )\right )^3}{(d+e x)^3} \, dx=-\frac {3 b (e f-d g) n x \left (a+b \log \left (c x^n\right )\right )^2}{2 d^2 e (d+e x)}+\frac {f^2 \left (a+b \log \left (c x^n\right )\right )^3}{2 d^2 (e f-d g)}-\frac {(f+g x)^2 \left (a+b \log \left (c x^n\right )\right )^3}{2 (e f-d g) (d+e x)^2}+\frac {3 b^2 (e f-d g) n^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x}{d}\right )}{d^2 e^2}-\frac {3 b (e f+d g) n \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )}{2 d^2 e^2}+\frac {3 b^3 (e f-d g) n^3 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2 e^2}-\frac {3 b^2 (e f+d g) n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{d^2 e^2}+\frac {3 b^3 (e f+d g) n^3 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )}{d^2 e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.15 \[ \int \frac {(f+g x) \left (a+b \log \left (c x^n\right )\right )^3}{(d+e x)^3} \, dx=\frac {-\frac {(e f-d g) \left (a+b \log \left (c x^n\right )\right )^3}{(d+e x)^2}-\frac {2 g \left (a+b \log \left (c x^n\right )\right )^3}{d+e x}+\frac {2 g \left (\left (a+b \log \left (c x^n\right )\right )^2 \left (a+b \log \left (c x^n\right )-3 b n \log \left (1+\frac {e x}{d}\right )\right )-6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )+6 b^3 n^3 \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right )}{d}+\frac {(e f-d g) \left (3 b d n \left (a+b \log \left (c x^n\right )\right )^2+(d+e x) \left (a+b \log \left (c x^n\right )\right )^3-3 b n (d+e x) \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {e x}{d}\right )-3 b n (d+e x) \left (\left (a+b \log \left (c x^n\right )\right ) \left (a+b \log \left (c x^n\right )-2 b n \log \left (1+\frac {e x}{d}\right )\right )-2 b^2 n^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )\right )-6 b^2 n^2 (d+e x) \left (\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )-b n \operatorname {PolyLog}\left (3,-\frac {e x}{d}\right )\right )\right )}{d^2 (d+e x)}}{2 e^2} \] Input:

Integrate[((f + g*x)*(a + b*Log[c*x^n])^3)/(d + e*x)^3,x]
 

Output:

(-(((e*f - d*g)*(a + b*Log[c*x^n])^3)/(d + e*x)^2) - (2*g*(a + b*Log[c*x^n 
])^3)/(d + e*x) + (2*g*((a + b*Log[c*x^n])^2*(a + b*Log[c*x^n] - 3*b*n*Log 
[1 + (e*x)/d]) - 6*b^2*n^2*(a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] + 6*b 
^3*n^3*PolyLog[3, -((e*x)/d)]))/d + ((e*f - d*g)*(3*b*d*n*(a + b*Log[c*x^n 
])^2 + (d + e*x)*(a + b*Log[c*x^n])^3 - 3*b*n*(d + e*x)*(a + b*Log[c*x^n]) 
^2*Log[1 + (e*x)/d] - 3*b*n*(d + e*x)*((a + b*Log[c*x^n])*(a + b*Log[c*x^n 
] - 2*b*n*Log[1 + (e*x)/d]) - 2*b^2*n^2*PolyLog[2, -((e*x)/d)]) - 6*b^2*n^ 
2*(d + e*x)*((a + b*Log[c*x^n])*PolyLog[2, -((e*x)/d)] - b*n*PolyLog[3, -( 
(e*x)/d)])))/(d^2*(d + e*x)))/(2*e^2)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2798, 2804, 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.

\(\displaystyle \int \frac {(f+g x) \left (a+b \log \left (c x^n\right )\right )^3}{(d+e x)^3} \, dx\)

\(\Big \downarrow \) 2798

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

\(\Big \downarrow \) 2804

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

\(\Big \downarrow \) 2009

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

Input:

Int[((f + g*x)*(a + b*Log[c*x^n])^3)/(d + e*x)^3,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2798

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_)*(( 
f_) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(f + g*x)^(m + 1)*(d + e*x)^(q + 
 1)*((a + b*Log[c*x^n])^p/((q + 1)*(e*f - d*g))), x] - Simp[b*n*(p/((q + 1) 
*(e*f - d*g)))   Int[(f + g*x)^(m + 1)*(d + e*x)^(q + 1)*((a + b*Log[c*x^n] 
)^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && NeQ[e*f 
 - d*g, 0] && EqQ[m + q + 2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

rule 2804

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{ 
u = ExpandIntegrand[(a + b*Log[c*x^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] / 
; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.44 (sec) , antiderivative size = 1652, normalized size of antiderivative = 5.60

method	result	size
risch	\(\text {Expression too large to display}\)	\(1652\)

Input:

int((g*x+f)*(a+b*ln(c*x^n))^3/(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

-3*b^3*n^3/e/d^2*ln(e*x+d)*ln(-e*x/d)*f-3*b^3*n^3/e/d^2*dilog(-e*x/d)*ln(x 
)*f-3*b^3*n^3/e/d^2*ln(e*x+d)*ln(-e*x/d)*ln(x)*f+3*b^3*n^2/e^2/d*ln(e*x+d) 
*ln(-e*x/d)*ln(x^n)*g-3*b^3*n^3/e^2/d*ln(e*x+d)*ln(-e*x/d)*ln(x)*g+3*b^3*n 
^2/e/d^2*ln(e*x+d)*ln(-e*x/d)*ln(x^n)*f+3/2*b^3*n^3/e/d^2*ln(e*x+d)*ln(x)^ 
2*f-3/2*b^3*n^3/e/d^2*ln(1+e*x/d)*ln(x)^2*f+3/2*b^3*n*ln(x^n)^2/e^2/d*ln(x 
)*g+3/2*b^3*n*ln(x^n)^2/e/d^2*ln(x)*f-3/2*b^3*n^2/e^2/d*ln(x)^2*ln(x^n)*g- 
3*b^3*n^2/e^2/d*ln(x^n)*ln(e*x+d)*g+3/2*b^3*n*ln(x^n)^2/e/d/(e*x+d)*f-3/2* 
b^3*n*ln(x^n)^2/e^2/d*ln(e*x+d)*g-3/2*b^3*n*ln(x^n)^2/e/d^2*ln(e*x+d)*f+3* 
b^3*n^2/e^2/d*ln(x^n)*ln(x)*g-b^3*ln(x^n)^3*g/e^2/(e*x+d)-1/2*b^3*ln(x^n)^ 
3/e/(e*x+d)^2*f-3*b^3*n^3/e^2/d*polylog(2,-e*x/d)*ln(x)*g-3/2*b^3*n^2/e/d^ 
2*ln(x)^2*ln(x^n)*f+3*b^3*n^2/e/d^2*ln(x^n)*ln(e*x+d)*f-3*b^3*n^2/e/d^2*ln 
(x^n)*ln(x)*f+3*b^3*n^3/e^2/d*ln(e*x+d)*ln(-e*x/d)*g-3*b^3*n^3/e/d^2*polyl 
og(2,-e*x/d)*ln(x)*f+3/2*b^3*n^3/e/d^2*ln(x)^2*f-3*b^3*n^3/e/d^2*dilog(-e* 
x/d)*f+1/2*b^3*ln(x^n)^3/e^2/(e*x+d)^2*d*g-3/2*b^3*n*ln(x^n)^2*g/e^2/(e*x+ 
d)+1/2*b^3*n^3/e^2/d*ln(x)^3*g-3/2*b^3*n^3/e^2/d*ln(x)^2*g+3*b^3*n^3/e^2/d 
*dilog(-e*x/d)*g+3*b^3*n^3/e^2/d*polylog(3,-e*x/d)*g+1/8*(I*Pi*b*csgn(I*x^ 
n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn( 
I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2*a)^3*(-g/e^2/(e*x+ 
d)-1/2*(-d*g+e*f)/e^2/(e*x+d)^2)+3*b^3*n^2/e^2/d*dilog(-e*x/d)*ln(x^n)*g-3 
*b^3*n^3/e^2/d*dilog(-e*x/d)*ln(x)*g+3*b^3*n^2/e/d^2*dilog(-e*x/d)*ln(x...
 

Fricas [F]

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

integrate((g*x+f)*(a+b*log(c*x^n))^3/(e*x+d)^3,x, algorithm="fricas")
 

Output:

integral((a^3*g*x + a^3*f + (b^3*g*x + b^3*f)*log(c*x^n)^3 + 3*(a*b^2*g*x 
+ a*b^2*f)*log(c*x^n)^2 + 3*(a^2*b*g*x + a^2*b*f)*log(c*x^n))/(e^3*x^3 + 3 
*d*e^2*x^2 + 3*d^2*e*x + d^3), x)
 

Sympy [F]

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

integrate((g*x+f)*(a+b*ln(c*x**n))**3/(e*x+d)**3,x)
 

Output:

Integral((a + b*log(c*x**n))**3*(f + g*x)/(d + e*x)**3, x)
 

Maxima [F]

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

integrate((g*x+f)*(a+b*log(c*x^n))^3/(e*x+d)^3,x, algorithm="maxima")
 

Output:

3/2*a^2*b*f*n*(1/(d*e^2*x + d^2*e) - log(e*x + d)/(d^2*e) + log(x)/(d^2*e) 
) - 3/2*a^2*b*g*n*(1/(e^3*x + d*e^2) + log(e*x + d)/(d*e^2) - log(x)/(d*e^ 
2)) - 3/2*(2*e*x + d)*a^2*b*g*log(c*x^n)/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) - 
 1/2*(2*e*x + d)*a^3*g/(e^4*x^2 + 2*d*e^3*x + d^2*e^2) - 3/2*a^2*b*f*log(c 
*x^n)/(e^3*x^2 + 2*d*e^2*x + d^2*e) - 1/2*a^3*f/(e^3*x^2 + 2*d*e^2*x + d^2 
*e) - 1/2*(2*b^3*e*g*x + (e*f + d*g)*b^3)*log(x^n)^3/(e^4*x^2 + 2*d*e^3*x 
+ d^2*e^2) + integrate(1/2*(2*(b^3*e^2*g*log(c)^3 + 3*a*b^2*e^2*g*log(c)^2 
)*x^2 + 3*((d*e*f*n + d^2*g*n)*b^3 + 2*(a*b^2*e^2*g + (e^2*g*n + e^2*g*log 
(c))*b^3)*x^2 + (2*a*b^2*e^2*f + (e^2*f*n + 3*d*e*g*n + 2*e^2*f*log(c))*b^ 
3)*x)*log(x^n)^2 + 2*(b^3*e^2*f*log(c)^3 + 3*a*b^2*e^2*f*log(c)^2)*x + 6*( 
(b^3*e^2*g*log(c)^2 + 2*a*b^2*e^2*g*log(c))*x^2 + (b^3*e^2*f*log(c)^2 + 2* 
a*b^2*e^2*f*log(c))*x)*log(x^n))/(e^5*x^4 + 3*d*e^4*x^3 + 3*d^2*e^3*x^2 + 
d^3*e^2*x), x)
 

Giac [F]

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

integrate((g*x+f)*(a+b*log(c*x^n))^3/(e*x+d)^3,x, algorithm="giac")
 

Output:

integrate((g*x + f)*(b*log(c*x^n) + a)^3/(e*x + d)^3, x)
 

Mupad [F(-1)]

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

int(((f + g*x)*(a + b*log(c*x^n))^3)/(d + e*x)^3,x)
 

Output:

int(((f + g*x)*(a + b*log(c*x^n))^3)/(d + e*x)^3, x)
 

Reduce [F]

\[ \int \frac {(f+g x) \left (a+b \log \left (c x^n\right )\right )^3}{(d+e x)^3} \, dx=\text {too large to display} \] Input:

int((g*x+f)*(a+b*log(c*x^n))^3/(e*x+d)^3,x)
 

Output:

(12*int(log(x**n*c)**2/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4 
),x)*b**3*d**6*g*n + 12*int(log(x**n*c)**2/(d**3*x + 3*d**2*e*x**2 + 3*d*e 
**2*x**3 + e**3*x**4),x)*b**3*d**5*e*f*n + 24*int(log(x**n*c)**2/(d**3*x + 
 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*b**3*d**5*e*g*n*x + 24*int( 
log(x**n*c)**2/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*b** 
3*d**4*e**2*f*n*x + 12*int(log(x**n*c)**2/(d**3*x + 3*d**2*e*x**2 + 3*d*e* 
*2*x**3 + e**3*x**4),x)*b**3*d**4*e**2*g*n*x**2 + 12*int(log(x**n*c)**2/(d 
**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*b**3*d**3*e**3*f*n*x 
**2 + 24*int(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x* 
*4),x)*a*b**2*d**6*g*n + 24*int(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d* 
e**2*x**3 + e**3*x**4),x)*a*b**2*d**5*e*f*n + 48*int(log(x**n*c)/(d**3*x + 
 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*a*b**2*d**5*e*g*n*x + 48*in 
t(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*a*b* 
*2*d**4*e**2*f*n*x + 24*int(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2 
*x**3 + e**3*x**4),x)*a*b**2*d**4*e**2*g*n*x**2 + 24*int(log(x**n*c)/(d**3 
*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*a*b**2*d**3*e**3*f*n*x* 
*2 + 60*int(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x** 
4),x)*b**3*d**6*g*n**2 + 12*int(log(x**n*c)/(d**3*x + 3*d**2*e*x**2 + 3*d* 
e**2*x**3 + e**3*x**4),x)*b**3*d**5*e*f*n**2 + 120*int(log(x**n*c)/(d**3*x 
 + 3*d**2*e*x**2 + 3*d*e**2*x**3 + e**3*x**4),x)*b**3*d**5*e*g*n**2*x +...