3.3.42 \(\int \frac {x^3 (a+b \log (c (d+e x)^n))}{f+g x} \, dx\) [242]

3.3.42.1 Optimal result
3.3.42.2 Mathematica [A] (verified)
3.3.42.3 Rubi [A] (verified)
3.3.42.4 Maple [C] (warning: unable to verify)
3.3.42.5 Fricas [F]
3.3.42.6 Sympy [F]
3.3.42.7 Maxima [F]
3.3.42.8 Giac [F]
3.3.42.9 Mupad [F(-1)]

3.3.42.1 Optimal result

Integrand size = 25, antiderivative size = 281 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {a f^2 x}{g^3}-\frac {b f^2 n x}{g^3}-\frac {b d f n x}{2 e g^2}-\frac {b d^2 n x}{3 e^2 g}+\frac {b f n x^2}{4 g^2}+\frac {b d n x^2}{6 e g}-\frac {b n x^3}{9 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {f^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{g^4}-\frac {b f^3 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4} \]

output
a*f^2*x/g^3-b*f^2*n*x/g^3-1/2*b*d*f*n*x/e/g^2-1/3*b*d^2*n*x/e^2/g+1/4*b*f* 
n*x^2/g^2+1/6*b*d*n*x^2/e/g-1/9*b*n*x^3/g+1/2*b*d^2*f*n*ln(e*x+d)/e^2/g^2+ 
1/3*b*d^3*n*ln(e*x+d)/e^3/g+b*f^2*(e*x+d)*ln(c*(e*x+d)^n)/e/g^3-1/2*f*x^2* 
(a+b*ln(c*(e*x+d)^n))/g^2+1/3*x^3*(a+b*ln(c*(e*x+d)^n))/g-f^3*(a+b*ln(c*(e 
*x+d)^n))*ln(e*(g*x+f)/(-d*g+e*f))/g^4-b*f^3*n*polylog(2,-g*(e*x+d)/(-d*g+ 
e*f))/g^4
 
3.3.42.2 Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.86 \[ \int \frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x} \, dx=\frac {6 b d^2 g^2 (3 e f+2 d g) n \log (d+e x)+e \left (g x \left (6 a e^2 \left (6 f^2-3 f g x+2 g^2 x^2\right )-b n \left (12 d^2 g^2-6 d e g (-3 f+g x)+e^2 \left (36 f^2-9 f g x+4 g^2 x^2\right )\right )\right )-36 a e^2 f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )+6 b e \log \left (c (d+e x)^n\right ) \left (6 d f^2 g+e g x \left (6 f^2-3 f g x+2 g^2 x^2\right )-6 e f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right )\right )\right )-36 b e^3 f^3 n \operatorname {PolyLog}\left (2,\frac {g (d+e x)}{-e f+d g}\right )}{36 e^3 g^4} \]

input
Integrate[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x),x]
 
output
(6*b*d^2*g^2*(3*e*f + 2*d*g)*n*Log[d + e*x] + e*(g*x*(6*a*e^2*(6*f^2 - 3*f 
*g*x + 2*g^2*x^2) - b*n*(12*d^2*g^2 - 6*d*e*g*(-3*f + g*x) + e^2*(36*f^2 - 
 9*f*g*x + 4*g^2*x^2))) - 36*a*e^2*f^3*Log[(e*(f + g*x))/(e*f - d*g)] + 6* 
b*e*Log[c*(d + e*x)^n]*(6*d*f^2*g + e*g*x*(6*f^2 - 3*f*g*x + 2*g^2*x^2) - 
6*e*f^3*Log[(e*(f + g*x))/(e*f - d*g)])) - 36*b*e^3*f^3*n*PolyLog[2, (g*(d 
 + e*x))/(-(e*f) + d*g)])/(36*e^3*g^4)
 
3.3.42.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2863, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {f^3 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^4}-\frac {f x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {x^3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {a f^2 x}{g^3}+\frac {b f^2 (d+e x) \log \left (c (d+e x)^n\right )}{e g^3}+\frac {b d^3 n \log (d+e x)}{3 e^3 g}+\frac {b d^2 f n \log (d+e x)}{2 e^2 g^2}-\frac {b d^2 n x}{3 e^2 g}-\frac {b f^3 n \operatorname {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{g^4}-\frac {b d f n x}{2 e g^2}+\frac {b d n x^2}{6 e g}-\frac {b f^2 n x}{g^3}+\frac {b f n x^2}{4 g^2}-\frac {b n x^3}{9 g}\)

input
Int[(x^3*(a + b*Log[c*(d + e*x)^n]))/(f + g*x),x]
 
output
(a*f^2*x)/g^3 - (b*f^2*n*x)/g^3 - (b*d*f*n*x)/(2*e*g^2) - (b*d^2*n*x)/(3*e 
^2*g) + (b*f*n*x^2)/(4*g^2) + (b*d*n*x^2)/(6*e*g) - (b*n*x^3)/(9*g) + (b*d 
^2*f*n*Log[d + e*x])/(2*e^2*g^2) + (b*d^3*n*Log[d + e*x])/(3*e^3*g) + (b*f 
^2*(d + e*x)*Log[c*(d + e*x)^n])/(e*g^3) - (f*x^2*(a + b*Log[c*(d + e*x)^n 
]))/(2*g^2) + (x^3*(a + b*Log[c*(d + e*x)^n]))/(3*g) - (f^3*(a + b*Log[c*( 
d + e*x)^n])*Log[(e*(f + g*x))/(e*f - d*g)])/g^4 - (b*f^3*n*PolyLog[2, -(( 
g*(d + e*x))/(e*f - d*g))])/g^4
 

3.3.42.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2863
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 
3.3.42.4 Maple [C] (warning: unable to verify)

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

Time = 0.94 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.80

method result size
risch \(\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x^{3}}{3 g}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f \,x^{2}}{2 g^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) x \,f^{2}}{g^{3}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) f^{3} \ln \left (g x +f \right )}{g^{4}}-\frac {b n \,x^{3}}{9 g}+\frac {b f n \,x^{2}}{4 g^{2}}-\frac {b \,f^{2} n x}{g^{3}}-\frac {49 b n \,f^{3}}{36 g^{4}}+\frac {b d n \,x^{2}}{6 e g}-\frac {b d f n x}{2 e \,g^{2}}-\frac {2 b n d \,f^{2}}{3 e \,g^{3}}-\frac {b \,d^{2} n x}{3 e^{2} g}-\frac {b n \,d^{2} f}{3 e^{2} g^{2}}+\frac {b n \,d^{3} \ln \left (\left (g x +f \right ) e +d g -e f \right )}{3 e^{3} g}+\frac {b n \,d^{2} \ln \left (\left (g x +f \right ) e +d g -e f \right ) f}{2 e^{2} g^{2}}+\frac {b n d \ln \left (\left (g x +f \right ) e +d g -e f \right ) f^{2}}{e \,g^{3}}+\frac {b n \,f^{3} \operatorname {dilog}\left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{4}}+\frac {b n \,f^{3} \ln \left (g x +f \right ) \ln \left (\frac {\left (g x +f \right ) e +d g -e f}{d g -e f}\right )}{g^{4}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{3} g^{2} x^{3}-\frac {1}{2} f g \,x^{2}+f^{2} x}{g^{3}}-\frac {f^{3} \ln \left (g x +f \right )}{g^{4}}\right )\) \(507\)

input
int(x^3*(a+b*ln(c*(e*x+d)^n))/(g*x+f),x,method=_RETURNVERBOSE)
 
output
1/3*b*ln((e*x+d)^n)/g*x^3-1/2*b*ln((e*x+d)^n)/g^2*f*x^2+b*ln((e*x+d)^n)/g^ 
3*x*f^2-b*ln((e*x+d)^n)*f^3/g^4*ln(g*x+f)-1/9*b*n*x^3/g+1/4*b*f*n*x^2/g^2- 
b*f^2*n*x/g^3-49/36*b*n/g^4*f^3+1/6*b*d*n*x^2/e/g-1/2*b*d*f*n*x/e/g^2-2/3* 
b/e*n/g^3*d*f^2-1/3*b*d^2*n*x/e^2/g-1/3*b/e^2*n/g^2*d^2*f+1/3*b/e^3*n/g*d^ 
3*ln((g*x+f)*e+d*g-e*f)+1/2*b/e^2*n/g^2*d^2*ln((g*x+f)*e+d*g-e*f)*f+b/e*n/ 
g^3*d*ln((g*x+f)*e+d*g-e*f)*f^2+b*n/g^4*f^3*dilog(((g*x+f)*e+d*g-e*f)/(d*g 
-e*f))+b*n/g^4*f^3*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+(-1/2*I*b*P 
i*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*b*Pi*csgn(I*c)*csg 
n(I*c*(e*x+d)^n)^2+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2* 
I*b*Pi*csgn(I*c*(e*x+d)^n)^3+b*ln(c)+a)*(1/g^3*(1/3*g^2*x^3-1/2*f*g*x^2+f^ 
2*x)-f^3/g^4*ln(g*x+f))
 
3.3.42.5 Fricas [F]

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

input
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="fricas")
 
output
integral((b*x^3*log((e*x + d)^n*c) + a*x^3)/(g*x + f), x)
 
3.3.42.6 Sympy [F]

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

input
integrate(x**3*(a+b*ln(c*(e*x+d)**n))/(g*x+f),x)
 
output
Integral(x**3*(a + b*log(c*(d + e*x)**n))/(f + g*x), x)
 
3.3.42.7 Maxima [F]

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

input
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="maxima")
 
output
-1/6*a*(6*f^3*log(g*x + f)/g^4 - (2*g^2*x^3 - 3*f*g*x^2 + 6*f^2*x)/g^3) + 
b*integrate((x^3*log((e*x + d)^n) + x^3*log(c))/(g*x + f), x)
 
3.3.42.8 Giac [F]

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

input
integrate(x^3*(a+b*log(c*(e*x+d)^n))/(g*x+f),x, algorithm="giac")
 
output
integrate((b*log((e*x + d)^n*c) + a)*x^3/(g*x + f), x)
 
3.3.42.9 Mupad [F(-1)]

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

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