\(\int \frac {x^3 \log (c (a+b x)^p)}{d+e x} \, dx\) [219]

Optimal result

Integrand size = 21, antiderivative size = 250 \[ \int \frac {x^3 \log \left (c (a+b x)^p\right )}{d+e x} \, dx=-\frac {d^2 p x}{e^3}-\frac {a d p x}{2 b e^2}-\frac {a^2 p x}{3 b^2 e}+\frac {d p x^2}{4 e^2}+\frac {a p x^2}{6 b e}-\frac {p x^3}{9 e}+\frac {a^2 d p \log (a+b x)}{2 b^2 e^2}+\frac {a^3 p \log (a+b x)}{3 b^3 e}-\frac {d x^2 \log \left (c (a+b x)^p\right )}{2 e^2}+\frac {x^3 \log \left (c (a+b x)^p\right )}{3 e}+\frac {d^2 (a+b x) \log \left (c (a+b x)^p\right )}{b e^3}-\frac {d^3 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^4}-\frac {d^3 p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^4} \] Output:

-d^2*p*x/e^3-1/2*a*d*p*x/b/e^2-1/3*a^2*p*x/b^2/e+1/4*d*p*x^2/e^2+1/6*a*p*x 
^2/b/e-1/9*p*x^3/e+1/2*a^2*d*p*ln(b*x+a)/b^2/e^2+1/3*a^3*p*ln(b*x+a)/b^3/e 
-1/2*d*x^2*ln(c*(b*x+a)^p)/e^2+1/3*x^3*ln(c*(b*x+a)^p)/e+d^2*(b*x+a)*ln(c* 
(b*x+a)^p)/b/e^3-d^3*ln(c*(b*x+a)^p)*ln(b*(e*x+d)/(-a*e+b*d))/e^4-d^3*p*po 
lylog(2,-e*(b*x+a)/(-a*e+b*d))/e^4
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {6 a^2 e^2 (3 b d+2 a e) p \log (a+b x)+b \left (-e p x \left (12 a^2 e^2-6 a b e (-3 d+e x)+b^2 \left (36 d^2-9 d e x+4 e^2 x^2\right )\right )+6 b \log \left (c (a+b x)^p\right ) \left (6 a d^2 e+b e x \left (6 d^2-3 d e x+2 e^2 x^2\right )-6 b d^3 \log \left (\frac {b (d+e x)}{b d-a e}\right )\right )\right )-36 b^3 d^3 p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )}{36 b^3 e^4} \] Input:

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

Output:

(6*a^2*e^2*(3*b*d + 2*a*e)*p*Log[a + b*x] + b*(-(e*p*x*(12*a^2*e^2 - 6*a*b 
*e*(-3*d + e*x) + b^2*(36*d^2 - 9*d*e*x + 4*e^2*x^2))) + 6*b*Log[c*(a + b* 
x)^p]*(6*a*d^2*e + b*e*x*(6*d^2 - 3*d*e*x + 2*e^2*x^2) - 6*b*d^3*Log[(b*(d 
 + e*x))/(b*d - a*e)])) - 36*b^3*d^3*p*PolyLog[2, (e*(a + b*x))/(-(b*d) + 
a*e)])/(36*b^3*e^4)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((d^2*p*x)/e^3) - (a*d*p*x)/(2*b*e^2) - (a^2*p*x)/(3*b^2*e) + (d*p*x^2)/( 
4*e^2) + (a*p*x^2)/(6*b*e) - (p*x^3)/(9*e) + (a^2*d*p*Log[a + b*x])/(2*b^2 
*e^2) + (a^3*p*Log[a + b*x])/(3*b^3*e) - (d*x^2*Log[c*(a + b*x)^p])/(2*e^2 
) + (x^3*Log[c*(a + b*x)^p])/(3*e) + (d^2*(a + b*x)*Log[c*(a + b*x)^p])/(b 
*e^3) - (d^3*Log[c*(a + b*x)^p]*Log[(b*(d + e*x))/(b*d - a*e)])/e^4 - (d^3 
*p*PolyLog[2, -((e*(a + b*x))/(b*d - a*e))])/e^4
 

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]
 

Maple [A] (verified)

Time = 2.85 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.19

method	result	size
parts	\(\frac {x^{3} \ln \left (c \left (b x +a \right )^{p}\right )}{3 e}-\frac {d \,x^{2} \ln \left (c \left (b x +a \right )^{p}\right )}{2 e^{2}}+\frac {\ln \left (c \left (b x +a \right )^{p}\right ) d^{2} x}{e^{3}}-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) d^{3} \ln \left (e x +d \right )}{e^{4}}-\frac {p b \left (-\frac {-\frac {\frac {2 \left (e x +d \right )^{3} b^{2}}{3}-\left (e x +d \right )^{2} a b e -\frac {7 \left (e x +d \right )^{2} b^{2} d}{2}+2 \left (e x +d \right ) a^{2} e^{2}+5 \left (e x +d \right ) a b d e +11 \left (e x +d \right ) b^{2} d^{2}}{b^{3}}+\frac {e a \left (2 a^{2} e^{2}+3 d e a b +6 d^{2} b^{2}\right ) \ln \left (\left (e x +d \right ) b +e a -b d \right )}{b^{4}}}{6 e^{3}}-\frac {d^{3} \left (\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}\right )}{e^{3}}\right )}{e}\)	\(297\)
risch	\(\frac {\ln \left (\left (b x +a \right )^{p}\right ) x^{3}}{3 e}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d \,x^{2}}{2 e^{2}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) x \,d^{2}}{e^{3}}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d^{3} \ln \left (e x +d \right )}{e^{4}}-\frac {p \,x^{3}}{9 e}+\frac {d p \,x^{2}}{4 e^{2}}-\frac {d^{2} p x}{e^{3}}-\frac {49 p \,d^{3}}{36 e^{4}}+\frac {a p \,x^{2}}{6 b e}-\frac {a d p x}{2 b \,e^{2}}-\frac {2 p a \,d^{2}}{3 b \,e^{3}}-\frac {a^{2} p x}{3 b^{2} e}-\frac {p \,a^{2} d}{3 b^{2} e^{2}}+\frac {p \,a^{3} \ln \left (\left (e x +d \right ) b +e a -b d \right )}{3 b^{3} e}+\frac {p \,a^{2} \ln \left (\left (e x +d \right ) b +e a -b d \right ) d}{2 b^{2} e^{2}}+\frac {p a \ln \left (\left (e x +d \right ) b +e a -b d \right ) d^{2}}{b \,e^{3}}+\frac {p \,d^{3} \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e^{4}}+\frac {p \,d^{3} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e^{4}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\frac {1}{3} e^{2} x^{3}-\frac {1}{2} e \,x^{2} d +d^{2} x}{e^{3}}-\frac {d^{3} \ln \left (e x +d \right )}{e^{4}}\right )\)	\(482\)

Input:

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

Output:

1/3*x^3*ln(c*(b*x+a)^p)/e-1/2*d*x^2*ln(c*(b*x+a)^p)/e^2+ln(c*(b*x+a)^p)/e^ 
3*d^2*x-ln(c*(b*x+a)^p)*d^3/e^4*ln(e*x+d)-p*b/e*(-1/6/e^3*(-1/b^3*(2/3*(e* 
x+d)^3*b^2-(e*x+d)^2*a*b*e-7/2*(e*x+d)^2*b^2*d+2*(e*x+d)*a^2*e^2+5*(e*x+d) 
*a*b*d*e+11*(e*x+d)*b^2*d^2)+e*a*(2*a^2*e^2+3*a*b*d*e+6*b^2*d^2)/b^4*ln((e 
*x+d)*b+e*a-b*d))-1/e^3*d^3*(dilog(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b+ln(e*x 
+d)*ln(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b))
 

Fricas [F]

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

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

Output:

integral(x^3*log((b*x + a)^p*c)/(e*x + d), x)
 

Sympy [F]

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

integrate(x**3*ln(c*(b*x+a)**p)/(e*x+d),x)
 

Output:

Integral(x**3*log(c*(a + b*x)**p)/(d + e*x), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {-36 \left (\int \frac {\mathrm {log}\left (\left (b x +a \right )^{p} c \right )}{b e \,x^{2}+a e x +b d x +a d}d x \right ) a \,b^{3} d^{3} e p +36 \left (\int \frac {\mathrm {log}\left (\left (b x +a \right )^{p} c \right )}{b e \,x^{2}+a e x +b d x +a d}d x \right ) b^{4} d^{4} p -18 \mathrm {log}\left (\left (b x +a \right )^{p} c \right )^{2} b^{3} d^{3}+12 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) a^{3} e^{3} p +18 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) a^{2} b d \,e^{2} p +36 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) a \,b^{2} d^{2} e p +36 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) b^{3} d^{2} e p x -18 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) b^{3} d \,e^{2} p \,x^{2}+12 \,\mathrm {log}\left (\left (b x +a \right )^{p} c \right ) b^{3} e^{3} p \,x^{3}-12 a^{2} b \,e^{3} p^{2} x -18 a \,b^{2} d \,e^{2} p^{2} x +6 a \,b^{2} e^{3} p^{2} x^{2}-36 b^{3} d^{2} e \,p^{2} x +9 b^{3} d \,e^{2} p^{2} x^{2}-4 b^{3} e^{3} p^{2} x^{3}}{36 b^{3} e^{4} p} \] Input:

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

Output:

( - 36*int(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)*a*b**3* 
d**3*e*p + 36*int(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)* 
b**4*d**4*p - 18*log((a + b*x)**p*c)**2*b**3*d**3 + 12*log((a + b*x)**p*c) 
*a**3*e**3*p + 18*log((a + b*x)**p*c)*a**2*b*d*e**2*p + 36*log((a + b*x)** 
p*c)*a*b**2*d**2*e*p + 36*log((a + b*x)**p*c)*b**3*d**2*e*p*x - 18*log((a 
+ b*x)**p*c)*b**3*d*e**2*p*x**2 + 12*log((a + b*x)**p*c)*b**3*e**3*p*x**3 
- 12*a**2*b*e**3*p**2*x - 18*a*b**2*d*e**2*p**2*x + 6*a*b**2*e**3*p**2*x** 
2 - 36*b**3*d**2*e*p**2*x + 9*b**3*d*e**2*p**2*x**2 - 4*b**3*e**3*p**2*x** 
3)/(36*b**3*e**4*p)