\(\int \frac {x \log (c (a+b x)^p)}{d+e x} \, dx\) [221]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 91 \[ \int \frac {x \log \left (c (a+b x)^p\right )}{d+e x} \, dx=-\frac {p x}{e}+\frac {(a+b x) \log \left (c (a+b x)^p\right )}{b e}-\frac {d \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{e^2}-\frac {d p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{e^2} \] Output:

-p*x/e+(b*x+a)*ln(c*(b*x+a)^p)/b/e-d*ln(c*(b*x+a)^p)*ln(b*(e*x+d)/(-a*e+b* 
d))/e^2-d*p*polylog(2,-e*(b*x+a)/(-a*e+b*d))/e^2
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.87 \[ \int \frac {x \log \left (c (a+b x)^p\right )}{d+e x} \, dx=\frac {-b e p x+\log \left (c (a+b x)^p\right ) \left (a e+b e x-b d \log \left (\frac {b (d+e x)}{b d-a e}\right )\right )-b d p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )}{b e^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 2.64 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.71

method result size
parts \(\frac {x \ln \left (c \left (b x +a \right )^{p}\right )}{e}-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {p b \left (\frac {e x +d}{e b}-\frac {a \ln \left (\left (e x +d \right ) b +e a -b d \right )}{b^{2}}-\frac {d \left (\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}\right )}{e}\right )}{e}\) \(156\)
risch \(\frac {\ln \left (\left (b x +a \right )^{p}\right ) x}{e}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {p x}{e}-\frac {p d}{e^{2}}+\frac {p a \ln \left (\left (e x +d \right ) b +e a -b d \right )}{b e}+\frac {p d \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e^{2}}+\frac {p d \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{e^{2}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\) \(270\)

Input:

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

Output:

x*ln(c*(b*x+a)^p)/e-ln(c*(b*x+a)^p)*d/e^2*ln(e*x+d)-p*b/e*(1/e*(e*x+d)/b-a 
/b^2*ln((e*x+d)*b+e*a-b*d)-1/e*d*(dilog(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b+l 
n(e*x+d)*ln(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*int(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)*a*b*d*e* 
p + 2*int(log((a + b*x)**p*c)/(a*d + a*e*x + b*d*x + b*e*x**2),x)*b**2*d** 
2*p - log((a + b*x)**p*c)**2*b*d + 2*log((a + b*x)**p*c)*a*e*p + 2*log((a 
+ b*x)**p*c)*b*e*p*x - 2*b*e*p**2*x)/(2*b*e**2*p)