\(\int \frac {\log (c (a+b x)^p)}{x^2 (d+e x)} \, dx\) [224]

Optimal result

Integrand size = 21, antiderivative size = 146 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^2 (d+e x)} \, dx=\frac {b p \log (x)}{a d}-\frac {b p \log (a+b x)}{a d}-\frac {\log \left (c (a+b x)^p\right )}{d x}-\frac {e \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac {e \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{d^2} \] Output:

b*p*ln(x)/a/d-b*p*ln(b*x+a)/a/d-ln(c*(b*x+a)^p)/d/x-e*ln(-b*x/a)*ln(c*(b*x 
+a)^p)/d^2+e*ln(c*(b*x+a)^p)*ln(b*(e*x+d)/(-a*e+b*d))/d^2+e*p*polylog(2,-e 
*(b*x+a)/(-a*e+b*d))/d^2-e*p*polylog(2,1+b*x/a)/d^2
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^2 (d+e x)} \, dx=\frac {b p \log (x)}{a d}-\frac {b p \log (a+b x)}{a d}-\frac {\log \left (c (a+b x)^p\right )}{d x}-\frac {e \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^2}+\frac {e \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {a+b x}{a}\right )}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 146, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b*p*Log[x])/(a*d) - (b*p*Log[a + b*x])/(a*d) - Log[c*(a + b*x)^p]/(d*x) - 
 (e*Log[-((b*x)/a)]*Log[c*(a + b*x)^p])/d^2 + (e*Log[c*(a + b*x)^p]*Log[(b 
*(d + e*x))/(b*d - a*e)])/d^2 + (e*p*PolyLog[2, -((e*(a + b*x))/(b*d - a*e 
))])/d^2 - (e*p*PolyLog[2, 1 + (b*x)/a])/d^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 = 1.21 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.38

method	result	size
parts	\(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (c \left (b x +a \right )^{p}\right )}{d x}-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e \ln \left (x \right )}{d^{2}}-p b \left (\frac {e \left (\frac {\operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}+\frac {\ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{b}\right )}{d^{2}}-\frac {\ln \left (x \right )}{d a}+\frac {\ln \left (b x +a \right )}{d a}-\frac {e \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{2} b}-\frac {e \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{2} b}\right )\)	\(201\)
risch	\(\frac {\ln \left (\left (b x +a \right )^{p}\right ) e \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{d x}-\frac {\ln \left (\left (b x +a \right )^{p}\right ) e \ln \left (x \right )}{d^{2}}-\frac {p e \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{2}}-\frac {p e \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{2}}+\frac {b p \ln \left (x \right )}{a d}-\frac {b p \ln \left (b x +a \right )}{a d}+\frac {p e \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{2}}+\frac {p e \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{2}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {e \ln \left (e x +d \right )}{d^{2}}-\frac {1}{d x}-\frac {e \ln \left (x \right )}{d^{2}}\right )\)	\(322\)

Input:

int(ln(c*(b*x+a)^p)/x^2/(e*x+d),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

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

integrate(log(c*(b*x+a)^p)/x^2/(e*x+d),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate(ln(c*(b*x+a)**p)/x**2/(e*x+d),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.07 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^2 (d+e x)} \, dx=b p {\left (\frac {{\left (\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} e}{b d^{2}} - \frac {{\left (\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )\right )} e}{b d^{2}} - \frac {\log \left (b x + a\right )}{a d} + \frac {\log \left (x\right )}{a d}\right )} + {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \] Input:

integrate(log(c*(b*x+a)^p)/x^2/(e*x+d),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(log(c*(b*x+a)^p)/x^2/(e*x+d),x, algorithm="giac")
 

Output:

integrate(log((b*x + a)^p*c)/((e*x + d)*x^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(log((a + b*x)**p*c)/(d*x**2 + e*x**3),x)