\(\int \frac {\log (c (a+b x)^p)}{x^3 (d+e x)} \, dx\) [225]

Optimal result

Integrand size = 21, antiderivative size = 227 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=-\frac {b p}{2 a d x}-\frac {b^2 p \log (x)}{2 a^2 d}-\frac {b e p \log (x)}{a d^2}+\frac {b^2 p \log (a+b x)}{2 a^2 d}+\frac {b e p \log (a+b x)}{a d^2}-\frac {\log \left (c (a+b x)^p\right )}{2 d x^2}+\frac {e \log \left (c (a+b x)^p\right )}{d^2 x}+\frac {e^2 \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )}{d^3}-\frac {e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )}{d^3}-\frac {e^2 p \operatorname {PolyLog}\left (2,-\frac {e (a+b x)}{b d-a e}\right )}{d^3}+\frac {e^2 p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.85 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=-\frac {\frac {2 b d e p \log (x)}{a}-\frac {2 b d e p \log (a+b x)}{a}+\frac {b d^2 p (a+b x \log (x)-b x \log (a+b x))}{a^2 x}+\frac {d^2 \log \left (c (a+b x)^p\right )}{x^2}-\frac {2 d e \log \left (c (a+b x)^p\right )}{x}-2 e^2 \log \left (-\frac {b x}{a}\right ) \log \left (c (a+b x)^p\right )+2 e^2 \log \left (c (a+b x)^p\right ) \log \left (\frac {b (d+e x)}{b d-a e}\right )+2 e^2 p \operatorname {PolyLog}\left (2,\frac {e (a+b x)}{-b d+a e}\right )-2 e^2 p \operatorname {PolyLog}\left (2,1+\frac {b x}{a}\right )}{2 d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 227, 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.

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

\(\Big \downarrow \) 2863

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 1.49 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.15

method	result	size
parts	\(-\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (c \left (b x +a \right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {e \ln \left (c \left (b x +a \right )^{p}\right )}{d^{2} x}-\frac {p b \left (\frac {2 e^{2} \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{3} b}+\frac {2 e^{2} \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{3} b}-\frac {2 e^{2} \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{3} b}-\frac {2 e^{2} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{3} b}-\frac {-\frac {d}{a x}+\frac {\left (-2 e a -b d \right ) \ln \left (x \right )}{a^{2}}+\frac {\left (2 e a +b d \right ) \ln \left (b x +a \right )}{a^{2}}}{d^{2}}\right )}{2}\)	\(260\)
risch	\(-\frac {\ln \left (\left (b x +a \right )^{p}\right ) e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{2 d \,x^{2}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) e^{2} \ln \left (x \right )}{d^{3}}+\frac {\ln \left (\left (b x +a \right )^{p}\right ) e}{d^{2} x}-\frac {b e p \ln \left (x \right )}{a \,d^{2}}-\frac {b^{2} p \ln \left (x \right )}{2 a^{2} d}-\frac {b p}{2 a d x}+\frac {b e p \ln \left (b x +a \right )}{a \,d^{2}}+\frac {b^{2} p \ln \left (b x +a \right )}{2 a^{2} d}-\frac {p \,e^{2} \operatorname {dilog}\left (\frac {b x +a}{a}\right )}{d^{3}}-\frac {p \,e^{2} \ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{d^{3}}+\frac {p \,e^{2} \operatorname {dilog}\left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{3}}+\frac {p \,e^{2} \ln \left (e x +d \right ) \ln \left (\frac {\left (e x +d \right ) b +e a -b d}{e a -b d}\right )}{d^{3}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {e^{2} \ln \left (e x +d \right )}{d^{3}}-\frac {1}{2 d \,x^{2}}+\frac {e^{2} \ln \left (x \right )}{d^{3}}+\frac {e}{d^{2} x}\right )\)	\(409\)

Input:

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

Output:

-ln(c*(b*x+a)^p)*e^2/d^3*ln(e*x+d)-1/2*ln(c*(b*x+a)^p)/d/x^2+ln(c*(b*x+a)^ 
p)*e^2/d^3*ln(x)+e*ln(c*(b*x+a)^p)/d^2/x-1/2*p*b*(2*e^2/d^3*dilog((b*x+a)/ 
a)/b+2*e^2/d^3*ln(x)*ln((b*x+a)/a)/b-2*e^2/d^3*dilog(((e*x+d)*b+e*a-b*d)/( 
a*e-b*d))/b-2*e^2/d^3*ln(e*x+d)*ln(((e*x+d)*b+e*a-b*d)/(a*e-b*d))/b-1/d^2* 
(-d/a/x+1/a^2*(-2*a*e-b*d)*ln(x)+(2*a*e+b*d)/a^2*ln(b*x+a)))
 

Fricas [F]

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

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

Output:

integral(log((b*x + a)^p*c)/(e*x^4 + d*x^3), x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.95 \[ \int \frac {\log \left (c (a+b x)^p\right )}{x^3 (d+e x)} \, dx=\frac {1}{2} \, {\left (2 \, e {\left (\frac {\log \left (b x + a\right )}{a d^{2}} - \frac {\log \left (x\right )}{a d^{2}}\right )} - \frac {2 \, {\left (\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )\right )} e^{2}}{b d^{3}} + \frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {b e x + b d}{b d - a e} + 1\right ) + {\rm Li}_2\left (\frac {b e x + b d}{b d - a e}\right )\right )} e^{2}}{b d^{3}} + \frac {b \log \left (b x + a\right )}{a^{2} d} - \frac {b \log \left (x\right )}{a^{2} d} - \frac {1}{a d x}\right )} b p - \frac {1}{2} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {2 \, e x - d}{d^{2} x^{2}}\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \] Input:

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

Output:

1/2*(2*e*(log(b*x + a)/(a*d^2) - log(x)/(a*d^2)) - 2*(log(b*x/a + 1)*log(x 
) + dilog(-b*x/a))*e^2/(b*d^3) + 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^2/(b*d^3) + b*log(b*x + a 
)/(a^2*d) - b*log(x)/(a^2*d) - 1/(a*d*x))*b*p - 1/2*(2*e^2*log(e*x + d)/d^ 
3 - 2*e^2*log(x)/d^3 - (2*e*x - d)/(d^2*x^2))*log((b*x + a)^p*c)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(log((a + b*x)**p*c)/(d*x**3 + e*x**4),x)