\(\int \frac {x^3 \log (c (a+\frac {b}{x})^p)}{d+e x} \, dx\) [240]

Optimal result

Integrand size = 23, antiderivative size = 297 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=-\frac {b d p x}{2 a e^2}-\frac {b^2 p x}{3 a^2 e}+\frac {b p x^2}{6 a e}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {b d^2 p \log (b+a x)}{a e^3}+\frac {b^2 d p \log (b+a x)}{2 a^2 e^2}+\frac {b^3 p \log (b+a x)}{3 a^3 e}-\frac {d^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4}-\frac {d^3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{d+e x} \, dx=\frac {b d^2 p \log \left (a+\frac {b}{x}\right )}{a e^3}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {b d^2 p \log (x)}{a e^3}-\frac {b p \left (\frac {2 b x}{a^2}-\frac {x^2}{a}-\frac {2 b^2 \log \left (a+\frac {b}{x}\right )}{a^3}-\frac {2 b^2 \log (x)}{a^3}\right )}{6 e}-\frac {b d p \left (\frac {x}{a}-\frac {b \log (b+a x)}{a^2}\right )}{2 e^2}-\frac {d^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \log (d+e x)}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}+\frac {d^3 p \log \left (-\frac {e (b+a x)}{a d-b e}\right ) \log (d+e x)}{e^4}-\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e^4}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2916, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^3 p \log (a x+b)}{3 a^3 e}+\frac {b^2 d p \log (a x+b)}{2 a^2 e^2}-\frac {b^2 p x}{3 a^2 e}-\frac {d^3 \log (d+e x) \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^4}+\frac {d^2 x \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{e^3}-\frac {d x^2 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{2 e^2}+\frac {x^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{3 e}+\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {a (d+e x)}{a d-b e}\right )}{e^4}+\frac {d^3 p \log (d+e x) \log \left (-\frac {e (a x+b)}{a d-b e}\right )}{e^4}+\frac {b d^2 p \log (a x+b)}{a e^3}-\frac {b d p x}{2 a e^2}+\frac {b p x^2}{6 a e}-\frac {d^3 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e^4}-\frac {d^3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^4}\)

Input:

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

Output:

-1/2*(b*d*p*x)/(a*e^2) - (b^2*p*x)/(3*a^2*e) + (b*p*x^2)/(6*a*e) + (d^2*x* 
Log[c*(a + b/x)^p])/e^3 - (d*x^2*Log[c*(a + b/x)^p])/(2*e^2) + (x^3*Log[c* 
(a + b/x)^p])/(3*e) + (b*d^2*p*Log[b + a*x])/(a*e^3) + (b^2*d*p*Log[b + a* 
x])/(2*a^2*e^2) + (b^3*p*Log[b + a*x])/(3*a^3*e) - (d^3*Log[c*(a + b/x)^p] 
*Log[d + e*x])/e^4 - (d^3*p*Log[-((e*x)/d)]*Log[d + e*x])/e^4 + (d^3*p*Log 
[-((e*(b + a*x))/(a*d - b*e))]*Log[d + e*x])/e^4 + (d^3*p*PolyLog[2, (a*(d 
 + e*x))/(a*d - b*e)])/e^4 - (d^3*p*PolyLog[2, 1 + (e*x)/d])/e^4
 

Defintions of rubi rules used

rule 2009

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

rule 2916

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log 
[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g 
, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
 

Maple [A] (verified)

Time = 3.10 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.01

method	result	size
parts	\(\frac {x^{3} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{3 e}-\frac {d \,x^{2} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{2 e^{2}}+\frac {d^{2} x \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )}{e^{3}}-\frac {d^{3} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{4}}+p b e \left (-\frac {\frac {5 a d \left (e x +d \right )-a \left (e x +d \right )^{2}+2 b e \left (e x +d \right )}{a^{2}}+\frac {\left (-6 a^{2} d^{2}-3 d e a b -2 b^{2} e^{2}\right ) \ln \left (d a -a \left (e x +d \right )-b e \right )}{a^{3}}}{6 e^{4}}-\frac {d^{3} \ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{e^{5} b}-\frac {d^{3} \operatorname {dilog}\left (-\frac {e x}{d}\right )}{e^{5} b}+\frac {d^{3} \operatorname {dilog}\left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{5} b}+\frac {d^{3} \ln \left (e x +d \right ) \ln \left (\frac {-d a +a \left (e x +d \right )+b e}{-d a +b e}\right )}{e^{5} b}\right )\)	\(301\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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