\(\int \frac {x \log (c (a+\frac {b}{x^3})^p)}{d+e x} \, dx\) [256]

Optimal result

Integrand size = 21, antiderivative size = 488 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=-\frac {\sqrt {3} \sqrt [3]{b} p \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt [3]{a} e}+\frac {x \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}+\frac {\sqrt [3]{b} p \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} e}-\frac {d \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{e^2}-\frac {3 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}-\frac {\sqrt [3]{b} p \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{2 \sqrt [3]{a} e}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{e^2}-\frac {3 d p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e^2} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.23 (sec) , antiderivative size = 403, normalized size of antiderivative = 0.83 \[ \int \frac {x \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=-\frac {3 b p \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b}{a x^3}\right )}{2 a e x^2}+\frac {x \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}-\frac {d \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{e^2}-\frac {3 d p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {(-1)^{2/3} e \left (\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{e^2}-\frac {3 d p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{e^2} \] Input:

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

Output:

(-3*b*p*Hypergeometric2F1[2/3, 1, 5/3, -(b/(a*x^3))])/(2*a*e*x^2) + (x*Log 
[c*(a + b/x^3)^p])/e - (d*Log[c*(a + b/x^3)^p]*Log[d + e*x])/e^2 - (3*d*p* 
Log[-((e*x)/d)]*Log[d + e*x])/e^2 + (d*p*Log[-((e*(b^(1/3) + a^(1/3)*x))/( 
a^(1/3)*d - b^(1/3)*e))]*Log[d + e*x])/e^2 + (d*p*Log[-(((-1)^(2/3)*e*(b^( 
1/3) - (-1)^(1/3)*a^(1/3)*x))/(a^(1/3)*d - (-1)^(2/3)*b^(1/3)*e))]*Log[d + 
 e*x])/e^2 + (d*p*Log[((-1)^(1/3)*e*(b^(1/3) + (-1)^(2/3)*a^(1/3)*x))/(a^( 
1/3)*d + (-1)^(1/3)*b^(1/3)*e)]*Log[d + e*x])/e^2 - (3*d*p*PolyLog[2, (d + 
 e*x)/d])/e^2 + (d*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - b^(1/3)*e 
)])/e^2 + (d*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d + (-1)^(1/3)*b^(1 
/3)*e)])/e^2 + (d*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - (-1)^(2/3) 
*b^(1/3)*e)])/e^2
 

Rubi [A] (verified)

Time = 1.49 (sec) , antiderivative size = 488, 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 = {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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.53 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.49

method	result	size
parts	\(\frac {x \ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right )}{e}-\frac {d \ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) \ln \left (e x +d \right )}{e^{2}}+3 p b \,e^{3} \left (-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a -3 \textit {\_Z}^{2} a d +3 \textit {\_Z} a \,d^{2}-a \,d^{3}+b \,e^{3}\right )}{\sum }\frac {\ln \left (e x -\textit {\_R} +d \right )}{-\textit {\_R}^{2}+2 \textit {\_R} d -d^{2}}}{3 e^{2} a}-\frac {d \left (-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a -3 \textit {\_Z}^{2} a d +3 \textit {\_Z} a \,d^{2}-a \,d^{3}+b \,e^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )}{3 b \,e^{3}}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{3}}\right )}{e^{2}}\right )\)	\(238\)

Input:

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

Output:

x*ln(c*(a+b/x^3)^p)/e-d*ln(c*(a+b/x^3)^p)*ln(e*x+d)/e^2+3*p*b*e^3*(-1/3/e^ 
2/a*sum(1/(-_R^2+2*_R*d-d^2)*ln(e*x-_R+d),_R=RootOf(_Z^3*a-3*_Z^2*a*d+3*_Z 
*a*d^2-a*d^3+b*e^3))-1/e^2*d*(-1/3*sum(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+dilo 
g((-e*x+_R1-d)/_R1),_R1=RootOf(_Z^3*a-3*_Z^2*a*d+3*_Z*a*d^2-a*d^3+b*e^3))/ 
b/e^3+(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))/b/e^3))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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