3.3.59 \(\int \frac {\log (c (a+\frac {b}{x^3})^p)}{x^2 (d+e x)} \, dx\) [259]

3.3.59.1 Optimal result
3.3.59.2 Mathematica [C] (verified)
3.3.59.3 Rubi [A] (verified)
3.3.59.4 Maple [C] (warning: unable to verify)
3.3.59.5 Fricas [F]
3.3.59.6 Sympy [F(-1)]
3.3.59.7 Maxima [F]
3.3.59.8 Giac [F]
3.3.59.9 Mupad [F(-1)]

3.3.59.1 Optimal result

Integrand size = 23, antiderivative size = 557 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {3 p}{d x}-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} x}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt [3]{b} d}-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d x}+\frac {e \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log \left (-\frac {b}{a x^3}\right )}{3 d^2}-\frac {\sqrt [3]{a} p \log \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{b} d}+\frac {e \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{d^2}+\frac {3 e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}+\frac {\sqrt [3]{a} p \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3} x^2\right )}{2 \sqrt [3]{b} d}+\frac {e p \operatorname {PolyLog}\left (2,1+\frac {b}{a x^3}\right )}{3 d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d^2}+\frac {3 e p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d^2} \]

output
3*p/d/x-ln(c*(a+b/x^3)^p)/d/x+1/3*e*ln(c*(a+b/x^3)^p)*ln(-b/a/x^3)/d^2-a^( 
1/3)*p*ln(b^(1/3)+a^(1/3)*x)/b^(1/3)/d+e*ln(c*(a+b/x^3)^p)*ln(e*x+d)/d^2+3 
*e*p*ln(-e*x/d)*ln(e*x+d)/d^2-e*p*ln(-e*(b^(1/3)+a^(1/3)*x)/(a^(1/3)*d-b^( 
1/3)*e))*ln(e*x+d)/d^2-e*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)/d^2-e*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)/d^2+1/2*a^(1/3) 
*p*ln(b^(2/3)-a^(1/3)*b^(1/3)*x+a^(2/3)*x^2)/b^(1/3)/d+1/3*e*p*polylog(2,1 
+b/a/x^3)/d^2-e*p*polylog(2,a^(1/3)*(e*x+d)/(a^(1/3)*d-b^(1/3)*e))/d^2-e*p 
*polylog(2,a^(1/3)*(e*x+d)/(a^(1/3)*d+(-1)^(1/3)*b^(1/3)*e))/d^2-e*p*polyl 
og(2,a^(1/3)*(e*x+d)/(a^(1/3)*d-(-1)^(2/3)*b^(1/3)*e))/d^2+3*e*p*polylog(2 
,1+e*x/d)/d^2-a^(1/3)*p*arctan(1/3*(b^(1/3)-2*a^(1/3)*x)/b^(1/3)*3^(1/2))* 
3^(1/2)/b^(1/3)/d
 
3.3.59.2 Mathematica [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 463, normalized size of antiderivative = 0.83 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x^2 (d+e x)} \, dx=\frac {3 b p \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {7}{3},-\frac {b}{a x^3}\right )}{4 a d x^4}-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d x}+\frac {e \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log \left (-\frac {b}{a x^3}\right )}{3 d^2}+\frac {e \log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{d^2}+\frac {3 e p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {(-1)^{2/3} e \left (\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{d^2}+\frac {e p \operatorname {PolyLog}\left (2,\frac {a+\frac {b}{x^3}}{a}\right )}{3 d^2}+\frac {3 e p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d^2}-\frac {e p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d^2} \]

input
Integrate[Log[c*(a + b/x^3)^p]/(x^2*(d + e*x)),x]
 
output
(3*b*p*Hypergeometric2F1[1, 4/3, 7/3, -(b/(a*x^3))])/(4*a*d*x^4) - Log[c*( 
a + b/x^3)^p]/(d*x) + (e*Log[c*(a + b/x^3)^p]*Log[-(b/(a*x^3))])/(3*d^2) + 
 (e*Log[c*(a + b/x^3)^p]*Log[d + e*x])/d^2 + (3*e*p*Log[-((e*x)/d)]*Log[d 
+ e*x])/d^2 - (e*p*Log[-((e*(b^(1/3) + a^(1/3)*x))/(a^(1/3)*d - b^(1/3)*e) 
)]*Log[d + e*x])/d^2 - (e*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])/d^2 - (e*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])/d^2 + (e*p*PolyLog[2, (a + b/x^3)/a])/(3*d^2) + (3 
*e*p*PolyLog[2, (d + e*x)/d])/d^2 - (e*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a 
^(1/3)*d - b^(1/3)*e)])/d^2 - (e*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3) 
*d + (-1)^(1/3)*b^(1/3)*e)])/d^2 - (e*p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^ 
(1/3)*d - (-1)^(2/3)*b^(1/3)*e)])/d^2
 
3.3.59.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 557, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

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

3.3.59.3.1 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]
 
3.3.59.4 Maple [C] (warning: unable to verify)

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

Time = 1.97 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.62

method result size
parts \(\frac {e \ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) \ln \left (e x +d \right )}{d^{2}}-\frac {\ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right )}{d x}-\frac {\ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) e \ln \left (x \right )}{d^{2}}+3 p b \left (\frac {e \left (-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a -3 \textit {\_Z}^{2} a d +3 \textit {\_Z} a \,d^{2}-a \,d^{3}+e^{3} b \right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )}{3 b}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b}\right )}{d^{2}}+\frac {1}{d b x}-\frac {\ln \left (x +\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 d b \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} x +\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 d b \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d b \left (\frac {b}{a}\right )^{\frac {1}{3}}}-\frac {e \ln \left (x \right )^{2}}{2 d^{2} b}+\frac {e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a +b \right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right )}{3 d^{2} b}\right )\) \(345\)

input
int(ln(c*(a+b/x^3)^p)/x^2/(e*x+d),x,method=_RETURNVERBOSE)
 
output
e*ln(c*(a+b/x^3)^p)*ln(e*x+d)/d^2-ln(c*(a+b/x^3)^p)/d/x-ln(c*(a+b/x^3)^p)* 
e/d^2*ln(x)+3*p*b*(e/d^2*(-1/3/b*sum(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+dilog( 
(-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))+1/ 
b*(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d)))+1/d/b/x-1/3/d/b/(1/a*b)^(1/3)*ln(x 
+(1/a*b)^(1/3))+1/6/d/b/(1/a*b)^(1/3)*ln(x^2-(1/a*b)^(1/3)*x+(1/a*b)^(2/3) 
)+1/3/d/b*3^(1/2)/(1/a*b)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/a*b)^(1/3)*x-1))- 
1/2*e/d^2/b*ln(x)^2+1/3*e/d^2/b*sum(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R 
1),_R1=RootOf(_Z^3*a+b)))
 
3.3.59.5 Fricas [F]

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

input
integrate(log(c*(a+b/x^3)^p)/x^2/(e*x+d),x, algorithm="fricas")
 
output
integral(log(c*((a*x^3 + b)/x^3)^p)/(e*x^3 + d*x^2), x)
 
3.3.59.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x^2 (d+e x)} \, dx=\text {Timed out} \]

input
integrate(ln(c*(a+b/x**3)**p)/x**2/(e*x+d),x)
 
output
Timed out
 
3.3.59.7 Maxima [F]

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

input
integrate(log(c*(a+b/x^3)^p)/x^2/(e*x+d),x, algorithm="maxima")
 
output
integrate(log((a + b/x^3)^p*c)/((e*x + d)*x^2), x)
 
3.3.59.8 Giac [F]

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

input
integrate(log(c*(a+b/x^3)^p)/x^2/(e*x+d),x, algorithm="giac")
 
output
integrate(log((a + b/x^3)^p*c)/((e*x + d)*x^2), x)
 
3.3.59.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x^2 (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^3}\right )}^p\right )}{x^2\,\left (d+e\,x\right )} \,d x \]

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