Integrand size = 23, antiderivative size = 388 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log \left (-\frac {b}{a x^3}\right )}{3 d}-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{d}-\frac {3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{d}-\frac {p \operatorname {PolyLog}\left (2,1+\frac {b}{a x^3}\right )}{3 d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d}-\frac {3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{d} \]
-1/3*ln(c*(a+b/x^3)^p)*ln(-b/a/x^3)/d-ln(c*(a+b/x^3)^p)*ln(e*x+d)/d-3*p*ln (-e*x/d)*ln(e*x+d)/d+p*ln(-e*(b^(1/3)+a^(1/3)*x)/(a^(1/3)*d-b^(1/3)*e))*ln (e*x+d)/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)/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)/d-1/3*p*polylog(2,1+b/a/x^3)/d+p*p olylog(2,a^(1/3)*(e*x+d)/(a^(1/3)*d-b^(1/3)*e))/d+p*polylog(2,a^(1/3)*(e*x +d)/(a^(1/3)*d+(-1)^(1/3)*b^(1/3)*e))/d+p*polylog(2,a^(1/3)*(e*x+d)/(a^(1/ 3)*d-(-1)^(2/3)*b^(1/3)*e))/d-3*p*polylog(2,1+e*x/d)/d
Time = 0.09 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.02 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log \left (-\frac {b}{a x^3}\right )}{3 d}-\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{d}-\frac {3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {(-1)^{2/3} e \left (\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{d}-\frac {p \operatorname {PolyLog}\left (2,\frac {a+\frac {b}{x^3}}{a}\right )}{3 d}-\frac {3 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d} \]
-1/3*(Log[c*(a + b/x^3)^p]*Log[-(b/(a*x^3))])/d - (Log[c*(a + b/x^3)^p]*Lo g[d + e*x])/d - (3*p*Log[-((e*x)/d)]*Log[d + e*x])/d + (p*Log[-((e*(b^(1/3 ) + a^(1/3)*x))/(a^(1/3)*d - b^(1/3)*e))]*Log[d + e*x])/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])/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])/d - (p*PolyLog[2, ( a + b/x^3)/a])/(3*d) - (3*p*PolyLog[2, (d + e*x)/d])/d + (p*PolyLog[2, (a^ (1/3)*(d + e*x))/(a^(1/3)*d - b^(1/3)*e)])/d + (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d + (-1)^(1/3)*b^(1/3)*e)])/d + (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - (-1)^(2/3)*b^(1/3)*e)])/d
Time = 0.72 (sec) , antiderivative size = 388, 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.
\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 2916 |
\(\displaystyle \int \left (\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d x}-\frac {e \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d}-\frac {\log \left (-\frac {b}{a x^3}\right ) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{3 d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt [3]{a} x+(-1)^{2/3} \sqrt [3]{b}\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{d}+\frac {p \log (d+e x) \log \left (\frac {\sqrt [3]{-1} e \left ((-1)^{2/3} \sqrt [3]{a} x+\sqrt [3]{b}\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{d}-\frac {p \operatorname {PolyLog}\left (2,\frac {b}{a x^3}+1\right )}{3 d}-\frac {3 p \operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{d}-\frac {3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{d}\) |
-1/3*(Log[c*(a + b/x^3)^p]*Log[-(b/(a*x^3))])/d - (Log[c*(a + b/x^3)^p]*Lo g[d + e*x])/d - (3*p*Log[-((e*x)/d)]*Log[d + e*x])/d + (p*Log[-((e*(b^(1/3 ) + a^(1/3)*x))/(a^(1/3)*d - b^(1/3)*e))]*Log[d + e*x])/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])/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])/d - (p*PolyLog[2, 1 + b/(a*x^3 )])/(3*d) + (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - b^(1/3)*e)])/d + (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d + (-1)^(1/3)*b^(1/3)*e)])/d + (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - (-1)^(2/3)*b^(1/3)*e)])/ d - (3*p*PolyLog[2, 1 + (e*x)/d])/d
3.3.58.3.1 Defintions of rubi rules used
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.55 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.57
method | result | size |
parts | \(-\frac {\ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) \ln \left (x \right )}{d}+3 p b \left (\frac {\ln \left (x \right )^{2}}{2 d b}-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a +b \right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{3 d b}+\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a -3 \textit {\_Z}^{2} a d +3 \textit {\_Z} a \,d^{2}-a \,d^{3}+e^{3} b \right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )}{3 d b}-\frac {\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{d b}-\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )}{d b}\right )\) | \(220\) |
-ln(c*(a+b/x^3)^p)*ln(e*x+d)/d+ln(c*(a+b/x^3)^p)/d*ln(x)+3*p*b*(1/2/d/b*ln (x)^2-1/3/d/b*sum(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1),_R1=RootOf(_Z^3 *a+b))+1/3/d/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/d/b*ln(e*x+d)*ln(- e*x/d)-1/d/b*dilog(-e*x/d))
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \]
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \]
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{x (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^3}\right )}^p\right )}{x\,\left (d+e\,x\right )} \,d x \]