Integrand size = 23, antiderivative size = 352 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\frac {p \log \left (-\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)}{d}+\frac {p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)}{d}+\frac {\log \left (-\frac {b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right )}{3 d}-\frac {\log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,1+\frac {b x^3}{a}\right )}{3 d} \] Output:
p*ln(-e*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*d-a^(1/3)*e))*ln(e*x+d)/d+p*ln(-e*((- 1)^(2/3)*a^(1/3)+b^(1/3)*x)/(b^(1/3)*d-(-1)^(2/3)*a^(1/3)*e))*ln(e*x+d)/d+ p*ln((-1)^(1/3)*e*(a^(1/3)+(-1)^(2/3)*b^(1/3)*x)/(b^(1/3)*d+(-1)^(1/3)*a^( 1/3)*e))*ln(e*x+d)/d+1/3*ln(-b*x^3/a)*ln(c*(b*x^3+a)^p)/d-ln(e*x+d)*ln(c*( b*x^3+a)^p)/d+p*polylog(2,b^(1/3)*(e*x+d)/(b^(1/3)*d-a^(1/3)*e))/d+p*polyl og(2,b^(1/3)*(e*x+d)/(b^(1/3)*d+(-1)^(1/3)*a^(1/3)*e))/d+p*polylog(2,b^(1/ 3)*(e*x+d)/(b^(1/3)*d-(-1)^(2/3)*a^(1/3)*e))/d+1/3*p*polylog(2,1+b*x^3/a)/ d
Time = 0.14 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.93 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\frac {3 p \log \left (\frac {e \left (\sqrt [3]{-1} \sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)+3 p \log \left (\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+\sqrt [3]{a} e}\right ) \log (d+e x)+3 p \log \left (\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)+\log \left (-\frac {b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right )-3 \log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )+3 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )+3 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )+3 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )+p \operatorname {PolyLog}\left (2,1+\frac {b x^3}{a}\right )}{3 d} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/(x*(d + e*x)),x]
Output:
(3*p*Log[(e*((-1)^(1/3)*a^(1/3) - b^(1/3)*x))/(b^(1/3)*d + (-1)^(1/3)*a^(1 /3)*e)]*Log[d + e*x] + 3*p*Log[(e*(a^(1/3) + b^(1/3)*x))/(-(b^(1/3)*d) + a ^(1/3)*e)]*Log[d + e*x] + 3*p*Log[(e*((-1)^(2/3)*a^(1/3) + b^(1/3)*x))/(-( b^(1/3)*d) + (-1)^(2/3)*a^(1/3)*e)]*Log[d + e*x] + Log[-((b*x^3)/a)]*Log[c *(a + b*x^3)^p] - 3*Log[d + e*x]*Log[c*(a + b*x^3)^p] + 3*p*PolyLog[2, (b^ (1/3)*(d + e*x))/(b^(1/3)*d - a^(1/3)*e)] + 3*p*PolyLog[2, (b^(1/3)*(d + e *x))/(b^(1/3)*d + (-1)^(1/3)*a^(1/3)*e)] + 3*p*PolyLog[2, (b^(1/3)*(d + e* x))/(b^(1/3)*d - (-1)^(2/3)*a^(1/3)*e)] + p*PolyLog[2, 1 + (b*x^3)/a])/(3* d)
Time = 1.18 (sec) , antiderivative size = 352, 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.
\(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx\) |
\(\Big \downarrow \) 2916 |
\(\displaystyle \int \left (\frac {\log \left (c \left (a+b x^3\right )^p\right )}{d x}-\frac {e \log \left (c \left (a+b x^3\right )^p\right )}{d (d+e x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )}{d}+\frac {\log \left (-\frac {b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right )}{3 d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )}{d}+\frac {p \log (d+e x) \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{d}+\frac {p \log (d+e x) \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{-1} \sqrt [3]{a} e+\sqrt [3]{b} d}\right )}{d}+\frac {p \operatorname {PolyLog}\left (2,\frac {b x^3}{a}+1\right )}{3 d}\) |
Input:
Int[Log[c*(a + b*x^3)^p]/(x*(d + e*x)),x]
Output:
(p*Log[-((e*(a^(1/3) + b^(1/3)*x))/(b^(1/3)*d - a^(1/3)*e))]*Log[d + e*x]) /d + (p*Log[-((e*((-1)^(2/3)*a^(1/3) + b^(1/3)*x))/(b^(1/3)*d - (-1)^(2/3) *a^(1/3)*e))]*Log[d + e*x])/d + (p*Log[((-1)^(1/3)*e*(a^(1/3) + (-1)^(2/3) *b^(1/3)*x))/(b^(1/3)*d + (-1)^(1/3)*a^(1/3)*e)]*Log[d + e*x])/d + (Log[-( (b*x^3)/a)]*Log[c*(a + b*x^3)^p])/(3*d) - (Log[d + e*x]*Log[c*(a + b*x^3)^ p])/d + (p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d - a^(1/3)*e)])/d + (p *PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d + (-1)^(1/3)*a^(1/3)*e)])/d + ( p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d - (-1)^(2/3)*a^(1/3)*e)])/d + (p*PolyLog[2, 1 + (b*x^3)/a])/(3*d)
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 = 1.38 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.48
method | result | size |
parts | \(-\frac {\ln \left (e x +d \right ) \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{d}+\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) \ln \left (x \right )}{d}-3 p b \left (\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )}{3 d b}-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )}{3 d b}\right )\) | \(170\) |
risch | \(-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) \ln \left (e x +d \right )}{d}+\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) \ln \left (x \right )}{d}-\frac {p \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right )}{d}+\frac {p \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{d}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-\frac {\ln \left (e x +d \right )}{d}+\frac {\ln \left (x \right )}{d}\right )\) | \(292\) |
Input:
int(ln(c*(b*x^3+a)^p)/x/(e*x+d),x,method=_RETURNVERBOSE)
Output:
-ln(e*x+d)*ln(c*(b*x^3+a)^p)/d+ln(c*(b*x^3+a)^p)/d*ln(x)-3*p*b*(1/3/d/b*su m(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1),_R1=RootOf(_Z^3*b+a))-1/3/d/b*s um(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+dilog((-e*x+_R1-d)/_R1),_R1=RootOf(_Z^3* b-3*_Z^2*b*d+3*_Z*b*d^2+a*e^3-b*d^3)))
\[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate(log(c*(b*x^3+a)^p)/x/(e*x+d),x, algorithm="fricas")
Output:
integral(log((b*x^3 + a)^p*c)/(e*x^2 + d*x), x)
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(b*x**3+a)**p)/x/(e*x+d),x)
Output:
Timed out
\[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate(log(c*(b*x^3+a)^p)/x/(e*x+d),x, algorithm="maxima")
Output:
integrate(log((b*x^3 + a)^p*c)/((e*x + d)*x), x)
\[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\int { \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x} \,d x } \] Input:
integrate(log(c*(b*x^3+a)^p)/x/(e*x+d),x, algorithm="giac")
Output:
integrate(log((b*x^3 + a)^p*c)/((e*x + d)*x), x)
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\int \frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{x\,\left (d+e\,x\right )} \,d x \] Input:
int(log(c*(a + b*x^3)^p)/(x*(d + e*x)),x)
Output:
int(log(c*(a + b*x^3)^p)/(x*(d + e*x)), x)
\[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{x (d+e x)} \, dx=\int \frac {\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )}{e \,x^{2}+d x}d x \] Input:
int(log(c*(b*x^3+a)^p)/x/(e*x+d),x)
Output:
int(log((a + b*x**3)**p*c)/(d*x + e*x**2),x)