Integrand size = 20, antiderivative size = 344 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{e}+\frac {3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{e}+\frac {3 p \operatorname {PolyLog}\left (2,1+\frac {e x}{d}\right )}{e} \] Output:
ln(c*(a+b/x^3)^p)*ln(e*x+d)/e+3*p*ln(-e*x/d)*ln(e*x+d)/e-p*ln(-e*(b^(1/3)+ a^(1/3)*x)/(a^(1/3)*d-b^(1/3)*e))*ln(e*x+d)/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)/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 )/e-p*polylog(2,a^(1/3)*(e*x+d)/(a^(1/3)*d-b^(1/3)*e))/e-p*polylog(2,a^(1/ 3)*(e*x+d)/(a^(1/3)*d+(-1)^(1/3)*b^(1/3)*e))/e-p*polylog(2,a^(1/3)*(e*x+d) /(a^(1/3)*d-(-1)^(2/3)*b^(1/3)*e))/e+3*p*polylog(2,1+e*x/d)/e
Time = 0.17 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.02 \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right ) \log (d+e x)}{e}+\frac {3 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {e \left (\sqrt [3]{b}+\sqrt [3]{a} x\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (-\frac {(-1)^{2/3} e \left (\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right ) \log (d+e x)}{e}-\frac {p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} x\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right ) \log (d+e x)}{e}+\frac {3 p \operatorname {PolyLog}\left (2,\frac {d+e x}{d}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{e}-\frac {p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{e} \] Input:
Integrate[Log[c*(a + b/x^3)^p]/(d + e*x),x]
Output:
(Log[c*(a + b/x^3)^p]*Log[d + e*x])/e + (3*p*Log[-((e*x)/d)]*Log[d + e*x]) /e - (p*Log[-((e*(b^(1/3) + a^(1/3)*x))/(a^(1/3)*d - b^(1/3)*e))]*Log[d + e*x])/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])/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])/e + (3*p*PolyLog[2, (d + e*x)/d])/e - (p*PolyLog[2, (a^(1/3)*(d + e*x ))/(a^(1/3)*d - b^(1/3)*e)])/e - (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3 )*d + (-1)^(1/3)*b^(1/3)*e)])/e - (p*PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/ 3)*d - (-1)^(2/3)*b^(1/3)*e)])/e
Time = 1.18 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2912, 2005, 2863, 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.
\(\displaystyle \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx\) |
\(\Big \downarrow \) 2912 |
\(\displaystyle \frac {3 b p \int \frac {\log (d+e x)}{\left (a+\frac {b}{x^3}\right ) x^4}dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle \frac {3 b p \int \frac {\log (d+e x)}{x \left (a x^3+b\right )}dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}\) |
\(\Big \downarrow \) 2863 |
\(\displaystyle \frac {3 b p \int \left (\frac {\log (d+e x)}{b x}-\frac {a x^2 \log (d+e x)}{b \left (a x^3+b\right )}\right )dx}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{e}+\frac {3 b p \left (-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{3 b}-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{3 b}-\frac {\operatorname {PolyLog}\left (2,\frac {\sqrt [3]{a} (d+e x)}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{3 b}-\frac {\log (d+e x) \log \left (-\frac {e \left (\sqrt [3]{a} x+\sqrt [3]{b}\right )}{\sqrt [3]{a} d-\sqrt [3]{b} e}\right )}{3 b}-\frac {\log (d+e x) \log \left (-\frac {e \left (\sqrt [3]{a} x+(-1)^{2/3} \sqrt [3]{b}\right )}{\sqrt [3]{a} d-(-1)^{2/3} \sqrt [3]{b} e}\right )}{3 b}-\frac {\log (d+e x) \log \left (\frac {\sqrt [3]{-1} e \left ((-1)^{2/3} \sqrt [3]{a} x+\sqrt [3]{b}\right )}{\sqrt [3]{a} d+\sqrt [3]{-1} \sqrt [3]{b} e}\right )}{3 b}+\frac {\operatorname {PolyLog}\left (2,\frac {e x}{d}+1\right )}{b}+\frac {\log \left (-\frac {e x}{d}\right ) \log (d+e x)}{b}\right )}{e}\) |
Input:
Int[Log[c*(a + b/x^3)^p]/(d + e*x),x]
Output:
(Log[c*(a + b/x^3)^p]*Log[d + e*x])/e + (3*b*p*((Log[-((e*x)/d)]*Log[d + e *x])/b - (Log[-((e*(b^(1/3) + a^(1/3)*x))/(a^(1/3)*d - b^(1/3)*e))]*Log[d + e*x])/(3*b) - (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])/(3*b) - (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])/ (3*b) - PolyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - b^(1/3)*e)]/(3*b) - Po lyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d + (-1)^(1/3)*b^(1/3)*e)]/(3*b) - P olyLog[2, (a^(1/3)*(d + e*x))/(a^(1/3)*d - (-1)^(2/3)*b^(1/3)*e)]/(3*b) + PolyLog[2, 1 + (e*x)/d]/b))/e
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) ^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c , d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_. )*(x_)), x_Symbol] :> Simp[Log[f + g*x]*((a + b*Log[c*(d + e*x^n)^p])/g), x ] - Simp[b*e*n*(p/g) Int[x^(n - 1)*(Log[f + g*x]/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.41
method | result | size |
parts | \(\frac {\ln \left (c \left (a +\frac {b}{x^{3}}\right )^{p}\right ) \ln \left (e x +d \right )}{e}+3 p b \,e^{2} \left (-\frac {\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} a -3 \textit {\_Z}^{2} a d +3 \textit {\_Z} a \,d^{2}-a \,d^{3}+b \,e^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )}{3 b \,e^{3}}+\frac {\operatorname {dilog}\left (-\frac {e x}{d}\right )+\ln \left (e x +d \right ) \ln \left (-\frac {e x}{d}\right )}{b \,e^{3}}\right )\) | \(142\) |
Input:
int(ln(c*(a+b/x^3)^p)/(e*x+d),x,method=_RETURNVERBOSE)
Output:
ln(c*(a+b/x^3)^p)*ln(e*x+d)/e+3*p*b*e^2*(-1/3*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))/b/e^3+(dilog(-e*x/d)+ln(e*x+d)*ln(-e*x/d))/b/e^3)
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{e x + d} \,d x } \] Input:
integrate(log(c*(a+b/x^3)^p)/(e*x+d),x, algorithm="fricas")
Output:
integral(log(c*((a*x^3 + b)/x^3)^p)/(e*x + d), x)
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(a+b/x**3)**p)/(e*x+d),x)
Output:
Timed out
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{e x + d} \,d x } \] Input:
integrate(log(c*(a+b/x^3)^p)/(e*x+d),x, algorithm="maxima")
Output:
integrate(log((a + b/x^3)^p*c)/(e*x + d), x)
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\int { \frac {\log \left ({\left (a + \frac {b}{x^{3}}\right )}^{p} c\right )}{e x + d} \,d x } \] Input:
integrate(log(c*(a+b/x^3)^p)/(e*x+d),x, algorithm="giac")
Output:
integrate(log((a + b/x^3)^p*c)/(e*x + d), x)
Timed out. \[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\int \frac {\ln \left (c\,{\left (a+\frac {b}{x^3}\right )}^p\right )}{d+e\,x} \,d x \] Input:
int(log(c*(a + b/x^3)^p)/(d + e*x),x)
Output:
int(log(c*(a + b/x^3)^p)/(d + e*x), x)
\[ \int \frac {\log \left (c \left (a+\frac {b}{x^3}\right )^p\right )}{d+e x} \, dx=\int \frac {\mathrm {log}\left (\frac {\left (a \,x^{3}+b \right )^{p} c}{x^{3 p}}\right )}{e x +d}d x \] Input:
int(log(c*(a+b/x^3)^p)/(e*x+d),x)
Output:
int(log(((a*x**3 + b)**p*c)/x**(3*p))/(d + e*x),x)