\(\int \frac {\log ^2(c (d+e x^3)^p)}{x^4} \, dx\) [132]

Optimal result

Integrand size = 18, antiderivative size = 86 \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d}-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d x^3}+\frac {2 e p^2 \operatorname {PolyLog}\left (2,1+\frac {e x^3}{d}\right )}{3 d} \] Output:

2/3*e*p*ln(-e*x^3/d)*ln(c*(e*x^3+d)^p)/d-1/3*(e*x^3+d)*ln(c*(e*x^3+d)^p)^2 
/d/x^3+2/3*e*p^2*polylog(2,1+e*x^3/d)/d
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d}-\frac {e \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d}-\frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{3 x^3}+\frac {2 e p^2 \operatorname {PolyLog}\left (2,\frac {d+e x^3}{d}\right )}{3 d} \] Input:

Integrate[Log[c*(d + e*x^3)^p]^2/x^4,x]
 

Output:

(2*e*p*Log[-((e*x^3)/d)]*Log[c*(d + e*x^3)^p])/(3*d) - (e*Log[c*(d + e*x^3 
)^p]^2)/(3*d) - Log[c*(d + e*x^3)^p]^2/(3*x^3) + (2*e*p^2*PolyLog[2, (d + 
e*x^3)/d])/(3*d)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2904, 2844, 2841, 2752}

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.

\(\displaystyle \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx\)

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{3} \int \frac {\log ^2\left (c \left (e x^3+d\right )^p\right )}{x^6}dx^3\)

\(\Big \downarrow \) 2844

\(\displaystyle \frac {1}{3} \left (\frac {2 e p \int \frac {\log \left (c \left (e x^3+d\right )^p\right )}{x^3}dx^3}{d}-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{d x^3}\right )\)

\(\Big \downarrow \) 2841

\(\displaystyle \frac {1}{3} \left (\frac {2 e p \left (\log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )-e p \int \frac {\log \left (-\frac {e x^3}{d}\right )}{e x^3+d}dx^3\right )}{d}-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{d x^3}\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle \frac {1}{3} \left (\frac {2 e p \left (\log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {e x^3}{d}+1\right )\right )}{d}-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{d x^3}\right )\)

Input:

Int[Log[c*(d + e*x^3)^p]^2/x^4,x]
 

Output:

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

Defintions of rubi rules used

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2841

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_ 
)), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x 
)^n])/g), x] - Simp[b*e*(n/g)   Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d + e*x 
), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]
 

rule 2844

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_. 
)*(x_))^2, x_Symbol] :> Simp[(d + e*x)*((a + b*Log[c*(d + e*x)^n])^p/((e*f 
- d*g)*(f + g*x))), x] - Simp[b*e*n*(p/(e*f - d*g))   Int[(a + b*Log[c*(d + 
 e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] & 
& NeQ[e*f - d*g, 0] && GtQ[p, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, 
 x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & 
&  !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
 

Maple [C] (warning: unable to verify)

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

Time = 1.16 (sec) , antiderivative size = 411, normalized size of antiderivative = 4.78

method	result	size
risch	\(-\frac {{\ln \left (\left (e \,x^{3}+d \right )^{p}\right )}^{2}}{3 x^{3}}+\frac {2 e p \ln \left (\left (e \,x^{3}+d \right )^{p}\right ) \ln \left (x \right )}{d}-\frac {2 e p \ln \left (\left (e \,x^{3}+d \right )^{p}\right ) \ln \left (e \,x^{3}+d \right )}{3 d}-\frac {2 e \,p^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (e \,\textit {\_Z}^{3}+d \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 {e \,p^{2} \ln \left (e \,x^{3}+d \right )^{2}}{3 d}+\left (i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right ) \left (-\frac {\ln \left (\left (e \,x^{3}+d \right )^{p}\right )}{3 x^{3}}+e p \left (\frac {\ln \left (x \right )}{d}-\frac {\ln \left (e \,x^{3}+d \right )}{3 d}\right )\right )-\frac {{\left (i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (e \,x^{3}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )\right )}^{2}}{12 x^{3}}\)	\(411\)

Input:

int(ln(c*(e*x^3+d)^p)^2/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*ln((e*x^3+d)^p)^2/x^3+2*e*p*ln((e*x^3+d)^p)/d*ln(x)-2/3*e*p*ln((e*x^3 
+d)^p)/d*ln(e*x^3+d)-2*e*p^2/d*sum(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1 
),_R1=RootOf(_Z^3*e+d))+1/3*e*p^2/d*ln(e*x^3+d)^2+(I*Pi*csgn(I*(e*x^3+d)^p 
)*csgn(I*c*(e*x^3+d)^p)^2-I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*c 
sgn(I*c)-I*Pi*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*csgn(I*c*(e*x^3+d)^p)^2*csgn(I* 
c)+2*ln(c))*(-1/3*ln((e*x^3+d)^p)/x^3+e*p*(1/d*ln(x)-1/3/d*ln(e*x^3+d)))-1 
/12*(I*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-I*Pi*csgn(I*(e*x^3+d 
)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-I*Pi*csgn(I*c*(e*x^3+d)^p)^3+I*Pi*csg 
n(I*c*(e*x^3+d)^p)^2*csgn(I*c)+2*ln(c))^2/x^3
 

Fricas [F]

\[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\int { \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{x^{4}} \,d x } \] Input:

integrate(log(c*(e*x^3+d)^p)^2/x^4,x, algorithm="fricas")
 

Output:

integral(log((e*x^3 + d)^p*c)^2/x^4, x)
 

Sympy [F]

\[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\int \frac {\log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{x^{4}}\, dx \] Input:

integrate(ln(c*(e*x**3+d)**p)**2/x**4,x)
 

Output:

Integral(log(c*(d + e*x**3)**p)**2/x**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.37 \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {1}{3} \, e^{2} p^{2} {\left (\frac {\log \left (e x^{3} + d\right )^{2}}{d e} - \frac {2 \, {\left (3 \, \log \left (\frac {e x^{3}}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x^{3}}{d}\right )\right )}}{d e}\right )} - \frac {2}{3} \, e p {\left (\frac {\log \left (e x^{3} + d\right )}{d} - \frac {\log \left (x^{3}\right )}{d}\right )} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) - \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{3 \, x^{3}} \] Input:

integrate(log(c*(e*x^3+d)^p)^2/x^4,x, algorithm="maxima")
 

Output:

1/3*e^2*p^2*(log(e*x^3 + d)^2/(d*e) - 2*(3*log(e*x^3/d + 1)*log(x) + dilog 
(-e*x^3/d))/(d*e)) - 2/3*e*p*(log(e*x^3 + d)/d - log(x^3)/d)*log((e*x^3 + 
d)^p*c) - 1/3*log((e*x^3 + d)^p*c)^2/x^3
 

Giac [F]

\[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\int { \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{x^{4}} \,d x } \] Input:

integrate(log(c*(e*x^3+d)^p)^2/x^4,x, algorithm="giac")
 

Output:

integrate(log((e*x^3 + d)^p*c)^2/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\int \frac {{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2}{x^4} \,d x \] Input:

int(log(c*(d + e*x^3)^p)^2/x^4,x)
 

Output:

int(log(c*(d + e*x^3)^p)^2/x^4, x)
 

Reduce [F]

\[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {-6 \left (\int \frac {\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right )}{e \,x^{7}+d \,x^{4}}d x \right ) d^{2} p \,x^{3}-{\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right )}^{2} d -2 \,\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) d p -2 \,\mathrm {log}\left (\left (e \,x^{3}+d \right )^{p} c \right ) e p \,x^{3}+6 \,\mathrm {log}\left (x \right ) e \,p^{2} x^{3}}{3 d \,x^{3}} \] Input:

int(log(c*(e*x^3+d)^p)^2/x^4,x)
 

Output:

( - 6*int(log((d + e*x**3)**p*c)/(d*x**4 + e*x**7),x)*d**2*p*x**3 - log((d 
 + e*x**3)**p*c)**2*d - 2*log((d + e*x**3)**p*c)*d*p - 2*log((d + e*x**3)* 
*p*c)*e*p*x**3 + 6*log(x)*e*p**2*x**3)/(3*d*x**3)