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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2504, 2444, 2441, 2352} \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d x^3}+\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d}+\frac {2 e p^2 \operatorname {PolyLog}\left (2,\frac {e x^3}{d}+1\right )}{3 d} \]

[In]

Int[Log[c*(d + e*x^3)^p]^2/x^4,x]

[Out]

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

Rule 2352

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

Rule 2441

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] - Dist[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 2444

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] - Dist[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 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {\log ^2\left (c (d+e x)^p\right )}{x^2} \, dx,x,x^3\right ) \\ & = -\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) \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^3\right )}{3 d} \\ & = \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 {\left (2 e^2 p^2\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^3\right )}{3 d} \\ & = \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 \text {Li}_2\left (1+\frac {e x^3}{d}\right )}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d}-\frac {e \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d}-\frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{3 x^3}+\frac {2 e p^2 \operatorname {PolyLog}\left (2,\frac {d+e x^3}{d}\right )}{3 d} \]

[In]

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

[Out]

(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)

Maple [C] (warning: unable to verify)

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

Time = 0.82 (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 p e \ln \left (\left (e \,x^{3}+d \right )^{p}\right ) \ln \left (x \right )}{d}-\frac {2 p e \ln \left (\left (e \,x^{3}+d \right )^{p}\right ) \ln \left (e \,x^{3}+d \right )}{3 d}-\frac {2 p^{2} e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} e +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 {p^{2} e \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}}+p e \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}}\)	\(411\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.37 \[ \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx=\frac {1}{3} \, e^{2} p^{2} {\left (\frac {\log \left (e x^{3} + d\right )^{2}}{d e} - \frac {2 \, {\left (3 \, \log \left (\frac {e x^{3}}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x^{3}}{d}\right )\right )}}{d e}\right )} - \frac {2}{3} \, e p {\left (\frac {\log \left (e x^{3} + d\right )}{d} - \frac {\log \left (x^{3}\right )}{d}\right )} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) - \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(log(c*(d + e*x^3)^p)^2/x^4,x)

[Out]

int(log(c*(d + e*x^3)^p)^2/x^4, x)