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

Optimal. Leaf size=86 \[ -\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d x^3}+\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d}+\frac {2 e p^2 \text {Li}_2\left (\frac {e x^3}{d}+1\right )}{3 d} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2454, 2397, 2394, 2315} \[ \frac {2 e p^2 \text {PolyLog}\left (2,\frac {e x^3}{d}+1\right )}{3 d}-\frac {\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )}{3 d x^3}+\frac {2 e p \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )}{3 d} \]

Antiderivative was successfully verified.

[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 2315

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

Rule 2394

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 2397

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 2454

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*} \int \frac {\log ^2\left (c \left (d+e x^3\right )^p\right )}{x^4} \, dx &=\frac {1}{3} \operatorname {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) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.04, size = 84, normalized size = 0.98 \[ \frac {-\left (d+e x^3\right ) \log ^2\left (c \left (d+e x^3\right )^p\right )+2 e p x^3 \log \left (-\frac {e x^3}{d}\right ) \log \left (c \left (d+e x^3\right )^p\right )+2 e p^2 x^3 \text {Li}_2\left (\frac {e x^3}{d}+1\right )}{3 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.53, size = 771, normalized size = 8.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/x^3*ln((e*x^3+d)^p)^2+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)^2-1/3*I*p*e/d*ln(e*x^3+d
)*Pi*csgn(I*c*(e*x^3+d)^p)^2*csgn(I*c)+I*p*e/d*ln(x)*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2+1/3*I*p*e/
d*ln(e*x^3+d)*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)+1/3*I/x^3*ln((e*x^3+d)^p)*Pi*csgn(I*(e*x^
3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)-2/3/x^3*ln((e*x^3+d)^p)*ln(c)-1/3*I/x^3*ln((e*x^3+d)^p)*Pi*csgn(I*c*(e
*x^3+d)^p)^2*csgn(I*c)+1/3*I*p*e/d*ln(e*x^3+d)*Pi*csgn(I*c*(e*x^3+d)^p)^3+I*p*e/d*ln(x)*Pi*csgn(I*c*(e*x^3+d)^
p)^2*csgn(I*c)-I*p*e/d*ln(x)*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)*csgn(I*c)+2*p*e/d*ln(x)*ln(c)-I*p*e/
d*ln(x)*Pi*csgn(I*c*(e*x^3+d)^p)^3+1/3*I/x^3*ln((e*x^3+d)^p)*Pi*csgn(I*c*(e*x^3+d)^p)^3-1/3*I*p*e/d*ln(e*x^3+d
)*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*x^3+d)^p)^2-1/3*I/x^3*ln((e*x^3+d)^p)*Pi*csgn(I*(e*x^3+d)^p)*csgn(I*c*(e*
x^3+d)^p)^2-2/3*p*e/d*ln(e*x^3+d)*ln(c)-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

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maxima [A]  time = 0.82, size = 118, normalized size = 1.37 \[ \frac {1}{3} \, e^{2} p^{2} {\left (\frac {\log \left (e x^{3} + d\right )^{2}}{d e} - \frac {2 \, {\left (3 \, \log \left (\frac {e x^{3}}{d} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {e x^{3}}{d}\right )\right )}}{d e}\right )} - \frac {2}{3} \, e p {\left (\frac {\log \left (e x^{3} + d\right )}{d} - \frac {\log \left (x^{3}\right )}{d}\right )} \log \left ({\left (e x^{3} + d\right )}^{p} c\right ) - \frac {\log \left ({\left (e x^{3} + d\right )}^{p} c\right )^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\ln \left (c\,{\left (e\,x^3+d\right )}^p\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (d + e x^{3}\right )^{p} \right )}^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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