3.32.75 \(\int \frac {-2 x^3-2 x^2 \log (3)+(8 x^3+7 x^2 \log (3)) \log (x)+(-16 x^3-12 x^2 \log (3)) \log ^2(x)+(-24 x+16 x^3+(-12+12 x^2) \log (3)) \log ^3(x)}{4 \log ^3(x)} \, dx\)

Optimal. Leaf size=21 \[ x (x+\log (3)) \left (-3+\left (x-\frac {x}{2 \log (x)}\right )^2\right ) \]

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Rubi [C]  time = 0.53, antiderivative size = 85, normalized size of antiderivative = 4.05, number of steps used = 28, number of rules used = 6, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {12, 6742, 2353, 2306, 2309, 2178} \begin {gather*} -\log (27) \text {Ei}(3 \log (x))+3 \log (3) \text {Ei}(3 \log (x))+x^4+\frac {x^4}{4 \log ^2(x)}-\frac {x^4}{\log (x)}+\frac {x^3 \log (3)}{4 \log ^2(x)}-\frac {x^3 \log (3)}{\log (x)}+\frac {1}{3} x^3 \log (27)-3 x^2-3 x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x^3 - 2*x^2*Log[3] + (8*x^3 + 7*x^2*Log[3])*Log[x] + (-16*x^3 - 12*x^2*Log[3])*Log[x]^2 + (-24*x + 16*
x^3 + (-12 + 12*x^2)*Log[3])*Log[x]^3)/(4*Log[x]^3),x]

[Out]

-3*x^2 + x^4 - 3*x*Log[3] + 3*ExpIntegralEi[3*Log[x]]*Log[3] + (x^3*Log[27])/3 - ExpIntegralEi[3*Log[x]]*Log[2
7] + x^4/(4*Log[x]^2) + (x^3*Log[3])/(4*Log[x]^2) - x^4/Log[x] - (x^3*Log[3])/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-2 x^3-2 x^2 \log (3)+\left (8 x^3+7 x^2 \log (3)\right ) \log (x)+\left (-16 x^3-12 x^2 \log (3)\right ) \log ^2(x)+\left (-24 x+16 x^3+\left (-12+12 x^2\right ) \log (3)\right ) \log ^3(x)}{\log ^3(x)} \, dx\\ &=\frac {1}{4} \int \left (4 \left (-6 x+4 x^3-3 \log (3)+x^2 \log (27)\right )-\frac {2 x^2 (x+\log (3))}{\log ^3(x)}+\frac {x^2 (8 x+7 \log (3))}{\log ^2(x)}-\frac {4 x^2 (4 x+\log (27))}{\log (x)}\right ) \, dx\\ &=\frac {1}{4} \int \frac {x^2 (8 x+7 \log (3))}{\log ^2(x)} \, dx-\frac {1}{2} \int \frac {x^2 (x+\log (3))}{\log ^3(x)} \, dx+\int \left (-6 x+4 x^3-3 \log (3)+x^2 \log (27)\right ) \, dx-\int \frac {x^2 (4 x+\log (27))}{\log (x)} \, dx\\ &=-3 x^2+x^4-3 x \log (3)+\frac {1}{3} x^3 \log (27)+\frac {1}{4} \int \left (\frac {8 x^3}{\log ^2(x)}+\frac {7 x^2 \log (3)}{\log ^2(x)}\right ) \, dx-\frac {1}{2} \int \left (\frac {x^3}{\log ^3(x)}+\frac {x^2 \log (3)}{\log ^3(x)}\right ) \, dx-\int \left (\frac {4 x^3}{\log (x)}+\frac {x^2 \log (27)}{\log (x)}\right ) \, dx\\ &=-3 x^2+x^4-3 x \log (3)+\frac {1}{3} x^3 \log (27)-\frac {1}{2} \int \frac {x^3}{\log ^3(x)} \, dx+2 \int \frac {x^3}{\log ^2(x)} \, dx-4 \int \frac {x^3}{\log (x)} \, dx-\frac {1}{2} \log (3) \int \frac {x^2}{\log ^3(x)} \, dx+\frac {1}{4} (7 \log (3)) \int \frac {x^2}{\log ^2(x)} \, dx-\log (27) \int \frac {x^2}{\log (x)} \, dx\\ &=-3 x^2+x^4-3 x \log (3)+\frac {1}{3} x^3 \log (27)+\frac {x^4}{4 \log ^2(x)}+\frac {x^3 \log (3)}{4 \log ^2(x)}-\frac {2 x^4}{\log (x)}-\frac {7 x^3 \log (3)}{4 \log (x)}-4 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )+8 \int \frac {x^3}{\log (x)} \, dx-\frac {1}{4} (3 \log (3)) \int \frac {x^2}{\log ^2(x)} \, dx+\frac {1}{4} (21 \log (3)) \int \frac {x^2}{\log (x)} \, dx-\log (27) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )-\int \frac {x^3}{\log ^2(x)} \, dx\\ &=-3 x^2+x^4-4 \text {Ei}(4 \log (x))-3 x \log (3)+\frac {1}{3} x^3 \log (27)-\text {Ei}(3 \log (x)) \log (27)+\frac {x^4}{4 \log ^2(x)}+\frac {x^3 \log (3)}{4 \log ^2(x)}-\frac {x^4}{\log (x)}-\frac {x^3 \log (3)}{\log (x)}-4 \int \frac {x^3}{\log (x)} \, dx+8 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{4} (9 \log (3)) \int \frac {x^2}{\log (x)} \, dx+\frac {1}{4} (21 \log (3)) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=-3 x^2+x^4+4 \text {Ei}(4 \log (x))-3 x \log (3)+\frac {21}{4} \text {Ei}(3 \log (x)) \log (3)+\frac {1}{3} x^3 \log (27)-\text {Ei}(3 \log (x)) \log (27)+\frac {x^4}{4 \log ^2(x)}+\frac {x^3 \log (3)}{4 \log ^2(x)}-\frac {x^4}{\log (x)}-\frac {x^3 \log (3)}{\log (x)}-4 \operatorname {Subst}\left (\int \frac {e^{4 x}}{x} \, dx,x,\log (x)\right )-\frac {1}{4} (9 \log (3)) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (x)\right )\\ &=-3 x^2+x^4-3 x \log (3)+3 \text {Ei}(3 \log (x)) \log (3)+\frac {1}{3} x^3 \log (27)-\text {Ei}(3 \log (x)) \log (27)+\frac {x^4}{4 \log ^2(x)}+\frac {x^3 \log (3)}{4 \log ^2(x)}-\frac {x^4}{\log (x)}-\frac {x^3 \log (3)}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 35, normalized size = 1.67 \begin {gather*} \frac {x (x+\log (3)) \left (x^2-4 x^2 \log (x)+4 \left (-3+x^2\right ) \log ^2(x)\right )}{4 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x^3 - 2*x^2*Log[3] + (8*x^3 + 7*x^2*Log[3])*Log[x] + (-16*x^3 - 12*x^2*Log[3])*Log[x]^2 + (-24*x
+ 16*x^3 + (-12 + 12*x^2)*Log[3])*Log[x]^3)/(4*Log[x]^3),x]

[Out]

(x*(x + Log[3])*(x^2 - 4*x^2*Log[x] + 4*(-3 + x^2)*Log[x]^2))/(4*Log[x]^2)

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fricas [B]  time = 0.55, size = 55, normalized size = 2.62 \begin {gather*} \frac {x^{4} + x^{3} \log \relax (3) + 4 \, {\left (x^{4} - 3 \, x^{2} + {\left (x^{3} - 3 \, x\right )} \log \relax (3)\right )} \log \relax (x)^{2} - 4 \, {\left (x^{4} + x^{3} \log \relax (3)\right )} \log \relax (x)}{4 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((12*x^2-12)*log(3)+16*x^3-24*x)*log(x)^3+(-12*x^2*log(3)-16*x^3)*log(x)^2+(7*x^2*log(3)+8*x^3)
*log(x)-2*x^2*log(3)-2*x^3)/log(x)^3,x, algorithm="fricas")

[Out]

1/4*(x^4 + x^3*log(3) + 4*(x^4 - 3*x^2 + (x^3 - 3*x)*log(3))*log(x)^2 - 4*(x^4 + x^3*log(3))*log(x))/log(x)^2

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giac [B]  time = 0.28, size = 60, normalized size = 2.86 \begin {gather*} x^{4} + x^{3} \log \relax (3) - \frac {x^{4}}{\log \relax (x)} - \frac {x^{3} \log \relax (3)}{\log \relax (x)} - 3 \, x^{2} - 3 \, x \log \relax (3) + \frac {x^{4}}{4 \, \log \relax (x)^{2}} + \frac {x^{3} \log \relax (3)}{4 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((12*x^2-12)*log(3)+16*x^3-24*x)*log(x)^3+(-12*x^2*log(3)-16*x^3)*log(x)^2+(7*x^2*log(3)+8*x^3)
*log(x)-2*x^2*log(3)-2*x^3)/log(x)^3,x, algorithm="giac")

[Out]

x^4 + x^3*log(3) - x^4/log(x) - x^3*log(3)/log(x) - 3*x^2 - 3*x*log(3) + 1/4*x^4/log(x)^2 + 1/4*x^3*log(3)/log
(x)^2

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maple [B]  time = 0.04, size = 49, normalized size = 2.33

method	result	size
risch	\(x^{3} \ln \relax (3)+x^{4}-3 x \ln \relax (3)-3 x^{2}-\frac {x^{3} \left (4 \ln \relax (3) \ln \relax (x )+4 x \ln \relax (x )-\ln \relax (3)-x \right )}{4 \ln \relax (x )^{2}}\)	\(49\)
norman	\(\frac {x^{4} \ln \relax (x )^{2}+x^{3} \ln \relax (3) \ln \relax (x )^{2}+\frac {x^{4}}{4}-3 x^{2} \ln \relax (x )^{2}+\frac {x^{3} \ln \relax (3)}{4}-x^{4} \ln \relax (x )-3 x \ln \relax (3) \ln \relax (x )^{2}-x^{3} \ln \relax (3) \ln \relax (x )}{\ln \relax (x )^{2}}\)	\(71\)
default	\(x^{3} \ln \relax (3)+x^{4}+3 \ln \relax (3) \expIntegralEi \left (1, -3 \ln \relax (x )\right )-3 x \ln \relax (3)-3 x^{2}+\frac {7 \ln \relax (3) \left (-\frac {x^{3}}{\ln \relax (x )}-3 \expIntegralEi \left (1, -3 \ln \relax (x )\right )\right )}{4}-\frac {x^{4}}{\ln \relax (x )}-\frac {\ln \relax (3) \left (-\frac {x^{3}}{2 \ln \relax (x )^{2}}-\frac {3 x^{3}}{2 \ln \relax (x )}-\frac {9 \expIntegralEi \left (1, -3 \ln \relax (x )\right )}{2}\right )}{2}+\frac {x^{4}}{4 \ln \relax (x )^{2}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(((12*x^2-12)*ln(3)+16*x^3-24*x)*ln(x)^3+(-12*x^2*ln(3)-16*x^3)*ln(x)^2+(7*x^2*ln(3)+8*x^3)*ln(x)-2*x^
2*ln(3)-2*x^3)/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x^3*ln(3)+x^4-3*x*ln(3)-3*x^2-1/4*x^3*(4*ln(3)*ln(x)+4*x*ln(x)-ln(3)-x)/ln(x)^2

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maxima [C]  time = 0.74, size = 72, normalized size = 3.43 \begin {gather*} x^{4} + x^{3} \log \relax (3) - 3 \, x^{2} - 3 \, x \log \relax (3) - 3 \, {\rm Ei}\left (3 \, \log \relax (x)\right ) \log \relax (3) + \frac {21}{4} \, \Gamma \left (-1, -3 \, \log \relax (x)\right ) \log \relax (3) + \frac {9}{2} \, \Gamma \left (-2, -3 \, \log \relax (x)\right ) \log \relax (3) - 4 \, {\rm Ei}\left (4 \, \log \relax (x)\right ) + 8 \, \Gamma \left (-1, -4 \, \log \relax (x)\right ) + 8 \, \Gamma \left (-2, -4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((12*x^2-12)*log(3)+16*x^3-24*x)*log(x)^3+(-12*x^2*log(3)-16*x^3)*log(x)^2+(7*x^2*log(3)+8*x^3)
*log(x)-2*x^2*log(3)-2*x^3)/log(x)^3,x, algorithm="maxima")

[Out]

x^4 + x^3*log(3) - 3*x^2 - 3*x*log(3) - 3*Ei(3*log(x))*log(3) + 21/4*gamma(-1, -3*log(x))*log(3) + 9/2*gamma(-
2, -3*log(x))*log(3) - 4*Ei(4*log(x)) + 8*gamma(-1, -4*log(x)) + 8*gamma(-2, -4*log(x))

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mupad [B]  time = 1.94, size = 60, normalized size = 2.86 \begin {gather*} \frac {\frac {x\,\left (x^3+\ln \relax (3)\,x^2\right )}{4}-\frac {x\,\ln \relax (x)\,\left (4\,x^3+\ln \left (81\right )\,x^2\right )}{4}}{{\ln \relax (x)}^2}-\frac {x\,\left (-4\,x^3-\ln \left (81\right )\,x^2+12\,x+12\,\ln \relax (3)\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x^2*log(3))/2 - (log(x)*(7*x^2*log(3) + 8*x^3))/4 - (log(x)^3*(log(3)*(12*x^2 - 12) - 24*x + 16*x^3))/4
+ x^3/2 + (log(x)^2*(12*x^2*log(3) + 16*x^3))/4)/log(x)^3,x)

[Out]

((x*(x^2*log(3) + x^3))/4 - (x*log(x)*(x^2*log(81) + 4*x^3))/4)/log(x)^2 - (x*(12*x + 12*log(3) - x^2*log(81)
- 4*x^3))/4

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sympy [B]  time = 0.13, size = 56, normalized size = 2.67 \begin {gather*} x^{4} + x^{3} \log {\relax (3 )} - 3 x^{2} - 3 x \log {\relax (3 )} + \frac {x^{4} + x^{3} \log {\relax (3 )} + \left (- 4 x^{4} - 4 x^{3} \log {\relax (3 )}\right ) \log {\relax (x )}}{4 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(((12*x**2-12)*ln(3)+16*x**3-24*x)*ln(x)**3+(-12*x**2*ln(3)-16*x**3)*ln(x)**2+(7*x**2*ln(3)+8*x*
*3)*ln(x)-2*x**2*ln(3)-2*x**3)/ln(x)**3,x)

[Out]

x**4 + x**3*log(3) - 3*x**2 - 3*x*log(3) + (x**4 + x**3*log(3) + (-4*x**4 - 4*x**3*log(3))*log(x))/(4*log(x)**
2)

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