3.72.88 \(\int \frac {(1296-432 x+1296 e x) \log ^3(x)+(1296-432 x+1296 e x) \log ^2(x) \log (4-x+3 e x)+(432-144 x+432 e x) \log (x) \log ^2(4-x+3 e x)+(48-16 x+48 e x) \log ^3(4-x+3 e x)}{4 x-x^2+3 e x^2} \, dx\)

Optimal. Leaf size=17 \[ (3 \log (x)+\log (4-x+3 e x))^4 \]

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Rubi [A]  time = 0.39, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 103, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 1593, 6688, 12, 6686} \begin {gather*} (3 \log (x)+\log (3 e x-x+4))^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1296 - 432*x + 1296*E*x)*Log[x]^3 + (1296 - 432*x + 1296*E*x)*Log[x]^2*Log[4 - x + 3*E*x] + (432 - 144*x
+ 432*E*x)*Log[x]*Log[4 - x + 3*E*x]^2 + (48 - 16*x + 48*E*x)*Log[4 - x + 3*E*x]^3)/(4*x - x^2 + 3*E*x^2),x]

[Out]

(3*Log[x] + Log[4 - x + 3*E*x])^4

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(1296-432 x+1296 e x) \log ^3(x)+(1296-432 x+1296 e x) \log ^2(x) \log (4-x+3 e x)+(432-144 x+432 e x) \log (x) \log ^2(4-x+3 e x)+(48-16 x+48 e x) \log ^3(4-x+3 e x)}{4 x+(-1+3 e) x^2} \, dx\\ &=\int \frac {(1296-432 x+1296 e x) \log ^3(x)+(1296-432 x+1296 e x) \log ^2(x) \log (4-x+3 e x)+(432-144 x+432 e x) \log (x) \log ^2(4-x+3 e x)+(48-16 x+48 e x) \log ^3(4-x+3 e x)}{x (4+(-1+3 e) x)} \, dx\\ &=\int \frac {16 (3+(-1+3 e) x) (3 \log (x)+\log (4-x+3 e x))^3}{x (4+(-1+3 e) x)} \, dx\\ &=16 \int \frac {(3+(-1+3 e) x) (3 \log (x)+\log (4-x+3 e x))^3}{x (4+(-1+3 e) x)} \, dx\\ &=(3 \log (x)+\log (4-x+3 e x))^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 1.00 \begin {gather*} (3 \log (x)+\log (4-x+3 e x))^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1296 - 432*x + 1296*E*x)*Log[x]^3 + (1296 - 432*x + 1296*E*x)*Log[x]^2*Log[4 - x + 3*E*x] + (432 -
144*x + 432*E*x)*Log[x]*Log[4 - x + 3*E*x]^2 + (48 - 16*x + 48*E*x)*Log[4 - x + 3*E*x]^3)/(4*x - x^2 + 3*E*x^
2),x]

[Out]

(3*Log[x] + Log[4 - x + 3*E*x])^4

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fricas [B]  time = 1.14, size = 73, normalized size = 4.29 \begin {gather*} \log \left (3 \, x e - x + 4\right )^{4} + 12 \, \log \left (3 \, x e - x + 4\right )^{3} \log \relax (x) + 54 \, \log \left (3 \, x e - x + 4\right )^{2} \log \relax (x)^{2} + 108 \, \log \left (3 \, x e - x + 4\right ) \log \relax (x)^{3} + 81 \, \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*exp(1)-16*x+48)*log(3*x*exp(1)-x+4)^3+(432*x*exp(1)-144*x+432)*log(x)*log(3*x*exp(1)-x+4)^2+(
1296*x*exp(1)-432*x+1296)*log(x)^2*log(3*x*exp(1)-x+4)+(1296*x*exp(1)-432*x+1296)*log(x)^3)/(3*x^2*exp(1)-x^2+
4*x),x, algorithm="fricas")

[Out]

log(3*x*e - x + 4)^4 + 12*log(3*x*e - x + 4)^3*log(x) + 54*log(3*x*e - x + 4)^2*log(x)^2 + 108*log(3*x*e - x +
4)*log(x)^3 + 81*log(x)^4

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giac [B]  time = 0.18, size = 75, normalized size = 4.41 \begin {gather*} 4 \, \log \left (3 \, x e - x + 4\right )^{4} + 12 \, \log \left (3 \, x e - x + 4\right )^{3} \log \relax (x) + 54 \, \log \left (3 \, x e - x + 4\right )^{2} \log \relax (x)^{2} + 108 \, \log \left (3 \, x e - x + 4\right ) \log \relax (x)^{3} + 81 \, \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*exp(1)-16*x+48)*log(3*x*exp(1)-x+4)^3+(432*x*exp(1)-144*x+432)*log(x)*log(3*x*exp(1)-x+4)^2+(
1296*x*exp(1)-432*x+1296)*log(x)^2*log(3*x*exp(1)-x+4)+(1296*x*exp(1)-432*x+1296)*log(x)^3)/(3*x^2*exp(1)-x^2+
4*x),x, algorithm="giac")

[Out]

4*log(3*x*e - x + 4)^4 + 12*log(3*x*e - x + 4)^3*log(x) + 54*log(3*x*e - x + 4)^2*log(x)^2 + 108*log(3*x*e - x
+ 4)*log(x)^3 + 81*log(x)^4

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maple [B]  time = 0.14, size = 74, normalized size = 4.35

method	result	size
risch	\(81 \ln \relax (x )^{4}+108 \ln \relax (x )^{3} \ln \left (3 x \,{\mathrm e}-x +4\right )+54 \ln \relax (x )^{2} \ln \left (3 x \,{\mathrm e}-x +4\right )^{2}+12 \ln \relax (x ) \ln \left (3 x \,{\mathrm e}-x +4\right )^{3}+\ln \left (3 x \,{\mathrm e}-x +4\right )^{4}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((48*x*exp(1)-16*x+48)*ln(3*x*exp(1)-x+4)^3+(432*x*exp(1)-144*x+432)*ln(x)*ln(3*x*exp(1)-x+4)^2+(1296*x*ex
p(1)-432*x+1296)*ln(x)^2*ln(3*x*exp(1)-x+4)+(1296*x*exp(1)-432*x+1296)*ln(x)^3)/(3*x^2*exp(1)-x^2+4*x),x,metho
d=_RETURNVERBOSE)

[Out]

81*ln(x)^4+108*ln(x)^3*ln(3*x*exp(1)-x+4)+54*ln(x)^2*ln(3*x*exp(1)-x+4)^2+12*ln(x)*ln(3*x*exp(1)-x+4)^3+ln(3*x
*exp(1)-x+4)^4

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maxima [B]  time = 0.42, size = 73, normalized size = 4.29 \begin {gather*} \log \left (x {\left (3 \, e - 1\right )} + 4\right )^{4} + 12 \, \log \left (x {\left (3 \, e - 1\right )} + 4\right )^{3} \log \relax (x) + 54 \, \log \left (x {\left (3 \, e - 1\right )} + 4\right )^{2} \log \relax (x)^{2} + 108 \, \log \left (x {\left (3 \, e - 1\right )} + 4\right ) \log \relax (x)^{3} + 81 \, \log \relax (x)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*exp(1)-16*x+48)*log(3*x*exp(1)-x+4)^3+(432*x*exp(1)-144*x+432)*log(x)*log(3*x*exp(1)-x+4)^2+(
1296*x*exp(1)-432*x+1296)*log(x)^2*log(3*x*exp(1)-x+4)+(1296*x*exp(1)-432*x+1296)*log(x)^3)/(3*x^2*exp(1)-x^2+
4*x),x, algorithm="maxima")

[Out]

log(x*(3*e - 1) + 4)^4 + 12*log(x*(3*e - 1) + 4)^3*log(x) + 54*log(x*(3*e - 1) + 4)^2*log(x)^2 + 108*log(x*(3*
e - 1) + 4)*log(x)^3 + 81*log(x)^4

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mupad [B]  time = 4.76, size = 73, normalized size = 4.29 \begin {gather*} {\ln \left (3\,x\,\mathrm {e}-x+4\right )}^4+12\,{\ln \left (3\,x\,\mathrm {e}-x+4\right )}^3\,\ln \relax (x)+54\,{\ln \left (3\,x\,\mathrm {e}-x+4\right )}^2\,{\ln \relax (x)}^2+108\,\ln \left (3\,x\,\mathrm {e}-x+4\right )\,{\ln \relax (x)}^3+81\,{\ln \relax (x)}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3*x*exp(1) - x + 4)^3*(48*x*exp(1) - 16*x + 48) + log(x)^3*(1296*x*exp(1) - 432*x + 1296) + log(3*x*e
xp(1) - x + 4)^2*log(x)*(432*x*exp(1) - 144*x + 432) + log(3*x*exp(1) - x + 4)*log(x)^2*(1296*x*exp(1) - 432*x
+ 1296))/(4*x + 3*x^2*exp(1) - x^2),x)

[Out]

log(3*x*exp(1) - x + 4)^4 + 81*log(x)^4 + 108*log(3*x*exp(1) - x + 4)*log(x)^3 + 12*log(3*x*exp(1) - x + 4)^3*
log(x) + 54*log(3*x*exp(1) - x + 4)^2*log(x)^2

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sympy [B]  time = 4.92, size = 76, normalized size = 4.47 \begin {gather*} 81 \log {\relax (x )}^{4} + 108 \log {\relax (x )}^{3} \log {\left (- x + 3 e x + 4 \right )} + 54 \log {\relax (x )}^{2} \log {\left (- x + 3 e x + 4 \right )}^{2} + 12 \log {\relax (x )} \log {\left (- x + 3 e x + 4 \right )}^{3} + \log {\left (- x + 3 e x + 4 \right )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*exp(1)-16*x+48)*ln(3*x*exp(1)-x+4)**3+(432*x*exp(1)-144*x+432)*ln(x)*ln(3*x*exp(1)-x+4)**2+(1
296*x*exp(1)-432*x+1296)*ln(x)**2*ln(3*x*exp(1)-x+4)+(1296*x*exp(1)-432*x+1296)*ln(x)**3)/(3*x**2*exp(1)-x**2+
4*x),x)

[Out]

81*log(x)**4 + 108*log(x)**3*log(-x + 3*E*x + 4) + 54*log(x)**2*log(-x + 3*E*x + 4)**2 + 12*log(x)*log(-x + 3*
E*x + 4)**3 + log(-x + 3*E*x + 4)**4

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