3.76.92 \(\int \frac {e^4 (4-4 x+9 x^2-9 x^3)+(36 x^3-36 x^2 \log (x)) \log (48+108 x^2)+(4-4 x+9 x^2-9 x^3) \log ^2(48+108 x^2)}{e^8 (4 x+9 x^3)+e^4 (-4 x^2-9 x^4)+e^4 (4 x+9 x^3) \log (x)+(-4 x^2-9 x^4+e^4 (8 x+18 x^3)+(4 x+9 x^3) \log (x)) \log ^2(48+108 x^2)+(4 x+9 x^3) \log ^4(48+108 x^2)} \, dx\)

Optimal. Leaf size=28 \[ \log \left (-1+\frac {x-\log (x)}{e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )}\right ) \]

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Rubi [A]  time = 3.87, antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 5, integrand size = 182, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 6725, 6684, 6696, 203} \begin {gather*} \log \left (\log ^2\left (12 \left (9 x^2+4\right )\right )-x+\log (x)+e^4\right )-\log \left (\log ^2\left (12 \left (9 x^2+4\right )\right )+e^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^4*(4 - 4*x + 9*x^2 - 9*x^3) + (36*x^3 - 36*x^2*Log[x])*Log[48 + 108*x^2] + (4 - 4*x + 9*x^2 - 9*x^3)*Lo
g[48 + 108*x^2]^2)/(E^8*(4*x + 9*x^3) + E^4*(-4*x^2 - 9*x^4) + E^4*(4*x + 9*x^3)*Log[x] + (-4*x^2 - 9*x^4 + E^
4*(8*x + 18*x^3) + (4*x + 9*x^3)*Log[x])*Log[48 + 108*x^2]^2 + (4*x + 9*x^3)*Log[48 + 108*x^2]^4),x]

[Out]

-Log[E^4 + Log[12*(4 + 9*x^2)]^2] + Log[E^4 - x + Log[x] + Log[12*(4 + 9*x^2)]^2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6696

Int[(u_.)*((a_.) + (b_.)*(y_)^(n_))^(p_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Dist[q, Subst[In
t[(a + b*x^n)^p, x], x, y], x] /;  !FalseQ[q]] /; FreeQ[{a, b, n, p}, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^4 \left (-4+4 x-9 x^2+9 x^3\right )+36 x^2 (x-\log (x)) \log \left (12 \left (4+9 x^2\right )\right )-\left (-4+4 x-9 x^2+9 x^3\right ) \log ^2\left (12 \left (4+9 x^2\right )\right )}{x \left (4+9 x^2\right ) \left (e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )\right ) \left (e^4-x+\log (x)+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )} \, dx\\ &=\int \left (\frac {-4+4 x-9 x^2+9 x^3-36 x^2 \log \left (12 \left (4+9 x^2\right )\right )}{x \left (4+9 x^2\right ) \left (-e^4+x-\log (x)-\log ^2\left (12 \left (4+9 x^2\right )\right )\right )}+\frac {36 x \log \left (48+108 x^2\right )}{\left (-4-9 x^2\right ) \left (e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )}\right ) \, dx\\ &=36 \int \frac {x \log \left (48+108 x^2\right )}{\left (-4-9 x^2\right ) \left (e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )} \, dx+\int \frac {-4+4 x-9 x^2+9 x^3-36 x^2 \log \left (12 \left (4+9 x^2\right )\right )}{x \left (4+9 x^2\right ) \left (-e^4+x-\log (x)-\log ^2\left (12 \left (4+9 x^2\right )\right )\right )} \, dx\\ &=-\log \left (e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )+\log \left (e^4-x+\log (x)+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 42, normalized size = 1.50 \begin {gather*} -\log \left (e^4+\log ^2\left (12 \left (4+9 x^2\right )\right )\right )+\log \left (e^4-x+\log (x)+\log ^2\left (12 \left (4+9 x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(4 - 4*x + 9*x^2 - 9*x^3) + (36*x^3 - 36*x^2*Log[x])*Log[48 + 108*x^2] + (4 - 4*x + 9*x^2 - 9*x
^3)*Log[48 + 108*x^2]^2)/(E^8*(4*x + 9*x^3) + E^4*(-4*x^2 - 9*x^4) + E^4*(4*x + 9*x^3)*Log[x] + (-4*x^2 - 9*x^
4 + E^4*(8*x + 18*x^3) + (4*x + 9*x^3)*Log[x])*Log[48 + 108*x^2]^2 + (4*x + 9*x^3)*Log[48 + 108*x^2]^4),x]

[Out]

-Log[E^4 + Log[12*(4 + 9*x^2)]^2] + Log[E^4 - x + Log[x] + Log[12*(4 + 9*x^2)]^2]

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fricas [A]  time = 2.46, size = 36, normalized size = 1.29 \begin {gather*} \log \left (\log \left (108 \, x^{2} + 48\right )^{2} - x + e^{4} + \log \relax (x)\right ) - \log \left (\log \left (108 \, x^{2} + 48\right )^{2} + e^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^3+9*x^2-4*x+4)*log(108*x^2+48)^2+(-36*x^2*log(x)+36*x^3)*log(108*x^2+48)+(-9*x^3+9*x^2-4*x+4)
*exp(4))/((9*x^3+4*x)*log(108*x^2+48)^4+((9*x^3+4*x)*log(x)+(18*x^3+8*x)*exp(4)-9*x^4-4*x^2)*log(108*x^2+48)^2
+(9*x^3+4*x)*exp(4)*log(x)+(9*x^3+4*x)*exp(4)^2+(-9*x^4-4*x^2)*exp(4)),x, algorithm="fricas")

[Out]

log(log(108*x^2 + 48)^2 - x + e^4 + log(x)) - log(log(108*x^2 + 48)^2 + e^4)

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giac [A]  time = 29.61, size = 36, normalized size = 1.29 \begin {gather*} \log \left (\log \left (108 \, x^{2} + 48\right )^{2} - x + e^{4} + \log \relax (x)\right ) - \log \left (\log \left (108 \, x^{2} + 48\right )^{2} + e^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^3+9*x^2-4*x+4)*log(108*x^2+48)^2+(-36*x^2*log(x)+36*x^3)*log(108*x^2+48)+(-9*x^3+9*x^2-4*x+4)
*exp(4))/((9*x^3+4*x)*log(108*x^2+48)^4+((9*x^3+4*x)*log(x)+(18*x^3+8*x)*exp(4)-9*x^4-4*x^2)*log(108*x^2+48)^2
+(9*x^3+4*x)*exp(4)*log(x)+(9*x^3+4*x)*exp(4)^2+(-9*x^4-4*x^2)*exp(4)),x, algorithm="giac")

[Out]

log(log(108*x^2 + 48)^2 - x + e^4 + log(x)) - log(log(108*x^2 + 48)^2 + e^4)

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maple [A]  time = 0.06, size = 37, normalized size = 1.32

method	result	size
risch	\(\ln \left (\ln \left (108 x^{2}+48\right )^{2}+{\mathrm e}^{4}-x +\ln \relax (x )\right )-\ln \left (\ln \left (108 x^{2}+48\right )^{2}+{\mathrm e}^{4}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x^3+9*x^2-4*x+4)*ln(108*x^2+48)^2+(-36*x^2*ln(x)+36*x^3)*ln(108*x^2+48)+(-9*x^3+9*x^2-4*x+4)*exp(4))/
((9*x^3+4*x)*ln(108*x^2+48)^4+((9*x^3+4*x)*ln(x)+(18*x^3+8*x)*exp(4)-9*x^4-4*x^2)*ln(108*x^2+48)^2+(9*x^3+4*x)
*exp(4)*ln(x)+(9*x^3+4*x)*exp(4)^2+(-9*x^4-4*x^2)*exp(4)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(108*x^2+48)^2+exp(4)-x+ln(x))-ln(ln(108*x^2+48)^2+exp(4))

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maxima [B]  time = 0.60, size = 102, normalized size = 3.64 \begin {gather*} \log \left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2} + 2 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \left (9 \, x^{2} + 4\right ) + \log \left (9 \, x^{2} + 4\right )^{2} - x + e^{4} + \log \relax (x)\right ) - \log \left (\log \relax (3)^{2} + 4 \, \log \relax (3) \log \relax (2) + 4 \, \log \relax (2)^{2} + 2 \, {\left (\log \relax (3) + 2 \, \log \relax (2)\right )} \log \left (9 \, x^{2} + 4\right ) + \log \left (9 \, x^{2} + 4\right )^{2} + e^{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^3+9*x^2-4*x+4)*log(108*x^2+48)^2+(-36*x^2*log(x)+36*x^3)*log(108*x^2+48)+(-9*x^3+9*x^2-4*x+4)
*exp(4))/((9*x^3+4*x)*log(108*x^2+48)^4+((9*x^3+4*x)*log(x)+(18*x^3+8*x)*exp(4)-9*x^4-4*x^2)*log(108*x^2+48)^2
+(9*x^3+4*x)*exp(4)*log(x)+(9*x^3+4*x)*exp(4)^2+(-9*x^4-4*x^2)*exp(4)),x, algorithm="maxima")

[Out]

log(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 + 2*(log(3) + 2*log(2))*log(9*x^2 + 4) + log(9*x^2 + 4)^2 - x + e^
4 + log(x)) - log(log(3)^2 + 4*log(3)*log(2) + 4*log(2)^2 + 2*(log(3) + 2*log(2))*log(9*x^2 + 4) + log(9*x^2 +
4)^2 + e^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {\left (9\,x^3-9\,x^2+4\,x-4\right )\,{\ln \left (108\,x^2+48\right )}^2+\left (36\,x^2\,\ln \relax (x)-36\,x^3\right )\,\ln \left (108\,x^2+48\right )+{\mathrm {e}}^4\,\left (9\,x^3-9\,x^2+4\,x-4\right )}{\left (9\,x^3+4\,x\right )\,{\ln \left (108\,x^2+48\right )}^4+\left ({\mathrm {e}}^4\,\left (18\,x^3+8\,x\right )+\ln \relax (x)\,\left (9\,x^3+4\,x\right )-4\,x^2-9\,x^4\right )\,{\ln \left (108\,x^2+48\right )}^2+{\mathrm {e}}^8\,\left (9\,x^3+4\,x\right )-{\mathrm {e}}^4\,\left (9\,x^4+4\,x^2\right )+{\mathrm {e}}^4\,\ln \relax (x)\,\left (9\,x^3+4\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(108*x^2 + 48)*(36*x^2*log(x) - 36*x^3) + log(108*x^2 + 48)^2*(4*x - 9*x^2 + 9*x^3 - 4) + exp(4)*(4*x
- 9*x^2 + 9*x^3 - 4))/(log(108*x^2 + 48)^4*(4*x + 9*x^3) + exp(8)*(4*x + 9*x^3) + log(108*x^2 + 48)^2*(exp(4)
*(8*x + 18*x^3) + log(x)*(4*x + 9*x^3) - 4*x^2 - 9*x^4) - exp(4)*(4*x^2 + 9*x^4) + exp(4)*log(x)*(4*x + 9*x^3)
),x)

[Out]

-int((log(108*x^2 + 48)*(36*x^2*log(x) - 36*x^3) + log(108*x^2 + 48)^2*(4*x - 9*x^2 + 9*x^3 - 4) + exp(4)*(4*x
- 9*x^2 + 9*x^3 - 4))/(log(108*x^2 + 48)^4*(4*x + 9*x^3) + exp(8)*(4*x + 9*x^3) + log(108*x^2 + 48)^2*(exp(4)
*(8*x + 18*x^3) + log(x)*(4*x + 9*x^3) - 4*x^2 - 9*x^4) - exp(4)*(4*x^2 + 9*x^4) + exp(4)*log(x)*(4*x + 9*x^3)
), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x**3+9*x**2-4*x+4)*ln(108*x**2+48)**2+(-36*x**2*ln(x)+36*x**3)*ln(108*x**2+48)+(-9*x**3+9*x**2-
4*x+4)*exp(4))/((9*x**3+4*x)*ln(108*x**2+48)**4+((9*x**3+4*x)*ln(x)+(18*x**3+8*x)*exp(4)-9*x**4-4*x**2)*ln(108
*x**2+48)**2+(9*x**3+4*x)*exp(4)*ln(x)+(9*x**3+4*x)*exp(4)**2+(-9*x**4-4*x**2)*exp(4)),x)

[Out]

Exception raised: PolynomialError

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