3.47.14 \(\int \frac {(-3-3 x) \log (3)+(-3-3 x) \log (3) \log (x^2)+((9+9 x-9 x^2) \log (3)+(3+3 x-3 x^2) \log (3) \log (x^2)) \log (\frac {9+3 \log (x^2)}{x})}{3 e^x+e^x \log (x^2)} \, dx\) [4614]

Optimal. Leaf size=26 \[ 3 e^{-x} \left (x+x^2\right ) \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right ) \]

[Out]

3*(x^2+x)/exp(x)*ln(3*(ln(x^2)+3)/x)*ln(3)

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Rubi [A]
time = 1.31, antiderivative size = 47, normalized size of antiderivative = 1.81, number of steps used = 18, number of rules used = 6, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {6873, 6874, 2225, 2207, 2227, 2635} \begin {gather*} 3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (\log \left (x^2\right )+3\right )}{x}\right )+3 e^{-x} x \log (3) \log \left (\frac {3 \left (\log \left (x^2\right )+3\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-3 - 3*x)*Log[3] + (-3 - 3*x)*Log[3]*Log[x^2] + ((9 + 9*x - 9*x^2)*Log[3] + (3 + 3*x - 3*x^2)*Log[3]*Log
[x^2])*Log[(9 + 3*Log[x^2])/x])/(3*E^x + E^x*Log[x^2]),x]

[Out]

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

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rule 2635

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6873

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

Rule 6874

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left ((-3-3 x) \log (3)+(-3-3 x) \log (3) \log \left (x^2\right )+\left (\left (9+9 x-9 x^2\right ) \log (3)+\left (3+3 x-3 x^2\right ) \log (3) \log \left (x^2\right )\right ) \log \left (\frac {9+3 \log \left (x^2\right )}{x}\right )\right )}{3+\log \left (x^2\right )} \, dx\\ &=\int \left (-\frac {3 e^{-x} (1+x) \log (3) \left (1+\log \left (x^2\right )\right )}{3+\log \left (x^2\right )}-3 e^{-x} \left (-1-x+x^2\right ) \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )\right ) \, dx\\ &=-\left ((3 \log (3)) \int \frac {e^{-x} (1+x) \left (1+\log \left (x^2\right )\right )}{3+\log \left (x^2\right )} \, dx\right )-(3 \log (3)) \int e^{-x} \left (-1-x+x^2\right ) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=3 e^{-x} x \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+(3 \log (3)) \int \frac {e^{-x} (1+x) \left (1+\log \left (x^2\right )\right )}{3+\log \left (x^2\right )} \, dx-(3 \log (3)) \int \left (e^{-x}+e^{-x} x-\frac {2 e^{-x} (1+x)}{3+\log \left (x^2\right )}\right ) \, dx\\ &=3 e^{-x} x \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )-(3 \log (3)) \int e^{-x} \, dx-(3 \log (3)) \int e^{-x} x \, dx+(3 \log (3)) \int \left (e^{-x}+e^{-x} x-\frac {2 e^{-x} (1+x)}{3+\log \left (x^2\right )}\right ) \, dx+(6 \log (3)) \int \frac {e^{-x} (1+x)}{3+\log \left (x^2\right )} \, dx\\ &=3 e^{-x} \log (3)+3 e^{-x} x \log (3)+3 e^{-x} x \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+(3 \log (3)) \int e^{-x} x \, dx-(6 \log (3)) \int \frac {e^{-x} (1+x)}{3+\log \left (x^2\right )} \, dx+(6 \log (3)) \int \left (\frac {e^{-x}}{3+\log \left (x^2\right )}+\frac {e^{-x} x}{3+\log \left (x^2\right )}\right ) \, dx\\ &=3 e^{-x} \log (3)+3 e^{-x} x \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+(3 \log (3)) \int e^{-x} \, dx+(6 \log (3)) \int \frac {e^{-x}}{3+\log \left (x^2\right )} \, dx+(6 \log (3)) \int \frac {e^{-x} x}{3+\log \left (x^2\right )} \, dx-(6 \log (3)) \int \left (\frac {e^{-x}}{3+\log \left (x^2\right )}+\frac {e^{-x} x}{3+\log \left (x^2\right )}\right ) \, dx\\ &=3 e^{-x} x \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )+3 e^{-x} x^2 \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.17, size = 25, normalized size = 0.96 \begin {gather*} 3 e^{-x} x (1+x) \log (3) \log \left (\frac {3 \left (3+\log \left (x^2\right )\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3 - 3*x)*Log[3] + (-3 - 3*x)*Log[3]*Log[x^2] + ((9 + 9*x - 9*x^2)*Log[3] + (3 + 3*x - 3*x^2)*Log[
3]*Log[x^2])*Log[(9 + 3*Log[x^2])/x])/(3*E^x + E^x*Log[x^2]),x]

[Out]

(3*x*(1 + x)*Log[3]*Log[(3*(3 + Log[x^2]))/x])/E^x

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.05, size = 1616, normalized size = 62.15

method	result	size
risch	\(\text {Expression too large to display}\)	\(1616\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x^2+3*x+3)*ln(3)*ln(x^2)+(-9*x^2+9*x+9)*ln(3))*ln((3*ln(x^2)+9)/x)+(-3*x-3)*ln(3)*ln(x^2)+(-3*x-3)*l
n(3))/(exp(x)*ln(x^2)+3*exp(x)),x,method=_RETURNVERBOSE)

[Out]

3*ln(3)*x*(x+1)*exp(-x)*ln(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2
)^3)+3/2*ln(3)*x*(I*x*Pi*csgn(I*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn
(I*x^2)^3))*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))
^2+I*x*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^2
*csgn(I/x)+I*x*Pi*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^
2)^3))^2-I*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3
))^3-I*Pi+I*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^
3))*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^2-I*x*P
i*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^3-I*Pi*cs
gn(I*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(I/x*(6*I+4
*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(I/x)-I*Pi*x-I*Pi*csgn
(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^3-I*x*Pi*csgn(I
/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(1/x*(6*I+4*I
*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))-I*x*Pi*csgn(I*(6*I+4*I*ln(x)
+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*
x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(I/x)+I*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csg
n(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^2*csgn(I/x)+I*Pi*csgn(I*(6*I+4*I*ln(x)+Pi
*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^
2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^2-I*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*cs
gn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2
*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))+I*x*Pi*csgn(I/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*
csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3))*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*c
sgn(I*x^2)^2+Pi*csgn(I*x^2)^3))^2+I*Pi*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(
I*x^2)^2+Pi*csgn(I*x^2)^3))^2-I*x*Pi*csgn(1/x*(6*I+4*I*ln(x)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*
x^2)^2+Pi*csgn(I*x^2)^3))^3+2*x*ln(3/2)-2*x*ln(x)+2*ln(3/2)-2*ln(x))*exp(-x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (25) = 50\).
time = 0.52, size = 61, normalized size = 2.35 \begin {gather*} 3 \, {\left (x^{2} \log \left (3\right ) + x \log \left (3\right )\right )} e^{\left (-x\right )} \log \left (2 \, \log \left (x\right ) + 3\right ) + 3 \, {\left (x^{2} \log \left (3\right )^{2} + x \log \left (3\right )^{2} - {\left (x^{2} \log \left (3\right ) + x \log \left (3\right )\right )} \log \left (x\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2+3*x+3)*log(3)*log(x^2)+(-9*x^2+9*x+9)*log(3))*log((3*log(x^2)+9)/x)+(-3*x-3)*log(3)*log(x^
2)+(-3*x-3)*log(3))/(exp(x)*log(x^2)+3*exp(x)),x, algorithm="maxima")

[Out]

3*(x^2*log(3) + x*log(3))*e^(-x)*log(2*log(x) + 3) + 3*(x^2*log(3)^2 + x*log(3)^2 - (x^2*log(3) + x*log(3))*lo
g(x))*e^(-x)

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Fricas [A]
time = 0.35, size = 25, normalized size = 0.96 \begin {gather*} 3 \, {\left (x^{2} + x\right )} e^{\left (-x\right )} \log \left (3\right ) \log \left (\frac {3 \, {\left (\log \left (x^{2}\right ) + 3\right )}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2+3*x+3)*log(3)*log(x^2)+(-9*x^2+9*x+9)*log(3))*log((3*log(x^2)+9)/x)+(-3*x-3)*log(3)*log(x^
2)+(-3*x-3)*log(3))/(exp(x)*log(x^2)+3*exp(x)),x, algorithm="fricas")

[Out]

3*(x^2 + x)*e^(-x)*log(3)*log(3*(log(x^2) + 3)/x)

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Sympy [A]
time = 2.30, size = 41, normalized size = 1.58 \begin {gather*} \left (3 x^{2} \log {\left (3 \right )} \log {\left (\frac {3 \log {\left (x^{2} \right )} + 9}{x} \right )} + 3 x \log {\left (3 \right )} \log {\left (\frac {3 \log {\left (x^{2} \right )} + 9}{x} \right )}\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x**2+3*x+3)*ln(3)*ln(x**2)+(-9*x**2+9*x+9)*ln(3))*ln((3*ln(x**2)+9)/x)+(-3*x-3)*ln(3)*ln(x**2)
+(-3*x-3)*ln(3))/(exp(x)*ln(x**2)+3*exp(x)),x)

[Out]

(3*x**2*log(3)*log((3*log(x**2) + 9)/x) + 3*x*log(3)*log((3*log(x**2) + 9)/x))*exp(-x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (25) = 50\).
time = 0.44, size = 83, normalized size = 3.19 \begin {gather*} 3 \, x^{2} e^{\left (-x\right )} \log \left (3\right )^{2} - 3 \, x^{2} e^{\left (-x\right )} \log \left (3\right ) \log \left (x\right ) + 3 \, x^{2} e^{\left (-x\right )} \log \left (3\right ) \log \left (\log \left (x^{2}\right ) + 3\right ) + 3 \, x e^{\left (-x\right )} \log \left (3\right )^{2} - 3 \, x e^{\left (-x\right )} \log \left (3\right ) \log \left (x\right ) + 3 \, x e^{\left (-x\right )} \log \left (3\right ) \log \left (\log \left (x^{2}\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2+3*x+3)*log(3)*log(x^2)+(-9*x^2+9*x+9)*log(3))*log((3*log(x^2)+9)/x)+(-3*x-3)*log(3)*log(x^
2)+(-3*x-3)*log(3))/(exp(x)*log(x^2)+3*exp(x)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \left (3\right )\,\left (3\,x+3\right )-\ln \left (\frac {3\,\ln \left (x^2\right )+9}{x}\right )\,\left (\ln \left (3\right )\,\left (-9\,x^2+9\,x+9\right )+\ln \left (x^2\right )\,\ln \left (3\right )\,\left (-3\,x^2+3\,x+3\right )\right )+\ln \left (x^2\right )\,\ln \left (3\right )\,\left (3\,x+3\right )}{3\,{\mathrm {e}}^x+\ln \left (x^2\right )\,{\mathrm {e}}^x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(3*x + 3) - log((3*log(x^2) + 9)/x)*(log(3)*(9*x - 9*x^2 + 9) + log(x^2)*log(3)*(3*x - 3*x^2 + 3)
) + log(x^2)*log(3)*(3*x + 3))/(3*exp(x) + log(x^2)*exp(x)),x)

[Out]

int(-(log(3)*(3*x + 3) - log((3*log(x^2) + 9)/x)*(log(3)*(9*x - 9*x^2 + 9) + log(x^2)*log(3)*(3*x - 3*x^2 + 3)
) + log(x^2)*log(3)*(3*x + 3))/(3*exp(x) + log(x^2)*exp(x)), x)

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