3.52.4 \(\int \frac {1}{9} (108 x+e^{\frac {\log ^2(x)}{9}} (18+4 \log (x))-72 \log (x^2)-18 \log ^2(x^2)) \, dx\)

Optimal. Leaf size=25 \[ 2 x \left (e^{\frac {\log ^2(x)}{9}}+3 x-\log ^2\left (x^2\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2288, 2295, 2296} \begin {gather*} 6 x^2-2 x \log ^2\left (x^2\right )+2 x e^{\frac {\log ^2(x)}{9}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(108*x + E^(Log[x]^2/9)*(18 + 4*Log[x]) - 72*Log[x^2] - 18*Log[x^2]^2)/9,x]

[Out]

2*E^(Log[x]^2/9)*x + 6*x^2 - 2*x*Log[x^2]^2

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \left (108 x+e^{\frac {\log ^2(x)}{9}} (18+4 \log (x))-72 \log \left (x^2\right )-18 \log ^2\left (x^2\right )\right ) \, dx\\ &=6 x^2+\frac {1}{9} \int e^{\frac {\log ^2(x)}{9}} (18+4 \log (x)) \, dx-2 \int \log ^2\left (x^2\right ) \, dx-8 \int \log \left (x^2\right ) \, dx\\ &=16 x+2 e^{\frac {\log ^2(x)}{9}} x+6 x^2-8 x \log \left (x^2\right )-2 x \log ^2\left (x^2\right )+8 \int \log \left (x^2\right ) \, dx\\ &=2 e^{\frac {\log ^2(x)}{9}} x+6 x^2-2 x \log ^2\left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 28, normalized size = 1.12 \begin {gather*} 2 e^{\frac {\log ^2(x)}{9}} x+6 x^2-2 x \log ^2\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(108*x + E^(Log[x]^2/9)*(18 + 4*Log[x]) - 72*Log[x^2] - 18*Log[x^2]^2)/9,x]

[Out]

2*E^(Log[x]^2/9)*x + 6*x^2 - 2*x*Log[x^2]^2

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fricas [A]  time = 0.61, size = 23, normalized size = 0.92 \begin {gather*} -8 \, x \log \relax (x)^{2} + 6 \, x^{2} + 2 \, x e^{\left (\frac {1}{9} \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*log(x)+18)*exp(1/9*log(x)^2)-2*log(x^2)^2-8*log(x^2)+12*x,x, algorithm="fricas")

[Out]

-8*x*log(x)^2 + 6*x^2 + 2*x*e^(1/9*log(x)^2)

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giac [A]  time = 0.17, size = 25, normalized size = 1.00 \begin {gather*} -2 \, x \log \left (x^{2}\right )^{2} + 6 \, x^{2} + 2 \, x e^{\left (\frac {1}{9} \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*log(x)+18)*exp(1/9*log(x)^2)-2*log(x^2)^2-8*log(x^2)+12*x,x, algorithm="giac")

[Out]

-2*x*log(x^2)^2 + 6*x^2 + 2*x*e^(1/9*log(x)^2)

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maple [A]  time = 0.03, size = 26, normalized size = 1.04

method	result	size
default	\(6 x^{2}-2 x \ln \left (x^{2}\right )^{2}+2 \,{\mathrm e}^{\frac {\ln \relax (x )^{2}}{9}} x\)	\(26\)
risch	\(6 x^{2}-2 x \ln \left (x^{2}\right )^{2}+2 \,{\mathrm e}^{\frac {\ln \relax (x )^{2}}{9}} x\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(4*ln(x)+18)*exp(1/9*ln(x)^2)-2*ln(x^2)^2-8*ln(x^2)+12*x,x,method=_RETURNVERBOSE)

[Out]

6*x^2-2*x*ln(x^2)^2+2*exp(1/9*ln(x)^2)*x

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maxima [A]  time = 0.34, size = 25, normalized size = 1.00 \begin {gather*} -2 \, x \log \left (x^{2}\right )^{2} + 6 \, x^{2} + 2 \, x e^{\left (\frac {1}{9} \, \log \relax (x)^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*log(x)+18)*exp(1/9*log(x)^2)-2*log(x^2)^2-8*log(x^2)+12*x,x, algorithm="maxima")

[Out]

-2*x*log(x^2)^2 + 6*x^2 + 2*x*e^(1/9*log(x)^2)

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mupad [B]  time = 3.26, size = 22, normalized size = 0.88 \begin {gather*} 2\,x\,\left (-{\ln \left (x^2\right )}^2+3\,x+{\mathrm {e}}^{\frac {{\ln \relax (x)}^2}{9}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(12*x - 8*log(x^2) + (exp(log(x)^2/9)*(4*log(x) + 18))/9 - 2*log(x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.28, size = 24, normalized size = 0.96 \begin {gather*} 6 x^{2} + 2 x e^{\frac {\log {\relax (x )}^{2}}{9}} - 8 x \log {\relax (x )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(4*ln(x)+18)*exp(1/9*ln(x)**2)-2*ln(x**2)**2-8*ln(x**2)+12*x,x)

[Out]

6*x**2 + 2*x*exp(log(x)**2/9) - 8*x*log(x)**2

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