3.64.86 \(\int \frac {2 x+(5-2 x) \log (3)+e^{2 \log ^2(x)} (\log (3)+4 \log (3) \log (x))}{\log (3)} \, dx\)

Optimal. Leaf size=25 \[ x-(-4+x) x+x \left (e^{2 \log ^2(x)}+\frac {x}{\log (3)}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 2288} \begin {gather*} \frac {x^2}{\log (3)}-\frac {1}{4} (5-2 x)^2+x e^{2 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x + (5 - 2*x)*Log[3] + E^(2*Log[x]^2)*(Log[3] + 4*Log[3]*Log[x]))/Log[3],x]

[Out]

-1/4*(5 - 2*x)^2 + E^(2*Log[x]^2)*x + x^2/Log[3]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (2 x+(5-2 x) \log (3)+e^{2 \log ^2(x)} (\log (3)+4 \log (3) \log (x))\right ) \, dx}{\log (3)}\\ &=-\frac {1}{4} (5-2 x)^2+\frac {x^2}{\log (3)}+\frac {\int e^{2 \log ^2(x)} (\log (3)+4 \log (3) \log (x)) \, dx}{\log (3)}\\ &=-\frac {1}{4} (5-2 x)^2+e^{2 \log ^2(x)} x+\frac {x^2}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 27, normalized size = 1.08 \begin {gather*} 5 x+e^{2 \log ^2(x)} x-x^2+\frac {x^2}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x + (5 - 2*x)*Log[3] + E^(2*Log[x]^2)*(Log[3] + 4*Log[3]*Log[x]))/Log[3],x]

[Out]

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

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fricas [A]  time = 0.88, size = 31, normalized size = 1.24 \begin {gather*} \frac {x e^{\left (2 \, \log \relax (x)^{2}\right )} \log \relax (3) + x^{2} - {\left (x^{2} - 5 \, x\right )} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)*log(x)+log(3))*exp(log(x)^2)^2+(-2*x+5)*log(3)+2*x)/log(3),x, algorithm="fricas")

[Out]

(x*e^(2*log(x)^2)*log(3) + x^2 - (x^2 - 5*x)*log(3))/log(3)

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giac [A]  time = 0.24, size = 31, normalized size = 1.24 \begin {gather*} \frac {x e^{\left (2 \, \log \relax (x)^{2}\right )} \log \relax (3) + x^{2} - {\left (x^{2} - 5 \, x\right )} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)*log(x)+log(3))*exp(log(x)^2)^2+(-2*x+5)*log(3)+2*x)/log(3),x, algorithm="giac")

[Out]

(x*e^(2*log(x)^2)*log(3) + x^2 - (x^2 - 5*x)*log(3))/log(3)

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maple [A]  time = 0.10, size = 27, normalized size = 1.08

method	result	size
norman	\(x \,{\mathrm e}^{2 \ln \relax (x )^{2}}+5 x -\frac {\left (\ln \relax (3)-1\right ) x^{2}}{\ln \relax (3)}\)	\(27\)
risch	\(-x^{2}+5 x +x \,{\mathrm e}^{2 \ln \relax (x )^{2}}+\frac {x^{2}}{\ln \relax (3)}\)	\(27\)
default	\(\frac {-x^{2} \ln \relax (3)+5 x \ln \relax (3)+x \ln \relax (3) {\mathrm e}^{2 \ln \relax (x )^{2}}+x^{2}}{\ln \relax (3)}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(3)*ln(x)+ln(3))*exp(ln(x)^2)^2+(-2*x+5)*ln(3)+2*x)/ln(3),x,method=_RETURNVERBOSE)

[Out]

x*exp(ln(x)^2)^2+5*x-(ln(3)-1)/ln(3)*x^2

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maxima [A]  time = 0.43, size = 31, normalized size = 1.24 \begin {gather*} \frac {x e^{\left (2 \, \log \relax (x)^{2}\right )} \log \relax (3) + x^{2} - {\left (x^{2} - 5 \, x\right )} \log \relax (3)}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(3)*log(x)+log(3))*exp(log(x)^2)^2+(-2*x+5)*log(3)+2*x)/log(3),x, algorithm="maxima")

[Out]

(x*e^(2*log(x)^2)*log(3) + x^2 - (x^2 - 5*x)*log(3))/log(3)

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mupad [B]  time = 4.24, size = 26, normalized size = 1.04 \begin {gather*} 5\,x+\frac {x^2}{\ln \relax (3)}-x^2+x\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - log(3)*(2*x - 5) + exp(2*log(x)^2)*(log(3) + 4*log(3)*log(x)))/log(3),x)

[Out]

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

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sympy [A]  time = 0.29, size = 24, normalized size = 0.96 \begin {gather*} \frac {x^{2} \left (1 - \log {\relax (3 )}\right )}{\log {\relax (3 )}} + x e^{2 \log {\relax (x )}^{2}} + 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(3)*ln(x)+ln(3))*exp(ln(x)**2)**2+(-2*x+5)*ln(3)+2*x)/ln(3),x)

[Out]

x**2*(1 - log(3))/log(3) + x*exp(2*log(x)**2) + 5*x

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