3.48.94 \(\int (-1+e^{2 x} (3+x)+e^{2 x} (3+8 x+2 x^2) \log (x)+(e^{2 x}+e^{2 x} (1+2 x) \log (x)) \log (\frac {\log (4)}{2})) \, dx\)

Optimal. Leaf size=22 \[ \left (-1+e^{2 x} x \log (x)\right ) \left (3+x+\log \left (\frac {\log (4)}{2}\right )\right ) \]

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Rubi [B]  time = 0.13, antiderivative size = 58, normalized size of antiderivative = 2.64, number of steps used = 10, number of rules used = 4, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2176, 2194, 2196, 2554} \begin {gather*} e^{2 x} x^2 \log (x)-x+3 e^{2 x} x \log (x)-\frac {1}{2} e^{2 x} \log (\log (2)) \log (x)+\frac {1}{2} e^{2 x} (2 x+1) \log (\log (2)) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + E^(2*x)*(3 + x) + E^(2*x)*(3 + 8*x + 2*x^2)*Log[x] + (E^(2*x) + E^(2*x)*(1 + 2*x)*Log[x])*Log[Log[4]/
2],x]

[Out]

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

Rule 2176

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] &&  !$UseGamma === True

Rule 2194

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 2196

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] &&  !$UseGamma === True

Rule 2554

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x+\log (\log (2)) \int \left (e^{2 x}+e^{2 x} (1+2 x) \log (x)\right ) \, dx+\int e^{2 x} (3+x) \, dx+\int e^{2 x} \left (3+8 x+2 x^2\right ) \log (x) \, dx\\ &=-x+\frac {1}{2} e^{2 x} (3+x)+3 e^{2 x} x \log (x)+e^{2 x} x^2 \log (x)-\frac {1}{2} \int e^{2 x} \, dx+\log (\log (2)) \int e^{2 x} \, dx+\log (\log (2)) \int e^{2 x} (1+2 x) \log (x) \, dx-\int e^{2 x} (3+x) \, dx\\ &=-\frac {e^{2 x}}{4}-x+3 e^{2 x} x \log (x)+e^{2 x} x^2 \log (x)+\frac {1}{2} e^{2 x} \log (\log (2))-\frac {1}{2} e^{2 x} \log (x) \log (\log (2))+\frac {1}{2} e^{2 x} (1+2 x) \log (x) \log (\log (2))+\frac {1}{2} \int e^{2 x} \, dx-\log (\log (2)) \int e^{2 x} \, dx\\ &=-x+3 e^{2 x} x \log (x)+e^{2 x} x^2 \log (x)-\frac {1}{2} e^{2 x} \log (x) \log (\log (2))+\frac {1}{2} e^{2 x} (1+2 x) \log (x) \log (\log (2))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 18, normalized size = 0.82 \begin {gather*} x \left (-1+e^{2 x} \log (x) (3+x+\log (\log (2)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + E^(2*x)*(3 + x) + E^(2*x)*(3 + 8*x + 2*x^2)*Log[x] + (E^(2*x) + E^(2*x)*(1 + 2*x)*Log[x])*Log[L
og[4]/2],x]

[Out]

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

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fricas [A]  time = 0.53, size = 29, normalized size = 1.32 \begin {gather*} x e^{\left (2 \, x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) + {\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \relax (x) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(x)^2*log(x)+exp(x)^2)*log(log(2))+(2*x^2+8*x+3)*exp(x)^2*log(x)+(3+x)*exp(x)^2-1,x, alg
orithm="fricas")

[Out]

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

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giac [B]  time = 0.13, size = 57, normalized size = 2.59 \begin {gather*} x e^{\left (2 \, x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) + {\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \relax (x) + \frac {1}{4} \, {\left (2 \, x + 5\right )} e^{\left (2 \, x\right )} - \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - x - \frac {3}{2} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(x)^2*log(x)+exp(x)^2)*log(log(2))+(2*x^2+8*x+3)*exp(x)^2*log(x)+(3+x)*exp(x)^2-1,x, alg
orithm="giac")

[Out]

x*e^(2*x)*log(x)*log(log(2)) + (x^2 + 3*x)*e^(2*x)*log(x) + 1/4*(2*x + 5)*e^(2*x) - 1/4*(2*x - 1)*e^(2*x) - x
- 3/2*e^(2*x)

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maple [A]  time = 0.09, size = 28, normalized size = 1.27

method	result	size
norman	\(x^{2} {\mathrm e}^{2 x} \ln \relax (x )+\left (\ln \left (\ln \relax (2)\right )+3\right ) x \,{\mathrm e}^{2 x} \ln \relax (x )-x\)	\(28\)
default	\(-x +\ln \relax (x ) {\mathrm e}^{2 x} \ln \left (\ln \relax (2)\right ) x +x^{2} {\mathrm e}^{2 x} \ln \relax (x )+3 x \,{\mathrm e}^{2 x} \ln \relax (x )\)	\(35\)
risch	\(-x +\ln \relax (x ) {\mathrm e}^{2 x} \ln \left (\ln \relax (2)\right ) x +x^{2} {\mathrm e}^{2 x} \ln \relax (x )+3 x \,{\mathrm e}^{2 x} \ln \relax (x )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+1)*exp(x)^2*ln(x)+exp(x)^2)*ln(ln(2))+(2*x^2+8*x+3)*exp(x)^2*ln(x)+(3+x)*exp(x)^2-1,x,method=_RETURN
VERBOSE)

[Out]

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

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maxima [B]  time = 0.38, size = 57, normalized size = 2.59 \begin {gather*} x e^{\left (2 \, x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) + {\left (x^{2} + 3 \, x\right )} e^{\left (2 \, x\right )} \log \relax (x) - \frac {1}{4} \, {\left (2 \, x + 5\right )} e^{\left (2 \, x\right )} + \frac {1}{4} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - x + \frac {3}{2} \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(x)^2*log(x)+exp(x)^2)*log(log(2))+(2*x^2+8*x+3)*exp(x)^2*log(x)+(3+x)*exp(x)^2-1,x, alg
orithm="maxima")

[Out]

x*e^(2*x)*log(x)*log(log(2)) + (x^2 + 3*x)*e^(2*x)*log(x) - 1/4*(2*x + 5)*e^(2*x) + 1/4*(2*x - 1)*e^(2*x) - x
+ 3/2*e^(2*x)

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mupad [B]  time = 3.37, size = 27, normalized size = 1.23 \begin {gather*} x\,\left ({\mathrm {e}}^{2\,x}\,\ln \relax (x)\,\left (\ln \left (\ln \relax (2)\right )+3\right )-1\right )+x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*(x + 3) + log(log(2))*(exp(2*x) + exp(2*x)*log(x)*(2*x + 1)) + exp(2*x)*log(x)*(8*x + 2*x^2 + 3)
- 1,x)

[Out]

x*(exp(2*x)*log(x)*(log(log(2)) + 3) - 1) + x^2*exp(2*x)*log(x)

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sympy [A]  time = 0.36, size = 29, normalized size = 1.32 \begin {gather*} - x + \left (x^{2} \log {\relax (x )} + x \log {\relax (x )} \log {\left (\log {\relax (2 )} \right )} + 3 x \log {\relax (x )}\right ) e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(x)**2*ln(x)+exp(x)**2)*ln(ln(2))+(2*x**2+8*x+3)*exp(x)**2*ln(x)+(3+x)*exp(x)**2-1,x)

[Out]

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

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