3.15.69 \(\int \frac {e^{x^2} (-4+2 x-e^x \log (3)+(-e^5+x) \log (3)) (-2+8 x-4 x^2+e^x (1+2 x) \log (3)+(-1+2 e^5 x-2 x^2) \log (3))}{4-2 x+e^x \log (3)+(e^5-x) \log (3)} \, dx\)

Optimal. Leaf size=26 \[ e^{x^2} \left (-4+2 x+\left (-e^5-e^x+x\right ) \log (3)\right ) \]

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Rubi [A]  time = 2.29, antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6688, 2288} \begin {gather*} \frac {e^{x^2} \left (x^2 (4+\log (9))-x \left (e^x \log (9)+8+e^5 \log (9)\right )\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(-4 + 2*x - E^x*Log[3] + (-E^5 + x)*Log[3])*(-2 + 8*x - 4*x^2 + E^x*(1 + 2*x)*Log[3] + (-1 + 2*E^5*
x - 2*x^2)*Log[3]))/(4 - 2*x + E^x*Log[3] + (E^5 - x)*Log[3]),x]

[Out]

(E^x^2*(x^2*(4 + Log[9]) - x*(8 + E^5*Log[9] + E^x*Log[9])))/(2*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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{x^2} \left (2 \left (1+\frac {\log (3)}{2}\right )-e^x \log (3)+x^2 (4+\log (9))-x \left (8+e^5 \log (9)+e^x \log (9)\right )\right ) \, dx\\ &=\frac {e^{x^2} \left (x^2 (4+\log (9))-x \left (8+e^5 \log (9)+e^x \log (9)\right )\right )}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 31, normalized size = 1.19 \begin {gather*} \frac {1}{2} e^{x^2} \left (-8-e^5 \log (9)-e^x \log (9)+x (4+\log (9))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(-4 + 2*x - E^x*Log[3] + (-E^5 + x)*Log[3])*(-2 + 8*x - 4*x^2 + E^x*(1 + 2*x)*Log[3] + (-1 +
2*E^5*x - 2*x^2)*Log[3]))/(4 - 2*x + E^x*Log[3] + (E^5 - x)*Log[3]),x]

[Out]

(E^x^2*(-8 - E^5*Log[9] - E^x*Log[9] + x*(4 + Log[9])))/2

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fricas [A]  time = 0.78, size = 26, normalized size = 1.00 \begin {gather*} e^{\left (x^{2} + \log \left ({\left (x - e^{5}\right )} \log \relax (3) - e^{x} \log \relax (3) + 2 \, x - 4\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(3)*exp(x)+(2*x*exp(5)-2*x^2-1)*log(3)-4*x^2+8*x-2)*exp(log(-log(3)*exp(x)+(-exp(5)+x)*l
og(3)+2*x-4)+x^2)/(log(3)*exp(x)+(exp(5)-x)*log(3)+4-2*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 42, normalized size = 1.62 \begin {gather*} x e^{\left (x^{2}\right )} \log \relax (3) + 2 \, x e^{\left (x^{2}\right )} - e^{\left (x^{2} + x\right )} \log \relax (3) - e^{\left (x^{2} + 5\right )} \log \relax (3) - 4 \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(3)*exp(x)+(2*x*exp(5)-2*x^2-1)*log(3)-4*x^2+8*x-2)*exp(log(-log(3)*exp(x)+(-exp(5)+x)*l
og(3)+2*x-4)+x^2)/(log(3)*exp(x)+(exp(5)-x)*log(3)+4-2*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.26, size = 26, normalized size = 1.00

method	result	size
risch	\(\left (-\ln \relax (3) {\mathrm e}^{x}+\left (-{\mathrm e}^{5}+x \right ) \ln \relax (3)+2 x -4\right ) {\mathrm e}^{x^{2}}\)	\(26\)
norman	\({\mathrm e}^{\ln \left (-\ln \relax (3) {\mathrm e}^{x}+\left (-{\mathrm e}^{5}+x \right ) \ln \relax (3)+2 x -4\right )+x^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+1)*ln(3)*exp(x)+(2*x*exp(5)-2*x^2-1)*ln(3)-4*x^2+8*x-2)*exp(ln(-ln(3)*exp(x)+(-exp(5)+x)*ln(3)+2*x-4
)+x^2)/(ln(3)*exp(x)+(exp(5)-x)*ln(3)+4-2*x),x,method=_RETURNVERBOSE)

[Out]

(-ln(3)*exp(x)+(-exp(5)+x)*ln(3)+2*x-4)*exp(x^2)

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maxima [C]  time = 0.49, size = 126, normalized size = 4.85 \begin {gather*} \frac {1}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {1}{2} i\right ) e^{\left (-\frac {1}{4}\right )} \log \relax (3) + \frac {1}{2} \, {\left (\frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} \log \relax (3) - \frac {1}{2} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) \log \relax (3) + 2 \, x e^{\left (x^{2}\right )} + \frac {1}{2} \, {\left (2 \, x e^{\left (x^{2}\right )} + i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right )\right )} \log \relax (3) - e^{\left (x^{2} + 5\right )} \log \relax (3) - 4 \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*log(3)*exp(x)+(2*x*exp(5)-2*x^2-1)*log(3)-4*x^2+8*x-2)*exp(log(-log(3)*exp(x)+(-exp(5)+x)*l
og(3)+2*x-4)+x^2)/(log(3)*exp(x)+(exp(5)-x)*log(3)+4-2*x),x, algorithm="maxima")

[Out]

1/2*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4)*log(3) + 1/2*(sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sq
rt(-(2*x + 1)^2) - 2*e^(1/4*(2*x + 1)^2))*e^(-1/4)*log(3) - 1/2*I*sqrt(pi)*erf(I*x)*log(3) + 2*x*e^(x^2) + 1/2
*(2*x*e^(x^2) + I*sqrt(pi)*erf(I*x))*log(3) - e^(x^2 + 5)*log(3) - 4*e^(x^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\ln \left (2\,x+\ln \relax (3)\,\left (x-{\mathrm {e}}^5\right )-{\mathrm {e}}^x\,\ln \relax (3)-4\right )+x^2}\,\left (\ln \relax (3)\,\left (2\,x^2-2\,{\mathrm {e}}^5\,x+1\right )-8\,x+4\,x^2-{\mathrm {e}}^x\,\ln \relax (3)\,\left (2\,x+1\right )+2\right )}{2\,x+\ln \relax (3)\,\left (x-{\mathrm {e}}^5\right )-{\mathrm {e}}^x\,\ln \relax (3)-4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(2*x + log(3)*(x - exp(5)) - exp(x)*log(3) - 4) + x^2)*(log(3)*(2*x^2 - 2*x*exp(5) + 1) - 8*x + 4*
x^2 - exp(x)*log(3)*(2*x + 1) + 2))/(2*x + log(3)*(x - exp(5)) - exp(x)*log(3) - 4),x)

[Out]

int((exp(log(2*x + log(3)*(x - exp(5)) - exp(x)*log(3) - 4) + x^2)*(log(3)*(2*x^2 - 2*x*exp(5) + 1) - 8*x + 4*
x^2 - exp(x)*log(3)*(2*x + 1) + 2))/(2*x + log(3)*(x - exp(5)) - exp(x)*log(3) - 4), x)

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sympy [A]  time = 0.30, size = 27, normalized size = 1.04 \begin {gather*} \left (x \log {\relax (3 )} + 2 x - e^{x} \log {\relax (3 )} - e^{5} \log {\relax (3 )} - 4\right ) e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*ln(3)*exp(x)+(2*x*exp(5)-2*x**2-1)*ln(3)-4*x**2+8*x-2)*exp(ln(-ln(3)*exp(x)+(-exp(5)+x)*ln(
3)+2*x-4)+x**2)/(ln(3)*exp(x)+(exp(5)-x)*ln(3)+4-2*x),x)

[Out]

(x*log(3) + 2*x - exp(x)*log(3) - exp(5)*log(3) - 4)*exp(x**2)

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