3.94.54 \(\int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} (e^{20+4 x} (-4-24 x)+8 x+8 x^2)+(1-4 e^{20+4 x}+e^{x^2} (4-16 e^{20+4 x})) \log (\frac {1}{5} (3+12 e^{x^2}))}{1+4 e^{x^2}} \, dx\)

Optimal. Leaf size=30 \[ 2+\left (-e^{4 (5+x)}+x\right ) \left (x+\log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )\right ) \]

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Rubi [F]  time = 1.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{20+4 x} (-1-4 x)+2 x+e^{x^2} \left (e^{20+4 x} (-4-24 x)+8 x+8 x^2\right )+\left (1-4 e^{20+4 x}+e^{x^2} \left (4-16 e^{20+4 x}\right )\right ) \log \left (\frac {1}{5} \left (3+12 e^{x^2}\right )\right )}{1+4 e^{x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(20 + 4*x)*(-1 - 4*x) + 2*x + E^x^2*(E^(20 + 4*x)*(-4 - 24*x) + 8*x + 8*x^2) + (1 - 4*E^(20 + 4*x) + E^
x^2*(4 - 16*E^(20 + 4*x)))*Log[(3 + 12*E^x^2)/5])/(1 + 4*E^x^2),x]

[Out]

E^(20 + 4*x)/8 - (3*E^(20 + 4*x)*x)/2 + x^2 + (2*x^3)/3 - E^(20 + 4*x)*Log[(3*(1 + 4*E^x^2))/5] + x*Log[(3*(1
+ 4*E^x^2))/5] + 2*Defer[Int][(E^(20 + 4*x)*x)/(1 + 4*E^x^2), x] + 8*Defer[Int][(E^(20 + 4*x + x^2)*x)/(1 + 4*
E^x^2), x] - 2*Defer[Int][x^2/(1 + 4*E^x^2), x] - 8*Defer[Int][(E^x^2*x^2)/(1 + 4*E^x^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^{20+4 x}+2 x-6 e^{20+4 x} x+\frac {2 \left (e^{20+4 x}-x\right ) x}{1+4 e^{x^2}}+2 x^2+\log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )-4 e^{20+4 x} \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )\right ) \, dx\\ &=x^2+\frac {2 x^3}{3}+2 \int \frac {\left (e^{20+4 x}-x\right ) x}{1+4 e^{x^2}} \, dx-4 \int e^{20+4 x} \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right ) \, dx-6 \int e^{20+4 x} x \, dx-\int e^{20+4 x} \, dx+\int \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right ) \, dx\\ &=-\frac {1}{4} e^{20+4 x}-\frac {3}{2} e^{20+4 x} x+x^2+\frac {2 x^3}{3}-e^{20+4 x} \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )+x \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )+\frac {3}{2} \int e^{20+4 x} \, dx+2 \int \left (\frac {e^{20+4 x} x}{1+4 e^{x^2}}-\frac {x^2}{1+4 e^{x^2}}\right ) \, dx+4 \int \frac {2 e^{20+4 x+x^2} x}{1+4 e^{x^2}} \, dx-\int \frac {8 e^{x^2} x^2}{1+4 e^{x^2}} \, dx\\ &=\frac {1}{8} e^{20+4 x}-\frac {3}{2} e^{20+4 x} x+x^2+\frac {2 x^3}{3}-e^{20+4 x} \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )+x \log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )+2 \int \frac {e^{20+4 x} x}{1+4 e^{x^2}} \, dx-2 \int \frac {x^2}{1+4 e^{x^2}} \, dx+8 \int \frac {e^{20+4 x+x^2} x}{1+4 e^{x^2}} \, dx-8 \int \frac {e^{x^2} x^2}{1+4 e^{x^2}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 29, normalized size = 0.97 \begin {gather*} -\left (\left (e^{4 (5+x)}-x\right ) \left (x+\log \left (\frac {3}{5} \left (1+4 e^{x^2}\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(20 + 4*x)*(-1 - 4*x) + 2*x + E^x^2*(E^(20 + 4*x)*(-4 - 24*x) + 8*x + 8*x^2) + (1 - 4*E^(20 + 4*x
) + E^x^2*(4 - 16*E^(20 + 4*x)))*Log[(3 + 12*E^x^2)/5])/(1 + 4*E^x^2),x]

[Out]

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

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fricas [A]  time = 0.54, size = 33, normalized size = 1.10 \begin {gather*} x^{2} - x e^{\left (4 \, x + 20\right )} + {\left (x - e^{\left (4 \, x + 20\right )}\right )} \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+
8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*exp(x^2)+1),x, algorithm="fricas")

[Out]

x^2 - x*e^(4*x + 20) + (x - e^(4*x + 20))*log(12/5*e^(x^2) + 3/5)

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giac [A]  time = 0.31, size = 41, normalized size = 1.37 \begin {gather*} x^{2} - x e^{\left (4 \, x + 20\right )} + x \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) - e^{\left (4 \, x + 20\right )} \log \left (\frac {12}{5} \, e^{\left (x^{2}\right )} + \frac {3}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+
8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*exp(x^2)+1),x, algorithm="giac")

[Out]

x^2 - x*e^(4*x + 20) + x*log(12/5*e^(x^2) + 3/5) - e^(4*x + 20)*log(12/5*e^(x^2) + 3/5)

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maple [A]  time = 0.05, size = 34, normalized size = 1.13

method	result	size
risch	\(\left (x -{\mathrm e}^{20+4 x}\right ) \ln \left (\frac {12 \,{\mathrm e}^{x^{2}}}{5}+\frac {3}{5}\right )+x^{2}-{\mathrm e}^{20+4 x} x\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*ln(12/5*exp(x^2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+8*x)*ex
p(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*exp(x^2)+1),x,method=_RETURNVERBOSE)

[Out]

(x-exp(20+4*x))*ln(12/5*exp(x^2)+3/5)+x^2-exp(20+4*x)*x

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maxima [B]  time = 0.49, size = 68, normalized size = 2.27 \begin {gather*} x^{2} - x {\left (\log \relax (5) - \log \relax (3)\right )} - {\left (x e^{20} - {\left (\log \relax (5) - \log \relax (3)\right )} e^{20}\right )} e^{\left (4 \, x\right )} + {\left (x - e^{\left (4 \, x + 20\right )} + 1\right )} \log \left (4 \, e^{\left (x^{2}\right )} + 1\right ) - \log \left (4 \, e^{\left (x^{2}\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*exp(20+4*x)+4)*exp(x^2)-4*exp(20+4*x)+1)*log(12/5*exp(x^2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x^2+
8*x)*exp(x^2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*exp(x^2)+1),x, algorithm="maxima")

[Out]

x^2 - x*(log(5) - log(3)) - (x*e^20 - (log(5) - log(3))*e^20)*e^(4*x) + (x - e^(4*x + 20) + 1)*log(4*e^(x^2) +
1) - log(4*e^(x^2) + 1)

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mupad [B]  time = 7.75, size = 22, normalized size = 0.73 \begin {gather*} \left (x+\ln \left (\frac {12\,{\mathrm {e}}^{x^2}}{5}+\frac {3}{5}\right )\right )\,\left (x-{\mathrm {e}}^{4\,x+20}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - log((12*exp(x^2))/5 + 3/5)*(4*exp(4*x + 20) + exp(x^2)*(16*exp(4*x + 20) - 4) - 1) - exp(4*x + 20)*
(4*x + 1) + exp(x^2)*(8*x - exp(4*x + 20)*(24*x + 4) + 8*x^2))/(4*exp(x^2) + 1),x)

[Out]

(x + log((12*exp(x^2))/5 + 3/5))*(x - exp(4*x + 20))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*exp(20+4*x)+4)*exp(x**2)-4*exp(20+4*x)+1)*ln(12/5*exp(x**2)+3/5)+((-24*x-4)*exp(20+4*x)+8*x**
2+8*x)*exp(x**2)+(-4*x-1)*exp(20+4*x)+2*x)/(4*exp(x**2)+1),x)

[Out]

Timed out

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