3.85.23 \(\int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} (-10 x^3+e^5 (1+2 x^2))-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+(-8 x^2+2 x^3) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} (8 x^2-2 x^3-2 x^2 (i \pi +\log (25)))} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^5-5 x}{x \left (-4-e^{x^2}+i \pi +x+\log (25)\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 3.26, 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^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{16 x^2+e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+x^2 (i \pi +\log (25))^2+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(4 - 2*x) + 5*x^2 + E^x^2*(-10*x^3 + E^5*(1 + 2*x^2)) - E^5*(I*Pi + Log[25]))/(16*x^2 + E^(2*x^2)*x^2
- 8*x^3 + x^4 + (-8*x^2 + 2*x^3)*(I*Pi + Log[25]) + x^2*(I*Pi + Log[25])^2 + E^x^2*(8*x^2 - 2*x^3 - 2*x^2*(I*
Pi + Log[25]))),x]

[Out]

10*Defer[Int][x/(-E^x^2 + x - 4*(1 - (I/4)*(Pi - (2*I)*Log[5]))), x] - (5 - E^5*(8 - (2*I)*Pi - Log[625]))*Def
er[Int][(I*E^x^2 - I*x + (4*I)*(1 - (I/4)*(Pi - (2*I)*Log[5])))^(-2), x] - E^5*Defer[Int][1/(x*(E^x^2 - x + 4*
(1 - (I/4)*(Pi - (2*I)*Log[5])))^2), x] + (40 + 2*E^5 - (10*I)*Pi - 5*Log[625])*Defer[Int][x/(E^x^2 - x + 4*(1
- (I/4)*(Pi - (2*I)*Log[5])))^2, x] - 10*Defer[Int][x^2/(E^x^2 - x + 4*(1 - (I/4)*(Pi - (2*I)*Log[5])))^2, x]
+ 2*E^5*Defer[Int][(E^x^2 - x + 4*(1 - (I/4)*(Pi - (2*I)*Log[5])))^(-1), x] + E^5*Defer[Int][1/(x^2*(E^x^2 -
x + 4*(1 - (I/4)*(Pi - (2*I)*Log[5])))), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^5 (4-2 x)+5 x^2+e^{x^2} \left (-10 x^3+e^5 \left (1+2 x^2\right )\right )-e^5 (i \pi +\log (25))}{e^{2 x^2} x^2-8 x^3+x^4+\left (-8 x^2+2 x^3\right ) (i \pi +\log (25))+e^{x^2} \left (8 x^2-2 x^3-2 x^2 (i \pi +\log (25))\right )+x^2 \left (16+(i \pi +\log (25))^2\right )} \, dx\\ &=\int \frac {5 x^2-10 e^{x^2} x^3+e^{5+x^2} \left (1+2 x^2\right )+e^5 (4-i \pi -2 x-\log (25))}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx\\ &=\int \left (\frac {e^5+2 e^5 x^2-10 x^3}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )}+\frac {\left (e^5-5 x\right ) \left (-1+2 x^2-x (8-2 i \pi -\log (625))\right )}{x \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}\right ) \, dx\\ &=\int \frac {e^5+2 e^5 x^2-10 x^3}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )} \, dx+\int \frac {\left (e^5-5 x\right ) \left (-1+2 x^2-x (8-2 i \pi -\log (625))\right )}{x \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx\\ &=\int \left (\frac {10 x}{-e^{x^2}+x-4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )}+\frac {2 e^5}{e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )}+\frac {e^5}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )}\right ) \, dx+\int \left (-\frac {e^5}{x \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}-\frac {10 x^2}{\left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}-\frac {2 i e^5 \pi \left (1-\frac {i \left (5+4 e^5 (-2+\log (5))\right )}{2 e^5 \pi }\right )}{\left (i e^{x^2}-i x+4 i \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}+\frac {x \left (40+2 e^5-10 i \pi -5 \log (625)\right )}{\left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2}\right ) \, dx\\ &=10 \int \frac {x}{-e^{x^2}+x-4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )} \, dx-10 \int \frac {x^2}{\left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx-e^5 \int \frac {1}{x \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx+e^5 \int \frac {1}{x^2 \left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )} \, dx+\left (2 e^5\right ) \int \frac {1}{e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )} \, dx-\left (5-e^5 (8-2 i \pi -\log (625))\right ) \int \frac {1}{\left (i e^{x^2}-i x+4 i \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx+\left (40+2 e^5-10 i \pi -5 \log (625)\right ) \int \frac {x}{\left (e^{x^2}-x+4 \left (1-\frac {1}{4} i (\pi -2 i \log (5))\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.83, size = 30, normalized size = 1.00 \begin {gather*} \frac {e^5-5 x}{x \left (-4-e^{x^2}+i \pi +x+\log (25)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(4 - 2*x) + 5*x^2 + E^x^2*(-10*x^3 + E^5*(1 + 2*x^2)) - E^5*(I*Pi + Log[25]))/(16*x^2 + E^(2*x^
2)*x^2 - 8*x^3 + x^4 + (-8*x^2 + 2*x^3)*(I*Pi + Log[25]) + x^2*(I*Pi + Log[25])^2 + E^x^2*(8*x^2 - 2*x^3 - 2*x
^2*(I*Pi + Log[25]))),x]

[Out]

(E^5 - 5*x)/(x*(-4 - E^x^2 + I*Pi + x + Log[25]))

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 35, normalized size = 1.17 \begin {gather*} \frac {5 \, x - e^{5}}{{\left (-i \, \pi + 4\right )} x - x^{2} + x e^{\left (x^{2}\right )} - 2 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2*x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2
*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp(x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^
2),x, algorithm="fricas")

[Out]

(5*x - e^5)/((-I*pi + 4)*x - x^2 + x*e^(x^2) - 2*x*log(5))

________________________________________________________________________________________

giac [A]  time = 0.17, size = 35, normalized size = 1.17 \begin {gather*} \frac {5 i \, x - i \, e^{5}}{\pi x - i \, x^{2} + i \, x e^{\left (x^{2}\right )} - 2 i \, x \log \relax (5) + 4 i \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2*x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2
*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp(x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^
2),x, algorithm="giac")

[Out]

(5*I*x - I*e^5)/(pi*x - I*x^2 + I*x*e^(x^2) - 2*I*x*log(5) + 4*I*x)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 30, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{5}-5 x}{x \left (2 \ln \relax (5)+i \pi -4-{\mathrm e}^{x^{2}}+x \right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*ln(5)+I*Pi)+(4-2*x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2*x^2*(2
*ln(5)+I*Pi)-2*x^3+8*x^2)*exp(x^2)+x^2*(2*ln(5)+I*Pi)^2+(2*x^3-8*x^2)*(2*ln(5)+I*Pi)+x^4-8*x^3+16*x^2),x,metho
d=_RETURNVERBOSE)

[Out]

(exp(5)-5*x)/x/(2*ln(5)+I*Pi-4-exp(x^2)+x)

________________________________________________________________________________________

maxima [A]  time = 2.11, size = 34, normalized size = 1.13 \begin {gather*} \frac {5 \, x - e^{5}}{{\left (-i \, \pi - 2 \, \log \relax (5) + 4\right )} x - x^{2} + x e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+1)*exp(5)-10*x^3)*exp(x^2)-exp(5)*(2*log(5)+I*pi)+(4-2*x)*exp(5)+5*x^2)/(x^2*exp(x^2)^2+(-2
*x^2*(2*log(5)+I*pi)-2*x^3+8*x^2)*exp(x^2)+x^2*(2*log(5)+I*pi)^2+(2*x^3-8*x^2)*(2*log(5)+I*pi)+x^4-8*x^3+16*x^
2),x, algorithm="maxima")

[Out]

(5*x - e^5)/((-I*pi - 2*log(5) + 4)*x - x^2 + x*e^(x^2))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^5\,\left (2\,x^2+1\right )-10\,x^3\right )-{\mathrm {e}}^5\,\left (2\,\ln \relax (5)+\Pi \,1{}\mathrm {i}\right )+5\,x^2-{\mathrm {e}}^5\,\left (2\,x-4\right )}{x^2\,{\mathrm {e}}^{2\,x^2}+x^2\,{\left (2\,\ln \relax (5)+\Pi \,1{}\mathrm {i}\right )}^2-{\mathrm {e}}^{x^2}\,\left (2\,x^2\,\left (2\,\ln \relax (5)+\Pi \,1{}\mathrm {i}\right )-8\,x^2+2\,x^3\right )-\left (8\,x^2-2\,x^3\right )\,\left (2\,\ln \relax (5)+\Pi \,1{}\mathrm {i}\right )+16\,x^2-8\,x^3+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2)*(exp(5)*(2*x^2 + 1) - 10*x^3) - exp(5)*(Pi*1i + 2*log(5)) + 5*x^2 - exp(5)*(2*x - 4))/(x^2*(Pi*1
i + 2*log(5))^2 - (8*x^2 - 2*x^3)*(Pi*1i + 2*log(5)) - exp(x^2)*(2*x^2*(Pi*1i + 2*log(5)) - 8*x^2 + 2*x^3) + x
^2*exp(2*x^2) + 16*x^2 - 8*x^3 + x^4),x)

[Out]

int((exp(x^2)*(exp(5)*(2*x^2 + 1) - 10*x^3) - exp(5)*(Pi*1i + 2*log(5)) + 5*x^2 - exp(5)*(2*x - 4))/(x^2*(Pi*1
i + 2*log(5))^2 - (8*x^2 - 2*x^3)*(Pi*1i + 2*log(5)) - exp(x^2)*(2*x^2*(Pi*1i + 2*log(5)) - 8*x^2 + 2*x^3) + x
^2*exp(2*x^2) + 16*x^2 - 8*x^3 + x^4), x)

________________________________________________________________________________________

sympy [A]  time = 21.92, size = 31, normalized size = 1.03 \begin {gather*} \frac {- 5 x + e^{5}}{x^{2} - x e^{x^{2}} - 4 x + 2 x \log {\relax (5 )} + i \pi x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+1)*exp(5)-10*x**3)*exp(x**2)-exp(5)*(2*ln(5)+I*pi)+(4-2*x)*exp(5)+5*x**2)/(x**2*exp(x**2)*
*2+(-2*x**2*(2*ln(5)+I*pi)-2*x**3+8*x**2)*exp(x**2)+x**2*(2*ln(5)+I*pi)**2+(2*x**3-8*x**2)*(2*ln(5)+I*pi)+x**4
-8*x**3+16*x**2),x)

[Out]

(-5*x + exp(5))/(x**2 - x*exp(x**2) - 4*x + 2*x*log(5) + I*pi*x)

________________________________________________________________________________________