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

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

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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 = {} $$\begin{array}{c}\int \frac{e^5(4-2x)+5x^2+e^{x^2}(-10x^3+e^5(1+2x^2))-e^5(i\pi +\log (25))}{16x^2+e^{2x^2}{x}^2-8x^3+x^4+(-8x^2+2x^3)(i\pi +\log (25))+x^2(i\pi +\log (25){)}^2+e^{x^2}(8x^2-2x^3-2x^2(i\pi +\log (25)))}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{e^5(4-2x)+5x^2+e^{x^2}(-10x^3+e^5(1+2x^2))-e^5(i\pi +\log (25))}{e^{2x^2}{x}^2-8x^3+x^4+(-8x^2+2x^3)(i\pi +\log (25))+e^{x^2}(8x^2-2x^3-2x^2(i\pi +\log (25)))+x^2(16+(i\pi +\log (25){)}^2)}dx\\ & =\int \frac{5x^2-10e^{x^2}{x}^3+e^{5+x^2}(1+2x^2)+e^5(4-i\pi -2x-\log (25))}{x^2{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx\\ & =\int (\frac{e^5+2e^5{x}^2-10x^3}{x^2(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}+\frac{(e^5-5x)(-1+2x^2-x(8-2i\pi -\log (625)))}{x{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2})dx\\ & =\int \frac{e^5+2e^5{x}^2-10x^3}{x^2(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}dx+\int \frac{(e^5-5x)(-1+2x^2-x(8-2i\pi -\log (625)))}{x{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx\\ & =\int (\frac{10x}{-e^{x^2}+x-4(1-\frac{1}{4}i(\pi -2i\log (5)))}+\frac{2e^5}{e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5)))}+\frac{e^5}{x^2(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))})dx+\int (-\frac{e^5}{x{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}-\frac{10x^2}{{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}-\frac{2i{e}^5\pi (1-\frac{i(5+4e^5(-2+\log (5)))}{2e^5\pi })}{{(i{e}^{x^2}-ix+4i(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}+\frac{x(40+2e^5-10i\pi -5\log (625))}{{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2})dx\\ & =10\int \frac{x}{-e^{x^2}+x-4(1-\frac{1}{4}i(\pi -2i\log (5)))}dx-10\int \frac{x^2}{{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx-e^5\int \frac{1}{x{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx+e^5\int \frac{1}{x^2(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}dx+\left(2e^5\right)\int \frac{1}{e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5)))}dx-(5-e^5(8-2i\pi -\log (625)))\int \frac{1}{{(i{e}^{x^2}-ix+4i(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx+(40+2e^5-10i\pi -5\log (625))\int \frac{x}{{(e^{x^2}-x+4(1-\frac{1}{4}i(\pi -2i\log (5))))}^2}dx\end{array}\end{array}$$

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

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]))

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

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))

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

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)

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

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

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)

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

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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int \frac{{\mathrm{e}}^{x^2}({\mathrm{e}}^5(2x^2+1)-10x^3)-{\mathrm{e}}^5(2\ln (5)+\mathrm{\Pi }1\mathrm{i})+5x^2-{\mathrm{e}}^5(2x-4)}{x^2{\mathrm{e}}^{2x^2}+x^2{(2\ln (5)+\mathrm{\Pi }1\mathrm{i})}^2-{\mathrm{e}}^{x^2}(2x^2(2\ln (5)+\mathrm{\Pi }1\mathrm{i})-8x^2+2x^3)-(8x^2-2x^3)(2\ln (5)+\mathrm{\Pi }1\mathrm{i})+16x^2-8x^3+x^4}dx\end{array}$$

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)

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

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)

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