3.89.67 $\int \frac{31+e^{x^2}(-25+6x+50x^2)+(2+2e^{x^2}x)\log (\frac{e}{2})}{9+150x+625x^2+(6+50x)\log (\frac{e}{2})+{\log }^2(\frac{e}{2})}dx$

Optimal. Leaf size=30 $$\frac{-1+e^{x^2}+2x}{x(25+\frac{3+\log \left(\frac{e}{2}\right)}{x})}$$

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Rubi [A]  time = 0.35, antiderivative size = 59, normalized size of antiderivative = 1.97, number of steps used = 4, number of rules used = 3, integrand size = 68, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.044, Rules used = {6688, 6742, 2288} $$\begin{array}{c}\frac{e^{x^2}(50x^2+x(8-\log (4)))}{2x(25x+4-\log (2){)}^2}-\frac{33-\log (4)}{25(25x+4-\log (2))}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(31 + E^x^2*(-25 + 6*x + 50*x^2) + (2 + 2*E^x^2*x)*Log[E/2])/(9 + 150*x + 625*x^2 + (6 + 50*x)*Log[E/2] +
Log[E/2]^2),x]

[Out]

(E^x^2*(50*x^2 + x*(8 - Log[4])))/(2*x*(4 + 25*x - Log[2])^2) - (33 - Log[4])/(25*(4 + 25*x - Log[2]))

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]]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{33(1-\frac{2\log (2)}{33})+e^{x^2}(-25+50x^2-x(-8+\log (4)))}{(4+25x-\log (2){)}^2}dx\\ & =\int (\frac{e^{x^2}(-25+50x^2+x(8-\log (4)))}{(4+25x-\log (2){)}^2}+\frac{33-\log (4)}{(4+25x-\log (2){)}^2})dx\\ & =-\frac{33-\log (4)}{25(4+25x-\log (2))}+\int \frac{e^{x^2}(-25+50x^2+x(8-\log (4)))}{(4+25x-\log (2){)}^2}dx\\ & =\frac{e^{x^2}(50x^2+x(8-\log (4)))}{2x(4+25x-\log (2){)}^2}-\frac{33-\log (4)}{25(4+25x-\log (2))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.13, size = 45, normalized size = 1.50 $$\begin{array}{c}\frac{25e^{x^2}(8+50x-\log (4))+2(4+25x-\log (2))(-33+\log (4))}{50(-4-25x+\log (2){)}^2}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(31 + E^x^2*(-25 + 6*x + 50*x^2) + (2 + 2*E^x^2*x)*Log[E/2])/(9 + 150*x + 625*x^2 + (6 + 50*x)*Log[E
/2] + Log[E/2]^2),x]

[Out]

(25*E^x^2*(8 + 50*x - Log[4]) + 2*(4 + 25*x - Log[2])*(-33 + Log[4]))/(50*(-4 - 25*x + Log[2])^2)

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fricas [A]  time = 0.49, size = 25, normalized size = 0.83 $$\begin{array}{c}\frac{25e^{\left(x^2\right)}+2\log (2)-33}{25(25x-\log (2)+4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(log(1/2*exp(1))^2+(50*x+6)*log(1/2*e
xp(1))+625*x^2+150*x+9),x, algorithm="fricas")

[Out]

1/25*(25*e^(x^2) + 2*log(2) - 33)/(25*x - log(2) + 4)

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giac [A]  time = 0.16, size = 25, normalized size = 0.83 $$\begin{array}{c}\frac{25e^{\left(x^2\right)}+2\log (2)-33}{25(25x-\log (2)+4)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(log(1/2*exp(1))^2+(50*x+6)*log(1/2*e
xp(1))+625*x^2+150*x+9),x, algorithm="giac")

[Out]

1/25*(25*e^(x^2) + 2*log(2) - 33)/(25*x - log(2) + 4)

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maple [A]  time = 0.52, size = 23, normalized size = 0.77

method	result	size
norman	$\frac{-{\mathrm{e}}^{x^2}+\frac{33}{25}-\frac{2\ln (2)}{25}}{\ln (2)-25x-4}$	$23$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x^2)*x+2)*ln(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(ln(1/2*exp(1))^2+(50*x+6)*ln(1/2*exp(1))+62
5*x^2+150*x+9),x,method=_RETURNVERBOSE)

[Out]

(-exp(x^2)+33/25-2/25*ln(2))/(ln(2)-25*x-4)

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maxima [A]  time = 0.52, size = 50, normalized size = 1.67 $$\begin{array}{c}\frac{e^{\left(x^2\right)}}{25x-\log (2)+4}-\frac{2\log \left(\frac{1}{2}e\right)}{25(25x+\log \left(\frac{1}{2}e\right)+3)}-\frac{31}{25(25x+\log \left(\frac{1}{2}e\right)+3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(log(1/2*exp(1))^2+(50*x+6)*log(1/2*e
xp(1))+625*x^2+150*x+9),x, algorithm="maxima")

[Out]

e^(x^2)/(25*x - log(2) + 4) - 2/25*log(1/2*e)/(25*x + log(1/2*e) + 3) - 31/25/(25*x + log(1/2*e) + 3)

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mupad [B]  time = 0.25, size = 29, normalized size = 0.97 $$\begin{array}{c}-\frac{\frac{\ln \left(\frac{{\mathrm{e}}^2}{4}\right)}{25}-{\mathrm{e}}^{x^2}+\frac{31}{25}}{25x+\ln \left(\frac{\mathrm{e}}{2}\right)+3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(1)/2)*(2*x*exp(x^2) + 2) + exp(x^2)*(6*x + 50*x^2 - 25) + 31)/(150*x + log(exp(1)/2)^2 + log(exp(
1)/2)*(50*x + 6) + 625*x^2 + 9),x)

[Out]

-(log(exp(2)/4)/25 - exp(x^2) + 31/25)/(25*x + log(exp(1)/2) + 3)

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sympy [A]  time = 0.26, size = 29, normalized size = 0.97 $$\begin{array}{c}-\frac{33-2\log (2)}{625x-25\log (2)+100}+\frac{e^{x^2}}{25x-\log (2)+4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x**2)*x+2)*ln(1/2*exp(1))+(50*x**2+6*x-25)*exp(x**2)+31)/(ln(1/2*exp(1))**2+(50*x+6)*ln(1/2*
exp(1))+625*x**2+150*x+9),x)

[Out]

-(33 - 2*log(2))/(625*x - 25*log(2) + 100) + exp(x**2)/(25*x - log(2) + 4)

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