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

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

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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 {gather*} \frac {e^{x^2} \left (50 x^2+x (8-\log (4))\right )}{2 x (25 x+4-\log (2))^2}-\frac {33-\log (4)}{25 (25 x+4-\log (2))} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {33 \left (1-\frac {2 \log (2)}{33}\right )+e^{x^2} \left (-25+50 x^2-x (-8+\log (4))\right )}{(4+25 x-\log (2))^2} \, dx\\ &=\int \left (\frac {e^{x^2} \left (-25+50 x^2+x (8-\log (4))\right )}{(4+25 x-\log (2))^2}+\frac {33-\log (4)}{(4+25 x-\log (2))^2}\right ) \, dx\\ &=-\frac {33-\log (4)}{25 (4+25 x-\log (2))}+\int \frac {e^{x^2} \left (-25+50 x^2+x (8-\log (4))\right )}{(4+25 x-\log (2))^2} \, dx\\ &=\frac {e^{x^2} \left (50 x^2+x (8-\log (4))\right )}{2 x (4+25 x-\log (2))^2}-\frac {33-\log (4)}{25 (4+25 x-\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 45, normalized size = 1.50 \begin {gather*} \frac {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} \end {gather*}

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 {gather*} \frac {25 \, e^{\left (x^{2}\right )} + 2 \, \log \relax (2) - 33}{25 \, {\left (25 \, x - \log \relax (2) + 4\right )}} \end {gather*}

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 {gather*} \frac {25 \, e^{\left (x^{2}\right )} + 2 \, \log \relax (2) - 33}{25 \, {\left (25 \, x - \log \relax (2) + 4\right )}} \end {gather*}

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	\(\frac {-{\mathrm e}^{x^{2}}+\frac {33}{25}-\frac {2 \ln \relax (2)}{25}}{\ln \relax (2)-25 x -4}\)	\(23\)

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 {gather*} \frac {e^{\left (x^{2}\right )}}{25 \, x - \log \relax (2) + 4} - \frac {2 \, \log \left (\frac {1}{2} \, e\right )}{25 \, {\left (25 \, x + \log \left (\frac {1}{2} \, e\right ) + 3\right )}} - \frac {31}{25 \, {\left (25 \, x + \log \left (\frac {1}{2} \, e\right ) + 3\right )}} \end {gather*}

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 {gather*} -\frac {\frac {\ln \left (\frac {{\mathrm {e}}^2}{4}\right )}{25}-{\mathrm {e}}^{x^2}+\frac {31}{25}}{25\,x+\ln \left (\frac {\mathrm {e}}{2}\right )+3} \end {gather*}

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 {gather*} - \frac {33 - 2 \log {\relax (2 )}}{625 x - 25 \log {\relax (2 )} + 100} + \frac {e^{x^{2}}}{25 x - \log {\relax (2 )} + 4} \end {gather*}

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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