3.63.43 \(\int \frac {x^2+4 x \log (2)+4 \log ^2(2)+e^{x+x^2} (2-2 x-4 x^2+(-4-8 x) \log (2))}{x^2+4 x \log (2)+4 \log ^2(2)} \, dx\)

Optimal. Leaf size=21 \[ x-\frac {e^{x+x^2}}{\frac {x}{2}+\log (2)} \]

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Rubi [A]  time = 0.32, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 4, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {27, 6688, 2288} \begin {gather*} x-\frac {2 e^{x^2+x} \left (2 x^2+x (1+\log (16))+\log (4)\right )}{(2 x+1) (x+\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + 4*x*Log[2] + 4*Log[2]^2 + E^(x + x^2)*(2 - 2*x - 4*x^2 + (-4 - 8*x)*Log[2]))/(x^2 + 4*x*Log[2] + 4*
Log[2]^2),x]

[Out]

x - (2*E^(x + x^2)*(2*x^2 + Log[4] + x*(1 + Log[16])))/((1 + 2*x)*(x + Log[4])^2)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2+4 x \log (2)+4 \log ^2(2)+e^{x+x^2} \left (2-2 x-4 x^2+(-4-8 x) \log (2)\right )}{(x+2 \log (2))^2} \, dx\\ &=\int \left (1-\frac {2 e^{x+x^2} \left (-1+2 x^2+\log (4)+x (1+\log (16))\right )}{(x+\log (4))^2}\right ) \, dx\\ &=x-2 \int \frac {e^{x+x^2} \left (-1+2 x^2+\log (4)+x (1+\log (16))\right )}{(x+\log (4))^2} \, dx\\ &=x-\frac {2 e^{x+x^2} \left (2 x^2+\log (4)+x (1+\log (16))\right )}{(1+2 x) (x+\log (4))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 38, normalized size = 1.81 \begin {gather*} x-\frac {2 e^{x+x^2} \left (2 x^2+\log (4)+x (1+\log (16))\right )}{(1+2 x) (x+\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 4*x*Log[2] + 4*Log[2]^2 + E^(x + x^2)*(2 - 2*x - 4*x^2 + (-4 - 8*x)*Log[2]))/(x^2 + 4*x*Log[2
] + 4*Log[2]^2),x]

[Out]

x - (2*E^(x + x^2)*(2*x^2 + Log[4] + x*(1 + Log[16])))/((1 + 2*x)*(x + Log[4])^2)

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fricas [A]  time = 0.57, size = 26, normalized size = 1.24 \begin {gather*} \frac {x^{2} + 2 \, x \log \relax (2) - 2 \, e^{\left (x^{2} + x\right )}}{x + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-4)*log(2)-4*x^2-2*x+2)*exp(x^2+x)+4*log(2)^2+4*x*log(2)+x^2)/(4*log(2)^2+4*x*log(2)+x^2),x,
algorithm="fricas")

[Out]

(x^2 + 2*x*log(2) - 2*e^(x^2 + x))/(x + 2*log(2))

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giac [A]  time = 0.14, size = 26, normalized size = 1.24 \begin {gather*} \frac {x^{2} + 2 \, x \log \relax (2) - 2 \, e^{\left (x^{2} + x\right )}}{x + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-4)*log(2)-4*x^2-2*x+2)*exp(x^2+x)+4*log(2)^2+4*x*log(2)+x^2)/(4*log(2)^2+4*x*log(2)+x^2),x,
algorithm="giac")

[Out]

(x^2 + 2*x*log(2) - 2*e^(x^2 + x))/(x + 2*log(2))

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maple [A]  time = 0.41, size = 19, normalized size = 0.90

method	result	size
risch	\(x -\frac {2 \,{\mathrm e}^{\left (x +1\right ) x}}{x +2 \ln \relax (2)}\)	\(19\)
norman	\(\frac {x^{2}-2 \,{\mathrm e}^{x^{2}+x}-4 \ln \relax (2)^{2}}{x +2 \ln \relax (2)}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*x-4)*ln(2)-4*x^2-2*x+2)*exp(x^2+x)+4*ln(2)^2+4*x*ln(2)+x^2)/(4*ln(2)^2+4*x*ln(2)+x^2),x,method=_RETU
RNVERBOSE)

[Out]

x-2/(x+2*ln(2))*exp((x+1)*x)

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maxima [B]  time = 0.64, size = 67, normalized size = 3.19 \begin {gather*} 4 \, {\left (\frac {2 \, \log \relax (2)}{x + 2 \, \log \relax (2)} + \log \left (x + 2 \, \log \relax (2)\right )\right )} \log \relax (2) - 4 \, \log \relax (2) \log \left (x + 2 \, \log \relax (2)\right ) + x - \frac {8 \, \log \relax (2)^{2}}{x + 2 \, \log \relax (2)} - \frac {2 \, e^{\left (x^{2} + x\right )}}{x + 2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-4)*log(2)-4*x^2-2*x+2)*exp(x^2+x)+4*log(2)^2+4*x*log(2)+x^2)/(4*log(2)^2+4*x*log(2)+x^2),x,
algorithm="maxima")

[Out]

4*(2*log(2)/(x + 2*log(2)) + log(x + 2*log(2)))*log(2) - 4*log(2)*log(x + 2*log(2)) + x - 8*log(2)^2/(x + 2*lo
g(2)) - 2*e^(x^2 + x)/(x + 2*log(2))

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mupad [B]  time = 0.20, size = 16, normalized size = 0.76 \begin {gather*} x-\frac {2\,{\mathrm {e}}^{x^2+x}}{x+\ln \relax (4)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*log(2) + 4*log(2)^2 + x^2 - exp(x + x^2)*(2*x + log(2)*(8*x + 4) + 4*x^2 - 2))/(4*x*log(2) + 4*log(2)
^2 + x^2),x)

[Out]

x - (2*exp(x + x^2))/(x + log(4))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.71 \begin {gather*} x - \frac {2 e^{x^{2} + x}}{x + 2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*x-4)*ln(2)-4*x**2-2*x+2)*exp(x**2+x)+4*ln(2)**2+4*x*ln(2)+x**2)/(4*ln(2)**2+4*x*ln(2)+x**2),x)

[Out]

x - 2*exp(x**2 + x)/(x + 2*log(2))

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