3.58.47 \(\int \frac {e^{-2-\log ^2(-2 x-x^2)} (-2-x+e^{2+\log ^2(-2 x-x^2)} (2+x+e^5 (2+x))+(4+4 x) \log (x) \log (-2 x-x^2))}{2 x+x^2} \, dx\)

Optimal. Leaf size=26 \[ \left (1+e^5-e^{-2-\log ^2((-2-x) x)}\right ) \log (x) \]

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Rubi [A]  time = 1.72, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 5, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {1593, 6742, 6688, 29, 2288} \begin {gather*} \left (1+e^5\right ) \log (x)-e^{-\log ^2(-x (x+2))-2} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 - Log[-2*x - x^2]^2)*(-2 - x + E^(2 + Log[-2*x - x^2]^2)*(2 + x + E^5*(2 + x)) + (4 + 4*x)*Log[x]*L
og[-2*x - x^2]))/(2*x + x^2),x]

[Out]

-(E^(-2 - Log[-(x*(2 + x))]^2)*Log[x]) + (1 + E^5)*Log[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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]]

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 {e^{-2-\log ^2\left (-2 x-x^2\right )} \left (-2-x+e^{2+\log ^2\left (-2 x-x^2\right )} \left (2+x+e^5 (2+x)\right )+(4+4 x) \log (x) \log \left (-2 x-x^2\right )\right )}{x (2+x)} \, dx\\ &=\int \left (\frac {e^{\log ^2(-x (2+x))-\log ^2\left (-2 x-x^2\right )} \left (1+e^5\right )}{x}+\frac {e^{-2-\log ^2\left (-2 x-x^2\right )} (-2-x+4 \log (x) \log (-x (2+x))+4 x \log (x) \log (-x (2+x)))}{x (2+x)}\right ) \, dx\\ &=\left (1+e^5\right ) \int \frac {e^{\log ^2(-x (2+x))-\log ^2\left (-2 x-x^2\right )}}{x} \, dx+\int \frac {e^{-2-\log ^2\left (-2 x-x^2\right )} (-2-x+4 \log (x) \log (-x (2+x))+4 x \log (x) \log (-x (2+x)))}{x (2+x)} \, dx\\ &=\left (1+e^5\right ) \int \frac {1}{x} \, dx+\int \frac {e^{-2-\log ^2(-x (2+x))} (-2-x+4 (1+x) \log (x) \log (-x (2+x)))}{x (2+x)} \, dx\\ &=-e^{-2-\log ^2(-x (2+x))} \log (x)+\left (1+e^5\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 25, normalized size = 0.96 \begin {gather*} \left (1+e^5-e^{-2-\log ^2(-x (2+x))}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 - Log[-2*x - x^2]^2)*(-2 - x + E^(2 + Log[-2*x - x^2]^2)*(2 + x + E^5*(2 + x)) + (4 + 4*x)*Lo
g[x]*Log[-2*x - x^2]))/(2*x + x^2),x]

[Out]

(1 + E^5 - E^(-2 - Log[-(x*(2 + x))]^2))*Log[x]

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fricas [A]  time = 0.49, size = 45, normalized size = 1.73 \begin {gather*} {\left ({\left (e^{5} + 1\right )} e^{\left (\log \left (-x^{2} - 2 \, x\right )^{2} + 2\right )} \log \relax (x) - \log \relax (x)\right )} e^{\left (-\log \left (-x^{2} - 2 \, x\right )^{2} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4*x+4)*log(-x^2-2*x)*log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-
2*x)^2+2),x, algorithm="fricas")

[Out]

((e^5 + 1)*e^(log(-x^2 - 2*x)^2 + 2)*log(x) - log(x))*e^(-log(-x^2 - 2*x)^2 - 2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, {\left (x + 1\right )} \log \left (-x^{2} - 2 \, x\right ) \log \relax (x) + {\left ({\left (x + 2\right )} e^{5} + x + 2\right )} e^{\left (\log \left (-x^{2} - 2 \, x\right )^{2} + 2\right )} - x - 2\right )} e^{\left (-\log \left (-x^{2} - 2 \, x\right )^{2} - 2\right )}}{x^{2} + 2 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4*x+4)*log(-x^2-2*x)*log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-
2*x)^2+2),x, algorithm="giac")

[Out]

integrate((4*(x + 1)*log(-x^2 - 2*x)*log(x) + ((x + 2)*e^5 + x + 2)*e^(log(-x^2 - 2*x)^2 + 2) - x - 2)*e^(-log
(-x^2 - 2*x)^2 - 2)/(x^2 + 2*x), x)

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maple [C]  time = 0.36, size = 632, normalized size = 24.31

method	result	size
risch	\({\mathrm e}^{5} \ln \relax (x )+\ln \relax (x )-\ln \relax (x ) x^{-i \pi \,\mathrm {csgn}\left (i x \right )} x^{-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right )} \left (2+x \right )^{-i \pi \,\mathrm {csgn}\left (i x \right )} \left (2+x \right )^{-i \pi \,\mathrm {csgn}\left (i \left (2+x \right )\right )} \left (2+x \right )^{-2 \ln \relax (x )} \left (2+x \right )^{-2 i \pi } \left (2+x \right )^{i \pi \,\mathrm {csgn}\left (i x \left (2+x \right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right )} x^{i \pi \,\mathrm {csgn}\left (i x \left (2+x \right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right )} \left (2+x \right )^{-i \pi \,\mathrm {csgn}\left (i x \left (2+x \right )\right )} x^{2 i \pi } \left (2+x \right )^{2 i \pi } x^{-i \pi \,\mathrm {csgn}\left (i x \left (2+x \right )\right )} x^{-2 i \pi } {\mathrm e}^{-2+\pi ^{2}-\ln \relax (x )^{2}-\ln \left (2+x \right )^{2}} {\mathrm e}^{-\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right )} {\mathrm e}^{-\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{3} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (2+x \right )\right )}{2}} {\mathrm e}^{-\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{3} \mathrm {csgn}\left (i \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i \left (2+x \right )\right )^{2}}{4}} {\mathrm e}^{\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (2+x \right )\right )} {\mathrm e}^{-\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{4} \mathrm {csgn}\left (i x \right )} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{4} \mathrm {csgn}\left (i \left (2+x \right )\right )^{2}}{4}} {\mathrm e}^{-\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{4} \mathrm {csgn}\left (i \left (2+x \right )\right )} {\mathrm e}^{\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i x \right )} {\mathrm e}^{\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{2} \mathrm {csgn}\left (i \left (2+x \right )\right )} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{5} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{5} \mathrm {csgn}\left (i \left (2+x \right )\right )}{2}} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{4} \mathrm {csgn}\left (i x \right )^{2}}{4}} {\mathrm e}^{\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{3}} {\mathrm e}^{-2 \pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{2}} {\mathrm e}^{\frac {\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{6}}{4}} {\mathrm e}^{-\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{5}} {\mathrm e}^{\pi ^{2} \mathrm {csgn}\left (i x \left (2+x \right )\right )^{4}}\)	\(632\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2+x)*exp(5)+2+x)*exp(ln(-x^2-2*x)^2+2)+(4*x+4)*ln(-x^2-2*x)*ln(x)-x-2)/(x^2+2*x)/exp(ln(-x^2-2*x)^2+2),
x,method=_RETURNVERBOSE)

[Out]

exp(5)*ln(x)+ln(x)-ln(x)/(x^(I*Pi*csgn(I*x)))/(x^(I*Pi*csgn(I*(2+x))))/((2+x)^(I*Pi*csgn(I*x)))/((2+x)^(I*Pi*c
sgn(I*(2+x))))/((2+x)^(2*ln(x)))/((2+x)^(2*I*Pi))/((2+x)^(-I*Pi*csgn(I*x*(2+x))*csgn(I*x)*csgn(I*(2+x))))/(x^(
-I*Pi*csgn(I*x*(2+x))*csgn(I*x)*csgn(I*(2+x))))/((2+x)^(I*Pi*csgn(I*x*(2+x))))/(x^(-2*I*Pi))/((2+x)^(-2*I*Pi))
/(x^(I*Pi*csgn(I*x*(2+x))))/(x^(2*I*Pi))*exp(-2+Pi^2-ln(x)^2-ln(2+x)^2)*exp(-Pi^2*csgn(I*x*(2+x))*csgn(I*x)*cs
gn(I*(2+x)))*exp(-1/2*Pi^2*csgn(I*x*(2+x))^3*csgn(I*x)^2*csgn(I*(2+x)))*exp(-1/2*Pi^2*csgn(I*x*(2+x))^3*csgn(I
*(2+x))^2*csgn(I*x))*exp(1/4*Pi^2*csgn(I*x*(2+x))^2*csgn(I*x)^2*csgn(I*(2+x))^2)*exp(Pi^2*csgn(I*x*(2+x))^3*cs
gn(I*x)*csgn(I*(2+x)))*exp(-Pi^2*csgn(I*x*(2+x))^4*csgn(I*x))*exp(1/4*Pi^2*csgn(I*x*(2+x))^4*csgn(I*(2+x))^2)*
exp(-Pi^2*csgn(I*x*(2+x))^4*csgn(I*(2+x)))*exp(Pi^2*csgn(I*x*(2+x))^2*csgn(I*x))*exp(Pi^2*csgn(I*x*(2+x))^2*cs
gn(I*(2+x)))*exp(1/2*Pi^2*csgn(I*x*(2+x))^5*csgn(I*x))*exp(1/2*Pi^2*csgn(I*x*(2+x))^5*csgn(I*(2+x)))*exp(1/4*P
i^2*csgn(I*x*(2+x))^4*csgn(I*x)^2)*exp(Pi^2*csgn(I*x*(2+x))^3)*exp(-2*Pi^2*csgn(I*x*(2+x))^2)*exp(1/4*Pi^2*csg
n(I*x*(2+x))^6)*exp(-Pi^2*csgn(I*x*(2+x))^5)*exp(Pi^2*csgn(I*x*(2+x))^4)

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maxima [B]  time = 0.58, size = 56, normalized size = 2.15 \begin {gather*} -{\left (\log \left (x + 2\right ) - \log \relax (x)\right )} e^{5} + e^{5} \log \left (x + 2\right ) - e^{\left (-\log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (-x - 2\right ) - \log \left (-x - 2\right )^{2} - 2\right )} \log \relax (x) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(5)+2+x)*exp(log(-x^2-2*x)^2+2)+(4*x+4)*log(-x^2-2*x)*log(x)-x-2)/(x^2+2*x)/exp(log(-x^2-
2*x)^2+2),x, algorithm="maxima")

[Out]

-(log(x + 2) - log(x))*e^5 + e^5*log(x + 2) - e^(-log(x)^2 - 2*log(x)*log(-x - 2) - log(-x - 2)^2 - 2)*log(x)
+ log(x)

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mupad [B]  time = 3.96, size = 29, normalized size = 1.12 \begin {gather*} \ln \relax (x)\,\left ({\mathrm {e}}^5+1\right )-{\mathrm {e}}^{-{\ln \left (-x^2-2\,x\right )}^2-2}\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- log(- 2*x - x^2)^2 - 2)*(x - exp(log(- 2*x - x^2)^2 + 2)*(x + exp(5)*(x + 2) + 2) - log(- 2*x - x^
2)*log(x)*(4*x + 4) + 2))/(2*x + x^2),x)

[Out]

log(x)*(exp(5) + 1) - exp(- log(- 2*x - x^2)^2 - 2)*log(x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2+x)*exp(5)+2+x)*exp(ln(-x**2-2*x)**2+2)+(4*x+4)*ln(-x**2-2*x)*ln(x)-x-2)/(x**2+2*x)/exp(ln(-x**2
-2*x)**2+2),x)

[Out]

Timed out

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