3.90.70 \(\int \frac {-48+64 e+32 x+e^{x^2} (48-64 e-32 x+128 x^2)+(-64+64 e^{x^2}) \log (-x+e^{x^2} x)}{-x^2-16 e^2 x^2-2 x^3-x^4+e (-8 x^2-8 x^3)+e^{x^2} (x^2+16 e^2 x^2+2 x^3+x^4+e (8 x^2+8 x^3))+(8 x^2+32 e x^2+8 x^3+e^{x^2} (-8 x^2-32 e x^2-8 x^3)) \log (-x+e^{x^2} x)+(-16 x^2+16 e^{x^2} x^2) \log ^2(-x+e^{x^2} x)} \, dx\)

Optimal. Leaf size=30 \[ \frac {4}{x \left (e+\frac {1+x}{4}-\log \left (-x+e^{x^2} x\right )\right )} \]

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Rubi [A]  time = 0.94, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, integrand size = 212, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6, 6688, 12, 6687} \begin {gather*} \frac {16}{x \left (-4 \log \left (-\left (\left (1-e^{x^2}\right ) x\right )\right )+x+4 e+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48 + 64*E + 32*x + E^x^2*(48 - 64*E - 32*x + 128*x^2) + (-64 + 64*E^x^2)*Log[-x + E^x^2*x])/(-x^2 - 16*E
^2*x^2 - 2*x^3 - x^4 + E*(-8*x^2 - 8*x^3) + E^x^2*(x^2 + 16*E^2*x^2 + 2*x^3 + x^4 + E*(8*x^2 + 8*x^3)) + (8*x^
2 + 32*E*x^2 + 8*x^3 + E^x^2*(-8*x^2 - 32*E*x^2 - 8*x^3))*Log[-x + E^x^2*x] + (-16*x^2 + 16*E^x^2*x^2)*Log[-x
+ E^x^2*x]^2),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, x] && NeQ[m, -1]

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 {-48+64 e+32 x+e^{x^2} \left (48-64 e-32 x+128 x^2\right )+\left (-64+64 e^{x^2}\right ) \log \left (-x+e^{x^2} x\right )}{\left (-1-16 e^2\right ) x^2-2 x^3-x^4+e \left (-8 x^2-8 x^3\right )+e^{x^2} \left (x^2+16 e^2 x^2+2 x^3+x^4+e \left (8 x^2+8 x^3\right )\right )+\left (8 x^2+32 e x^2+8 x^3+e^{x^2} \left (-8 x^2-32 e x^2-8 x^3\right )\right ) \log \left (-x+e^{x^2} x\right )+\left (-16 x^2+16 e^{x^2} x^2\right ) \log ^2\left (-x+e^{x^2} x\right )} \, dx\\ &=\int \frac {16 \left (3 \left (1-\frac {4 e}{3}\right )+4 e^{1+x^2}-2 x-e^{x^2} \left (3-2 x+8 x^2\right )-4 \left (-1+e^{x^2}\right ) \log \left (\left (-1+e^{x^2}\right ) x\right )\right )}{\left (1-e^{x^2}\right ) x^2 \left (1+4 e+x-4 \log \left (\left (-1+e^{x^2}\right ) x\right )\right )^2} \, dx\\ &=16 \int \frac {3 \left (1-\frac {4 e}{3}\right )+4 e^{1+x^2}-2 x-e^{x^2} \left (3-2 x+8 x^2\right )-4 \left (-1+e^{x^2}\right ) \log \left (\left (-1+e^{x^2}\right ) x\right )}{\left (1-e^{x^2}\right ) x^2 \left (1+4 e+x-4 \log \left (\left (-1+e^{x^2}\right ) x\right )\right )^2} \, dx\\ &=\frac {16}{x \left (1+4 e+x-4 \log \left (-\left (\left (1-e^{x^2}\right ) x\right )\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 26, normalized size = 0.87 \begin {gather*} \frac {16}{x+4 e x+x^2-4 x \log \left (\left (-1+e^{x^2}\right ) x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48 + 64*E + 32*x + E^x^2*(48 - 64*E - 32*x + 128*x^2) + (-64 + 64*E^x^2)*Log[-x + E^x^2*x])/(-x^2
- 16*E^2*x^2 - 2*x^3 - x^4 + E*(-8*x^2 - 8*x^3) + E^x^2*(x^2 + 16*E^2*x^2 + 2*x^3 + x^4 + E*(8*x^2 + 8*x^3)) +
(8*x^2 + 32*E*x^2 + 8*x^3 + E^x^2*(-8*x^2 - 32*E*x^2 - 8*x^3))*Log[-x + E^x^2*x] + (-16*x^2 + 16*E^x^2*x^2)*L
og[-x + E^x^2*x]^2),x]

[Out]

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

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fricas [A]  time = 0.48, size = 28, normalized size = 0.93 \begin {gather*} \frac {16}{x^{2} + 4 \, x e - 4 \, x \log \left (x e^{\left (x^{2}\right )} - x\right ) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*exp(x^2)-64)*log(exp(x^2)*x-x)+(-64*exp(1)+128*x^2-32*x+48)*exp(x^2)+64*exp(1)+32*x-48)/((16*x^
2*exp(x^2)-16*x^2)*log(exp(x^2)*x-x)^2+((-32*x^2*exp(1)-8*x^3-8*x^2)*exp(x^2)+32*x^2*exp(1)+8*x^3+8*x^2)*log(e
xp(x^2)*x-x)+(16*x^2*exp(1)^2+(8*x^3+8*x^2)*exp(1)+x^4+2*x^3+x^2)*exp(x^2)-16*x^2*exp(1)^2+(-8*x^3-8*x^2)*exp(
1)-x^4-2*x^3-x^2),x, algorithm="fricas")

[Out]

16/(x^2 + 4*x*e - 4*x*log(x*e^(x^2) - x) + x)

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giac [A]  time = 0.40, size = 28, normalized size = 0.93 \begin {gather*} \frac {16}{x^{2} + 4 \, x e - 4 \, x \log \left (x e^{\left (x^{2}\right )} - x\right ) + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*exp(x^2)-64)*log(exp(x^2)*x-x)+(-64*exp(1)+128*x^2-32*x+48)*exp(x^2)+64*exp(1)+32*x-48)/((16*x^
2*exp(x^2)-16*x^2)*log(exp(x^2)*x-x)^2+((-32*x^2*exp(1)-8*x^3-8*x^2)*exp(x^2)+32*x^2*exp(1)+8*x^3+8*x^2)*log(e
xp(x^2)*x-x)+(16*x^2*exp(1)^2+(8*x^3+8*x^2)*exp(1)+x^4+2*x^3+x^2)*exp(x^2)-16*x^2*exp(1)^2+(-8*x^3-8*x^2)*exp(
1)-x^4-2*x^3-x^2),x, algorithm="giac")

[Out]

16/(x^2 + 4*x*e - 4*x*log(x*e^(x^2) - x) + x)

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maple [C]  time = 0.18, size = 128, normalized size = 4.27

method	result	size
risch	\(-\frac {16 i}{x \left (2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}-1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x^{2}}-1\right )\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x^{2}}-1\right )\right )^{2}-2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}-1\right )\right ) \mathrm {csgn}\left (i x \left ({\mathrm e}^{x^{2}}-1\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i x \left ({\mathrm e}^{x^{2}}-1\right )\right )^{3}-4 i {\mathrm e}-i x +4 i \ln \relax (x )+4 i \ln \left ({\mathrm e}^{x^{2}}-1\right )-i\right )}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*exp(x^2)-64)*ln(exp(x^2)*x-x)+(-64*exp(1)+128*x^2-32*x+48)*exp(x^2)+64*exp(1)+32*x-48)/((16*x^2*exp(x
^2)-16*x^2)*ln(exp(x^2)*x-x)^2+((-32*x^2*exp(1)-8*x^3-8*x^2)*exp(x^2)+32*x^2*exp(1)+8*x^3+8*x^2)*ln(exp(x^2)*x
-x)+(16*x^2*exp(1)^2+(8*x^3+8*x^2)*exp(1)+x^4+2*x^3+x^2)*exp(x^2)-16*x^2*exp(1)^2+(-8*x^3-8*x^2)*exp(1)-x^4-2*
x^3-x^2),x,method=_RETURNVERBOSE)

[Out]

-16*I/x/(2*Pi*csgn(I*x)*csgn(I*(exp(x^2)-1))*csgn(I*x*(exp(x^2)-1))-2*Pi*csgn(I*x)*csgn(I*x*(exp(x^2)-1))^2-2*
Pi*csgn(I*(exp(x^2)-1))*csgn(I*x*(exp(x^2)-1))^2+2*Pi*csgn(I*x*(exp(x^2)-1))^3-4*I*exp(1)-I*x+4*I*ln(x)+4*I*ln
(exp(x^2)-1)-I)

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maxima [A]  time = 0.41, size = 31, normalized size = 1.03 \begin {gather*} \frac {16}{x^{2} + x {\left (4 \, e + 1\right )} - 4 \, x \log \relax (x) - 4 \, x \log \left (e^{\left (x^{2}\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*exp(x^2)-64)*log(exp(x^2)*x-x)+(-64*exp(1)+128*x^2-32*x+48)*exp(x^2)+64*exp(1)+32*x-48)/((16*x^
2*exp(x^2)-16*x^2)*log(exp(x^2)*x-x)^2+((-32*x^2*exp(1)-8*x^3-8*x^2)*exp(x^2)+32*x^2*exp(1)+8*x^3+8*x^2)*log(e
xp(x^2)*x-x)+(16*x^2*exp(1)^2+(8*x^3+8*x^2)*exp(1)+x^4+2*x^3+x^2)*exp(x^2)-16*x^2*exp(1)^2+(-8*x^3-8*x^2)*exp(
1)-x^4-2*x^3-x^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {32\,x+64\,\mathrm {e}+\ln \left (x\,{\mathrm {e}}^{x^2}-x\right )\,\left (64\,{\mathrm {e}}^{x^2}-64\right )-{\mathrm {e}}^{x^2}\,\left (-128\,x^2+32\,x+64\,\mathrm {e}-48\right )-48}{\mathrm {e}\,\left (8\,x^3+8\,x^2\right )-{\ln \left (x\,{\mathrm {e}}^{x^2}-x\right )}^2\,\left (16\,x^2\,{\mathrm {e}}^{x^2}-16\,x^2\right )-\ln \left (x\,{\mathrm {e}}^{x^2}-x\right )\,\left (32\,x^2\,\mathrm {e}-{\mathrm {e}}^{x^2}\,\left (32\,x^2\,\mathrm {e}+8\,x^2+8\,x^3\right )+8\,x^2+8\,x^3\right )-{\mathrm {e}}^{x^2}\,\left (\mathrm {e}\,\left (8\,x^3+8\,x^2\right )+16\,x^2\,{\mathrm {e}}^2+x^2+2\,x^3+x^4\right )+16\,x^2\,{\mathrm {e}}^2+x^2+2\,x^3+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x + 64*exp(1) + log(x*exp(x^2) - x)*(64*exp(x^2) - 64) - exp(x^2)*(32*x + 64*exp(1) - 128*x^2 - 48) -
48)/(exp(1)*(8*x^2 + 8*x^3) - log(x*exp(x^2) - x)^2*(16*x^2*exp(x^2) - 16*x^2) - log(x*exp(x^2) - x)*(32*x^2*
exp(1) - exp(x^2)*(32*x^2*exp(1) + 8*x^2 + 8*x^3) + 8*x^2 + 8*x^3) - exp(x^2)*(exp(1)*(8*x^2 + 8*x^3) + 16*x^2
*exp(2) + x^2 + 2*x^3 + x^4) + 16*x^2*exp(2) + x^2 + 2*x^3 + x^4),x)

[Out]

-int((32*x + 64*exp(1) + log(x*exp(x^2) - x)*(64*exp(x^2) - 64) - exp(x^2)*(32*x + 64*exp(1) - 128*x^2 - 48) -
48)/(exp(1)*(8*x^2 + 8*x^3) - log(x*exp(x^2) - x)^2*(16*x^2*exp(x^2) - 16*x^2) - log(x*exp(x^2) - x)*(32*x^2*
exp(1) - exp(x^2)*(32*x^2*exp(1) + 8*x^2 + 8*x^3) + 8*x^2 + 8*x^3) - exp(x^2)*(exp(1)*(8*x^2 + 8*x^3) + 16*x^2
*exp(2) + x^2 + 2*x^3 + x^4) + 16*x^2*exp(2) + x^2 + 2*x^3 + x^4), x)

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sympy [A]  time = 0.46, size = 27, normalized size = 0.90 \begin {gather*} - \frac {16}{- x^{2} + 4 x \log {\left (x e^{x^{2}} - x \right )} - 4 e x - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*exp(x**2)-64)*ln(exp(x**2)*x-x)+(-64*exp(1)+128*x**2-32*x+48)*exp(x**2)+64*exp(1)+32*x-48)/((16
*x**2*exp(x**2)-16*x**2)*ln(exp(x**2)*x-x)**2+((-32*x**2*exp(1)-8*x**3-8*x**2)*exp(x**2)+32*x**2*exp(1)+8*x**3
+8*x**2)*ln(exp(x**2)*x-x)+(16*x**2*exp(1)**2+(8*x**3+8*x**2)*exp(1)+x**4+2*x**3+x**2)*exp(x**2)-16*x**2*exp(1
)**2+(-8*x**3-8*x**2)*exp(1)-x**4-2*x**3-x**2),x)

[Out]

-16/(-x**2 + 4*x*log(x*exp(x**2) - x) - 4*E*x - x)

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