3.85.3 \(\int \frac {8 e^x+e^{\frac {e^{x^2} x}{8}} (-8 e^x+e^{x^2} (-1-2 x^2))}{-8+8 e^{\frac {e^{x^2} x}{8}}} \, dx\)

Optimal. Leaf size=25 \[ -e^x-\log \left (-4+4 e^{\frac {e^{x^2} x}{8}}\right ) \]

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Rubi [A]  time = 0.61, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6741, 12, 6742, 2194, 6684} \begin {gather*} -\log \left (1-e^{\frac {e^{x^2} x}{8}}\right )-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^x + E^((E^x^2*x)/8)*(-8*E^x + E^x^2*(-1 - 2*x^2)))/(-8 + 8*E^((E^x^2*x)/8)),x]

[Out]

-E^x - Log[1 - E^((E^x^2*x)/8)]

Rule 12

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

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

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Mathematica [A]  time = 0.36, size = 25, normalized size = 1.00 \begin {gather*} -e^x-\log \left (1-e^{\frac {e^{x^2} x}{8}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^x + E^((E^x^2*x)/8)*(-8*E^x + E^x^2*(-1 - 2*x^2)))/(-8 + 8*E^((E^x^2*x)/8)),x]

[Out]

-E^x - Log[1 - E^((E^x^2*x)/8)]

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fricas [A]  time = 0.64, size = 18, normalized size = 0.72 \begin {gather*} -e^{x} - \log \left (e^{\left (\frac {1}{8} \, x e^{\left (x^{2}\right )}\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-1)*exp(x^2)-8*exp(x))*exp(1/16*exp(x^2)*x)^2+8*exp(x))/(8*exp(1/16*exp(x^2)*x)^2-8),x, alg
orithm="fricas")

[Out]

-e^x - log(e^(1/8*x*e^(x^2)) - 1)

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giac [A]  time = 0.16, size = 30, normalized size = 1.20 \begin {gather*} x^{2} - e^{x} - \log \left (e^{\left (x^{2} + \frac {1}{8} \, x e^{\left (x^{2}\right )}\right )} - e^{\left (x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-1)*exp(x^2)-8*exp(x))*exp(1/16*exp(x^2)*x)^2+8*exp(x))/(8*exp(1/16*exp(x^2)*x)^2-8),x, alg
orithm="giac")

[Out]

x^2 - e^x - log(e^(x^2 + 1/8*x*e^(x^2)) - e^(x^2))

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maple [A]  time = 0.06, size = 19, normalized size = 0.76

method	result	size
risch	\(-{\mathrm e}^{x}-\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x^{2}} x}{8}}-1\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^2-1)*exp(x^2)-8*exp(x))*exp(1/16*exp(x^2)*x)^2+8*exp(x))/(8*exp(1/16*exp(x^2)*x)^2-8),x,method=_RE
TURNVERBOSE)

[Out]

-exp(x)-ln(exp(1/8*exp(x^2)*x)-1)

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maxima [A]  time = 0.40, size = 18, normalized size = 0.72 \begin {gather*} -e^{x} - \log \left (e^{\left (\frac {1}{8} \, x e^{\left (x^{2}\right )}\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^2-1)*exp(x^2)-8*exp(x))*exp(1/16*exp(x^2)*x)^2+8*exp(x))/(8*exp(1/16*exp(x^2)*x)^2-8),x, alg
orithm="maxima")

[Out]

-e^x - log(e^(1/8*x*e^(x^2)) - 1)

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mupad [B]  time = 0.18, size = 18, normalized size = 0.72 \begin {gather*} -\ln \left ({\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{x^2}}{8}}-1\right )-{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(x) - exp((x*exp(x^2))/8)*(8*exp(x) + exp(x^2)*(2*x^2 + 1)))/(8*exp((x*exp(x^2))/8) - 8),x)

[Out]

- log(exp((x*exp(x^2))/8) - 1) - exp(x)

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sympy [A]  time = 0.28, size = 29, normalized size = 1.16 \begin {gather*} - \frac {7 x e^{x^{2}}}{64} - e^{x} - \frac {\log {\left (e^{\frac {x e^{x^{2}}}{8}} - 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**2-1)*exp(x**2)-8*exp(x))*exp(1/16*exp(x**2)*x)**2+8*exp(x))/(8*exp(1/16*exp(x**2)*x)**2-8),
x)

[Out]

-7*x*exp(x**2)/64 - exp(x) - log(exp(x*exp(x**2)/8) - 1)/8

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