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

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

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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} $$\begin{array}{c}-\log (1-e^{\frac{e^{x^2}x}{8}})-e^x\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-8e^x-e^{\frac{e^{x^2}x}{8}}(-8e^x+e^{x^2}(-1-2x^2))}{8(1-e^{\frac{e^{x^2}x}{8}})}dx\\ & =\frac{1}{8}\int \frac{-8e^x-e^{\frac{e^{x^2}x}{8}}(-8e^x+e^{x^2}(-1-2x^2))}{1-e^{\frac{e^{x^2}x}{8}}}dx\\ & =\frac{1}{8}\int (-8e^x+\frac{e^{\frac{1}{8}x(e^{x^2}+8x)}(1+2x^2)}{1-e^{\frac{e^{x^2}x}{8}}})dx\\ & =\frac{1}{8}\int \frac{e^{\frac{1}{8}x(e^{x^2}+8x)}(1+2x^2)}{1-e^{\frac{e^{x^2}x}{8}}}dx-\int e^{x}dx\\ & =-e^x-\log (1-e^{\frac{e^{x^2}x}{8}})\end{array}\end{array}$$

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

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 $$\begin{array}{c}-e^x-\log (e^{\left(\frac{1}{8}x{e}^{\left(x^2\right)}\right)}-1)\end{array}$$

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 $$\begin{array}{c}{x}^2-e^x-\log (e^{(x^2+\frac{1}{8}x{e}^{\left(x^2\right)})}-e^{\left(x^2\right)})\end{array}$$

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	$-{\mathrm{e}}^x-\ln ({\mathrm{e}}^{\frac{{\mathrm{e}}^{x^2}x}{8}}-1)$	$19$

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 $$\begin{array}{c}-e^x-\log (e^{\left(\frac{1}{8}x{e}^{\left(x^2\right)}\right)}-1)\end{array}$$

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 $$\begin{array}{c}-\ln ({\mathrm{e}}^{\frac{x{\mathrm{e}}^{x^2}}{8}}-1)-{\mathrm{e}}^x\end{array}$$

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 $$\begin{array}{c}-\frac{7x{e}^{x^2}}{64}-e^x-\frac{\log (e^{\frac{x{e}^{x^2}}{8}}-1)}{8}\end{array}$$

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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