3.17.17 $\int \frac{1}{2}(20+e^{\frac{1}{2}(-x+2e^{2}x-4x^2)}(1-2e^2+8x))dx$

Optimal. Leaf size=30 $$-e^{-x+x(e^2-(2-\frac{1}{2x})x)}+10x$$

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Rubi [A]  time = 0.06, antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 38, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.079, Rules used = {12, 2244, 2236} $$\begin{array}{c}10x-e^{-2x^2-\frac{1}{2}(1-2e^2)x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(20 + E^((-x + 2*E^2*x - 4*x^2)/2)*(1 - 2*E^2 + 8*x))/2,x]

[Out]

-E^(-1/2*((1 - 2*E^2)*x) - 2*x^2) + 10*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 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int (20+e^{\frac{1}{2}(-x+2e^{2}x-4x^2)}(1-2e^2+8x))dx\\ & =10x+\frac{1}{2}\int e^{\frac{1}{2}(-x+2e^{2}x-4x^2)}(1-2e^2+8x)dx\\ & =10x+\frac{1}{2}\int e^{\frac{1}{2}(-1+2e^2)x-2x^2}(1-2e^2+8x)dx\\ & =-e^{-\frac{1}{2}(1-2e^2)x-2x^2}+10x\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 26, normalized size = 0.87 $$\begin{array}{c}-e^{\frac{1}{2}(-1+2e^2)x-2x^2}+10x\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(20 + E^((-x + 2*E^2*x - 4*x^2)/2)*(1 - 2*E^2 + 8*x))/2,x]

[Out]

-E^(((-1 + 2*E^2)*x)/2 - 2*x^2) + 10*x

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fricas [A]  time = 0.87, size = 20, normalized size = 0.67 $$\begin{array}{c}10x-e^{(-2x^2+x{e}^2-\frac{1}{2}x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*exp(2)+8*x+1)*exp(exp(2)*x-2*x^2-1/2*x)+10,x, algorithm="fricas")

[Out]

10*x - e^(-2*x^2 + x*e^2 - 1/2*x)

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giac [A]  time = 0.26, size = 20, normalized size = 0.67 $$\begin{array}{c}10x-e^{(-2x^2+x{e}^2-\frac{1}{2}x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*exp(2)+8*x+1)*exp(exp(2)*x-2*x^2-1/2*x)+10,x, algorithm="giac")

[Out]

10*x - e^(-2*x^2 + x*e^2 - 1/2*x)

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maple [A]  time = 0.04, size = 20, normalized size = 0.67

method	result	size
risch	$10x-{\mathrm{e}}^{\frac{x(2{\mathrm{e}}^2-4x-1)}{2}}$	$20$
norman	$10x-{\mathrm{e}}^{{\mathrm{e}}^{2}x-2x^2-\frac{x}{2}}$	$21$
default	$10x+\frac{\sqrt{\pi }{\mathrm{e}}^{\frac{{({\mathrm{e}}^2-\frac{1}{2})}^2}{8}}\sqrt{2}\mathrm{erf}(\sqrt{2}x-\frac{({\mathrm{e}}^2-\frac{1}{2})\sqrt{2}}{4})}{8}-{\mathrm{e}}^{-2x^2+({\mathrm{e}}^2-\frac{1}{2})x}+\frac{({\mathrm{e}}^2-\frac{1}{2})\sqrt{\pi }{\mathrm{e}}^{\frac{{({\mathrm{e}}^2-\frac{1}{2})}^2}{8}}\sqrt{2}\mathrm{erf}(\sqrt{2}x-\frac{({\mathrm{e}}^2-\frac{1}{2})\sqrt{2}}{4})}{4}-\frac{\sqrt{\pi }{\mathrm{e}}^{2+\frac{{({\mathrm{e}}^2-\frac{1}{2})}^2}{8}}\sqrt{2}\mathrm{erf}(\sqrt{2}x-\frac{({\mathrm{e}}^2-\frac{1}{2})\sqrt{2}}{4})}{4}$	$125$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-2*exp(2)+8*x+1)*exp(exp(2)*x-2*x^2-1/2*x)+10,x,method=_RETURNVERBOSE)

[Out]

10*x-exp(1/2*x*(2*exp(2)-4*x-1))

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maxima [A]  time = 0.42, size = 20, normalized size = 0.67 $$\begin{array}{c}10x-e^{(-2x^2+x{e}^2-\frac{1}{2}x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*exp(2)+8*x+1)*exp(exp(2)*x-2*x^2-1/2*x)+10,x, algorithm="maxima")

[Out]

10*x - e^(-2*x^2 + x*e^2 - 1/2*x)

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mupad [B]  time = 1.16, size = 20, normalized size = 0.67 $$\begin{array}{c}10x-{\mathrm{e}}^{x{\mathrm{e}}^2-\frac{x}{2}-2x^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*exp(2) - x/2 - 2*x^2)*(8*x - 2*exp(2) + 1))/2 + 10,x)

[Out]

10*x - exp(x*exp(2) - x/2 - 2*x^2)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.57 $$\begin{array}{c}10x-e^{-2x^2-\frac{x}{2}+x{e}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-2*exp(2)+8*x+1)*exp(exp(2)*x-2*x**2-1/2*x)+10,x)

[Out]

10*x - exp(-2*x**2 - x/2 + x*exp(2))

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