3.16.35 \(\int \frac {1}{4} (121-44 e^3+4 e^6+e^{-6+2 x} (1+2 x)+e^{-3+x} (22+e^3 (-4-4 x)+22 x)) \, dx\)

Optimal. Leaf size=31 \[ x \left (e^3+\frac {1}{2} \left (-1-e^{-3+x}+2 (-5-x)+2 x\right )\right )^2 \]

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Rubi [B]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 2.68, number of steps used = 7, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 2176, 2194, 2187} \begin {gather*} \frac {1}{4} \left (11-2 e^3\right )^2 x-\frac {1}{8} e^{2 x-6}+\frac {1}{8} e^{2 x-6} (2 x+1)+\frac {1}{2} e^{x-3} \left (\left (11-2 e^3\right ) x-2 e^3+11\right )-\frac {1}{2} \left (11-2 e^3\right ) e^{x-3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(121 - 44*E^3 + 4*E^6 + E^(-6 + 2*x)*(1 + 2*x) + E^(-3 + x)*(22 + E^3*(-4 - 4*x) + 22*x))/4,x]

[Out]

-1/8*E^(-6 + 2*x) - (E^(-3 + x)*(11 - 2*E^3))/2 + ((11 - 2*E^3)^2*x)/4 + (E^(-6 + 2*x)*(1 + 2*x))/8 + (E^(-3 +
x)*(11 - 2*E^3 + (11 - 2*E^3)*x))/2

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \left (121-44 e^3+4 e^6+e^{-6+2 x} (1+2 x)+e^{-3+x} \left (22+e^3 (-4-4 x)+22 x\right )\right ) \, dx\\ &=\frac {1}{4} \left (11-2 e^3\right )^2 x+\frac {1}{4} \int e^{-6+2 x} (1+2 x) \, dx+\frac {1}{4} \int e^{-3+x} \left (22+e^3 (-4-4 x)+22 x\right ) \, dx\\ &=\frac {1}{4} \left (11-2 e^3\right )^2 x+\frac {1}{8} e^{-6+2 x} (1+2 x)-\frac {1}{4} \int e^{-6+2 x} \, dx+\frac {1}{4} \int e^{-3+x} \left (2 \left (11-2 e^3\right )+2 \left (11-2 e^3\right ) x\right ) \, dx\\ &=-\frac {1}{8} e^{-6+2 x}+\frac {1}{4} \left (11-2 e^3\right )^2 x+\frac {1}{8} e^{-6+2 x} (1+2 x)+\frac {1}{2} e^{-3+x} \left (11-2 e^3+\left (11-2 e^3\right ) x\right )+\frac {1}{2} \left (-11+2 e^3\right ) \int e^{-3+x} \, dx\\ &=-\frac {1}{8} e^{-6+2 x}-\frac {1}{2} e^{-3+x} \left (11-2 e^3\right )+\frac {1}{4} \left (11-2 e^3\right )^2 x+\frac {1}{8} e^{-6+2 x} (1+2 x)+\frac {1}{2} e^{-3+x} \left (11-2 e^3+\left (11-2 e^3\right ) x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 24, normalized size = 0.77 \begin {gather*} \frac {\left (11 e^3-2 e^6+e^x\right )^2 x}{4 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(121 - 44*E^3 + 4*E^6 + E^(-6 + 2*x)*(1 + 2*x) + E^(-3 + x)*(22 + E^3*(-4 - 4*x) + 22*x))/4,x]

[Out]

((11*E^3 - 2*E^6 + E^x)^2*x)/(4*E^6)

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fricas [B]  time = 0.75, size = 37, normalized size = 1.19 \begin {gather*} x e^{6} - 11 \, x e^{3} + \frac {1}{4} \, x e^{\left (2 \, x - 6\right )} - \frac {1}{2} \, {\left (2 \, x e^{3} - 11 \, x\right )} e^{\left (x - 3\right )} + \frac {121}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)*exp(x-3)^2+1/4*((-4*x-4)*exp(3)+22*x+22)*exp(x-3)+exp(3)^2-11*exp(3)+121/4,x, algorithm=
"fricas")

[Out]

x*e^6 - 11*x*e^3 + 1/4*x*e^(2*x - 6) - 1/2*(2*x*e^3 - 11*x)*e^(x - 3) + 121/4*x

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giac [A]  time = 0.16, size = 34, normalized size = 1.10 \begin {gather*} x e^{6} - 11 \, x e^{3} + \frac {1}{4} \, x e^{\left (2 \, x - 6\right )} + \frac {11}{2} \, x e^{\left (x - 3\right )} - x e^{x} + \frac {121}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)*exp(x-3)^2+1/4*((-4*x-4)*exp(3)+22*x+22)*exp(x-3)+exp(3)^2-11*exp(3)+121/4,x, algorithm=
"giac")

[Out]

x*e^6 - 11*x*e^3 + 1/4*x*e^(2*x - 6) + 11/2*x*e^(x - 3) - x*e^x + 121/4*x

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maple [B]  time = 0.07, size = 35, normalized size = 1.13

method	result	size
norman	\(\left ({\mathrm e}^{6}-11 \,{\mathrm e}^{3}+\frac {121}{4}\right ) x +\left (-{\mathrm e}^{3}+\frac {11}{2}\right ) x \,{\mathrm e}^{x -3}+\frac {x \,{\mathrm e}^{2 x -6}}{4}\)	\(35\)
risch	\(\frac {x \,{\mathrm e}^{2 x -6}}{4}-\frac {\left (2 \,{\mathrm e}^{3}-11\right ) x \,{\mathrm e}^{x -3}}{2}+x \,{\mathrm e}^{6}-11 x \,{\mathrm e}^{3}+\frac {121 x}{4}\)	\(36\)
default	\(\frac {121 x}{4}+x \,{\mathrm e}^{6}+\frac {3 \,{\mathrm e}^{2 x -6}}{4}+\frac {{\mathrm e}^{2 x -6} \left (x -3\right )}{4}+\frac {11 \,{\mathrm e}^{x -3} \left (x -3\right )}{2}+\frac {33 \,{\mathrm e}^{x -3}}{2}-4 \,{\mathrm e}^{3} {\mathrm e}^{x -3}-{\mathrm e}^{3} \left ({\mathrm e}^{x -3} \left (x -3\right )-{\mathrm e}^{x -3}\right )-11 x \,{\mathrm e}^{3}\)	\(77\)
derivativedivides	\(\frac {121 x}{4}-\frac {363}{4}+{\mathrm e}^{6} \left (x -3\right )+\frac {3 \,{\mathrm e}^{2 x -6}}{4}+\frac {{\mathrm e}^{2 x -6} \left (x -3\right )}{4}+\frac {11 \,{\mathrm e}^{x -3} \left (x -3\right )}{2}+\frac {33 \,{\mathrm e}^{x -3}}{2}-4 \,{\mathrm e}^{3} {\mathrm e}^{x -3}-{\mathrm e}^{3} \left ({\mathrm e}^{x -3} \left (x -3\right )-{\mathrm e}^{x -3}\right )-11 \,{\mathrm e}^{3} \left (x -3\right )\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(2*x+1)*exp(x-3)^2+1/4*((-4*x-4)*exp(3)+22*x+22)*exp(x-3)+exp(3)^2-11*exp(3)+121/4,x,method=_RETURNVER
BOSE)

[Out]

(exp(3)^2-11*exp(3)+121/4)*x+(-exp(3)+11/2)*x*exp(x-3)+1/4*x*exp(x-3)^2

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maxima [B]  time = 0.46, size = 35, normalized size = 1.13 \begin {gather*} -\frac {1}{2} \, x {\left (2 \, e^{3} - 11\right )} e^{\left (x - 3\right )} + x e^{6} - 11 \, x e^{3} + \frac {1}{4} \, x e^{\left (2 \, x - 6\right )} + \frac {121}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)*exp(x-3)^2+1/4*((-4*x-4)*exp(3)+22*x+22)*exp(x-3)+exp(3)^2-11*exp(3)+121/4,x, algorithm=
"maxima")

[Out]

-1/2*x*(2*e^3 - 11)*e^(x - 3) + x*e^6 - 11*x*e^3 + 1/4*x*e^(2*x - 6) + 121/4*x

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mupad [B]  time = 0.36, size = 15, normalized size = 0.48 \begin {gather*} \frac {x\,{\left ({\mathrm {e}}^{x-3}-2\,{\mathrm {e}}^3+11\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(6) - 11*exp(3) + (exp(2*x - 6)*(2*x + 1))/4 + (exp(x - 3)*(22*x - exp(3)*(4*x + 4) + 22))/4 + 121/4,x)

[Out]

(x*(exp(x - 3) - 2*exp(3) + 11)^2)/4

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sympy [B]  time = 0.15, size = 39, normalized size = 1.26 \begin {gather*} \frac {x e^{2 x - 6}}{4} + x \left (- 11 e^{3} + \frac {121}{4} + e^{6}\right ) + \frac {\left (- 8 x e^{3} + 44 x\right ) e^{x - 3}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(2*x+1)*exp(x-3)**2+1/4*((-4*x-4)*exp(3)+22*x+22)*exp(x-3)+exp(3)**2-11*exp(3)+121/4,x)

[Out]

x*exp(2*x - 6)/4 + x*(-11*exp(3) + 121/4 + exp(6)) + (-8*x*exp(3) + 44*x)*exp(x - 3)/8

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