3.59.88 \(\int \frac {1}{3} (125-375 e^3+e^{2 x} (-85+e^3 (255-30 x)+10 x)+e^x (-400+e^3 (1200-150 x)+50 x)) \, dx\)

Optimal. Leaf size=26 \[ -\frac {5 \left (5+e^x\right )^2 (-9+x) \left (-\frac {x}{3}+e^3 x\right )}{x} \]

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Rubi [B]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 3.50, number of steps used = 8, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 2187, 2176, 2194} \begin {gather*} -\frac {5}{6} \left (1-3 e^3\right ) e^{2 x} (17-2 x)+\frac {125}{3} \left (1-3 e^3\right ) x-\frac {50}{3} e^x \left (8 \left (1-3 e^3\right )-\left (1-3 e^3\right ) x\right )-\frac {50}{3} \left (1-3 e^3\right ) e^x-\frac {5}{6} \left (1-3 e^3\right ) e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(125 - 375*E^3 + E^(2*x)*(-85 + E^3*(255 - 30*x) + 10*x) + E^x*(-400 + E^3*(1200 - 150*x) + 50*x))/3,x]

[Out]

(-50*E^x*(1 - 3*E^3))/3 - (5*E^(2*x)*(1 - 3*E^3))/6 - (5*E^(2*x)*(1 - 3*E^3)*(17 - 2*x))/6 + (125*(1 - 3*E^3)*
x)/3 - (50*E^x*(8*(1 - 3*E^3) - (1 - 3*E^3)*x))/3

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}{3} \int \left (125-375 e^3+e^{2 x} \left (-85+e^3 (255-30 x)+10 x\right )+e^x \left (-400+e^3 (1200-150 x)+50 x\right )\right ) \, dx\\ &=\frac {125}{3} \left (1-3 e^3\right ) x+\frac {1}{3} \int e^{2 x} \left (-85+e^3 (255-30 x)+10 x\right ) \, dx+\frac {1}{3} \int e^x \left (-400+e^3 (1200-150 x)+50 x\right ) \, dx\\ &=\frac {125}{3} \left (1-3 e^3\right ) x+\frac {1}{3} \int e^{2 x} \left (-85 \left (1-3 e^3\right )+10 \left (1-3 e^3\right ) x\right ) \, dx+\frac {1}{3} \int e^x \left (-400 \left (1-3 e^3\right )+50 \left (1-3 e^3\right ) x\right ) \, dx\\ &=-\frac {5}{6} e^{2 x} \left (1-3 e^3\right ) (17-2 x)+\frac {125}{3} \left (1-3 e^3\right ) x-\frac {50}{3} e^x \left (8 \left (1-3 e^3\right )-\left (1-3 e^3\right ) x\right )-\frac {1}{3} \left (5 \left (1-3 e^3\right )\right ) \int e^{2 x} \, dx-\frac {1}{3} \left (50 \left (1-3 e^3\right )\right ) \int e^x \, dx\\ &=-\frac {50}{3} e^x \left (1-3 e^3\right )-\frac {5}{6} e^{2 x} \left (1-3 e^3\right )-\frac {5}{6} e^{2 x} \left (1-3 e^3\right ) (17-2 x)+\frac {125}{3} \left (1-3 e^3\right ) x-\frac {50}{3} e^x \left (8 \left (1-3 e^3\right )-\left (1-3 e^3\right ) x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 33, normalized size = 1.27 \begin {gather*} -\frac {5}{3} \left (-1+3 e^3\right ) \left (e^{2 x} (-9+x)+25 x+e^x (-90+10 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(125 - 375*E^3 + E^(2*x)*(-85 + E^3*(255 - 30*x) + 10*x) + E^x*(-400 + E^3*(1200 - 150*x) + 50*x))/3
,x]

[Out]

(-5*(-1 + 3*E^3)*(E^(2*x)*(-9 + x) + 25*x + E^x*(-90 + 10*x)))/3

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fricas [A]  time = 0.55, size = 43, normalized size = 1.65 \begin {gather*} -125 \, x e^{3} - \frac {5}{3} \, {\left (3 \, {\left (x - 9\right )} e^{3} - x + 9\right )} e^{\left (2 \, x\right )} - \frac {50}{3} \, {\left (3 \, {\left (x - 9\right )} e^{3} - x + 9\right )} e^{x} + \frac {125}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-30*x+255)*exp(3)+10*x-85)*exp(x)^2+1/3*((-150*x+1200)*exp(3)+50*x-400)*exp(x)-125*exp(3)+125/
3,x, algorithm="fricas")

[Out]

-125*x*e^3 - 5/3*(3*(x - 9)*e^3 - x + 9)*e^(2*x) - 50/3*(3*(x - 9)*e^3 - x + 9)*e^x + 125/3*x

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giac [A]  time = 0.15, size = 45, normalized size = 1.73 \begin {gather*} -125 \, x e^{3} + \frac {5}{3} \, {\left (x - 9\right )} e^{\left (2 \, x\right )} - 5 \, {\left (x - 9\right )} e^{\left (2 \, x + 3\right )} - 50 \, {\left (x - 9\right )} e^{\left (x + 3\right )} + \frac {50}{3} \, {\left (x - 9\right )} e^{x} + \frac {125}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-30*x+255)*exp(3)+10*x-85)*exp(x)^2+1/3*((-150*x+1200)*exp(3)+50*x-400)*exp(x)-125*exp(3)+125/
3,x, algorithm="giac")

[Out]

-125*x*e^3 + 5/3*(x - 9)*e^(2*x) - 5*(x - 9)*e^(2*x + 3) - 50*(x - 9)*e^(x + 3) + 50/3*(x - 9)*e^x + 125/3*x

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maple [B]  time = 0.04, size = 48, normalized size = 1.85

method	result	size
risch	\(\frac {\left (-15 x \,{\mathrm e}^{3}+135 \,{\mathrm e}^{3}+5 x -45\right ) {\mathrm e}^{2 x}}{3}+\frac {\left (-150 x \,{\mathrm e}^{3}+1350 \,{\mathrm e}^{3}+50 x -450\right ) {\mathrm e}^{x}}{3}-125 x \,{\mathrm e}^{3}+\frac {125 x}{3}\)	\(48\)
norman	\(\left (-150+450 \,{\mathrm e}^{3}\right ) {\mathrm e}^{x}+\left (-15+45 \,{\mathrm e}^{3}\right ) {\mathrm e}^{2 x}+\left (-125 \,{\mathrm e}^{3}+\frac {125}{3}\right ) x +\left (-50 \,{\mathrm e}^{3}+\frac {50}{3}\right ) x \,{\mathrm e}^{x}+\left (-5 \,{\mathrm e}^{3}+\frac {5}{3}\right ) x \,{\mathrm e}^{2 x}\)	\(52\)
default	\(\frac {125 x}{3}+400 \,{\mathrm e}^{x} {\mathrm e}^{3}+\frac {50 \,{\mathrm e}^{x} x}{3}-150 \,{\mathrm e}^{x}-50 \,{\mathrm e}^{3} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-15 \,{\mathrm e}^{2 x}+\frac {5 x \,{\mathrm e}^{2 x}}{3}+\frac {85 \,{\mathrm e}^{3} {\mathrm e}^{2 x}}{2}-10 \,{\mathrm e}^{3} \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {{\mathrm e}^{2 x}}{4}\right )-125 x \,{\mathrm e}^{3}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-30*x+255)*exp(3)+10*x-85)*exp(x)^2+1/3*((-150*x+1200)*exp(3)+50*x-400)*exp(x)-125*exp(3)+125/3,x,me
thod=_RETURNVERBOSE)

[Out]

1/3*(-15*x*exp(3)+135*exp(3)+5*x-45)*exp(2*x)+1/3*(-150*x*exp(3)+1350*exp(3)+50*x-450)*exp(x)-125*x*exp(3)+125
/3*x

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maxima [B]  time = 0.37, size = 47, normalized size = 1.81 \begin {gather*} -125 \, x e^{3} - \frac {5}{3} \, {\left (x {\left (3 \, e^{3} - 1\right )} - 27 \, e^{3} + 9\right )} e^{\left (2 \, x\right )} - \frac {50}{3} \, {\left (x {\left (3 \, e^{3} - 1\right )} - 27 \, e^{3} + 9\right )} e^{x} + \frac {125}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-30*x+255)*exp(3)+10*x-85)*exp(x)^2+1/3*((-150*x+1200)*exp(3)+50*x-400)*exp(x)-125*exp(3)+125/
3,x, algorithm="maxima")

[Out]

-125*x*e^3 - 5/3*(x*(3*e^3 - 1) - 27*e^3 + 9)*e^(2*x) - 50/3*(x*(3*e^3 - 1) - 27*e^3 + 9)*e^x + 125/3*x

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mupad [B]  time = 0.11, size = 33, normalized size = 1.27 \begin {gather*} -\left (5\,{\mathrm {e}}^3-\frac {5}{3}\right )\,\left (25\,x-9\,{\mathrm {e}}^{2\,x}-90\,{\mathrm {e}}^x+x\,{\mathrm {e}}^{2\,x}+10\,x\,{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(125/3 - (exp(x)*(exp(3)*(150*x - 1200) - 50*x + 400))/3 - (exp(2*x)*(exp(3)*(30*x - 255) - 10*x + 85))/3 -
125*exp(3),x)

[Out]

-(5*exp(3) - 5/3)*(25*x - 9*exp(2*x) - 90*exp(x) + x*exp(2*x) + 10*x*exp(x))

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sympy [B]  time = 0.17, size = 54, normalized size = 2.08 \begin {gather*} x \left (\frac {125}{3} - 125 e^{3}\right ) + \frac {\left (- 450 x e^{3} + 150 x - 1350 + 4050 e^{3}\right ) e^{x}}{9} + \frac {\left (- 45 x e^{3} + 15 x - 135 + 405 e^{3}\right ) e^{2 x}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-30*x+255)*exp(3)+10*x-85)*exp(x)**2+1/3*((-150*x+1200)*exp(3)+50*x-400)*exp(x)-125*exp(3)+125
/3,x)

[Out]

x*(125/3 - 125*exp(3)) + (-450*x*exp(3) + 150*x - 1350 + 4050*exp(3))*exp(x)/9 + (-45*x*exp(3) + 15*x - 135 +
405*exp(3))*exp(2*x)/9

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