3.28.47 \(\int (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x (50-50 x-50 x^2)+e^{\frac {2 e^{2+x}}{5}} (50 x+10 e^{2+x} x^2)+e^{\frac {e^{2+x}}{5}} (100 x-150 x^2+e^x (50+50 x)+e^{2+x} (10 e^x x+10 x^2-10 x^3))) \, dx\)

Optimal. Leaf size=34 \[ -x+\left (5 e^x+5 \left (x+e^{\frac {e^{2+x}}{5}} x-x^2\right )\right )^2 \]

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Rubi [B]  time = 0.12, antiderivative size = 83, normalized size of antiderivative = 2.44, number of steps used = 12, number of rules used = 4, integrand size = 115, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {2194, 2196, 2176, 2288} \begin {gather*} 25 x^4-50 x^3+25 e^{\frac {2 e^{x+2}}{5}} x^2-50 e^x x^2+25 x^2+50 e^{\frac {e^{x+2}}{5}} \left (-x^3+x^2+e^x x\right )+50 e^x x-x+25 e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + 50*E^(2*x) + 50*x - 150*x^2 + 100*x^3 + E^x*(50 - 50*x - 50*x^2) + E^((2*E^(2 + x))/5)*(50*x + 10*E^(
2 + x)*x^2) + E^(E^(2 + x)/5)*(100*x - 150*x^2 + E^x*(50 + 50*x) + E^(2 + x)*(10*E^x*x + 10*x^2 - 10*x^3)),x]

[Out]

25*E^(2*x) - x + 50*E^x*x + 25*x^2 + 25*E^((2*E^(2 + x))/5)*x^2 - 50*E^x*x^2 - 50*x^3 + 25*x^4 + 50*E^(E^(2 +
x)/5)*(E^x*x + x^2 - x^3)

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 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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-x+25 x^2-50 x^3+25 x^4+50 \int e^{2 x} \, dx+\int e^x \left (50-50 x-50 x^2\right ) \, dx+\int e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right ) \, dx+\int e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right ) \, dx\\ &=25 e^{2 x}-x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} \left (e^x x+x^2-x^3\right )+\int \left (50 e^x-50 e^x x-50 e^x x^2\right ) \, dx\\ &=25 e^{2 x}-x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} \left (e^x x+x^2-x^3\right )+50 \int e^x \, dx-50 \int e^x x \, dx-50 \int e^x x^2 \, dx\\ &=50 e^x+25 e^{2 x}-x-50 e^x x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 e^x x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} \left (e^x x+x^2-x^3\right )+50 \int e^x \, dx+100 \int e^x x \, dx\\ &=100 e^x+25 e^{2 x}-x+50 e^x x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 e^x x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} \left (e^x x+x^2-x^3\right )-100 \int e^x \, dx\\ &=25 e^{2 x}-x+50 e^x x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 e^x x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} \left (e^x x+x^2-x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.09, size = 78, normalized size = 2.29 \begin {gather*} 25 e^{2 x}-x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} x \left (e^x+x-x^2\right )-50 e^x \left (-x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + 50*E^(2*x) + 50*x - 150*x^2 + 100*x^3 + E^x*(50 - 50*x - 50*x^2) + E^((2*E^(2 + x))/5)*(50*x +
10*E^(2 + x)*x^2) + E^(E^(2 + x)/5)*(100*x - 150*x^2 + E^x*(50 + 50*x) + E^(2 + x)*(10*E^x*x + 10*x^2 - 10*x^3
)),x]

[Out]

25*E^(2*x) - x + 25*x^2 + 25*E^((2*E^(2 + x))/5)*x^2 - 50*x^3 + 25*x^4 + 50*E^(E^(2 + x)/5)*x*(E^x + x - x^2)
- 50*E^x*(-x + x^2)

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fricas [B]  time = 0.78, size = 90, normalized size = 2.65 \begin {gather*} {\left (25 \, x^{2} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )} + 4\right )} + {\left (25 \, x^{4} - 50 \, x^{3} + 25 \, x^{2} - x\right )} e^{4} - 50 \, {\left (x^{2} - x\right )} e^{\left (x + 4\right )} - 50 \, {\left ({\left (x^{3} - x^{2}\right )} e^{4} - x e^{\left (x + 4\right )}\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} + 25 \, e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2)*exp(2+x)+(50*x+50)*exp(x)-15
0*x^2+100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="frica
s")

[Out]

(25*x^2*e^(2/5*e^(x + 2) + 4) + (25*x^4 - 50*x^3 + 25*x^2 - x)*e^4 - 50*(x^2 - x)*e^(x + 4) - 50*((x^3 - x^2)*
e^4 - x*e^(x + 4))*e^(1/5*e^(x + 2)) + 25*e^(2*x + 4))*e^(-4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 100 \, x^{3} - 150 \, x^{2} - 50 \, {\left (x^{2} + x - 1\right )} e^{x} + 10 \, {\left (x^{2} e^{\left (x + 2\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )}\right )} - 10 \, {\left (15 \, x^{2} + {\left (x^{3} - x^{2} - x e^{x}\right )} e^{\left (x + 2\right )} - 5 \, {\left (x + 1\right )} e^{x} - 10 \, x\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} + 50 \, x + 50 \, e^{\left (2 \, x\right )} - 1\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2)*exp(2+x)+(50*x+50)*exp(x)-15
0*x^2+100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="giac"
)

[Out]

integrate(100*x^3 - 150*x^2 - 50*(x^2 + x - 1)*e^x + 10*(x^2*e^(x + 2) + 5*x)*e^(2/5*e^(x + 2)) - 10*(15*x^2 +
(x^3 - x^2 - x*e^x)*e^(x + 2) - 5*(x + 1)*e^x - 10*x)*e^(1/5*e^(x + 2)) + 50*x + 50*e^(2*x) - 1, x)

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maple [B]  time = 0.14, size = 71, normalized size = 2.09

method	result	size
risch	\(25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}-50 x \left (x^{2}-x -{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}}+25 \,{\mathrm e}^{2 x}+\left (-50 x^{2}+50 x \right ) {\mathrm e}^{x}+25 x^{4}-50 x^{3}+25 x^{2}-x\)	\(71\)
default	\(-x +25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}-50 x^{3} {\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}} x^{2}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}} x \,{\mathrm e}^{x}+50 \,{\mathrm e}^{x} x -50 \,{\mathrm e}^{x} x^{2}+25 x^{2}-50 x^{3}+25 x^{4}+25 \,{\mathrm e}^{2 x}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2)*exp(2+x)+(50*x+50)*exp(x)-150*x^2+
100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x,method=_RETURNVERBOSE)

[Out]

25*x^2*exp(2/5*exp(2+x))-50*x*(x^2-x-exp(x))*exp(1/5*exp(2+x))+25*exp(2*x)+(-50*x^2+50*x)*exp(x)+25*x^4-50*x^3
+25*x^2-x

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maxima [B]  time = 0.40, size = 71, normalized size = 2.09 \begin {gather*} 25 \, x^{4} - 50 \, x^{3} + 25 \, x^{2} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )}\right )} + 25 \, x^{2} - 50 \, {\left (x^{2} - x\right )} e^{x} - 50 \, {\left (x^{3} - x^{2} - x e^{x}\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} - x + 25 \, e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2)*exp(2+x)+(50*x+50)*exp(x)-15
0*x^2+100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="maxim
a")

[Out]

25*x^4 - 50*x^3 + 25*x^2*e^(2/5*e^(x + 2)) + 25*x^2 - 50*(x^2 - x)*e^x - 50*(x^3 - x^2 - x*e^x)*e^(1/5*e^(x +
2)) - x + 25*e^(2*x)

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mupad [B]  time = 1.84, size = 85, normalized size = 2.50 \begin {gather*} 25\,{\mathrm {e}}^{2\,x}-x-50\,x^2\,{\mathrm {e}}^x+50\,x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+25\,x^2\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}-50\,x^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+50\,x\,{\mathrm {e}}^x+50\,x\,{\mathrm {e}}^{x+\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+25\,x^2-50\,x^3+25\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(50*x + 50*exp(2*x) + exp((2*exp(x + 2))/5)*(50*x + 10*x^2*exp(x + 2)) + exp(exp(x + 2)/5)*(100*x + exp(x +
2)*(10*x*exp(x) + 10*x^2 - 10*x^3) + exp(x)*(50*x + 50) - 150*x^2) - exp(x)*(50*x + 50*x^2 - 50) - 150*x^2 +
100*x^3 - 1,x)

[Out]

25*exp(2*x) - x - 50*x^2*exp(x) + 50*x^2*exp((exp(2)*exp(x))/5) + 25*x^2*exp((2*exp(2)*exp(x))/5) - 50*x^3*exp
((exp(2)*exp(x))/5) + 50*x*exp(x) + 50*x*exp(x + (exp(2)*exp(x))/5) + 25*x^2 - 50*x^3 + 25*x^4

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sympy [B]  time = 0.32, size = 78, normalized size = 2.29 \begin {gather*} 25 x^{4} - 50 x^{3} + 25 x^{2} e^{\frac {2 e^{2} e^{x}}{5}} + 25 x^{2} - x + \left (- 50 x^{2} + 50 x\right ) e^{x} + \left (- 50 x^{3} + 50 x^{2} + 50 x e^{x}\right ) e^{\frac {e^{2} e^{x}}{5}} + 25 e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*x**2*exp(2+x)+50*x)*exp(1/5*exp(2+x))**2+((10*exp(x)*x-10*x**3+10*x**2)*exp(2+x)+(50*x+50)*exp(x
)-150*x**2+100*x)*exp(1/5*exp(2+x))+50*exp(x)**2+(-50*x**2-50*x+50)*exp(x)+100*x**3-150*x**2+50*x-1,x)

[Out]

25*x**4 - 50*x**3 + 25*x**2*exp(2*exp(2)*exp(x)/5) + 25*x**2 - x + (-50*x**2 + 50*x)*exp(x) + (-50*x**3 + 50*x
**2 + 50*x*exp(x))*exp(exp(2)*exp(x)/5) + 25*exp(2*x)

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