\(\int e^{-2 x} (150 x+20 e^{2 x} x-150 x^2+e^{3 x} (1+10 x+5 x^2)) \, dx\) [9285]

Optimal result

Integrand size = 39, antiderivative size = 22 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=-6+e^x+5 \left (2+15 e^{-2 x}+e^x\right ) x^2 \]

[Out]

exp(x)-6+5*(15/exp(x)^2+exp(x)+2)*x^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23, number of steps used = 17, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6820, 2227, 2207, 2225} \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=75 e^{-2 x} x^2+5 e^x x^2+10 x^2+e^x \]

[In]

Int[(150*x + 20*E^(2*x)*x - 150*x^2 + E^(3*x)*(1 + 10*x + 5*x^2))/E^(2*x),x]

[Out]

E^x + 10*x^2 + (75*x^2)/E^(2*x) + 5*E^x*x^2

Rule 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2227

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] &&  !TrueQ[$UseGamma]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (20 x-150 e^{-2 x} (-1+x) x+e^x \left (1+10 x+5 x^2\right )\right ) \, dx \\ & = 10 x^2-150 \int e^{-2 x} (-1+x) x \, dx+\int e^x \left (1+10 x+5 x^2\right ) \, dx \\ & = 10 x^2-150 \int \left (-e^{-2 x} x+e^{-2 x} x^2\right ) \, dx+\int \left (e^x+10 e^x x+5 e^x x^2\right ) \, dx \\ & = 10 x^2+5 \int e^x x^2 \, dx+10 \int e^x x \, dx+150 \int e^{-2 x} x \, dx-150 \int e^{-2 x} x^2 \, dx+\int e^x \, dx \\ & = e^x-75 e^{-2 x} x+10 e^x x+10 x^2+75 e^{-2 x} x^2+5 e^x x^2-10 \int e^x \, dx-10 \int e^x x \, dx+75 \int e^{-2 x} \, dx-150 \int e^{-2 x} x \, dx \\ & = -\frac {75}{2} e^{-2 x}-9 e^x+10 x^2+75 e^{-2 x} x^2+5 e^x x^2+10 \int e^x \, dx-75 \int e^{-2 x} \, dx \\ & = e^x+10 x^2+75 e^{-2 x} x^2+5 e^x x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=10 x^2+75 e^{-2 x} x^2+e^x \left (1+5 x^2\right ) \]

[In]

Integrate[(150*x + 20*E^(2*x)*x - 150*x^2 + E^(3*x)*(1 + 10*x + 5*x^2))/E^(2*x),x]

[Out]

10*x^2 + (75*x^2)/E^(2*x) + E^x*(1 + 5*x^2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14

method	result	size
default	\(5 \,{\mathrm e}^{x} x^{2}+75 \,{\mathrm e}^{-2 x} x^{2}+10 x^{2}+{\mathrm e}^{x}\)	\(25\)
parts	\(5 \,{\mathrm e}^{x} x^{2}+75 \,{\mathrm e}^{-2 x} x^{2}+10 x^{2}+{\mathrm e}^{x}\)	\(25\)
risch	\(10 x^{2}+\left (5 x^{2}+1\right ) {\mathrm e}^{x}+75 \,{\mathrm e}^{-2 x} x^{2}\)	\(26\)
norman	\(\left ({\mathrm e}^{3 x}+75 x^{2}+5 x^{2} {\mathrm e}^{3 x}+10 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-2 x}\)	\(34\)
parallelrisch	\(\left ({\mathrm e}^{3 x}+75 x^{2}+5 x^{2} {\mathrm e}^{3 x}+10 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-2 x}\)	\(34\)

[In]

int(((5*x^2+10*x+1)*exp(x)^3+20*x*exp(x)^2-150*x^2+150*x)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

10*x^2+75*x^2/exp(x)^2+5*exp(x)*x^2+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx={\left (10 \, x^{2} e^{\left (2 \, x\right )} + 75 \, x^{2} + {\left (5 \, x^{2} + 1\right )} e^{\left (3 \, x\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate(((5*x^2+10*x+1)*exp(x)^3+20*x*exp(x)^2-150*x^2+150*x)/exp(x)^2,x, algorithm="fricas")

[Out]

(10*x^2*e^(2*x) + 75*x^2 + (5*x^2 + 1)*e^(3*x))*e^(-2*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=10 x^{2} + 75 x^{2} e^{- 2 x} + \left (5 x^{2} + 1\right ) e^{x} \]

[In]

integrate(((5*x**2+10*x+1)*exp(x)**3+20*x*exp(x)**2-150*x**2+150*x)/exp(x)**2,x)

[Out]

10*x**2 + 75*x**2*exp(-2*x) + (5*x**2 + 1)*exp(x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=10 \, x^{2} + \frac {75}{2} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {75}{2} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + 5 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 10 \, {\left (x - 1\right )} e^{x} + e^{x} \]

[In]

integrate(((5*x^2+10*x+1)*exp(x)^3+20*x*exp(x)^2-150*x^2+150*x)/exp(x)^2,x, algorithm="maxima")

[Out]

10*x^2 + 75/2*(2*x^2 + 2*x + 1)*e^(-2*x) - 75/2*(2*x + 1)*e^(-2*x) + 5*(x^2 - 2*x + 2)*e^x + 10*(x - 1)*e^x +
e^x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=75 \, x^{2} e^{\left (-2 \, x\right )} + 10 \, x^{2} + {\left (5 \, x^{2} + 1\right )} e^{x} \]

[In]

integrate(((5*x^2+10*x+1)*exp(x)^3+20*x*exp(x)^2-150*x^2+150*x)/exp(x)^2,x, algorithm="giac")

[Out]

75*x^2*e^(-2*x) + 10*x^2 + (5*x^2 + 1)*e^x

Mupad [B] (verification not implemented)

Time = 13.51 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int e^{-2 x} \left (150 x+20 e^{2 x} x-150 x^2+e^{3 x} \left (1+10 x+5 x^2\right )\right ) \, dx=75\,x^2\,{\mathrm {e}}^{-2\,x}+{\mathrm {e}}^x\,\left (5\,x^2+1\right )+10\,x^2 \]

[In]

int(exp(-2*x)*(150*x + exp(3*x)*(10*x + 5*x^2 + 1) + 20*x*exp(2*x) - 150*x^2),x)

[Out]

75*x^2*exp(-2*x) + exp(x)*(5*x^2 + 1) + 10*x^2