\(\int \frac {1}{4} e^{-x} (30 e^x x+40 x^3-10 x^4+e^5 (15 x^2-5 x^3)) \, dx\) [6721]

Optimal result

Integrand size = 41, antiderivative size = 28 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=5 x \left (\frac {3 x}{4}+\frac {1}{4} e^{-x} x^2 \left (e^5+2 x\right )\right ) \]

[Out]

x*(15/4*x+5/4*x^2*(2*x+exp(5))/exp(x))

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(163\) vs. \(2(28)=56\).

Time = 0.33 (sec) , antiderivative size = 163, normalized size of antiderivative = 5.82, number of steps used = 20, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {12, 6873, 6820, 6874, 2227, 2207, 2225} \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {5}{2} e^{-x} x^4+10 e^{-x} x^3-\frac {5}{4} \left (8-e^5\right ) e^{-x} x^3-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} \left (8-e^5\right ) e^{-x} x^2+\frac {15 x^2}{4}-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} \left (8-e^5\right ) e^{-x} x-\frac {15 e^{5-x}}{2}+60 e^{-x}-\frac {15}{2} \left (8-e^5\right ) e^{-x} \]

[In]

Int[(30*E^x*x + 40*x^3 - 10*x^4 + E^5*(15*x^2 - 5*x^3))/(4*E^x),x]

[Out]

(-15*E^(5 - x))/2 + 60/E^x - (15*(8 - E^5))/(2*E^x) - (15*E^(5 - x)*x)/2 + (60*x)/E^x - (15*(8 - E^5)*x)/(2*E^
x) + (15*x^2)/4 - (15*E^(5 - x)*x^2)/4 + (30*x^2)/E^x - (15*(8 - E^5)*x^2)/(4*E^x) + (10*x^3)/E^x - (5*(8 - E^
5)*x^3)/(4*E^x) + (5*x^4)/(2*E^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 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]]

Rule 6873

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx \\ & = \frac {1}{4} \int 5 e^{-x} x \left (6 e^x+3 e^5 x+8 \left (1-\frac {e^5}{8}\right ) x^2-2 x^3\right ) \, dx \\ & = \frac {5}{4} \int e^{-x} x \left (6 e^x+3 e^5 x+8 \left (1-\frac {e^5}{8}\right ) x^2-2 x^3\right ) \, dx \\ & = \frac {5}{4} \int e^{-x} x \left (6 e^x-e^5 (-3+x) x-2 (-4+x) x^2\right ) \, dx \\ & = \frac {5}{4} \int \left (6 x+e^{-x} x^2 \left (3 e^5+\left (8-e^5\right ) x-2 x^2\right )\right ) \, dx \\ & = \frac {15 x^2}{4}+\frac {5}{4} \int e^{-x} x^2 \left (3 e^5+\left (8-e^5\right ) x-2 x^2\right ) \, dx \\ & = \frac {15 x^2}{4}+\frac {5}{4} \int \left (3 e^{5-x} x^2-e^{-x} \left (-8+e^5\right ) x^3-2 e^{-x} x^4\right ) \, dx \\ & = \frac {15 x^2}{4}-\frac {5}{2} \int e^{-x} x^4 \, dx+\frac {15}{4} \int e^{5-x} x^2 \, dx+\frac {1}{4} \left (5 \left (8-e^5\right )\right ) \int e^{-x} x^3 \, dx \\ & = \frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4+\frac {15}{2} \int e^{5-x} x \, dx-10 \int e^{-x} x^3 \, dx+\frac {1}{4} \left (15 \left (8-e^5\right )\right ) \int e^{-x} x^2 \, dx \\ & = -\frac {15}{2} e^{5-x} x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4+\frac {15}{2} \int e^{5-x} \, dx-30 \int e^{-x} x^2 \, dx+\frac {1}{2} \left (15 \left (8-e^5\right )\right ) \int e^{-x} x \, dx \\ & = -\frac {15 e^{5-x}}{2}-\frac {15}{2} e^{5-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4-60 \int e^{-x} x \, dx+\frac {1}{2} \left (15 \left (8-e^5\right )\right ) \int e^{-x} \, dx \\ & = -\frac {15 e^{5-x}}{2}-\frac {15}{2} e^{-x} \left (8-e^5\right )-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4-60 \int e^{-x} \, dx \\ & = -\frac {15 e^{5-x}}{2}+60 e^{-x}-\frac {15}{2} e^{-x} \left (8-e^5\right )-\frac {15}{2} e^{5-x} x+60 e^{-x} x-\frac {15}{2} e^{-x} \left (8-e^5\right ) x+\frac {15 x^2}{4}-\frac {15}{4} e^{5-x} x^2+30 e^{-x} x^2-\frac {15}{4} e^{-x} \left (8-e^5\right ) x^2+10 e^{-x} x^3-\frac {5}{4} e^{-x} \left (8-e^5\right ) x^3+\frac {5}{2} e^{-x} x^4 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {5}{4} \left (3 x^2+e^{-x} x^2 \left (e^5 x+2 x^2\right )\right ) \]

[In]

Integrate[(30*E^x*x + 40*x^3 - 10*x^4 + E^5*(15*x^2 - 5*x^3))/(4*E^x),x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
norman	\(\left (\frac {5 x^{4}}{2}+\frac {5 x^{3} {\mathrm e}^{5}}{4}+\frac {15 \,{\mathrm e}^{x} x^{2}}{4}\right ) {\mathrm e}^{-x}\)	\(26\)
risch	\(\frac {15 x^{2}}{4}+\frac {\left (5 x^{3} {\mathrm e}^{5}+10 x^{4}\right ) {\mathrm e}^{-x}}{4}\)	\(26\)
parallelrisch	\(\frac {\left (5 x^{3} {\mathrm e}^{5}+10 x^{4}+15 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}}{4}\)	\(27\)
default	\(\frac {15 x^{2}}{4}+\frac {5 \,{\mathrm e}^{-x} x^{4}}{2}+\frac {15 \,{\mathrm e}^{5} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )}{4}-\frac {5 \,{\mathrm e}^{5} \left (-x^{3} {\mathrm e}^{-x}-3 x^{2} {\mathrm e}^{-x}-6 x \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-x}\right )}{4}\)	\(79\)
parts	\(\frac {15 x^{2}}{4}+\frac {5 \,{\mathrm e}^{-x} x^{4}}{2}+\frac {15 \,{\mathrm e}^{5} \left (-x^{2} {\mathrm e}^{-x}-2 x \,{\mathrm e}^{-x}-2 \,{\mathrm e}^{-x}\right )}{4}-\frac {5 \,{\mathrm e}^{5} \left (-x^{3} {\mathrm e}^{-x}-3 x^{2} {\mathrm e}^{-x}-6 x \,{\mathrm e}^{-x}-6 \,{\mathrm e}^{-x}\right )}{4}\)	\(79\)

[In]

int(1/4*(30*exp(x)*x+(-5*x^3+15*x^2)*exp(5)-10*x^4+40*x^3)/exp(x),x,method=_RETURNVERBOSE)

[Out]

(5/2*x^4+5/4*x^3*exp(5)+15/4*exp(x)*x^2)/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.51 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {5}{4} \, {\left (2 \, x^{4} + x^{3} e^{5} + 3 \, x^{2} e^{x}\right )} e^{\left (-x\right )} \]

[In]

integrate(1/4*(30*exp(x)*x+(-5*x^3+15*x^2)*exp(5)-10*x^4+40*x^3)/exp(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {15 x^{2}}{4} + \frac {\left (10 x^{4} + 5 x^{3} e^{5}\right ) e^{- x}}{4} \]

[In]

integrate(1/4*(30*exp(x)*x+(-5*x**3+15*x**2)*exp(5)-10*x**4+40*x**3)/exp(x),x)

[Out]

15*x**2/4 + (10*x**4 + 5*x**3*exp(5))*exp(-x)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (21) = 42\).

Time = 0.21 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.57 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {15}{4} \, x^{2} + \frac {5}{2} \, {\left (x^{4} + 4 \, x^{3} + 12 \, x^{2} + 24 \, x + 24\right )} e^{\left (-x\right )} + \frac {5}{4} \, {\left (x^{3} e^{5} + 3 \, x^{2} e^{5} + 6 \, x e^{5} + 6 \, e^{5}\right )} e^{\left (-x\right )} - 10 \, {\left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right )} e^{\left (-x\right )} - \frac {15}{4} \, {\left (x^{2} e^{5} + 2 \, x e^{5} + 2 \, e^{5}\right )} e^{\left (-x\right )} \]

[In]

integrate(1/4*(30*exp(x)*x+(-5*x^3+15*x^2)*exp(5)-10*x^4+40*x^3)/exp(x),x, algorithm="maxima")

[Out]

15/4*x^2 + 5/2*(x^4 + 4*x^3 + 12*x^2 + 24*x + 24)*e^(-x) + 5/4*(x^3*e^5 + 3*x^2*e^5 + 6*x*e^5 + 6*e^5)*e^(-x)
- 10*(x^3 + 3*x^2 + 6*x + 6)*e^(-x) - 15/4*(x^2*e^5 + 2*x*e^5 + 2*e^5)*e^(-x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {5}{2} \, x^{4} e^{\left (-x\right )} + \frac {5}{4} \, x^{3} e^{\left (-x + 5\right )} + \frac {15}{4} \, x^{2} \]

[In]

integrate(1/4*(30*exp(x)*x+(-5*x^3+15*x^2)*exp(5)-10*x^4+40*x^3)/exp(x),x, algorithm="giac")

[Out]

5/2*x^4*e^(-x) + 5/4*x^3*e^(-x + 5) + 15/4*x^2

Mupad [B] (verification not implemented)

Time = 12.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{4} e^{-x} \left (30 e^x x+40 x^3-10 x^4+e^5 \left (15 x^2-5 x^3\right )\right ) \, dx=\frac {5\,x^4\,{\mathrm {e}}^{-x}}{2}+\frac {15\,x^2}{4}+\frac {5\,x^3\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^5}{4} \]

[In]

int(exp(-x)*((exp(5)*(15*x^2 - 5*x^3))/4 + (15*x*exp(x))/2 + 10*x^3 - (5*x^4)/2),x)

[Out]

(5*x^4*exp(-x))/2 + (15*x^2)/4 + (5*x^3*exp(-x)*exp(5))/4