\(\int \frac {1}{4} e^x (-2 x+47 x^2+16 x^3) \, dx\) [5452]

Optimal result

Integrand size = 21, antiderivative size = 17 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {1}{4} e^x x \left (-x+16 x^2\right ) \]

[Out]

1/4*exp(x)*x*(16*x^2-x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {12, 1608, 2227, 2207, 2225} \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=4 e^x x^3-\frac {e^x x^2}{4} \]

[In]

Int[(E^x*(-2*x + 47*x^2 + 16*x^3))/4,x]

[Out]

-1/4*(E^x*x^2) + 4*E^x*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 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx \\ & = \frac {1}{4} \int e^x x \left (-2+47 x+16 x^2\right ) \, dx \\ & = \frac {1}{4} \int \left (-2 e^x x+47 e^x x^2+16 e^x x^3\right ) \, dx \\ & = -\left (\frac {1}{2} \int e^x x \, dx\right )+4 \int e^x x^3 \, dx+\frac {47}{4} \int e^x x^2 \, dx \\ & = -\frac {e^x x}{2}+\frac {47 e^x x^2}{4}+4 e^x x^3+\frac {\int e^x \, dx}{2}-12 \int e^x x^2 \, dx-\frac {47}{2} \int e^x x \, dx \\ & = \frac {e^x}{2}-24 e^x x-\frac {e^x x^2}{4}+4 e^x x^3+\frac {47 \int e^x \, dx}{2}+24 \int e^x x \, dx \\ & = 24 e^x-\frac {e^x x^2}{4}+4 e^x x^3-24 \int e^x \, dx \\ & = -\frac {1}{4} e^x x^2+4 e^x x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {1}{4} e^x x^2 (-1+16 x) \]

[In]

Integrate[(E^x*(-2*x + 47*x^2 + 16*x^3))/4,x]

[Out]

(E^x*x^2*(-1 + 16*x))/4

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76

method	result	size
gosper	\(\frac {{\mathrm e}^{x} \left (16 x -1\right ) x^{2}}{4}\)	\(13\)
default	\(-\frac {{\mathrm e}^{x} x^{2}}{4}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
norman	\(-\frac {{\mathrm e}^{x} x^{2}}{4}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
risch	\(\frac {\left (16 x^{3}-x^{2}\right ) {\mathrm e}^{x}}{4}\)	\(16\)
parallelrisch	\(-\frac {{\mathrm e}^{x} x^{2}}{4}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
parts	\(-\frac {{\mathrm e}^{x} x^{2}}{4}+4 \,{\mathrm e}^{x} x^{3}\)	\(16\)
meijerg	\(-\left (-4 x^{3}+12 x^{2}-24 x +24\right ) {\mathrm e}^{x}+\frac {47 \left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{12}+\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{4}\)	\(44\)

[In]

int(1/4*(16*x^3+47*x^2-2*x)*exp(x),x,method=_RETURNVERBOSE)

[Out]

1/4*exp(x)*(16*x-1)*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {1}{4} \, {\left (16 \, x^{3} - x^{2}\right )} e^{x} \]

[In]

integrate(1/4*(16*x^3+47*x^2-2*x)*exp(x),x, algorithm="fricas")

[Out]

1/4*(16*x^3 - x^2)*e^x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {\left (16 x^{3} - x^{2}\right ) e^{x}}{4} \]

[In]

integrate(1/4*(16*x**3+47*x**2-2*x)*exp(x),x)

[Out]

(16*x**3 - x**2)*exp(x)/4

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (14) = 28\).

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=4 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} + \frac {47}{4} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} - \frac {1}{2} \, {\left (x - 1\right )} e^{x} \]

[In]

integrate(1/4*(16*x^3+47*x^2-2*x)*exp(x),x, algorithm="maxima")

[Out]

4*(x^3 - 3*x^2 + 6*x - 6)*e^x + 47/4*(x^2 - 2*x + 2)*e^x - 1/2*(x - 1)*e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {1}{4} \, {\left (16 \, x^{3} - x^{2}\right )} e^{x} \]

[In]

integrate(1/4*(16*x^3+47*x^2-2*x)*exp(x),x, algorithm="giac")

[Out]

1/4*(16*x^3 - x^2)*e^x

Mupad [B] (verification not implemented)

Time = 11.99 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {1}{4} e^x \left (-2 x+47 x^2+16 x^3\right ) \, dx=\frac {x^2\,{\mathrm {e}}^x\,\left (16\,x-1\right )}{4} \]

[In]

int((exp(x)*(47*x^2 - 2*x + 16*x^3))/4,x)

[Out]

(x^2*exp(x)*(16*x - 1))/4