\(\int \frac {1}{5} (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)) \, dx\) [5567]

Optimal result

Integrand size = 28, antiderivative size = 23 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=3 x \left (e^x-x-\left (-4+\frac {4 x^2}{5}\right )^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 2207, 2225} \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48 x^5}{25}+\frac {96 x^3}{5}-3 x^2-48 x-3 e^x+3 e^x (x+1) \]

[In]

Int[(-240 - 30*x + 288*x^2 - 48*x^4 + E^x*(15 + 15*x))/5,x]

[Out]

-3*E^x - 48*x - 3*x^2 + (96*x^3)/5 - (48*x^5)/25 + 3*E^x*(1 + 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx \\ & = -48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+\frac {1}{5} \int e^x (15+15 x) \, dx \\ & = -48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+3 e^x (1+x)-3 \int e^x \, dx \\ & = -3 e^x-48 x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25}+3 e^x (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-48 x+3 e^x x-3 x^2+\frac {96 x^3}{5}-\frac {48 x^5}{25} \]

[In]

Integrate[(-240 - 30*x + 288*x^2 - 48*x^4 + E^x*(15 + 15*x))/5,x]

[Out]

-48*x + 3*E^x*x - 3*x^2 + (96*x^3)/5 - (48*x^5)/25

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
default	\(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\)	\(25\)
norman	\(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\)	\(25\)
risch	\(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\)	\(25\)
parallelrisch	\(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\)	\(25\)
parts	\(-48 x +3 \,{\mathrm e}^{x} x -3 x^{2}+\frac {96 x^{3}}{5}-\frac {48 x^{5}}{25}\)	\(25\)

[In]

int(1/5*(15*x+15)*exp(x)-48/5*x^4+288/5*x^2-6*x-48,x,method=_RETURNVERBOSE)

[Out]

-48*x+3*exp(x)*x-3*x^2+96/5*x^3-48/25*x^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]

[In]

integrate(1/5*(15*x+15)*exp(x)-48/5*x^4+288/5*x^2-6*x-48,x, algorithm="fricas")

[Out]

-48/25*x^5 + 96/5*x^3 - 3*x^2 + 3*x*e^x - 48*x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=- \frac {48 x^{5}}{25} + \frac {96 x^{3}}{5} - 3 x^{2} + 3 x e^{x} - 48 x \]

[In]

integrate(1/5*(15*x+15)*exp(x)-48/5*x**4+288/5*x**2-6*x-48,x)

[Out]

-48*x**5/25 + 96*x**3/5 - 3*x**2 + 3*x*exp(x) - 48*x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]

[In]

integrate(1/5*(15*x+15)*exp(x)-48/5*x^4+288/5*x^2-6*x-48,x, algorithm="maxima")

[Out]

-48/25*x^5 + 96/5*x^3 - 3*x^2 + 3*x*e^x - 48*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {48}{25} \, x^{5} + \frac {96}{5} \, x^{3} - 3 \, x^{2} + 3 \, x e^{x} - 48 \, x \]

[In]

integrate(1/5*(15*x+15)*exp(x)-48/5*x^4+288/5*x^2-6*x-48,x, algorithm="giac")

[Out]

-48/25*x^5 + 96/5*x^3 - 3*x^2 + 3*x*e^x - 48*x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {1}{5} \left (-240-30 x+288 x^2-48 x^4+e^x (15+15 x)\right ) \, dx=-\frac {3\,x\,\left (25\,x-25\,{\mathrm {e}}^x-160\,x^2+16\,x^4+400\right )}{25} \]

[In]

int((exp(x)*(15*x + 15))/5 - 6*x + (288*x^2)/5 - (48*x^4)/5 - 48,x)

[Out]

-(3*x*(25*x - 25*exp(x) - 160*x^2 + 16*x^4 + 400))/25