\(\int \frac {1}{8} (16 x+e^{\frac {1}{8} (24 x+16 x^2-x^5)} (24+32 x-5 x^4)) \, dx\) [6663]

Optimal result

Integrand size = 39, antiderivative size = 22 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=e^{3 x+2 x \left (x-\frac {x^4}{16}\right )}+x^2 \]

[Out]

x^2+exp(3*x+2*x*(x-1/16*x^4))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6838} \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=x^2+e^{\frac {1}{8} \left (-x^5+16 x^2+24 x\right )} \]

[In]

Int[(16*x + E^((24*x + 16*x^2 - x^5)/8)*(24 + 32*x - 5*x^4))/8,x]

[Out]

E^((24*x + 16*x^2 - x^5)/8) + x^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx \\ & = x^2+\frac {1}{8} \int e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right ) \, dx \\ & = e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )}+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=e^{-\frac {1}{8} x \left (-24-16 x+x^4\right )}+x^2 \]

[In]

Integrate[(16*x + E^((24*x + 16*x^2 - x^5)/8)*(24 + 32*x - 5*x^4))/8,x]

[Out]

E^(-1/8*(x*(-24 - 16*x + x^4))) + x^2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
risch	\({\mathrm e}^{-\frac {x \left (x^{4}-16 x -24\right )}{8}}+x^{2}\)	\(17\)
default	\({\mathrm e}^{-\frac {1}{8} x^{5}+2 x^{2}+3 x}+x^{2}\)	\(20\)
norman	\({\mathrm e}^{-\frac {1}{8} x^{5}+2 x^{2}+3 x}+x^{2}\)	\(20\)
parallelrisch	\({\mathrm e}^{-\frac {1}{8} x^{5}+2 x^{2}+3 x}+x^{2}\)	\(20\)
parts	\({\mathrm e}^{-\frac {1}{8} x^{5}+2 x^{2}+3 x}+x^{2}\)	\(20\)

[In]

int(1/8*(-5*x^4+32*x+24)*exp(-1/8*x^5+2*x^2+3*x)+2*x,x,method=_RETURNVERBOSE)

[Out]

exp(-1/8*x*(x^4-16*x-24))+x^2

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=x^{2} + e^{\left (-\frac {1}{8} \, x^{5} + 2 \, x^{2} + 3 \, x\right )} \]

[In]

integrate(1/8*(-5*x^4+32*x+24)*exp(-1/8*x^5+2*x^2+3*x)+2*x,x, algorithm="fricas")

[Out]

x^2 + e^(-1/8*x^5 + 2*x^2 + 3*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=x^{2} + e^{- \frac {x^{5}}{8} + 2 x^{2} + 3 x} \]

[In]

integrate(1/8*(-5*x**4+32*x+24)*exp(-1/8*x**5+2*x**2+3*x)+2*x,x)

[Out]

x**2 + exp(-x**5/8 + 2*x**2 + 3*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=x^{2} + e^{\left (-\frac {1}{8} \, x^{5} + 2 \, x^{2} + 3 \, x\right )} \]

[In]

integrate(1/8*(-5*x^4+32*x+24)*exp(-1/8*x^5+2*x^2+3*x)+2*x,x, algorithm="maxima")

[Out]

x^2 + e^(-1/8*x^5 + 2*x^2 + 3*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx=x^{2} + e^{\left (-\frac {1}{8} \, x^{5} + 2 \, x^{2} + 3 \, x\right )} \]

[In]

integrate(1/8*(-5*x^4+32*x+24)*exp(-1/8*x^5+2*x^2+3*x)+2*x,x, algorithm="giac")

[Out]

x^2 + e^(-1/8*x^5 + 2*x^2 + 3*x)

Mupad [B] (verification not implemented)

Time = 12.71 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{8} \left (16 x+e^{\frac {1}{8} \left (24 x+16 x^2-x^5\right )} \left (24+32 x-5 x^4\right )\right ) \, dx={\mathrm {e}}^{-\frac {x^5}{8}+2\,x^2+3\,x}+x^2 \]

[In]

int(2*x + (exp(3*x + 2*x^2 - x^5/8)*(32*x - 5*x^4 + 24))/8,x)

[Out]

exp(3*x + 2*x^2 - x^5/8) + x^2