\(\int \frac {1}{16} (48+32 e^{2 x}+48 x+x^3+e^x (-192-8 x-4 x^2)+16 \log (4)) \, dx\) [308]

Optimal result

Integrand size = 37, antiderivative size = 25 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=-3+3 x+\left (6-e^x+\frac {x^2}{8}\right )^2+x \log (4) \]

[Out]

2*x*ln(2)-3+3*x+(1/8*x^2+6-exp(x))^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {12, 2225, 2227, 2207} \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=\frac {x^4}{64}-\frac {e^x x^2}{4}+\frac {3 x^2}{2}-12 e^x+e^{2 x}+x (3+\log (4)) \]

[In]

Int[(48 + 32*E^(2*x) + 48*x + x^3 + E^x*(-192 - 8*x - 4*x^2) + 16*Log[4])/16,x]

[Out]

-12*E^x + E^(2*x) + (3*x^2)/2 - (E^x*x^2)/4 + x^4/64 + x*(3 + Log[4])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx \\ & = \frac {3 x^2}{2}+\frac {x^4}{64}+x (3+\log (4))+\frac {1}{16} \int e^x \left (-192-8 x-4 x^2\right ) \, dx+2 \int e^{2 x} \, dx \\ & = e^{2 x}+\frac {3 x^2}{2}+\frac {x^4}{64}+x (3+\log (4))+\frac {1}{16} \int \left (-192 e^x-8 e^x x-4 e^x x^2\right ) \, dx \\ & = e^{2 x}+\frac {3 x^2}{2}+\frac {x^4}{64}+x (3+\log (4))-\frac {1}{4} \int e^x x^2 \, dx-\frac {1}{2} \int e^x x \, dx-12 \int e^x \, dx \\ & = -12 e^x+e^{2 x}-\frac {e^x x}{2}+\frac {3 x^2}{2}-\frac {e^x x^2}{4}+\frac {x^4}{64}+x (3+\log (4))+\frac {\int e^x \, dx}{2}+\frac {1}{2} \int e^x x \, dx \\ & = -\frac {23 e^x}{2}+e^{2 x}+\frac {3 x^2}{2}-\frac {e^x x^2}{4}+\frac {x^4}{64}+x (3+\log (4))-\frac {\int e^x \, dx}{2} \\ & = -12 e^x+e^{2 x}+\frac {3 x^2}{2}-\frac {e^x x^2}{4}+\frac {x^4}{64}+x (3+\log (4)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=e^{2 x}+3 x+\frac {3 x^2}{2}+\frac {x^4}{64}-\frac {1}{4} e^x \left (48+x^2\right )+x \log (4) \]

[In]

Integrate[(48 + 32*E^(2*x) + 48*x + x^3 + E^x*(-192 - 8*x - 4*x^2) + 16*Log[4])/16,x]

[Out]

E^(2*x) + 3*x + (3*x^2)/2 + x^4/64 - (E^x*(48 + x^2))/4 + x*Log[4]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40

method	result	size
default	\(3 x +\frac {x^{4}}{64}-\frac {{\mathrm e}^{x} x^{2}}{4}-12 \,{\mathrm e}^{x}+\frac {3 x^{2}}{2}+{\mathrm e}^{2 x}+2 x \ln \left (2\right )\)	\(35\)
norman	\({\mathrm e}^{2 x}+x \left (2 \ln \left (2\right )+3\right )+\frac {3 x^{2}}{2}+\frac {x^{4}}{64}-\frac {{\mathrm e}^{x} x^{2}}{4}-12 \,{\mathrm e}^{x}\)	\(35\)
risch	\(3 x +\frac {x^{4}}{64}-\frac {{\mathrm e}^{x} x^{2}}{4}-12 \,{\mathrm e}^{x}+\frac {3 x^{2}}{2}+{\mathrm e}^{2 x}+2 x \ln \left (2\right )\)	\(35\)
parallelrisch	\({\mathrm e}^{2 x}+x \left (2 \ln \left (2\right )+3\right )+\frac {3 x^{2}}{2}+\frac {x^{4}}{64}-\frac {{\mathrm e}^{x} x^{2}}{4}-12 \,{\mathrm e}^{x}\)	\(35\)
parts	\(3 x +\frac {x^{4}}{64}-\frac {{\mathrm e}^{x} x^{2}}{4}-12 \,{\mathrm e}^{x}+\frac {3 x^{2}}{2}+{\mathrm e}^{2 x}+2 x \ln \left (2\right )\)	\(35\)

[In]

int(2*exp(x)^2+1/16*(-4*x^2-8*x-192)*exp(x)+2*ln(2)+1/16*x^3+3*x+3,x,method=_RETURNVERBOSE)

[Out]

3*x+1/64*x^4-1/4*exp(x)*x^2-12*exp(x)+3/2*x^2+exp(x)^2+2*x*ln(2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=\frac {1}{64} \, x^{4} + \frac {3}{2} \, x^{2} - \frac {1}{4} \, {\left (x^{2} + 48\right )} e^{x} + 2 \, x \log \left (2\right ) + 3 \, x + e^{\left (2 \, x\right )} \]

[In]

integrate(2*exp(x)^2+1/16*(-4*x^2-8*x-192)*exp(x)+2*log(2)+1/16*x^3+3*x+3,x, algorithm="fricas")

[Out]

1/64*x^4 + 3/2*x^2 - 1/4*(x^2 + 48)*e^x + 2*x*log(2) + 3*x + e^(2*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=\frac {x^{4}}{64} + \frac {3 x^{2}}{2} + x \left (2 \log {\left (2 \right )} + 3\right ) + \frac {\left (- x^{2} - 48\right ) e^{x}}{4} + e^{2 x} \]

[In]

integrate(2*exp(x)**2+1/16*(-4*x**2-8*x-192)*exp(x)+2*ln(2)+1/16*x**3+3*x+3,x)

[Out]

x**4/64 + 3*x**2/2 + x*(2*log(2) + 3) + (-x**2 - 48)*exp(x)/4 + exp(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=\frac {1}{64} \, x^{4} + \frac {3}{2} \, x^{2} - \frac {1}{4} \, {\left (x^{2} + 48\right )} e^{x} + 2 \, x \log \left (2\right ) + 3 \, x + e^{\left (2 \, x\right )} \]

[In]

integrate(2*exp(x)^2+1/16*(-4*x^2-8*x-192)*exp(x)+2*log(2)+1/16*x^3+3*x+3,x, algorithm="maxima")

[Out]

1/64*x^4 + 3/2*x^2 - 1/4*(x^2 + 48)*e^x + 2*x*log(2) + 3*x + e^(2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx=\frac {1}{64} \, x^{4} + \frac {3}{2} \, x^{2} - \frac {1}{4} \, {\left (x^{2} + 48\right )} e^{x} + 2 \, x \log \left (2\right ) + 3 \, x + e^{\left (2 \, x\right )} \]

[In]

integrate(2*exp(x)^2+1/16*(-4*x^2-8*x-192)*exp(x)+2*log(2)+1/16*x^3+3*x+3,x, algorithm="giac")

[Out]

1/64*x^4 + 3/2*x^2 - 1/4*(x^2 + 48)*e^x + 2*x*log(2) + 3*x + e^(2*x)

Mupad [B] (verification not implemented)

Time = 7.59 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {1}{16} \left (48+32 e^{2 x}+48 x+x^3+e^x \left (-192-8 x-4 x^2\right )+16 \log (4)\right ) \, dx={\mathrm {e}}^{2\,x}-12\,{\mathrm {e}}^x+x\,\left (2\,\ln \left (2\right )+3\right )-\frac {x^2\,{\mathrm {e}}^x}{4}+\frac {3\,x^2}{2}+\frac {x^4}{64} \]

[In]

int(3*x + 2*exp(2*x) + 2*log(2) - (exp(x)*(8*x + 4*x^2 + 192))/16 + x^3/16 + 3,x)

[Out]

exp(2*x) - 12*exp(x) + x*(2*log(2) + 3) - (x^2*exp(x))/4 + (3*x^2)/2 + x^4/64