3.16.50 \(\int (-5+e^{-4+x}+4 x^2+\log (\frac {e^{2 x^2}}{16})) \, dx\)

Optimal. Leaf size=29 \[ -5+e^{-4+x}-x-x \left (4-\log \left (\frac {e^{2 x^2}}{16}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2194, 2548, 12, 30} \begin {gather*} x \log \left (\frac {e^{2 x^2}}{16}\right )-5 x+e^{x-4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-4 + x) - 5*x + x*Log[E^(2*x^2)/16]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2194

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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-5 x+\frac {4 x^3}{3}+\int e^{-4+x} \, dx+\int \log \left (\frac {e^{2 x^2}}{16}\right ) \, dx\\ &=e^{-4+x}-5 x+\frac {4 x^3}{3}+x \log \left (\frac {e^{2 x^2}}{16}\right )-\int 4 x^2 \, dx\\ &=e^{-4+x}-5 x+\frac {4 x^3}{3}+x \log \left (\frac {e^{2 x^2}}{16}\right )-4 \int x^2 \, dx\\ &=e^{-4+x}-5 x+x \log \left (\frac {e^{2 x^2}}{16}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 34, normalized size = 1.17 \begin {gather*} e^{-4+x}-5 x+2 x^3+x \left (-2 x^2+\log \left (\frac {e^{2 x^2}}{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-4 + x) - 5*x + 2*x^3 + x*(-2*x^2 + Log[E^(2*x^2)/16])

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 18, normalized size = 0.62 \begin {gather*} 2 \, x^{3} - 4 \, x \log \relax (2) - 5 \, x + e^{\left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1/16*exp(x^2)^2)+exp(x-4)+4*x^2-5,x, algorithm="fricas")

[Out]

2*x^3 - 4*x*log(2) - 5*x + e^(x - 4)

________________________________________________________________________________________

giac [A]  time = 0.15, size = 18, normalized size = 0.62 \begin {gather*} 2 \, x^{3} - 4 \, x \log \relax (2) - 5 \, x + e^{\left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1/16*exp(x^2)^2)+exp(x-4)+4*x^2-5,x, algorithm="giac")

[Out]

2*x^3 - 4*x*log(2) - 5*x + e^(x - 4)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 20, normalized size = 0.69

method	result	size
default	\(-5 x +{\mathrm e}^{x -4}+x \ln \left (\frac {{\mathrm e}^{2 x^{2}}}{16}\right )\)	\(20\)
risch	\(2 x \ln \left ({\mathrm e}^{x^{2}}\right )-\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x^{2}}\right ) x}{2}+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x^{2}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x^{2}}\right )^{2} x -\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x^{2}}\right )^{3} x}{2}-4 x \ln \relax (2)+{\mathrm e}^{x -4}-5 x\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(1/16*exp(x^2)^2)+exp(x-4)+4*x^2-5,x,method=_RETURNVERBOSE)

[Out]

-5*x+exp(x-4)+x*ln(1/16*exp(x^2)^2)

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 19, normalized size = 0.66 \begin {gather*} x \log \left (\frac {1}{16} \, e^{\left (2 \, x^{2}\right )}\right ) - 5 \, x + e^{\left (x - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1/16*exp(x^2)^2)+exp(x-4)+4*x^2-5,x, algorithm="maxima")

[Out]

x*log(1/16*e^(2*x^2)) - 5*x + e^(x - 4)

________________________________________________________________________________________

mupad [B]  time = 0.08, size = 17, normalized size = 0.59 \begin {gather*} {\mathrm {e}}^{x-4}-x\,\left (\ln \left (16\right )+5\right )+2\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x - 4) + log(exp(2*x^2)/16) + 4*x^2 - 5,x)

[Out]

exp(x - 4) - x*(log(16) + 5) + 2*x^3

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 19, normalized size = 0.66 \begin {gather*} 2 x^{3} + x \left (-5 - 4 \log {\relax (2 )}\right ) + e^{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(1/16*exp(x**2)**2)+exp(x-4)+4*x**2-5,x)

[Out]

2*x**3 + x*(-5 - 4*log(2)) + exp(x - 4)

________________________________________________________________________________________