\(\int \frac {1}{4} (16+e^{\frac {1}{4} (-100 x-4 x^2+x \log (4))} (4-100 x-8 x^2+x \log (4))) \, dx\) [1669]

Optimal result

Integrand size = 40, antiderivative size = 24 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=8 x+\left (-4+e^{x \left (-25+\frac {1}{4} (-4 x+\log (4))\right )}\right ) x \]

[Out]

x*(exp(x*(1/2*ln(2)-x-25))-4)+8*x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 6, 2326} \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=\frac {2^{x/2} e^{-x^2-25 x} \left (8 x^2+x (100-\log (4))\right )}{8 x+100-\log (4)}+4 x \]

[In]

Int[(16 + E^((-100*x - 4*x^2 + x*Log[4])/4)*(4 - 100*x - 8*x^2 + x*Log[4]))/4,x]

[Out]

4*x + (2^(x/2)*E^(-25*x - x^2)*(8*x^2 + x*(100 - Log[4])))/(100 + 8*x - Log[4])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx \\ & = 4 x+\frac {1}{4} \int e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right ) \, dx \\ & = 4 x+\frac {1}{4} \int e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-8 x^2+x (-100+\log (4))\right ) \, dx \\ & = 4 x+\frac {2^{x/2} e^{-25 x-x^2} \left (8 x^2+x (100-\log (4))\right )}{100+8 x-\log (4)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(24)=48\).

Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=4 x+\frac {e^{-x^2+x \left (-25+\frac {\log (2)}{2}\right )} \left (-8 x^2+x (-100+\log (4))\right )}{4 \left (-25-2 x+\frac {\log (2)}{2}\right )} \]

[In]

Integrate[(16 + E^((-100*x - 4*x^2 + x*Log[4])/4)*(4 - 100*x - 8*x^2 + x*Log[4]))/4,x]

[Out]

4*x + (E^(-x^2 + x*(-25 + Log[2]/2))*(-8*x^2 + x*(-100 + Log[4])))/(4*(-25 - 2*x + Log[2]/2))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(x \,{\mathrm e}^{\frac {x \left (\ln \left (2\right )-2 x -50\right )}{2}}+4 x\)	\(18\)
risch	\(2^{\frac {x}{2}} {\mathrm e}^{-x \left (x +25\right )} x +4 x\)	\(19\)
default	\(4 x +x \,{\mathrm e}^{-x^{2}+\left (\frac {\ln \left (2\right )}{2}-25\right ) x}\)	\(22\)
norman	\({\mathrm e}^{\frac {x \ln \left (2\right )}{2}-x^{2}-25 x} x +4 x\)	\(22\)
parts	\(4 x +x \,{\mathrm e}^{-x^{2}+\left (\frac {\ln \left (2\right )}{2}-25\right ) x}\)	\(22\)

[In]

int(1/4*(2*x*ln(2)-8*x^2-100*x+4)*exp(1/2*x*ln(2)-x^2-25*x)+4,x,method=_RETURNVERBOSE)

[Out]

x*exp(1/2*x*(ln(2)-2*x-50))+4*x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=x e^{\left (-x^{2} + \frac {1}{2} \, x \log \left (2\right ) - 25 \, x\right )} + 4 \, x \]

[In]

integrate(1/4*(2*x*log(2)-8*x^2-100*x+4)*exp(1/2*x*log(2)-x^2-25*x)+4,x, algorithm="fricas")

[Out]

x*e^(-x^2 + 1/2*x*log(2) - 25*x) + 4*x

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=x e^{- x^{2} - 25 x + \frac {x \log {\left (2 \right )}}{2}} + 4 x \]

[In]

integrate(1/4*(2*x*ln(2)-8*x**2-100*x+4)*exp(1/2*x*ln(2)-x**2-25*x)+4,x)

[Out]

x*exp(-x**2 - 25*x + x*log(2)/2) + 4*x

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.35 (sec) , antiderivative size = 306, normalized size of antiderivative = 12.75 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=-\frac {1}{16} i \, {\left (\frac {i \, \sqrt {\pi } {\left (4 \, x - \log \left (2\right ) + 50\right )} {\left (\operatorname {erf}\left (\frac {1}{4} \, \sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}\right ) - 1\right )} {\left (\log \left (2\right ) - 50\right )}}{\sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}} - 4 i \, e^{\left (-\frac {1}{16} \, {\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}\right )}\right )} e^{\left (\frac {1}{16} \, {\left (\log \left (2\right ) - 50\right )}^{2}\right )} \log \left (2\right ) + \frac {1}{2} \, \sqrt {\pi } \operatorname {erf}\left (x - \frac {1}{4} \, \log \left (2\right ) + \frac {25}{2}\right ) e^{\left (\frac {1}{16} \, {\left (\log \left (2\right ) - 50\right )}^{2}\right )} + \frac {1}{16} i \, {\left (\frac {i \, \sqrt {\pi } {\left (4 \, x - \log \left (2\right ) + 50\right )} {\left (\operatorname {erf}\left (\frac {1}{4} \, \sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}\right ) - 1\right )} {\left (\log \left (2\right ) - 50\right )}^{2}}{\sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}} - \frac {16 i \, {\left (4 \, x - \log \left (2\right ) + 50\right )}^{3} \Gamma \left (\frac {3}{2}, \frac {1}{16} \, {\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}\right )}{{\left ({\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}\right )}^{\frac {3}{2}}} - 8 i \, {\left (\log \left (2\right ) - 50\right )} e^{\left (-\frac {1}{16} \, {\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}\right )}\right )} e^{\left (\frac {1}{16} \, {\left (\log \left (2\right ) - 50\right )}^{2}\right )} + \frac {25}{8} i \, {\left (\frac {i \, \sqrt {\pi } {\left (4 \, x - \log \left (2\right ) + 50\right )} {\left (\operatorname {erf}\left (\frac {1}{4} \, \sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}\right ) - 1\right )} {\left (\log \left (2\right ) - 50\right )}}{\sqrt {{\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}}} - 4 i \, e^{\left (-\frac {1}{16} \, {\left (4 \, x - \log \left (2\right ) + 50\right )}^{2}\right )}\right )} e^{\left (\frac {1}{16} \, {\left (\log \left (2\right ) - 50\right )}^{2}\right )} + 4 \, x \]

[In]

integrate(1/4*(2*x*log(2)-8*x^2-100*x+4)*exp(1/2*x*log(2)-x^2-25*x)+4,x, algorithm="maxima")

[Out]

-1/16*I*(I*sqrt(pi)*(4*x - log(2) + 50)*(erf(1/4*sqrt((4*x - log(2) + 50)^2)) - 1)*(log(2) - 50)/sqrt((4*x - l
og(2) + 50)^2) - 4*I*e^(-1/16*(4*x - log(2) + 50)^2))*e^(1/16*(log(2) - 50)^2)*log(2) + 1/2*sqrt(pi)*erf(x - 1
/4*log(2) + 25/2)*e^(1/16*(log(2) - 50)^2) + 1/16*I*(I*sqrt(pi)*(4*x - log(2) + 50)*(erf(1/4*sqrt((4*x - log(2
) + 50)^2)) - 1)*(log(2) - 50)^2/sqrt((4*x - log(2) + 50)^2) - 16*I*(4*x - log(2) + 50)^3*gamma(3/2, 1/16*(4*x
- log(2) + 50)^2)/((4*x - log(2) + 50)^2)^(3/2) - 8*I*(log(2) - 50)*e^(-1/16*(4*x - log(2) + 50)^2))*e^(1/16*
(log(2) - 50)^2) + 25/8*I*(I*sqrt(pi)*(4*x - log(2) + 50)*(erf(1/4*sqrt((4*x - log(2) + 50)^2)) - 1)*(log(2) -
50)/sqrt((4*x - log(2) + 50)^2) - 4*I*e^(-1/16*(4*x - log(2) + 50)^2))*e^(1/16*(log(2) - 50)^2) + 4*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=x e^{\left (-x^{2} + \frac {1}{2} \, x \log \left (2\right ) - 25 \, x\right )} + 4 \, x \]

[In]

integrate(1/4*(2*x*log(2)-8*x^2-100*x+4)*exp(1/2*x*log(2)-x^2-25*x)+4,x, algorithm="giac")

[Out]

x*e^(-x^2 + 1/2*x*log(2) - 25*x) + 4*x

Mupad [B] (verification not implemented)

Time = 7.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{4} \left (16+e^{\frac {1}{4} \left (-100 x-4 x^2+x \log (4)\right )} \left (4-100 x-8 x^2+x \log (4)\right )\right ) \, dx=x\,\left (2^{x/2}\,{\mathrm {e}}^{-x^2-25\,x}+4\right ) \]

[In]

int(4 - (exp((x*log(2))/2 - 25*x - x^2)*(100*x - 2*x*log(2) + 8*x^2 - 4))/4,x)

[Out]

x*(2^(x/2)*exp(- 25*x - x^2) + 4)