\(\int e^{398+160 x+16 x^2+\frac {2 (e^2 x+e^4 x)}{e^2}} (18 e^4+e^2 (1458+288 x)) \, dx\) [9801]

Optimal result

Integrand size = 44, antiderivative size = 21 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 e^{2 x+2 e^2 x+16 (5+x)^2} \]

[Out]

9*exp(4*(5+x)*(20+4*x))*exp(x+x*exp(2)^2/exp(1)^2)^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2276, 2268} \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 e^{16 x^2+2 \left (81+e^2\right ) x+400} \]

[In]

Int[E^(398 + 160*x + 16*x^2 + (2*(E^2*x + E^4*x))/E^2)*(18*E^4 + E^2*(1458 + 288*x)),x]

[Out]

9*E^(400 + 2*(81 + E^2)*x + 16*x^2)

Rule 2268

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2276

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = \int e^{398+2 \left (81+e^2\right ) x+16 x^2} \left (18 e^2 \left (81+e^2\right )+288 e^2 x\right ) \, dx \\ & = 9 e^{400+2 \left (81+e^2\right ) x+16 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 e^{400+2 \left (81+e^2\right ) x+16 x^2} \]

[In]

Integrate[E^(398 + 160*x + 16*x^2 + (2*(E^2*x + E^4*x))/E^2)*(18*E^4 + E^2*(1458 + 288*x)),x]

[Out]

9*E^(400 + 2*(81 + E^2)*x + 16*x^2)

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33

method	result	size
risch	\(9 \,{\mathrm e}^{2 x \,{\mathrm e}^{-2} {\mathrm e}^{2}+2 x \,{\mathrm e}^{-2} {\mathrm e}^{4}+16 x^{2}+160 x +400}\)	\(28\)
gosper	\(9 \,{\mathrm e}^{16 x^{2}+160 x +400} {\mathrm e}^{2 x \left ({\mathrm e}^{2}+{\mathrm e}^{4}\right ) {\mathrm e}^{-2}}\)	\(32\)
parallelrisch	\(9 \,{\mathrm e}^{16 x^{2}+160 x +400} {\mathrm e}^{2 x \left ({\mathrm e}^{2}+{\mathrm e}^{4}\right ) {\mathrm e}^{-2}}\)	\(32\)
norman	\(9 \,{\mathrm e}^{16 x^{2}+160 x +400} {\mathrm e}^{2 \left (x \,{\mathrm e}^{4}+{\mathrm e}^{2} x \right ) {\mathrm e}^{-2}}\)	\(35\)
default	\({\mathrm e}^{-2} \left (-\frac {729 i {\mathrm e}^{400} {\mathrm e}^{2} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{2}+162\right )^{2}}{64}} \operatorname {erf}\left (4 i x +\frac {i \left (2 \,{\mathrm e}^{2}+162\right )}{8}\right )}{4}-\frac {9 i {\mathrm e}^{400} {\mathrm e}^{4} \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{2}+162\right )^{2}}{64}} \operatorname {erf}\left (4 i x +\frac {i \left (2 \,{\mathrm e}^{2}+162\right )}{8}\right )}{4}+288 \,{\mathrm e}^{400} {\mathrm e}^{2} \left (\frac {{\mathrm e}^{16 x^{2}+\left (2 \,{\mathrm e}^{2}+162\right ) x}}{32}+\frac {i \left (2 \,{\mathrm e}^{2}+162\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {\left (2 \,{\mathrm e}^{2}+162\right )^{2}}{64}} \operatorname {erf}\left (4 i x +\frac {i \left (2 \,{\mathrm e}^{2}+162\right )}{8}\right )}{256}\right )\right )\)	\(147\)

[In]

int((18*exp(2)^2+(288*x+1458)*exp(1)^2)*exp(16*x^2+160*x+400)*exp((x*exp(2)^2+x*exp(1)^2)/exp(1)^2)^2/exp(1)^2
,x,method=_RETURNVERBOSE)

[Out]

9*exp(2*x*exp(-2)*exp(2)+2*x*exp(-2)*exp(4)+16*x^2+160*x+400)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 \, e^{\left (16 \, x^{2} + 2 \, x e^{2} + 162 \, x + 400\right )} \]

[In]

integrate((18*exp(2)^2+(288*x+1458)*exp(1)^2)*exp(16*x^2+160*x+400)*exp((x*exp(2)^2+x*exp(1)^2)/exp(1)^2)^2/ex
p(1)^2,x, algorithm="fricas")

[Out]

9*e^(16*x^2 + 2*x*e^2 + 162*x + 400)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 e^{\frac {2 \left (x e^{2} + x e^{4}\right )}{e^{2}}} e^{16 x^{2} + 160 x + 400} \]

[In]

integrate((18*exp(2)**2+(288*x+1458)*exp(1)**2)*exp(16*x**2+160*x+400)*exp((x*exp(2)**2+x*exp(1)**2)/exp(1)**2
)**2/exp(1)**2,x)

[Out]

9*exp(2*(x*exp(2) + x*exp(4))*exp(-2))*exp(16*x**2 + 160*x + 400)

Maxima [C] (verification not implemented)

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

Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 6.05 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=-\frac {9}{4} i \, \sqrt {\pi } \operatorname {erf}\left (4 i \, x + \frac {1}{4} i \, e^{2} + \frac {81}{4} i\right ) e^{\left (-\frac {1}{16} \, {\left (e^{2} + 81\right )}^{2} + 402\right )} - \frac {729}{4} i \, \sqrt {\pi } \operatorname {erf}\left (4 i \, x + \frac {1}{4} i \, e^{2} + \frac {81}{4} i\right ) e^{\left (-\frac {1}{16} \, {\left (e^{2} + 81\right )}^{2} + 400\right )} - \frac {9}{4} \, {\left (\frac {\sqrt {\pi } {\left (16 \, x + e^{2} + 81\right )} {\left (\operatorname {erf}\left (\frac {1}{4} \, \sqrt {-{\left (16 \, x + e^{2} + 81\right )}^{2}}\right ) - 1\right )} {\left (e^{2} + 81\right )}}{\sqrt {-{\left (16 \, x + e^{2} + 81\right )}^{2}}} - 4 \, e^{\left (\frac {1}{16} \, {\left (16 \, x + e^{2} + 81\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{16} \, {\left (e^{2} + 81\right )}^{2} + 400\right )} \]

[In]

integrate((18*exp(2)^2+(288*x+1458)*exp(1)^2)*exp(16*x^2+160*x+400)*exp((x*exp(2)^2+x*exp(1)^2)/exp(1)^2)^2/ex
p(1)^2,x, algorithm="maxima")

[Out]

-9/4*I*sqrt(pi)*erf(4*I*x + 1/4*I*e^2 + 81/4*I)*e^(-1/16*(e^2 + 81)^2 + 402) - 729/4*I*sqrt(pi)*erf(4*I*x + 1/
4*I*e^2 + 81/4*I)*e^(-1/16*(e^2 + 81)^2 + 400) - 9/4*(sqrt(pi)*(16*x + e^2 + 81)*(erf(1/4*sqrt(-(16*x + e^2 +
81)^2)) - 1)*(e^2 + 81)/sqrt(-(16*x + e^2 + 81)^2) - 4*e^(1/16*(16*x + e^2 + 81)^2))*e^(-1/16*(e^2 + 81)^2 + 4
00)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9 \, e^{\left (16 \, x^{2} + 2 \, x e^{2} + 162 \, x + 400\right )} \]

[In]

integrate((18*exp(2)^2+(288*x+1458)*exp(1)^2)*exp(16*x^2+160*x+400)*exp((x*exp(2)^2+x*exp(1)^2)/exp(1)^2)^2/ex
p(1)^2,x, algorithm="giac")

[Out]

9*e^(16*x^2 + 2*x*e^2 + 162*x + 400)

Mupad [B] (verification not implemented)

Time = 15.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int e^{398+160 x+16 x^2+\frac {2 \left (e^2 x+e^4 x\right )}{e^2}} \left (18 e^4+e^2 (1458+288 x)\right ) \, dx=9\,{\mathrm {e}}^{162\,x}\,{\mathrm {e}}^{400}\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^2} \]

[In]

int(exp(2*exp(-2)*(x*exp(2) + x*exp(4)))*exp(-2)*exp(160*x + 16*x^2 + 400)*(18*exp(4) + exp(2)*(288*x + 1458))
,x)

[Out]

9*exp(162*x)*exp(400)*exp(16*x^2)*exp(2*x*exp(2))