3.4.64 \(\int \frac {32 e^4 x^3+2 x^4+e^{32-16 x+2 x^2} (-16 x^3+4 x^4+e^8 (-512-4096 x+1024 x^2)+e^4 (-32 x-512 x^2+128 x^3))+e^{16-8 x+x^2} (2 x^3-16 x^4+4 x^5+e^8 (-512 x-4096 x^2+1024 x^3)+e^4 (-512 x^3+128 x^4))}{e^8 x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (e^{(-4+x)^2}+x\right )^2 \left (16+\frac {x}{e^4}\right )^2}{x^2} \]

________________________________________________________________________________________

Rubi [C]  time = 0.84, antiderivative size = 209, normalized size of antiderivative = 8.71, number of steps used = 23, number of rules used = 11, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 14, 2288, 6742, 2227, 2204, 2242, 2234, 2226, 2209, 2212} \begin {gather*} -\frac {\left (1-256 e^4+512 e^8\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}+\frac {32 \left (1-8 e^4\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}-\frac {31 \sqrt {\pi } \text {erfi}(4-x)}{e^8}+512 \sqrt {\pi } \text {erfi}(4-x)+\frac {512 e^{x^2-8 x+16}}{x}+\frac {e^{2 (4-x)^2-8} \left (-x^3+4 \left (1-4 e^4\right ) x^2+64 e^4 x\right ) \left (x+16 e^4\right )}{(4-x) x^3}+\frac {\left (x+16 e^4\right )^2}{e^8}+16 e^{(x-4)^2-8}-2 e^{(x-4)^2-8} (4-x)-8 \left (1-8 e^4\right ) e^{(x-4)^2-8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*E^4*x^3 + 2*x^4 + E^(32 - 16*x + 2*x^2)*(-16*x^3 + 4*x^4 + E^8*(-512 - 4096*x + 1024*x^2) + E^4*(-32*x
- 512*x^2 + 128*x^3)) + E^(16 - 8*x + x^2)*(2*x^3 - 16*x^4 + 4*x^5 + E^8*(-512*x - 4096*x^2 + 1024*x^3) + E^4
*(-512*x^3 + 128*x^4)))/(E^8*x^3),x]

[Out]

16*E^(-8 + (-4 + x)^2) - 8*E^(-8 + (-4 + x)^2)*(1 - 8*E^4) - 2*E^(-8 + (-4 + x)^2)*(4 - x) + (512*E^(16 - 8*x
+ x^2))/x + (16*E^4 + x)^2/E^8 + (E^(-8 + 2*(4 - x)^2)*(16*E^4 + x)*(64*E^4*x + 4*(1 - 4*E^4)*x^2 - x^3))/((4
- x)*x^3) + 512*Sqrt[Pi]*Erfi[4 - x] - (31*Sqrt[Pi]*Erfi[4 - x])/E^8 + (32*(1 - 8*E^4)*Sqrt[Pi]*Erfi[4 - x])/E
^8 - ((1 - 256*E^4 + 512*E^8)*Sqrt[Pi]*Erfi[4 - x])/E^8

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2242

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*F
^(a + b*x + c*x^2))/(e*(m + 1)), x] + (-Dist[(2*c*Log[F])/(e^2*(m + 1)), Int[(d + e*x)^(m + 2)*F^(a + b*x + c*
x^2), x], x] - Dist[((b*e - 2*c*d)*Log[F])/(e^2*(m + 1)), Int[(d + e*x)^(m + 1)*F^(a + b*x + c*x^2), x], x]) /
; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && LtQ[m, -1]

Rule 2288

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {32 e^4 x^3+2 x^4+e^{32-16 x+2 x^2} \left (-16 x^3+4 x^4+e^8 \left (-512-4096 x+1024 x^2\right )+e^4 \left (-32 x-512 x^2+128 x^3\right )\right )+e^{16-8 x+x^2} \left (2 x^3-16 x^4+4 x^5+e^8 \left (-512 x-4096 x^2+1024 x^3\right )+e^4 \left (-512 x^3+128 x^4\right )\right )}{x^3} \, dx}{e^8}\\ &=\frac {\int \left (2 \left (16 e^4+x\right )+\frac {4 e^{2 (-4+x)^2} \left (16 e^4+x\right ) \left (-8 e^4-64 e^4 x-4 \left (1-4 e^4\right ) x^2+x^3\right )}{x^3}+\frac {2 e^{16-8 x+x^2} \left (16 e^4+x\right ) \left (-16 e^4+\left (1-128 e^4\right ) x-8 \left (1-4 e^4\right ) x^2+2 x^3\right )}{x^2}\right ) \, dx}{e^8}\\ &=\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {2 \int \frac {e^{16-8 x+x^2} \left (16 e^4+x\right ) \left (-16 e^4+\left (1-128 e^4\right ) x-8 \left (1-4 e^4\right ) x^2+2 x^3\right )}{x^2} \, dx}{e^8}+\frac {4 \int \frac {e^{2 (-4+x)^2} \left (16 e^4+x\right ) \left (-8 e^4-64 e^4 x-4 \left (1-4 e^4\right ) x^2+x^3\right )}{x^3} \, dx}{e^8}\\ &=\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+\frac {2 \int \left (e^{16-8 x+x^2} \left (1+256 e^4 \left (-1+2 e^4\right )\right )-\frac {256 e^{24-8 x+x^2}}{x^2}-\frac {2048 e^{24-8 x+x^2}}{x}+8 e^{16-8 x+x^2} \left (-1+8 e^4\right ) x+2 e^{16-8 x+x^2} x^2\right ) \, dx}{e^8}\\ &=\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+\frac {4 \int e^{16-8 x+x^2} x^2 \, dx}{e^8}-\frac {512 \int \frac {e^{24-8 x+x^2}}{x^2} \, dx}{e^8}-\frac {4096 \int \frac {e^{24-8 x+x^2}}{x} \, dx}{e^8}-\frac {\left (16 \left (1-8 e^4\right )\right ) \int e^{16-8 x+x^2} x \, dx}{e^8}+\frac {\left (2 \left (1-256 e^4+512 e^8\right )\right ) \int e^{16-8 x+x^2} \, dx}{e^8}\\ &=\frac {512 e^{16-8 x+x^2}}{x}+\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+\frac {4 \int e^{(-4+x)^2} x^2 \, dx}{e^8}-\frac {1024 \int e^{24-8 x+x^2} \, dx}{e^8}-\frac {\left (16 \left (1-8 e^4\right )\right ) \int e^{(-4+x)^2} x \, dx}{e^8}+\frac {\left (2 \left (1-256 e^4+512 e^8\right )\right ) \int e^{(-4+x)^2} \, dx}{e^8}\\ &=\frac {512 e^{16-8 x+x^2}}{x}+\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}-\frac {\left (1-256 e^4+512 e^8\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}-1024 \int e^{\frac {1}{4} (-8+2 x)^2} \, dx+\frac {4 \int \left (16 e^{(-4+x)^2}+8 e^{(-4+x)^2} (-4+x)+e^{(-4+x)^2} (-4+x)^2\right ) \, dx}{e^8}-\frac {\left (16 \left (1-8 e^4\right )\right ) \int \left (4 e^{(-4+x)^2}+e^{(-4+x)^2} (-4+x)\right ) \, dx}{e^8}\\ &=\frac {512 e^{16-8 x+x^2}}{x}+\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+512 \sqrt {\pi } \text {erfi}(4-x)-\frac {\left (1-256 e^4+512 e^8\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}+\frac {4 \int e^{(-4+x)^2} (-4+x)^2 \, dx}{e^8}+\frac {32 \int e^{(-4+x)^2} (-4+x) \, dx}{e^8}+\frac {64 \int e^{(-4+x)^2} \, dx}{e^8}-\frac {\left (16 \left (1-8 e^4\right )\right ) \int e^{(-4+x)^2} (-4+x) \, dx}{e^8}-\frac {\left (64 \left (1-8 e^4\right )\right ) \int e^{(-4+x)^2} \, dx}{e^8}\\ &=16 e^{-8+(-4+x)^2}-8 e^{-8+(-4+x)^2} \left (1-8 e^4\right )-2 e^{-8+(-4+x)^2} (4-x)+\frac {512 e^{16-8 x+x^2}}{x}+\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+512 \sqrt {\pi } \text {erfi}(4-x)-\frac {32 \sqrt {\pi } \text {erfi}(4-x)}{e^8}+\frac {32 \left (1-8 e^4\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}-\frac {\left (1-256 e^4+512 e^8\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}-\frac {2 \int e^{(-4+x)^2} \, dx}{e^8}\\ &=16 e^{-8+(-4+x)^2}-8 e^{-8+(-4+x)^2} \left (1-8 e^4\right )-2 e^{-8+(-4+x)^2} (4-x)+\frac {512 e^{16-8 x+x^2}}{x}+\frac {\left (16 e^4+x\right )^2}{e^8}+\frac {e^{-8+2 (4-x)^2} \left (16 e^4+x\right ) \left (64 e^4 x+4 \left (1-4 e^4\right ) x^2-x^3\right )}{(4-x) x^3}+512 \sqrt {\pi } \text {erfi}(4-x)-\frac {31 \sqrt {\pi } \text {erfi}(4-x)}{e^8}+\frac {32 \left (1-8 e^4\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}-\frac {\left (1-256 e^4+512 e^8\right ) \sqrt {\pi } \text {erfi}(4-x)}{e^8}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.20, size = 57, normalized size = 2.38 \begin {gather*} \frac {32 e^4 x^3+x^4+e^{2 (-4+x)^2} \left (16 e^4+x\right )^2+2 e^{(-4+x)^2} x \left (16 e^4+x\right )^2}{e^8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*E^4*x^3 + 2*x^4 + E^(32 - 16*x + 2*x^2)*(-16*x^3 + 4*x^4 + E^8*(-512 - 4096*x + 1024*x^2) + E^4*
(-32*x - 512*x^2 + 128*x^3)) + E^(16 - 8*x + x^2)*(2*x^3 - 16*x^4 + 4*x^5 + E^8*(-512*x - 4096*x^2 + 1024*x^3)
+ E^4*(-512*x^3 + 128*x^4)))/(E^8*x^3),x]

[Out]

(32*E^4*x^3 + x^4 + E^(2*(-4 + x)^2)*(16*E^4 + x)^2 + 2*E^(-4 + x)^2*x*(16*E^4 + x)^2)/(E^8*x^2)

________________________________________________________________________________________

fricas [B]  time = 1.21, size = 69, normalized size = 2.88 \begin {gather*} \frac {{\left (x^{4} + 32 \, x^{3} e^{4} + {\left (x^{2} + 32 \, x e^{4} + 256 \, e^{8}\right )} e^{\left (2 \, x^{2} - 16 \, x + 32\right )} + 2 \, {\left (x^{3} + 32 \, x^{2} e^{4} + 256 \, x e^{8}\right )} e^{\left (x^{2} - 8 \, x + 16\right )}\right )} e^{\left (-8\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1024*x^2-4096*x-512)*exp(2)^4+(128*x^3-512*x^2-32*x)*exp(2)^2+4*x^4-16*x^3)*exp(x^2-8*x+16)^2+((1
024*x^3-4096*x^2-512*x)*exp(2)^4+(128*x^4-512*x^3)*exp(2)^2+4*x^5-16*x^4+2*x^3)*exp(x^2-8*x+16)+32*x^3*exp(2)^
2+2*x^4)/x^3/exp(2)^4,x, algorithm="fricas")

[Out]

(x^4 + 32*x^3*e^4 + (x^2 + 32*x*e^4 + 256*e^8)*e^(2*x^2 - 16*x + 32) + 2*(x^3 + 32*x^2*e^4 + 256*x*e^8)*e^(x^2
- 8*x + 16))*e^(-8)/x^2

________________________________________________________________________________________

giac [B]  time = 0.48, size = 99, normalized size = 4.12 \begin {gather*} \frac {{\left (x^{4} + 32 \, x^{3} e^{4} + 2 \, x^{3} e^{\left (x^{2} - 8 \, x + 16\right )} + x^{2} e^{\left (2 \, x^{2} - 16 \, x + 32\right )} + 64 \, x^{2} e^{\left (x^{2} - 8 \, x + 20\right )} + 32 \, x e^{\left (2 \, x^{2} - 16 \, x + 36\right )} + 512 \, x e^{\left (x^{2} - 8 \, x + 24\right )} + 256 \, e^{\left (2 \, x^{2} - 16 \, x + 40\right )}\right )} e^{\left (-8\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1024*x^2-4096*x-512)*exp(2)^4+(128*x^3-512*x^2-32*x)*exp(2)^2+4*x^4-16*x^3)*exp(x^2-8*x+16)^2+((1
024*x^3-4096*x^2-512*x)*exp(2)^4+(128*x^4-512*x^3)*exp(2)^2+4*x^5-16*x^4+2*x^3)*exp(x^2-8*x+16)+32*x^3*exp(2)^
2+2*x^4)/x^3/exp(2)^4,x, algorithm="giac")

[Out]

(x^4 + 32*x^3*e^4 + 2*x^3*e^(x^2 - 8*x + 16) + x^2*e^(2*x^2 - 16*x + 32) + 64*x^2*e^(x^2 - 8*x + 20) + 32*x*e^
(2*x^2 - 16*x + 36) + 512*x*e^(x^2 - 8*x + 24) + 256*e^(2*x^2 - 16*x + 40))*e^(-8)/x^2

________________________________________________________________________________________

maple [B]  time = 0.17, size = 66, normalized size = 2.75

method	result	size
risch	\(32 x \,{\mathrm e}^{-4}+{\mathrm e}^{-8} x^{2}+\frac {\left (256 \,{\mathrm e}^{8}+32 x \,{\mathrm e}^{4}+x^{2}\right ) {\mathrm e}^{2 \left (x -2\right ) \left (x -6\right )}}{x^{2}}+\frac {2 \left (256 \,{\mathrm e}^{8}+32 x \,{\mathrm e}^{4}+x^{2}\right ) {\mathrm e}^{x^{2}-8 x +8}}{x}\)	\(66\)
norman	\(\frac {\left ({\mathrm e}^{-2} x^{4}+x^{2} {\mathrm e}^{-2} {\mathrm e}^{2 x^{2}-16 x +32}+32 x^{3} {\mathrm e}^{2}+256 \,{\mathrm e}^{6} {\mathrm e}^{2 x^{2}-16 x +32}+512 x \,{\mathrm e}^{6} {\mathrm e}^{x^{2}-8 x +16}+64 x^{2} {\mathrm e}^{2} {\mathrm e}^{x^{2}-8 x +16}+2 \,{\mathrm e}^{-2} x^{3} {\mathrm e}^{x^{2}-8 x +16}+32 \,{\mathrm e}^{2} x \,{\mathrm e}^{2 x^{2}-16 x +32}\right ) {\mathrm e}^{-6}}{x^{2}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((1024*x^2-4096*x-512)*exp(2)^4+(128*x^3-512*x^2-32*x)*exp(2)^2+4*x^4-16*x^3)*exp(x^2-8*x+16)^2+((1024*x^
3-4096*x^2-512*x)*exp(2)^4+(128*x^4-512*x^3)*exp(2)^2+4*x^5-16*x^4+2*x^3)*exp(x^2-8*x+16)+32*x^3*exp(2)^2+2*x^
4)/x^3/exp(2)^4,x,method=_RETURNVERBOSE)

[Out]

32*x*exp(-4)+exp(-8)*x^2+1/x^2*(256*exp(8)+32*x*exp(4)+x^2)*exp(2*(x-2)*(x-6))+2/x*(256*exp(8)+32*x*exp(4)+x^2
)*exp(x^2-8*x+8)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (512 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - 4 i\right ) e^{8} - 256 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - 4 i\right ) e^{4} + \frac {2 \, {\left (x - 4\right )}^{3} \Gamma \left (\frac {3}{2}, -{\left (x - 4\right )}^{2}\right )}{\left (-{\left (x - 4\right )}^{2}\right )^{\frac {3}{2}}} - x^{2} - 64 \, {\left (\frac {4 \, \sqrt {\pi } {\left (x - 4\right )} {\left (\operatorname {erf}\left (\sqrt {-{\left (x - 4\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x - 4\right )}^{2}}} + e^{\left ({\left (x - 4\right )}^{2}\right )}\right )} e^{4} - 32 \, x e^{4} + i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - 4 i\right ) - \frac {{\left (x^{2} e^{32} + 32 \, x e^{36} + 256 \, e^{40}\right )} e^{\left (2 \, x^{2} - 16 \, x\right )}}{x^{2}} - 8 \, e^{\left ({\left (x - 4\right )}^{2}\right )} + 2 \, \int \frac {256 \, {\left (8 \, x e^{24} + e^{24}\right )} e^{\left (x^{2} - 8 \, x\right )}}{x^{2}}\,{d x}\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1024*x^2-4096*x-512)*exp(2)^4+(128*x^3-512*x^2-32*x)*exp(2)^2+4*x^4-16*x^3)*exp(x^2-8*x+16)^2+((1
024*x^3-4096*x^2-512*x)*exp(2)^4+(128*x^4-512*x^3)*exp(2)^2+4*x^5-16*x^4+2*x^3)*exp(x^2-8*x+16)+32*x^3*exp(2)^
2+2*x^4)/x^3/exp(2)^4,x, algorithm="maxima")

[Out]

-(512*I*sqrt(pi)*erf(I*x - 4*I)*e^8 - 256*I*sqrt(pi)*erf(I*x - 4*I)*e^4 + 2*(x - 4)^3*gamma(3/2, -(x - 4)^2)/(
-(x - 4)^2)^(3/2) - x^2 - 64*(4*sqrt(pi)*(x - 4)*(erf(sqrt(-(x - 4)^2)) - 1)/sqrt(-(x - 4)^2) + e^((x - 4)^2))
*e^4 - 32*x*e^4 + I*sqrt(pi)*erf(I*x - 4*I) - (x^2*e^32 + 32*x*e^36 + 256*e^40)*e^(2*x^2 - 16*x)/x^2 - 8*e^((x
- 4)^2) + 2*integrate(256*(8*x*e^24 + e^24)*e^(x^2 - 8*x)/x^2, x))*e^(-8)

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 110, normalized size = 4.58 \begin {gather*} {\mathrm {e}}^{-16\,x-8}\,\left (64\,{\mathrm {e}}^{x^2+8\,x+20}+{\mathrm {e}}^{2\,x^2+32}\right )+x^2\,{\mathrm {e}}^{-8}+\frac {256\,{\mathrm {e}}^{2\,x^2-16\,x+32}+x\,{\mathrm {e}}^{-16\,x-8}\,\left (512\,{\mathrm {e}}^{x^2+8\,x+24}+32\,{\mathrm {e}}^{2\,x^2+36}\right )}{x^2}+x\,{\mathrm {e}}^{-16\,x-8}\,\left (2\,{\mathrm {e}}^{x^2+8\,x+16}+32\,{\mathrm {e}}^{16\,x+4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-8)*(exp(2*x^2 - 16*x + 32)*(exp(8)*(4096*x - 1024*x^2 + 512) + exp(4)*(32*x + 512*x^2 - 128*x^3) +
16*x^3 - 4*x^4) - 32*x^3*exp(4) + exp(x^2 - 8*x + 16)*(exp(8)*(512*x + 4096*x^2 - 1024*x^3) + exp(4)*(512*x^3
- 128*x^4) - 2*x^3 + 16*x^4 - 4*x^5) - 2*x^4))/x^3,x)

[Out]

exp(- 16*x - 8)*(64*exp(8*x + x^2 + 20) + exp(2*x^2 + 32)) + x^2*exp(-8) + (256*exp(2*x^2 - 16*x + 32) + x*exp
(- 16*x - 8)*(512*exp(8*x + x^2 + 24) + 32*exp(2*x^2 + 36)))/x^2 + x*exp(- 16*x - 8)*(2*exp(8*x + x^2 + 16) +
32*exp(16*x + 4))

________________________________________________________________________________________

sympy [B]  time = 0.27, size = 88, normalized size = 3.67 \begin {gather*} \frac {x^{2}}{e^{8}} + \frac {32 x}{e^{4}} + \frac {\left (x^{3} e^{8} + 32 x^{2} e^{12} + 256 x e^{16}\right ) e^{2 x^{2} - 16 x + 32} + \left (2 x^{4} e^{8} + 64 x^{3} e^{12} + 512 x^{2} e^{16}\right ) e^{x^{2} - 8 x + 16}}{x^{3} e^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((1024*x**2-4096*x-512)*exp(2)**4+(128*x**3-512*x**2-32*x)*exp(2)**2+4*x**4-16*x**3)*exp(x**2-8*x+1
6)**2+((1024*x**3-4096*x**2-512*x)*exp(2)**4+(128*x**4-512*x**3)*exp(2)**2+4*x**5-16*x**4+2*x**3)*exp(x**2-8*x
+16)+32*x**3*exp(2)**2+2*x**4)/x**3/exp(2)**4,x)

[Out]

x**2*exp(-8) + 32*x*exp(-4) + ((x**3*exp(8) + 32*x**2*exp(12) + 256*x*exp(16))*exp(2*x**2 - 16*x + 32) + (2*x*
*4*exp(8) + 64*x**3*exp(12) + 512*x**2*exp(16))*exp(x**2 - 8*x + 16))*exp(-16)/x**3

________________________________________________________________________________________