3.44.78 \(\int \frac {1}{5} e^{\frac {1}{5} (-16 x-2 e x-2 e^4 x+2 x^2)} (-48-6 e-6 e^4+5 e^{\frac {1}{5} (16 x+2 e x+2 e^4 x-2 x^2)}+12 x) \, dx\)

Optimal. Leaf size=25 \[ 3 e^{-\frac {16 x}{5}-\frac {2}{5} \left (e+e^4-x\right ) x}+x \]

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Rubi [A]  time = 1.11, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.145, Rules used = {12, 6741, 6742, 6688, 8, 2235, 2234, 2204, 2244, 2240} \begin {gather*} 3 e^{\frac {2 x^2}{5}-\frac {2}{5} \left (8+e+e^4\right ) x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-16*x - 2*E*x - 2*E^4*x + 2*x^2)/5)*(-48 - 6*E - 6*E^4 + 5*E^((16*x + 2*E*x + 2*E^4*x - 2*x^2)/5) + 1
2*x))/5,x]

[Out]

3*E^((-2*(8 + E + E^4)*x)/5 + (2*x^2)/5) + x

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 2235

Int[(F_)^(v_), x_Symbol] :> Int[F^ExpandToSum[v, x], x] /; FreeQ[F, x] && QuadraticQ[v, x] &&  !QuadraticMatch
Q[v, x]

Rule 2240

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] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2244

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])

Rule 6688

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

Rule 6741

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

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 {1}{5} \int e^{\frac {1}{5} \left (-16 x-2 e x-2 e^4 x+2 x^2\right )} \left (-48-6 e-6 e^4+5 e^{\frac {1}{5} \left (16 x+2 e x+2 e^4 x-2 x^2\right )}+12 x\right ) \, dx\\ &=\frac {1}{5} \int e^{\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )} \left (5 e^{\frac {1}{5} \left (16 x+2 e x+2 e^4 x-2 x^2\right )}-48 \left (1+\frac {1}{8} e \left (1+e^3\right )\right )+12 x\right ) \, dx\\ &=\frac {1}{5} \int \left (5 \exp \left (\frac {2}{5} \left (8+e+e^4-x\right ) x+\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )\right )-6 e^{\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )} \left (8+e+e^4\right )+12 e^{\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )} x\right ) \, dx\\ &=\frac {12}{5} \int e^{\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )} x \, dx-\frac {1}{5} \left (6 \left (8+e+e^4\right )\right ) \int e^{\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )} \, dx+\int \exp \left (\frac {2}{5} \left (8+e+e^4-x\right ) x+\frac {1}{5} x \left (-2 \left (8+e+e^4\right )+2 x\right )\right ) \, dx\\ &=\frac {12}{5} \int e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}} x \, dx-\frac {1}{5} \left (6 \left (8+e+e^4\right )\right ) \int e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}} \, dx+\int 1 \, dx\\ &=3 e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}}+x+\frac {1}{5} \left (6 \left (8+e+e^4\right )\right ) \int e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}} \, dx-\frac {1}{5} \left (6 e^{-\frac {1}{10} \left (8+e+e^4\right )^2} \left (8+e+e^4\right )\right ) \int e^{\frac {5}{8} \left (-\frac {2}{5} \left (8+e+e^4\right )+\frac {4 x}{5}\right )^2} \, dx\\ &=3 e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}}+x+3 e^{-\frac {1}{10} \left (8+e+e^4\right )^2} \left (8+e+e^4\right ) \sqrt {\frac {\pi }{10}} \text {erfi}\left (\frac {8+e+e^4-2 x}{\sqrt {10}}\right )+\frac {1}{5} \left (6 e^{-\frac {1}{10} \left (8+e+e^4\right )^2} \left (8+e+e^4\right )\right ) \int e^{\frac {5}{8} \left (-\frac {2}{5} \left (8+e+e^4\right )+\frac {4 x}{5}\right )^2} \, dx\\ &=3 e^{-\frac {2}{5} \left (8+e+e^4\right ) x+\frac {2 x^2}{5}}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 22, normalized size = 0.88 \begin {gather*} 3 e^{\frac {2}{5} x \left (-8-e-e^4+x\right )}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-16*x - 2*E*x - 2*E^4*x + 2*x^2)/5)*(-48 - 6*E - 6*E^4 + 5*E^((16*x + 2*E*x + 2*E^4*x - 2*x^2)/
5) + 12*x))/5,x]

[Out]

3*E^((2*x*(-8 - E - E^4 + x))/5) + x

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fricas [B]  time = 0.61, size = 45, normalized size = 1.80 \begin {gather*} {\left (x e^{\left (-\frac {2}{5} \, x^{2} + \frac {2}{5} \, x e^{4} + \frac {2}{5} \, x e + \frac {16}{5} \, x\right )} + 3\right )} e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{4} - \frac {2}{5} \, x e - \frac {16}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2/5*x*exp(4)+2/5*x*exp(1)-2/5*x^2+16/5*x)-6*exp(4)-6*exp(1)+12*x-48)/exp(2/5*x*exp(4)+2/5
*x*exp(1)-2/5*x^2+16/5*x),x, algorithm="fricas")

[Out]

(x*e^(-2/5*x^2 + 2/5*x*e^4 + 2/5*x*e + 16/5*x) + 3)*e^(2/5*x^2 - 2/5*x*e^4 - 2/5*x*e - 16/5*x)

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giac [C]  time = 0.38, size = 179, normalized size = 7.16 \begin {gather*} \frac {3}{10} i \, \sqrt {10} \sqrt {\pi } {\left (e^{4} + e\right )} \operatorname {erf}\left (-\frac {1}{10} i \, \sqrt {10} {\left (2 \, x - e^{4} - e - 8\right )}\right ) e^{\left (-\frac {1}{10} \, e^{8} - \frac {1}{5} \, e^{5} - \frac {8}{5} \, e^{4} - \frac {1}{10} \, e^{2} - \frac {8}{5} \, e - \frac {32}{5}\right )} - \frac {3}{10} i \, \sqrt {10} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{10} i \, \sqrt {10} {\left (2 \, x - e^{4} - e - 8\right )}\right ) e^{\left (-\frac {1}{10} \, e^{8} - \frac {1}{5} \, e^{5} - \frac {8}{5} \, e^{4} - \frac {1}{10} \, e^{2} - \frac {8}{5} \, e - \frac {12}{5}\right )} - \frac {3}{10} i \, \sqrt {10} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{10} i \, \sqrt {10} {\left (2 \, x - e^{4} - e - 8\right )}\right ) e^{\left (-\frac {1}{10} \, e^{8} - \frac {1}{5} \, e^{5} - \frac {8}{5} \, e^{4} - \frac {1}{10} \, e^{2} - \frac {8}{5} \, e - \frac {27}{5}\right )} + x + 3 \, e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{4} - \frac {2}{5} \, x e - \frac {16}{5} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2/5*x*exp(4)+2/5*x*exp(1)-2/5*x^2+16/5*x)-6*exp(4)-6*exp(1)+12*x-48)/exp(2/5*x*exp(4)+2/5
*x*exp(1)-2/5*x^2+16/5*x),x, algorithm="giac")

[Out]

3/10*I*sqrt(10)*sqrt(pi)*(e^4 + e)*erf(-1/10*I*sqrt(10)*(2*x - e^4 - e - 8))*e^(-1/10*e^8 - 1/5*e^5 - 8/5*e^4
- 1/10*e^2 - 8/5*e - 32/5) - 3/10*I*sqrt(10)*sqrt(pi)*erf(-1/10*I*sqrt(10)*(2*x - e^4 - e - 8))*e^(-1/10*e^8 -
1/5*e^5 - 8/5*e^4 - 1/10*e^2 - 8/5*e - 12/5) - 3/10*I*sqrt(10)*sqrt(pi)*erf(-1/10*I*sqrt(10)*(2*x - e^4 - e -
8))*e^(-1/10*e^8 - 1/5*e^5 - 8/5*e^4 - 1/10*e^2 - 8/5*e - 27/5) + x + 3*e^(2/5*x^2 - 2/5*x*e^4 - 2/5*x*e - 16
/5*x)

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maple [A]  time = 0.06, size = 18, normalized size = 0.72

method	result	size
risch	\(x +3 \,{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{4}+{\mathrm e}-x +8\right )}{5}}\)	\(18\)
norman	\(\left (3+x \,{\mathrm e}^{\frac {2 x \,{\mathrm e}^{4}}{5}+\frac {2 x \,{\mathrm e}}{5}-\frac {2 x^{2}}{5}+\frac {16 x}{5}}\right ) {\mathrm e}^{-\frac {2 x \,{\mathrm e}^{4}}{5}-\frac {2 x \,{\mathrm e}}{5}+\frac {2 x^{2}}{5}-\frac {16 x}{5}}\)	\(48\)
default	\(x +\frac {12 i \sqrt {\pi }\, {\mathrm e}^{-\frac {5 \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right )^{2}}{8}} \sqrt {10}\, \erf \left (\frac {i \sqrt {10}\, x}{5}+\frac {i \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) \sqrt {10}}{4}\right )}{5}+3 \,{\mathrm e}^{\frac {2 x^{2}}{5}+\left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) x}+\frac {3 i \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) \sqrt {\pi }\, {\mathrm e}^{-\frac {5 \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right )^{2}}{8}} \sqrt {10}\, \erf \left (\frac {i \sqrt {10}\, x}{5}+\frac {i \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) \sqrt {10}}{4}\right )}{4}+\frac {3 i \sqrt {\pi }\, {\mathrm e}^{1-\frac {5 \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right )^{2}}{8}} \sqrt {10}\, \erf \left (\frac {i \sqrt {10}\, x}{5}+\frac {i \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) \sqrt {10}}{4}\right )}{10}+\frac {3 i \sqrt {\pi }\, {\mathrm e}^{4-\frac {5 \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right )^{2}}{8}} \sqrt {10}\, \erf \left (\frac {i \sqrt {10}\, x}{5}+\frac {i \left (-\frac {2 \,{\mathrm e}^{4}}{5}-\frac {2 \,{\mathrm e}}{5}-\frac {16}{5}\right ) \sqrt {10}}{4}\right )}{10}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(5*exp(2/5*x*exp(4)+2/5*x*exp(1)-2/5*x^2+16/5*x)-6*exp(4)-6*exp(1)+12*x-48)/exp(2/5*x*exp(4)+2/5*x*exp
(1)-2/5*x^2+16/5*x),x,method=_RETURNVERBOSE)

[Out]

x+3*exp(-2/5*x*(exp(4)+exp(1)-x+8))

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maxima [C]  time = 0.57, size = 270, normalized size = 10.80 \begin {gather*} \frac {12}{5} i \, \sqrt {5} \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{5} i \, \sqrt {5} \sqrt {2} x - \frac {1}{10} i \, \sqrt {5} \sqrt {2} {\left (e^{4} + e + 8\right )}\right ) e^{\left (-\frac {1}{10} \, {\left (e^{4} + e + 8\right )}^{2}\right )} + \frac {3}{10} i \, \sqrt {5} \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{5} i \, \sqrt {5} \sqrt {2} x - \frac {1}{10} i \, \sqrt {5} \sqrt {2} {\left (e^{4} + e + 8\right )}\right ) e^{\left (-\frac {1}{10} \, {\left (e^{4} + e + 8\right )}^{2} + 4\right )} + \frac {3}{10} i \, \sqrt {5} \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{5} i \, \sqrt {5} \sqrt {2} x - \frac {1}{10} i \, \sqrt {5} \sqrt {2} {\left (e^{4} + e + 8\right )}\right ) e^{\left (-\frac {1}{10} \, {\left (e^{4} + e + 8\right )}^{2} + 1\right )} + \frac {3}{20} \, \sqrt {5} \sqrt {2} {\left (\frac {\sqrt {5} \sqrt {2} \sqrt {\frac {2}{5}} \sqrt {\pi } {\left (2 \, x - e^{4} - e - 8\right )} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{10}} \sqrt {-{\left (2 \, x - e^{4} - e - 8\right )}^{2}}\right ) - 1\right )} {\left (e^{4} + e + 8\right )}}{\sqrt {-{\left (2 \, x - e^{4} - e - 8\right )}^{2}}} + 2 \, \sqrt {5} \sqrt {2} e^{\left (\frac {1}{10} \, {\left (2 \, x - e^{4} - e - 8\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{10} \, {\left (e^{4} + e + 8\right )}^{2}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2/5*x*exp(4)+2/5*x*exp(1)-2/5*x^2+16/5*x)-6*exp(4)-6*exp(1)+12*x-48)/exp(2/5*x*exp(4)+2/5
*x*exp(1)-2/5*x^2+16/5*x),x, algorithm="maxima")

[Out]

12/5*I*sqrt(5)*sqrt(2)*sqrt(pi)*erf(1/5*I*sqrt(5)*sqrt(2)*x - 1/10*I*sqrt(5)*sqrt(2)*(e^4 + e + 8))*e^(-1/10*(
e^4 + e + 8)^2) + 3/10*I*sqrt(5)*sqrt(2)*sqrt(pi)*erf(1/5*I*sqrt(5)*sqrt(2)*x - 1/10*I*sqrt(5)*sqrt(2)*(e^4 +
e + 8))*e^(-1/10*(e^4 + e + 8)^2 + 4) + 3/10*I*sqrt(5)*sqrt(2)*sqrt(pi)*erf(1/5*I*sqrt(5)*sqrt(2)*x - 1/10*I*s
qrt(5)*sqrt(2)*(e^4 + e + 8))*e^(-1/10*(e^4 + e + 8)^2 + 1) + 3/20*sqrt(5)*sqrt(2)*(sqrt(5)*sqrt(2)*sqrt(2/5)*
sqrt(pi)*(2*x - e^4 - e - 8)*(erf(sqrt(1/10)*sqrt(-(2*x - e^4 - e - 8)^2)) - 1)*(e^4 + e + 8)/sqrt(-(2*x - e^4
- e - 8)^2) + 2*sqrt(5)*sqrt(2)*e^(1/10*(2*x - e^4 - e - 8)^2))*e^(-1/10*(e^4 + e + 8)^2) + x

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mupad [B]  time = 0.24, size = 24, normalized size = 0.96 \begin {gather*} x+3\,{\mathrm {e}}^{\frac {2\,x^2}{5}-\frac {2\,x\,\mathrm {e}}{5}-\frac {2\,x\,{\mathrm {e}}^4}{5}-\frac {16\,x}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp((2*x^2)/5 - (2*x*exp(1))/5 - (2*x*exp(4))/5 - (16*x)/5)*((6*exp(1))/5 - exp((16*x)/5 + (2*x*exp(1))/5
+ (2*x*exp(4))/5 - (2*x^2)/5) - (12*x)/5 + (6*exp(4))/5 + 48/5),x)

[Out]

x + 3*exp((2*x^2)/5 - (2*x*exp(1))/5 - (2*x*exp(4))/5 - (16*x)/5)

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sympy [A]  time = 0.16, size = 32, normalized size = 1.28 \begin {gather*} x + 3 e^{\frac {2 x^{2}}{5} - \frac {2 x e^{4}}{5} - \frac {16 x}{5} - \frac {2 e x}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(5*exp(2/5*x*exp(4)+2/5*x*exp(1)-2/5*x**2+16/5*x)-6*exp(4)-6*exp(1)+12*x-48)/exp(2/5*x*exp(4)+2/
5*x*exp(1)-2/5*x**2+16/5*x),x)

[Out]

x + 3*exp(2*x**2/5 - 2*x*exp(4)/5 - 16*x/5 - 2*E*x/5)

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