3.39.75 \(\int \frac {e^{9+\frac {4}{8 x+e x}} (-4+16 x+2 e x)}{8+e} \, dx\)

Optimal. Leaf size=18 \[ e^{9+\frac {4}{(8+e) x}} x^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {6, 12, 2227, 2226, 2206, 2210, 2214} \begin {gather*} e^{\frac {4}{(8+e) x}+9} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(9 + 4/(8*x + E*x))*(-4 + 16*x + 2*E*x))/(8 + E),x]

[Out]

E^(9 + 4/((8 + E)*x))*x^2

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 2206

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

Rule 2210

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

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{9+\frac {4}{8 x+e x}} (-4+(16+2 e) x)}{8+e} \, dx\\ &=\frac {\int e^{9+\frac {4}{8 x+e x}} (-4+(16+2 e) x) \, dx}{8+e}\\ &=\frac {\int e^{9+\frac {4}{(8+e) x}} (-4+(16+2 e) x) \, dx}{8+e}\\ &=\frac {\int \left (-4 e^{9+\frac {4}{(8+e) x}}+2 e^{9+\frac {4}{(8+e) x}} (8+e) x\right ) \, dx}{8+e}\\ &=2 \int e^{9+\frac {4}{(8+e) x}} x \, dx-\frac {4 \int e^{9+\frac {4}{(8+e) x}} \, dx}{8+e}\\ &=-\frac {4 e^{9+\frac {4}{(8+e) x}} x}{8+e}+e^{9+\frac {4}{(8+e) x}} x^2-\frac {16 \int \frac {e^{9+\frac {4}{(8+e) x}}}{x} \, dx}{(8+e)^2}+\frac {4 \int e^{9+\frac {4}{(8+e) x}} \, dx}{8+e}\\ &=e^{9+\frac {4}{(8+e) x}} x^2+\frac {16 e^9 \text {Ei}\left (\frac {4}{(8+e) x}\right )}{(8+e)^2}+\frac {16 \int \frac {e^{9+\frac {4}{(8+e) x}}}{x} \, dx}{(8+e)^2}\\ &=e^{9+\frac {4}{(8+e) x}} x^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 18, normalized size = 1.00 \begin {gather*} e^{9+\frac {4}{(8+e) x}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(9 + 4/(8*x + E*x))*(-4 + 16*x + 2*E*x))/(8 + E),x]

[Out]

E^(9 + 4/((8 + E)*x))*x^2

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 26, normalized size = 1.44 \begin {gather*} x^{2} e^{\left (\frac {9 \, x e + 72 \, x + 4}{x e + 8 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)+16*x-4)*exp(4)*exp(5)*exp(4/(x*exp(1)+8*x))/(exp(1)+8),x, algorithm="fricas")

[Out]

x^2*e^((9*x*e + 72*x + 4)/(x*e + 8*x))

________________________________________________________________________________________

giac [B]  time = 0.20, size = 181, normalized size = 10.06 \begin {gather*} \frac {16 \, e^{\left (\frac {9 \, x e + 72 \, x + 4}{x e + 8 \, x}\right )}}{\frac {{\left (9 \, x e + 72 \, x + 4\right )}^{2} e^{2}}{{\left (x e + 8 \, x\right )}^{2}} - \frac {18 \, {\left (9 \, x e + 72 \, x + 4\right )} e^{2}}{x e + 8 \, x} + \frac {16 \, {\left (9 \, x e + 72 \, x + 4\right )}^{2} e}{{\left (x e + 8 \, x\right )}^{2}} - \frac {288 \, {\left (9 \, x e + 72 \, x + 4\right )} e}{x e + 8 \, x} + \frac {64 \, {\left (9 \, x e + 72 \, x + 4\right )}^{2}}{{\left (x e + 8 \, x\right )}^{2}} - \frac {1152 \, {\left (9 \, x e + 72 \, x + 4\right )}}{x e + 8 \, x} + 81 \, e^{2} + 1296 \, e + 5184} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)+16*x-4)*exp(4)*exp(5)*exp(4/(x*exp(1)+8*x))/(exp(1)+8),x, algorithm="giac")

[Out]

16*e^((9*x*e + 72*x + 4)/(x*e + 8*x))/((9*x*e + 72*x + 4)^2*e^2/(x*e + 8*x)^2 - 18*(9*x*e + 72*x + 4)*e^2/(x*e
+ 8*x) + 16*(9*x*e + 72*x + 4)^2*e/(x*e + 8*x)^2 - 288*(9*x*e + 72*x + 4)*e/(x*e + 8*x) + 64*(9*x*e + 72*x +
4)^2/(x*e + 8*x)^2 - 1152*(9*x*e + 72*x + 4)/(x*e + 8*x) + 81*e^2 + 1296*e + 5184)

________________________________________________________________________________________

maple [A]  time = 0.10, size = 19, normalized size = 1.06

method	result	size
gosper	\({\mathrm e}^{9+\frac {4}{\left ({\mathrm e}+8\right ) x}} x^{2}\)	\(19\)
norman	\(x^{2} {\mathrm e}^{4} {\mathrm e}^{5} {\mathrm e}^{\frac {4}{x \,{\mathrm e}+8 x}}\)	\(22\)
risch	\(x^{2} {\mathrm e}^{\frac {9 x \,{\mathrm e}+72 x +4}{x \left ({\mathrm e}+8\right )}}\)	\(26\)
derivativedivides	\(-\frac {16 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}}{4}+\expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )+\frac {-\frac {{\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}} x^{2} \left ({\mathrm e}+8\right )^{2}}{2}-2 x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}-8 \expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )}{{\mathrm e}+8}+\frac {2 \,{\mathrm e} \left (-\frac {{\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}} x^{2} \left ({\mathrm e}+8\right )^{2}}{32}-\frac {x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}}{8}-\frac {\expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )}{2}\right )}{{\mathrm e}+8}\right )}{\left ({\mathrm e}+8\right )^{2}}\)	\(180\)
default	\(-\frac {16 \,{\mathrm e}^{4} {\mathrm e}^{5} \left (\frac {x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}}{4}+\expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )+\frac {-\frac {{\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}} x^{2} \left ({\mathrm e}+8\right )^{2}}{2}-2 x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}-8 \expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )}{{\mathrm e}+8}+\frac {2 \,{\mathrm e} \left (-\frac {{\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}} x^{2} \left ({\mathrm e}+8\right )^{2}}{32}-\frac {x \left ({\mathrm e}+8\right ) {\mathrm e}^{\frac {4}{\left ({\mathrm e}+8\right ) x}}}{8}-\frac {\expIntegralEi \left (1, -\frac {4}{\left ({\mathrm e}+8\right ) x}\right )}{2}\right )}{{\mathrm e}+8}\right )}{\left ({\mathrm e}+8\right )^{2}}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(1)+16*x-4)*exp(4)*exp(5)*exp(4/(x*exp(1)+8*x))/(exp(1)+8),x,method=_RETURNVERBOSE)

[Out]

exp(9+4/(exp(1)+8)/x)*x^2

________________________________________________________________________________________

maxima [C]  time = 0.38, size = 77, normalized size = 4.28 \begin {gather*} \frac {16 \, {\left (\frac {e^{9} \Gamma \left (-1, -\frac {4}{x {\left (e + 8\right )}}\right )}{e + 8} + \frac {2 \, e^{10} \Gamma \left (-2, -\frac {4}{x {\left (e + 8\right )}}\right )}{{\left (e + 8\right )}^{2}} + \frac {16 \, e^{9} \Gamma \left (-2, -\frac {4}{x {\left (e + 8\right )}}\right )}{{\left (e + 8\right )}^{2}}\right )}}{e + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)+16*x-4)*exp(4)*exp(5)*exp(4/(x*exp(1)+8*x))/(exp(1)+8),x, algorithm="maxima")

[Out]

16*(e^9*gamma(-1, -4/(x*(e + 8)))/(e + 8) + 2*e^10*gamma(-2, -4/(x*(e + 8)))/(e + 8)^2 + 16*e^9*gamma(-2, -4/(
x*(e + 8)))/(e + 8)^2)/(e + 8)

________________________________________________________________________________________

mupad [B]  time = 2.15, size = 19, normalized size = 1.06 \begin {gather*} x^2\,{\mathrm {e}}^9\,{\mathrm {e}}^{\frac {4}{8\,x+x\,\mathrm {e}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(9)*exp(4/(8*x + x*exp(1)))*(16*x + 2*x*exp(1) - 4))/(exp(1) + 8),x)

[Out]

x^2*exp(9)*exp(4/(8*x + x*exp(1)))

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 17, normalized size = 0.94 \begin {gather*} x^{2} e^{9} e^{\frac {4}{e x + 8 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(1)+16*x-4)*exp(4)*exp(5)*exp(4/(x*exp(1)+8*x))/(exp(1)+8),x)

[Out]

x**2*exp(9)*exp(4/(E*x + 8*x))

________________________________________________________________________________________