3.63.20 \(\int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} (5-99 x-123 x^2-100 x^3)}{2+4 x+2 x^2} \, dx\)

Optimal. Leaf size=27 \[ 4+e^{\frac {3}{2+2 x}} \left (-1+e^{e^4}+x-25 x^2\right ) \]

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Rubi [B]  time = 0.50, antiderivative size = 57, normalized size of antiderivative = 2.11, number of steps used = 16, number of rules used = 9, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.155, Rules used = {27, 12, 6741, 6742, 2206, 2210, 2226, 2214, 2209} \begin {gather*} -25 e^{\frac {3}{2 (x+1)}} (x+1)^2+51 e^{\frac {3}{2 (x+1)}} (x+1)-\left (27-e^{e^4}\right ) e^{\frac {3}{2 (x+1)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^(E^4 + 3/(2 + 2*x)) + E^(3/(2 + 2*x))*(5 - 99*x - 123*x^2 - 100*x^3))/(2 + 4*x + 2*x^2),x]

[Out]

-(E^(3/(2*(1 + x)))*(27 - E^E^4)) + 51*E^(3/(2*(1 + x)))*(1 + x) - 25*E^(3/(2*(1 + x)))*(1 + x)^2

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 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 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} &=\int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{2 (1+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {-3 e^{e^4+\frac {3}{2+2 x}}+e^{\frac {3}{2+2 x}} \left (5-99 x-123 x^2-100 x^3\right )}{(1+x)^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {3}{2+2 x}} \left (5-3 e^{e^4}-99 x-123 x^2-100 x^3\right )}{(1+x)^2} \, dx\\ &=\frac {1}{2} \int \left (77 e^{\frac {3}{2+2 x}}-100 e^{\frac {3}{2+2 x}} x-\frac {3 e^{\frac {3}{2+2 x}} \left (-27+e^{e^4}\right )}{(1+x)^2}-\frac {153 e^{\frac {3}{2+2 x}}}{1+x}\right ) \, dx\\ &=\frac {77}{2} \int e^{\frac {3}{2+2 x}} \, dx-50 \int e^{\frac {3}{2+2 x}} x \, dx-\frac {153}{2} \int \frac {e^{\frac {3}{2+2 x}}}{1+x} \, dx+\frac {1}{2} \left (3 \left (27-e^{e^4}\right )\right ) \int \frac {e^{\frac {3}{2+2 x}}}{(1+x)^2} \, dx\\ &=-e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {77}{2} e^{\frac {3}{2 (1+x)}} (1+x)+\frac {153}{2} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-50 \int \left (-e^{\frac {3}{2+2 x}}+\frac {1}{2} e^{\frac {3}{2+2 x}} (2+2 x)\right ) \, dx+\frac {231}{2} \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx\\ &=-e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {77}{2} e^{\frac {3}{2 (1+x)}} (1+x)+\frac {75}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-25 \int e^{\frac {3}{2+2 x}} (2+2 x) \, dx+50 \int e^{\frac {3}{2+2 x}} \, dx\\ &=-e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+\frac {177}{2} e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2+\frac {75}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-\frac {75}{2} \int e^{\frac {3}{2+2 x}} \, dx+150 \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx\\ &=-e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+51 e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2-\frac {225}{4} \text {Ei}\left (\frac {3}{2 (1+x)}\right )-\frac {225}{2} \int \frac {e^{\frac {3}{2+2 x}}}{2+2 x} \, dx\\ &=-e^{\frac {3}{2 (1+x)}} \left (27-e^{e^4}\right )+51 e^{\frac {3}{2 (1+x)}} (1+x)-25 e^{\frac {3}{2 (1+x)}} (1+x)^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 25, normalized size = 0.93 \begin {gather*} e^{\frac {3}{2 (1+x)}} \left (-1+e^{e^4}+x-25 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^(E^4 + 3/(2 + 2*x)) + E^(3/(2 + 2*x))*(5 - 99*x - 123*x^2 - 100*x^3))/(2 + 4*x + 2*x^2),x]

[Out]

E^(3/(2*(1 + x)))*(-1 + E^E^4 + x - 25*x^2)

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fricas [A]  time = 1.17, size = 39, normalized size = 1.44 \begin {gather*} -{\left (25 \, x^{2} - x - e^{\left (e^{4}\right )} + 1\right )} e^{\left (\frac {2 \, {\left (x + 1\right )} e^{4} + 3}{2 \, {\left (x + 1\right )}} - e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(3/(2*x+2))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2*x+2)))/(2*x^2+4*x+2),x, algorithm=
"fricas")

[Out]

-(25*x^2 - x - e^(e^4) + 1)*e^(1/2*(2*(x + 1)*e^4 + 3)/(x + 1) - e^4)

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giac [B]  time = 0.39, size = 64, normalized size = 2.37 \begin {gather*} {\left (x + 1\right )}^{2} {\left (\frac {51 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}}{x + 1} - \frac {27 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}}{{\left (x + 1\right )}^{2}} + \frac {e^{\left (\frac {3}{2 \, {\left (x + 1\right )}} + e^{4}\right )}}{{\left (x + 1\right )}^{2}} - 25 \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(3/(2*x+2))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2*x+2)))/(2*x^2+4*x+2),x, algorithm=
"giac")

[Out]

(x + 1)^2*(51*e^(3/2/(x + 1))/(x + 1) - 27*e^(3/2/(x + 1))/(x + 1)^2 + e^(3/2/(x + 1) + e^4)/(x + 1)^2 - 25*e^
(3/2/(x + 1)))

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maple [A]  time = 0.44, size = 21, normalized size = 0.78

method	result	size
gosper	\(\left ({\mathrm e}^{{\mathrm e}^{4}}-1-25 x^{2}+x \right ) {\mathrm e}^{\frac {3}{2 \left (x +1\right )}}\)	\(21\)
risch	\(\left ({\mathrm e}^{{\mathrm e}^{4}}-1-25 x^{2}+x \right ) {\mathrm e}^{\frac {3}{2 \left (x +1\right )}}\)	\(21\)
derivativedivides	\(-25 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} \left (x +1\right )^{2}+51 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} \left (x +1\right )+{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} {\mathrm e}^{{\mathrm e}^{4}}-27 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}}\)	\(52\)
default	\(-25 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} \left (x +1\right )^{2}+51 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} \left (x +1\right )+{\mathrm e}^{\frac {3}{2 \left (x +1\right )}} {\mathrm e}^{{\mathrm e}^{4}}-27 \,{\mathrm e}^{\frac {3}{2 \left (x +1\right )}}\)	\(52\)
norman	\(\frac {\left ({\mathrm e}^{{\mathrm e}^{4}}-1\right ) {\mathrm e}^{\frac {3}{2 x +2}}+x \,{\mathrm e}^{{\mathrm e}^{4}} {\mathrm e}^{\frac {3}{2 x +2}}-24 x^{2} {\mathrm e}^{\frac {3}{2 x +2}}-25 x^{3} {\mathrm e}^{\frac {3}{2 x +2}}}{x +1}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*exp(3/(2*x+2))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2*x+2)))/(2*x^2+4*x+2),x,method=_RETURNVER
BOSE)

[Out]

(exp(exp(4))-1-25*x^2+x)*exp(3/2/(x+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (25 \, x^{2} - x\right )} e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )} - \frac {1}{2} \, {\left (3 \, e^{\left (e^{4}\right )} + 2\right )} \int \frac {e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )}}{x^{2} + 2 \, x + 1}\,{d x} - \frac {5}{3} \, e^{\left (\frac {3}{2 \, {\left (x + 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(3/(2*x+2))*exp(exp(4))+(-100*x^3-123*x^2-99*x+5)*exp(3/(2*x+2)))/(2*x^2+4*x+2),x, algorithm=
"maxima")

[Out]

-(25*x^2 - x)*e^(3/2/(x + 1)) - 1/2*(3*e^(e^4) + 2)*integrate(e^(3/2/(x + 1))/(x^2 + 2*x + 1), x) - 5/3*e^(3/2
/(x + 1))

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mupad [B]  time = 0.16, size = 22, normalized size = 0.81 \begin {gather*} {\mathrm {e}}^{\frac {3}{2\,\left (x+1\right )}}\,\left (-25\,x^2+x+{\mathrm {e}}^{{\mathrm {e}}^4}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3/(2*x + 2))*(99*x + 123*x^2 + 100*x^3 - 5) + 3*exp(3/(2*x + 2))*exp(exp(4)))/(4*x + 2*x^2 + 2),x)

[Out]

exp(3/(2*(x + 1)))*(x + exp(exp(4)) - 25*x^2 - 1)

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sympy [A]  time = 0.26, size = 20, normalized size = 0.74 \begin {gather*} \left (- 25 x^{2} + x - 1 + e^{e^{4}}\right ) e^{\frac {3}{2 x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*exp(3/(2*x+2))*exp(exp(4))+(-100*x**3-123*x**2-99*x+5)*exp(3/(2*x+2)))/(2*x**2+4*x+2),x)

[Out]

(-25*x**2 + x - 1 + exp(exp(4)))*exp(3/(2*x + 2))

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