3.34.85 \(\int \frac {e^x (16+24 x+e^4 x+12 x^2+2 x^3+(-8-20 x-18 x^2-7 x^3-x^4) \log (7))}{8+12 x+6 x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ -3+e^x \left (2+x \left (\frac {e^4}{x (2+x)^2}-\log (7)\right )\right ) \]

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Rubi [A]  time = 0.18, antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 11, number of rules used = 6, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {6, 2199, 2177, 2178, 2176, 2194} \begin {gather*} \frac {e^{x+4}}{(x+2)^2}-e^x (x+2) \log (7)+e^x (2+\log (7))+e^x \log (7) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(16 + 24*x + E^4*x + 12*x^2 + 2*x^3 + (-8 - 20*x - 18*x^2 - 7*x^3 - x^4)*Log[7]))/(8 + 12*x + 6*x^2 +
x^3),x]

[Out]

E^(4 + x)/(2 + x)^2 + E^x*Log[7] - E^x*(2 + x)*Log[7] + E^x*(2 + Log[7])

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (16+\left (24+e^4\right ) x+12 x^2+2 x^3+\left (-8-20 x-18 x^2-7 x^3-x^4\right ) \log (7)\right )}{8+12 x+6 x^2+x^3} \, dx\\ &=\int \left (-\frac {2 e^{4+x}}{(2+x)^3}+\frac {e^{4+x}}{(2+x)^2}-e^x (2+x) \log (7)+e^x (2+\log (7))\right ) \, dx\\ &=-\left (2 \int \frac {e^{4+x}}{(2+x)^3} \, dx\right )-\log (7) \int e^x (2+x) \, dx+(2+\log (7)) \int e^x \, dx+\int \frac {e^{4+x}}{(2+x)^2} \, dx\\ &=\frac {e^{4+x}}{(2+x)^2}-\frac {e^{4+x}}{2+x}-e^x (2+x) \log (7)+e^x (2+\log (7))+\log (7) \int e^x \, dx-\int \frac {e^{4+x}}{(2+x)^2} \, dx+\int \frac {e^{4+x}}{2+x} \, dx\\ &=\frac {e^{4+x}}{(2+x)^2}+e^2 \text {Ei}(2+x)+e^x \log (7)-e^x (2+x) \log (7)+e^x (2+\log (7))-\int \frac {e^{4+x}}{2+x} \, dx\\ &=\frac {e^{4+x}}{(2+x)^2}+e^x \log (7)-e^x (2+x) \log (7)+e^x (2+\log (7))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 37, normalized size = 1.37 \begin {gather*} -\frac {e^x \left (-8-e^4+x^3 \log (7)+x (-8+\log (2401))+x^2 (-2+\log (2401))\right )}{(2+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(16 + 24*x + E^4*x + 12*x^2 + 2*x^3 + (-8 - 20*x - 18*x^2 - 7*x^3 - x^4)*Log[7]))/(8 + 12*x + 6
*x^2 + x^3),x]

[Out]

-((E^x*(-8 - E^4 + x^3*Log[7] + x*(-8 + Log[2401]) + x^2*(-2 + Log[2401])))/(2 + x)^2)

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fricas [A]  time = 0.57, size = 41, normalized size = 1.52 \begin {gather*} \frac {{\left (2 \, x^{2} - {\left (x^{3} + 4 \, x^{2} + 4 \, x\right )} \log \relax (7) + 8 \, x + e^{4} + 8\right )} e^{x}}{x^{2} + 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*exp(x)/(x^3+6*x^2+12*x+8),x, alg
orithm="fricas")

[Out]

(2*x^2 - (x^3 + 4*x^2 + 4*x)*log(7) + 8*x + e^4 + 8)*e^x/(x^2 + 4*x + 4)

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giac [B]  time = 0.18, size = 59, normalized size = 2.19 \begin {gather*} -\frac {x^{3} e^{x} \log \relax (7) + 4 \, x^{2} e^{x} \log \relax (7) - 2 \, x^{2} e^{x} + 4 \, x e^{x} \log \relax (7) - 8 \, x e^{x} - e^{\left (x + 4\right )} - 8 \, e^{x}}{x^{2} + 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*exp(x)/(x^3+6*x^2+12*x+8),x, alg
orithm="giac")

[Out]

-(x^3*e^x*log(7) + 4*x^2*e^x*log(7) - 2*x^2*e^x + 4*x*e^x*log(7) - 8*x*e^x - e^(x + 4) - 8*e^x)/(x^2 + 4*x + 4
)

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maple [A]  time = 0.06, size = 25, normalized size = 0.93

method	result	size
default	\(\frac {{\mathrm e}^{4} {\mathrm e}^{x}}{\left (2+x \right )^{2}}+2 \,{\mathrm e}^{x}-\ln \relax (7) {\mathrm e}^{x} x\)	\(25\)
risch	\(\frac {\left (-\ln \relax (7) x^{3}-4 x^{2} \ln \relax (7)+{\mathrm e}^{4}-4 x \ln \relax (7)+2 x^{2}+8 x +8\right ) {\mathrm e}^{x}}{\left (2+x \right )^{2}}\)	\(40\)
gosper	\(\frac {\left (-\ln \relax (7) x^{3}-4 x^{2} \ln \relax (7)+{\mathrm e}^{4}-4 x \ln \relax (7)+2 x^{2}+8 x +8\right ) {\mathrm e}^{x}}{x^{2}+4 x +4}\)	\(47\)
norman	\(\frac {\left ({\mathrm e}^{4}+8\right ) {\mathrm e}^{x}+\left (-4 \ln \relax (7)+2\right ) x^{2} {\mathrm e}^{x}+\left (-4 \ln \relax (7)+8\right ) x \,{\mathrm e}^{x}-\ln \relax (7) x^{3} {\mathrm e}^{x}}{\left (2+x \right )^{2}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^4-7*x^3-18*x^2-20*x-8)*ln(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*exp(x)/(x^3+6*x^2+12*x+8),x,method=_RET
URNVERBOSE)

[Out]

exp(2)^2*exp(x)/(2+x)^2+2*exp(x)-ln(7)*exp(x)*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {{\left (x^{4} \log \relax (7) + 2 \, x^{3} {\left (3 \, \log \relax (7) - 1\right )} + 12 \, x^{2} {\left (\log \relax (7) - 1\right )} - x {\left (e^{4} + 12 \, \log \relax (7)\right )}\right )} e^{x}}{x^{3} + 6 \, x^{2} + 12 \, x + 8} - \frac {20 \, e^{x} \log \relax (7)}{x^{2} + 4 \, x + 4} + \frac {8 \, e^{\left (-2\right )} E_{3}\left (-x - 2\right ) \log \relax (7)}{{\left (x + 2\right )}^{2}} + \frac {24 \, e^{x}}{x^{2} + 4 \, x + 4} - \frac {16 \, e^{\left (-2\right )} E_{3}\left (-x - 2\right )}{{\left (x + 2\right )}^{2}} + \int \frac {2 \, {\left (x {\left (e^{4} + 24 \, \log \relax (7) - 24\right )} - e^{4} - 12 \, \log \relax (7)\right )} e^{x}}{x^{4} + 8 \, x^{3} + 24 \, x^{2} + 32 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^4-7*x^3-18*x^2-20*x-8)*log(7)+x*exp(2)^2+2*x^3+12*x^2+24*x+16)*exp(x)/(x^3+6*x^2+12*x+8),x, alg
orithm="maxima")

[Out]

-(x^4*log(7) + 2*x^3*(3*log(7) - 1) + 12*x^2*(log(7) - 1) - x*(e^4 + 12*log(7)))*e^x/(x^3 + 6*x^2 + 12*x + 8)
- 20*e^x*log(7)/(x^2 + 4*x + 4) + 8*e^(-2)*exp_integral_e(3, -x - 2)*log(7)/(x + 2)^2 + 24*e^x/(x^2 + 4*x + 4)
- 16*e^(-2)*exp_integral_e(3, -x - 2)/(x + 2)^2 + integrate(2*(x*(e^4 + 24*log(7) - 24) - e^4 - 12*log(7))*e^
x/(x^4 + 8*x^3 + 24*x^2 + 32*x + 16), x)

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mupad [B]  time = 2.12, size = 21, normalized size = 0.78 \begin {gather*} \frac {{\mathrm {e}}^{x+4}}{{\left (x+2\right )}^2}-{\mathrm {e}}^x\,\left (x\,\ln \relax (7)-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(24*x + x*exp(4) - log(7)*(20*x + 18*x^2 + 7*x^3 + x^4 + 8) + 12*x^2 + 2*x^3 + 16))/(12*x + 6*x^2
+ x^3 + 8),x)

[Out]

exp(x + 4)/(x + 2)^2 - exp(x)*(x*log(7) - 2)

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sympy [B]  time = 0.20, size = 46, normalized size = 1.70 \begin {gather*} \frac {\left (- x^{3} \log {\relax (7 )} - 4 x^{2} \log {\relax (7 )} + 2 x^{2} - 4 x \log {\relax (7 )} + 8 x + 8 + e^{4}\right ) e^{x}}{x^{2} + 4 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**4-7*x**3-18*x**2-20*x-8)*ln(7)+x*exp(2)**2+2*x**3+12*x**2+24*x+16)*exp(x)/(x**3+6*x**2+12*x+8)
,x)

[Out]

(-x**3*log(7) - 4*x**2*log(7) + 2*x**2 - 4*x*log(7) + 8*x + 8 + exp(4))*exp(x)/(x**2 + 4*x + 4)

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