3.65.52 \(\int \frac {e^x (1792-832 x+80 x^2+4 x^3+e^4 (-8 x-3 x^2+x^3))}{64-32 x+4 x^2} \, dx\)

Optimal. Leaf size=24 \[ e^x \left (27+x-\frac {e^4 x^2}{4 (4-x)}\right ) \]

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Rubi [B]  time = 0.14, antiderivative size = 58, normalized size of antiderivative = 2.42, number of steps used = 10, number of rules used = 7, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {27, 12, 2199, 2194, 2177, 2178, 2176} \begin {gather*} -\frac {1}{4} \left (4+e^4\right ) e^x (4-x)+\frac {1}{4} \left (128+9 e^4\right ) e^x-\frac {1}{4} \left (4+e^4\right ) e^x-\frac {4 e^{x+4}}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(1792 - 832*x + 80*x^2 + 4*x^3 + E^4*(-8*x - 3*x^2 + x^3)))/(64 - 32*x + 4*x^2),x]

[Out]

-1/4*(E^x*(4 + E^4)) + (E^x*(128 + 9*E^4))/4 - (4*E^(4 + x))/(4 - x) - (E^x*(4 + E^4)*(4 - x))/4

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 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 (1792-832 x+80 x^2+4 x^3+e^4 \left (-8 x-3 x^2+x^3\right )\right )}{4 (-4+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {e^x \left (1792-832 x+80 x^2+4 x^3+e^4 \left (-8 x-3 x^2+x^3\right )\right )}{(-4+x)^2} \, dx\\ &=\frac {1}{4} \int \left (e^x \left (128+9 e^4\right )-\frac {16 e^{4+x}}{(-4+x)^2}+\frac {16 e^{4+x}}{-4+x}+e^x \left (4+e^4\right ) (-4+x)\right ) \, dx\\ &=-\left (4 \int \frac {e^{4+x}}{(-4+x)^2} \, dx\right )+4 \int \frac {e^{4+x}}{-4+x} \, dx+\frac {1}{4} \left (4+e^4\right ) \int e^x (-4+x) \, dx+\frac {1}{4} \left (128+9 e^4\right ) \int e^x \, dx\\ &=\frac {1}{4} e^x \left (128+9 e^4\right )-\frac {4 e^{4+x}}{4-x}-\frac {1}{4} e^x \left (4+e^4\right ) (4-x)+4 e^8 \text {Ei}(-4+x)-4 \int \frac {e^{4+x}}{-4+x} \, dx+\frac {1}{4} \left (-4-e^4\right ) \int e^x \, dx\\ &=-\frac {1}{4} e^x \left (4+e^4\right )+\frac {1}{4} e^x \left (128+9 e^4\right )-\frac {4 e^{4+x}}{4-x}-\frac {1}{4} e^x \left (4+e^4\right ) (4-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 26, normalized size = 1.08 \begin {gather*} \frac {e^x \left (-432+92 x+\left (4+e^4\right ) x^2\right )}{4 (-4+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1792 - 832*x + 80*x^2 + 4*x^3 + E^4*(-8*x - 3*x^2 + x^3)))/(64 - 32*x + 4*x^2),x]

[Out]

(E^x*(-432 + 92*x + (4 + E^4)*x^2))/(4*(-4 + x))

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fricas [A]  time = 0.72, size = 25, normalized size = 1.04 \begin {gather*} \frac {{\left (x^{2} e^{4} + 4 \, x^{2} + 92 \, x - 432\right )} e^{x}}{4 \, {\left (x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-3*x^2-8*x)*exp(4)+4*x^3+80*x^2-832*x+1792)*exp(x)/(4*x^2-32*x+64),x, algorithm="fricas")

[Out]

1/4*(x^2*e^4 + 4*x^2 + 92*x - 432)*e^x/(x - 4)

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giac [A]  time = 0.13, size = 32, normalized size = 1.33 \begin {gather*} \frac {x^{2} e^{\left (x + 4\right )} + 4 \, x^{2} e^{x} + 92 \, x e^{x} - 432 \, e^{x}}{4 \, {\left (x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-3*x^2-8*x)*exp(4)+4*x^3+80*x^2-832*x+1792)*exp(x)/(4*x^2-32*x+64),x, algorithm="giac")

[Out]

1/4*(x^2*e^(x + 4) + 4*x^2*e^x + 92*x*e^x - 432*e^x)/(x - 4)

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maple [A]  time = 0.50, size = 26, normalized size = 1.08

method	result	size
gosper	\(\frac {\left (x^{2} {\mathrm e}^{4}+4 x^{2}+92 x -432\right ) {\mathrm e}^{x}}{4 x -16}\)	\(26\)
risch	\(\frac {\left (x^{2} {\mathrm e}^{4}+4 x^{2}+92 x -432\right ) {\mathrm e}^{x}}{4 x -16}\)	\(26\)
norman	\(\frac {\left (1+\frac {{\mathrm e}^{4}}{4}\right ) x^{2} {\mathrm e}^{x}+23 \,{\mathrm e}^{x} x -108 \,{\mathrm e}^{x}}{x -4}\)	\(29\)
default	\(\frac {{\mathrm e}^{4} \left ({\mathrm e}^{x} x +7 \,{\mathrm e}^{x}-\frac {64 \,{\mathrm e}^{x}}{x -4}-112 \,{\mathrm e}^{4} \expIntegralEi \left (1, -x +4\right )\right )}{4}+27 \,{\mathrm e}^{x}+{\mathrm e}^{x} x -2 \,{\mathrm e}^{4} \left (-\frac {4 \,{\mathrm e}^{x}}{x -4}-5 \,{\mathrm e}^{4} \expIntegralEi \left (1, -x +4\right )\right )-\frac {3 \,{\mathrm e}^{4} \left ({\mathrm e}^{x}-\frac {16 \,{\mathrm e}^{x}}{x -4}-24 \,{\mathrm e}^{4} \expIntegralEi \left (1, -x +4\right )\right )}{4}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-3*x^2-8*x)*exp(4)+4*x^3+80*x^2-832*x+1792)*exp(x)/(4*x^2-32*x+64),x,method=_RETURNVERBOSE)

[Out]

1/4*(x^2*exp(4)+4*x^2+92*x-432)*exp(x)/(x-4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left (x^{3} {\left (e^{4} + 4\right )} - 4 \, x^{2} {\left (e^{4} - 19\right )} - 800 \, x\right )} e^{x}}{4 \, {\left (x^{2} - 8 \, x + 16\right )}} - \frac {448 \, e^{4} E_{2}\left (-x + 4\right )}{x - 4} - \frac {1}{4} \, \int \frac {64 \, {\left (x + 50\right )} e^{x}}{x^{3} - 12 \, x^{2} + 48 \, x - 64}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-3*x^2-8*x)*exp(4)+4*x^3+80*x^2-832*x+1792)*exp(x)/(4*x^2-32*x+64),x, algorithm="maxima")

[Out]

1/4*(x^3*(e^4 + 4) - 4*x^2*(e^4 - 19) - 800*x)*e^x/(x^2 - 8*x + 16) - 448*e^4*exp_integral_e(2, -x + 4)/(x - 4
) - 1/4*integrate(64*(x + 50)*e^x/(x^3 - 12*x^2 + 48*x - 64), x)

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mupad [B]  time = 4.08, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (92\,x+x^2\,{\mathrm {e}}^4+4\,x^2-432\right )}{4\,\left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(80*x^2 - exp(4)*(8*x + 3*x^2 - x^3) - 832*x + 4*x^3 + 1792))/(4*x^2 - 32*x + 64),x)

[Out]

(exp(x)*(92*x + x^2*exp(4) + 4*x^2 - 432))/(4*(x - 4))

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sympy [A]  time = 0.15, size = 24, normalized size = 1.00 \begin {gather*} \frac {\left (4 x^{2} + x^{2} e^{4} + 92 x - 432\right ) e^{x}}{4 x - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-3*x**2-8*x)*exp(4)+4*x**3+80*x**2-832*x+1792)*exp(x)/(4*x**2-32*x+64),x)

[Out]

(4*x**2 + x**2*exp(4) + 92*x - 432)*exp(x)/(4*x - 16)

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