3.33.50 \(\int \frac {e^{5+x} (-9+24 x-16 x^2)+e^5 (9-24 x+16 x^2)+12 e^4 \log (2)}{e^5 (9-24 x+16 x^2)} \, dx\)

Optimal. Leaf size=22 \[ 4-e^x+x-\frac {3 \log (2)}{e (-3+4 x)} \]

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Rubi [A]  time = 0.25, antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 27, 6742, 2194, 683} \begin {gather*} x-e^x+\frac {\log (4096)}{4 e (3-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 + x)*(-9 + 24*x - 16*x^2) + E^5*(9 - 24*x + 16*x^2) + 12*E^4*Log[2])/(E^5*(9 - 24*x + 16*x^2)),x]

[Out]

-E^x + x + Log[4096]/(4*E*(3 - 4*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 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 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{9-24 x+16 x^2} \, dx}{e^5}\\ &=\frac {\int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{(-3+4 x)^2} \, dx}{e^5}\\ &=\frac {\int \left (-e^{5+x}+\frac {e^4 \left (9 e-24 e x+16 e x^2+\log (4096)\right )}{(-3+4 x)^2}\right ) \, dx}{e^5}\\ &=-\frac {\int e^{5+x} \, dx}{e^5}+\frac {\int \frac {9 e-24 e x+16 e x^2+\log (4096)}{(-3+4 x)^2} \, dx}{e}\\ &=-e^x+\frac {\int \left (e+\frac {\log (4096)}{(-3+4 x)^2}\right ) \, dx}{e}\\ &=-e^x+x+\frac {\log (4096)}{4 e (3-4 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 25, normalized size = 1.14 \begin {gather*} \frac {-e^{1+x}+e x+\frac {\log (4096)}{12-16 x}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + x)*(-9 + 24*x - 16*x^2) + E^5*(9 - 24*x + 16*x^2) + 12*E^4*Log[2])/(E^5*(9 - 24*x + 16*x^2))
,x]

[Out]

(-E^(1 + x) + E*x + Log[4096]/(12 - 16*x))/E

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fricas [A]  time = 0.56, size = 40, normalized size = 1.82 \begin {gather*} \frac {{\left ({\left (4 \, x^{2} - 3 \, x\right )} e^{5} - {\left (4 \, x - 3\right )} e^{\left (x + 5\right )} - 3 \, e^{4} \log \relax (2)\right )} e^{\left (-5\right )}}{4 \, x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="fricas")

[Out]

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

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giac [B]  time = 0.28, size = 42, normalized size = 1.91 \begin {gather*} \frac {{\left (4 \, x^{2} e^{5} - 3 \, x e^{5} - 4 \, x e^{\left (x + 5\right )} - 3 \, e^{4} \log \relax (2) + 3 \, e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{4 \, x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="giac")

[Out]

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

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maple [A]  time = 0.55, size = 18, normalized size = 0.82

method	result	size
risch	\(x -\frac {3 \,{\mathrm e}^{-1} \ln \relax (2)}{4 \left (x -\frac {3}{4}\right )}-{\mathrm e}^{x}\)	\(18\)
norman	\(\frac {4 x^{2}-4 \,{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}-\frac {3 \left (4 \,{\mathrm e}^{4} \ln \relax (2)+3 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{4}}{4 x -3}\)	\(41\)
default	\({\mathrm e}^{-5} \left (x \,{\mathrm e}^{5}-\frac {3 \,{\mathrm e}^{4} \ln \relax (2)}{4 x -3}-9 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{16 \left (x -\frac {3}{4}\right )}-\frac {{\mathrm e}^{\frac {3}{4}} \expIntegralEi \left (1, -x +\frac {3}{4}\right )}{16}\right )+24 \,{\mathrm e}^{5} \left (-\frac {3 \,{\mathrm e}^{x}}{64 \left (x -\frac {3}{4}\right )}-\frac {7 \,{\mathrm e}^{\frac {3}{4}} \expIntegralEi \left (1, -x +\frac {3}{4}\right )}{64}\right )-16 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{x}}{16}-\frac {9 \,{\mathrm e}^{x}}{256 \left (x -\frac {3}{4}\right )}-\frac {33 \,{\mathrm e}^{\frac {3}{4}} \expIntegralEi \left (1, -x +\frac {3}{4}\right )}{256}\right )\right )\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*ln(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x,method=_R
ETURNVERBOSE)

[Out]

x-3/4*exp(-1)*ln(2)/(x-3/4)-exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{4} \, {\left ({\left (4 \, x - \frac {9}{4 \, x - 3} + 6 \, \log \left (4 \, x - 3\right )\right )} e^{5} + 6 \, {\left (\frac {3}{4 \, x - 3} - \log \left (4 \, x - 3\right )\right )} e^{5} - \frac {32 \, {\left (2 \, x^{2} e^{5} - 3 \, x e^{5}\right )} e^{x}}{16 \, x^{2} - 24 \, x + 9} + \frac {9 \, e^{\frac {23}{4}} E_{2}\left (-x + \frac {3}{4}\right )}{4 \, x - 3} - \frac {12 \, e^{4} \log \relax (2)}{4 \, x - 3} - \frac {9 \, e^{5}}{4 \, x - 3} + 288 \, \int \frac {e^{\left (x + 5\right )}}{64 \, x^{3} - 144 \, x^{2} + 108 \, x - 27}\,{d x}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x^2+24*x-9)*exp(5)*exp(x)+12*exp(4)*log(2)+(16*x^2-24*x+9)*exp(5))/(16*x^2-24*x+9)/exp(5),x, a
lgorithm="maxima")

[Out]

1/4*((4*x - 9/(4*x - 3) + 6*log(4*x - 3))*e^5 + 6*(3/(4*x - 3) - log(4*x - 3))*e^5 - 32*(2*x^2*e^5 - 3*x*e^5)*
e^x/(16*x^2 - 24*x + 9) + 9*e^(23/4)*exp_integral_e(2, -x + 3/4)/(4*x - 3) - 12*e^4*log(2)/(4*x - 3) - 9*e^5/(
4*x - 3) + 288*integrate(e^(x + 5)/(64*x^3 - 144*x^2 + 108*x - 27), x))*e^(-5)

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mupad [B]  time = 0.15, size = 24, normalized size = 1.09 \begin {gather*} x-{\mathrm {e}}^x+\frac {3\,{\mathrm {e}}^4\,\ln \relax (2)}{3\,{\mathrm {e}}^5-4\,x\,{\mathrm {e}}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-5)*(12*exp(4)*log(2) + exp(5)*(16*x^2 - 24*x + 9) - exp(5)*exp(x)*(16*x^2 - 24*x + 9)))/(16*x^2 - 24
*x + 9),x)

[Out]

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

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sympy [A]  time = 0.25, size = 20, normalized size = 0.91 \begin {gather*} x - e^{x} - \frac {3 \log {\relax (2 )}}{4 e x - 3 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*x**2+24*x-9)*exp(5)*exp(x)+12*exp(4)*ln(2)+(16*x**2-24*x+9)*exp(5))/(16*x**2-24*x+9)/exp(5),x)

[Out]

x - exp(x) - 3*log(2)/(4*E*x - 3*E)

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