\(\int \frac {e^{5+x} (-9+24 x-16 x^2)+e^5 (9-24 x+16 x^2)+12 e^4 \log (2)}{e^5 (9-24 x+16 x^2)} \, dx\) [3250]

Optimal result

Integrand size = 54, antiderivative size = 22 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=4-e^x+x-\frac {3 \log (2)}{e (-3+4 x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 27, 6874, 2225, 697} \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=x-e^x+\frac {\log (4096)}{4 e (3-4 x)} \]

[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 697

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 2225

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {-e^{1+x}+e x+\frac {\log (4096)}{12-16 x}}{e} \]

[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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
risch	\(x -\frac {3 \,{\mathrm e}^{-1} \ln \left (2\right )}{4 \left (x -\frac {3}{4}\right )}-{\mathrm e}^{x}\)	\(18\)
parts	\(x -\frac {3 \,{\mathrm e}^{-5} {\mathrm e}^{4} \ln \left (2\right )}{-3+4 x}-{\mathrm e}^{x}\)	\(24\)
norman	\(\frac {4 x^{2}-4 \,{\mathrm e}^{x} x +3 \,{\mathrm e}^{x}-\frac {3 \left (4 \,{\mathrm e}^{4} \ln \left (2\right )+3 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}{4}}{-3+4 x}\)	\(41\)
parallelrisch	\(\frac {{\mathrm e}^{-5} \left (16 x^{2} {\mathrm e}^{5}-16 x \,{\mathrm e}^{5} {\mathrm e}^{x}+12 \,{\mathrm e}^{5} {\mathrm e}^{x}-12 \,{\mathrm e}^{4} \ln \left (2\right )-9 \,{\mathrm e}^{5}\right )}{16 x -12}\)	\(45\)
default	\({\mathrm e}^{-5} \left (-\frac {9 \,{\mathrm e}^{5}}{4 \left (-3+4 x \right )}-24 \,{\mathrm e}^{5} \left (-\frac {3}{16 \left (-3+4 x \right )}+\frac {\ln \left (-3+4 x \right )}{16}\right )+16 \,{\mathrm e}^{5} \left (\frac {x}{16}-\frac {9}{64 \left (-3+4 x \right )}+\frac {3 \ln \left (-3+4 x \right )}{32}\right )-\frac {3 \,{\mathrm e}^{4} \ln \left (2\right )}{-3+4 x}-9 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{16 \left (x -\frac {3}{4}\right )}-\frac {{\mathrm e}^{\frac {3}{4}} \operatorname {Ei}_{1}\left (-x +\frac {3}{4}\right )}{16}\right )+24 \,{\mathrm e}^{5} \left (-\frac {3 \,{\mathrm e}^{x}}{64 \left (x -\frac {3}{4}\right )}-\frac {7 \,{\mathrm e}^{\frac {3}{4}} \operatorname {Ei}_{1}\left (-x +\frac {3}{4}\right )}{64}\right )-16 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{x}}{16}-\frac {9 \,{\mathrm e}^{x}}{256 \left (x -\frac {3}{4}\right )}-\frac {33 \,{\mathrm e}^{\frac {3}{4}} \operatorname {Ei}_{1}\left (-x +\frac {3}{4}\right )}{256}\right )\right )\)	\(157\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {{\left ({\left (4 \, x^{2} - 3 \, x\right )} e^{5} - {\left (4 \, x - 3\right )} e^{\left (x + 5\right )} - 3 \, e^{4} \log \left (2\right )\right )} e^{\left (-5\right )}}{4 \, x - 3} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=x - e^{x} - \frac {3 \log {\left (2 \right )}}{4 e x - 3 e} \]

[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)

Maxima [F]

\[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\int { \frac {{\left ({\left (16 \, x^{2} - 24 \, x + 9\right )} e^{5} - {\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (x + 5\right )} + 12 \, e^{4} \log \left (2\right )\right )} e^{\left (-5\right )}}{16 \, x^{2} - 24 \, x + 9} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {{\left (4 \, x^{2} e^{5} - 3 \, x e^{5} - 4 \, x e^{\left (x + 5\right )} - 3 \, e^{4} \log \left (2\right ) + 3 \, e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{4 \, x - 3} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=x-{\mathrm {e}}^x+\frac {3\,{\mathrm {e}}^4\,\ln \left (2\right )}{3\,{\mathrm {e}}^5-4\,x\,{\mathrm {e}}^5} \]

[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))