Integrand size = 54, antiderivative size = 22 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=4-e^x+x-\frac {3 \log (2)}{e (-3+4 x)} \] Output:
4-exp(x)+x-3/(-3+4*x)/exp(5)*exp(4)*ln(2)
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {-e^{1+x}+e x+\frac {\log (4096)}{12-16 x}}{e} \] Input:
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]
Output:
(-E^(1 + x) + E*x + Log[4096]/(12 - 16*x))/E
Time = 0.53 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 7277, 27, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{x+5} \left (-16 x^2+24 x-9\right )+e^5 \left (16 x^2-24 x+9\right )+12 e^4 \log (2)}{e^5 \left (16 x^2-24 x+9\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-e^{x+5} \left (16 x^2-24 x+9\right )+e^5 \left (16 x^2-24 x+9\right )+12 e^4 \log (2)}{16 x^2-24 x+9}dx}{e^5}\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle \frac {64 \int \frac {-e^{x+5} \left (16 x^2-24 x+9\right )+e^5 \left (16 x^2-24 x+9\right )+12 e^4 \log (2)}{64 (3-4 x)^2}dx}{e^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-e^{x+5} \left (16 x^2-24 x+9\right )+e^5 \left (16 x^2-24 x+9\right )+12 e^4 \log (2)}{(3-4 x)^2}dx}{e^5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {e^4 \left (16 e x^2-24 e x+\log (4096)+9 e\right )}{(4 x-3)^2}-e^{x+5}\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^5 x-e^{x+5}+\frac {e^4 \log (4096)}{4 (3-4 x)}}{e^5}\) |
Input:
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]
Output:
(-E^(5 + x) + E^5*x + (E^4*Log[4096])/(4*(3 - 4*x)))/E^5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.32 (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 | \(\frac {{\mathrm e}^{-5} \left (16 x^{2} {\mathrm e}^{5}-16 \,{\mathrm e}^{x} {\mathrm e}^{5} x +12 \,{\mathrm e}^{5} {\mathrm e}^{x}-12 \,{\mathrm e}^{4} \ln \left (2\right )-9 \,{\mathrm e}^{5}\right )}{-12+16 x}\) | \(45\) |
default | \({\mathrm e}^{-5} \left (-\frac {9 \,{\mathrm e}^{5}}{4 \left (-3+4 x \right )}-24 \,{\mathrm e}^{5} \left (\frac {\ln \left (-3+4 x \right )}{16}-\frac {3}{16 \left (-3+4 x \right )}\right )+16 \,{\mathrm e}^{5} \left (\frac {x}{16}+\frac {3 \ln \left (-3+4 x \right )}{32}-\frac {9}{64 \left (-3+4 x \right )}\right )-\frac {3 \,{\mathrm e}^{4} \ln \left (2\right )}{-3+4 x}-9 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{x}}{16 \left (x -\frac {3}{4}\right )}-\frac {{\mathrm e}^{\frac {3}{4}} \operatorname {expIntegral}_{1}\left (-x +\frac {3}{4}\right )}{16}\right )-16 \,{\mathrm e}^{5} \left (\frac {{\mathrm e}^{x}}{16}-\frac {33 \,{\mathrm e}^{\frac {3}{4}} \operatorname {expIntegral}_{1}\left (-x +\frac {3}{4}\right )}{256}-\frac {9 \,{\mathrm e}^{x}}{256 \left (x -\frac {3}{4}\right )}\right )+24 \,{\mathrm e}^{5} \left (-\frac {7 \,{\mathrm e}^{\frac {3}{4}} \operatorname {expIntegral}_{1}\left (-x +\frac {3}{4}\right )}{64}-\frac {3 \,{\mathrm e}^{x}}{64 \left (x -\frac {3}{4}\right )}\right )\right )\) | \(157\) |
Input:
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=_RETURNVERBOSE)
Output:
x-3/4*exp(-1)*ln(2)/(x-3/4)-exp(x)
Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {{\left ({\left (4 \, x^{2} - 3 \, x\right )} e^{5} - {\left (4 \, x - 3\right )} e^{\left (x + 5\right )} - 3 \, e^{4} \log \left (2\right )\right )} e^{\left (-5\right )}}{4 \, x - 3} \] Input:
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, algorithm="fricas")
Output:
((4*x^2 - 3*x)*e^5 - (4*x - 3)*e^(x + 5) - 3*e^4*log(2))*e^(-5)/(4*x - 3)
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=x - e^{x} - \frac {3 \log {\left (2 \right )}}{4 e x - 3 e} \] Input:
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)
Output:
x - exp(x) - 3*log(2)/(4*E*x - 3*E)
\[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\int { \frac {{\left ({\left (16 \, x^{2} - 24 \, x + 9\right )} e^{5} - {\left (16 \, x^{2} - 24 \, x + 9\right )} e^{\left (x + 5\right )} + 12 \, e^{4} \log \left (2\right )\right )} e^{\left (-5\right )}}{16 \, x^{2} - 24 \, x + 9} \,d x } \] Input:
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, algorithm="maxima")
Output:
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)*e xp_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)
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {{\left (4 \, x^{2} e^{5} - 3 \, x e^{5} - 4 \, x e^{\left (x + 5\right )} - 3 \, e^{4} \log \left (2\right ) + 3 \, e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}}{4 \, x - 3} \] Input:
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, algorithm="giac")
Output:
(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)
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=x-{\mathrm {e}}^x+\frac {3\,{\mathrm {e}}^4\,\ln \left (2\right )}{3\,{\mathrm {e}}^5-4\,x\,{\mathrm {e}}^5} \] Input:
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)
Output:
x - exp(x) + (3*exp(4)*log(2))/(3*exp(5) - 4*x*exp(5))
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {e^{5+x} \left (-9+24 x-16 x^2\right )+e^5 \left (9-24 x+16 x^2\right )+12 e^4 \log (2)}{e^5 \left (9-24 x+16 x^2\right )} \, dx=\frac {-4 e^{x} e x +3 e^{x} e -4 \,\mathrm {log}\left (2\right ) x +4 e \,x^{2}-3 e x}{e \left (4 x -3\right )} \] Input:
int(((-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)
Output:
( - 4*e**x*e*x + 3*e**x*e - 4*log(2)*x + 4*e*x**2 - 3*e*x)/(e*(4*x - 3))