Integrand size = 62, antiderivative size = 25 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=9+x+\left (e^{2 x}-\frac {4}{1+x}\right ) \left (x+\frac {2}{\log (4)}\right ) \] Output:
(x+1/ln(2))*(exp(x)^2-4/(1+x))+x+9
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {-16+x \log (16)+x^2 \log (16)+e^{2 x} (1+x) (4+x \log (16))+\log (65536)}{2 (1+x) \log (4)} \] Input:
Integrate[(8 + (-3 + 2*x + x^2)*Log[4] + E^(2*x)*(4 + 8*x + 4*x^2 + (1 + 4 *x + 5*x^2 + 2*x^3)*Log[4]))/((1 + 2*x + x^2)*Log[4]),x]
Output:
(-16 + x*Log[16] + x^2*Log[16] + E^(2*x)*(1 + x)*(4 + x*Log[16]) + Log[655 36])/(2*(1 + x)*Log[4])
Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).
Time = 0.47 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 2007, 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.
\(\displaystyle \int \frac {\left (x^2+2 x-3\right ) \log (4)+e^{2 x} \left (4 x^2+\left (2 x^3+5 x^2+4 x+1\right ) \log (4)+8 x+4\right )+8}{\left (x^2+2 x+1\right ) \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-\log (4) \left (-x^2-2 x+3\right )+e^{2 x} \left (4 x^2+8 x+\left (2 x^3+5 x^2+4 x+1\right ) \log (4)+4\right )+8}{x^2+2 x+1}dx}{\log (4)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \frac {\int \frac {-\log (4) \left (-x^2-2 x+3\right )+e^{2 x} \left (4 x^2+8 x+\left (2 x^3+5 x^2+4 x+1\right ) \log (4)+4\right )+8}{(x+1)^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (e^{2 x} (\log (16) x+\log (4)+4)+\frac {\log (4) x^2+\log (16) x-3 \log (4)+8}{(x+1)^2}\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \log (4)+\frac {1}{2} e^{2 x} (x \log (16)+4+\log (4))-\frac {1}{4} e^{2 x} \log (16)-\frac {2 (4-\log (16))}{x+1}}{\log (4)}\) |
Input:
Int[(8 + (-3 + 2*x + x^2)*Log[4] + E^(2*x)*(4 + 8*x + 4*x^2 + (1 + 4*x + 5 *x^2 + 2*x^3)*Log[4]))/((1 + 2*x + x^2)*Log[4]),x]
Output:
(x*Log[4] - (2*(4 - Log[16]))/(1 + x) - (E^(2*x)*Log[16])/4 + (E^(2*x)*(4 + Log[4] + x*Log[16]))/2)/Log[4]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44
method | result | size |
parts | \(x \,{\mathrm e}^{2 x}+\frac {{\mathrm e}^{2 x}}{\ln \left (2\right )}+x +\frac {4}{1+x}-\frac {4}{\ln \left (2\right ) \left (1+x \right )}\) | \(36\) |
risch | \(x -\frac {4}{\ln \left (2\right ) \left (1+x \right )}+\frac {4}{1+x}+\frac {\left (2 x \ln \left (2\right )+2\right ) {\mathrm e}^{2 x}}{2 \ln \left (2\right )}\) | \(38\) |
norman | \(\frac {x^{2}+{\mathrm e}^{2 x} x^{2}+\frac {{\mathrm e}^{2 x}}{\ln \left (2\right )}+\frac {\left (1+\ln \left (2\right )\right ) x \,{\mathrm e}^{2 x}}{\ln \left (2\right )}+\frac {3 \ln \left (2\right )-4}{\ln \left (2\right )}}{1+x}\) | \(53\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{2 x} \ln \left (2\right ) x^{2}+2 \,{\mathrm e}^{2 x} \ln \left (2\right ) x +2 x^{2} \ln \left (2\right )+2 x \,{\mathrm e}^{2 x}-8+2 \,{\mathrm e}^{2 x}+6 \ln \left (2\right )}{2 \ln \left (2\right ) \left (1+x \right )}\) | \(58\) |
default | \(\frac {-\frac {8}{1+x}+\frac {6 \ln \left (2\right )}{1+x}+4 \ln \left (2\right ) \left (\ln \left (1+x \right )+\frac {1}{1+x}\right )+2 \ln \left (2\right ) \left (x -2 \ln \left (1+x \right )-\frac {1}{1+x}\right )+2 \,{\mathrm e}^{2 x}+2 \ln \left (2\right ) \left (-\frac {{\mathrm e}^{2 x}}{1+x}-2 \,{\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-2 x \right )\right )+8 \ln \left (2\right ) \left ({\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-2 x \right )+\frac {{\mathrm e}^{2 x}}{1+x}\right )+5 \ln \left (2\right ) {\mathrm e}^{2 x}-\frac {10 \ln \left (2\right ) {\mathrm e}^{2 x}}{1+x}+4 \ln \left (2\right ) \left (\frac {x \,{\mathrm e}^{2 x}}{2}-\frac {5 \,{\mathrm e}^{2 x}}{4}-{\mathrm e}^{-2} \operatorname {expIntegral}_{1}\left (-2-2 x \right )+\frac {{\mathrm e}^{2 x}}{1+x}\right )}{2 \ln \left (2\right )}\) | \(175\) |
Input:
int(1/2*((2*(2*x^3+5*x^2+4*x+1)*ln(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3)* ln(2)+8)/(x^2+2*x+1)/ln(2),x,method=_RETURNVERBOSE)
Output:
x*exp(x)^2+1/ln(2)*exp(x)^2+x+4/(1+x)-4/ln(2)/(1+x)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {{\left ({\left (x^{2} + x\right )} \log \left (2\right ) + x + 1\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + x + 4\right )} \log \left (2\right ) - 4}{{\left (x + 1\right )} \log \left (2\right )} \] Input:
integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+ 2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),x, algorithm="fricas")
Output:
(((x^2 + x)*log(2) + x + 1)*e^(2*x) + (x^2 + x + 4)*log(2) - 4)/((x + 1)*l og(2))
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=x + \frac {\left (x \log {\left (2 \right )} + 1\right ) e^{2 x}}{\log {\left (2 \right )}} + \frac {-4 + 4 \log {\left (2 \right )}}{x \log {\left (2 \right )} + \log {\left (2 \right )}} \] Input:
integrate(1/2*((2*(2*x**3+5*x**2+4*x+1)*ln(2)+4*x**2+8*x+4)*exp(x)**2+2*(x **2+2*x-3)*ln(2)+8)/(x**2+2*x+1)/ln(2),x)
Output:
x + (x*log(2) + 1)*exp(2*x)/log(2) + (-4 + 4*log(2))/(x*log(2) + log(2))
\[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\int { \frac {{\left (2 \, x^{2} + {\left (2 \, x^{3} + 5 \, x^{2} + 4 \, x + 1\right )} \log \left (2\right ) + 4 \, x + 2\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 2 \, x - 3\right )} \log \left (2\right ) + 4}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (2\right )} \,d x } \] Input:
integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+ 2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),x, algorithm="maxima")
Output:
((x - 1/(x + 1) - 2*log(x + 1))*log(2) + 2*(1/(x + 1) + log(x + 1))*log(2) - e^(-2)*exp_integral_e(2, -2*x - 2)*log(2)/(x + 1) + (x^3*log(2) + x^2*( 2*log(2) + 1) + x*(log(2) + 2))*e^(2*x)/(x^2 + 2*x + 1) - 2*e^(-2)*exp_int egral_e(2, -2*x - 2)/(x + 1) + 3*log(2)/(x + 1) - 4/(x + 1) - integrate((x *log(2) + log(2) + 2)*e^(2*x)/(x^3 + 3*x^2 + 3*x + 1), x))/log(2)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (2\right ) + x^{2} \log \left (2\right ) + x e^{\left (2 \, x\right )} \log \left (2\right ) + x e^{\left (2 \, x\right )} + x \log \left (2\right ) + e^{\left (2 \, x\right )} + 4 \, \log \left (2\right ) - 4}{{\left (x + 1\right )} \log \left (2\right )} \] Input:
integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+ 2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),x, algorithm="giac")
Output:
(x^2*e^(2*x)*log(2) + x^2*log(2) + x*e^(2*x)*log(2) + x*e^(2*x) + x*log(2) + e^(2*x) + 4*log(2) - 4)/((x + 1)*log(2))
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {{\mathrm {e}}^{2\,x}+x\,\ln \left (2\right )+x\,{\mathrm {e}}^{2\,x}\,\ln \left (2\right )}{\ln \left (2\right )}+\frac {\ln \left (2\right )+\ln \left (8\right )-4}{\ln \left (2\right )\,\left (x+1\right )} \] Input:
int(((exp(2*x)*(8*x + 2*log(2)*(4*x + 5*x^2 + 2*x^3 + 1) + 4*x^2 + 4))/2 + log(2)*(2*x + x^2 - 3) + 4)/(log(2)*(2*x + x^2 + 1)),x)
Output:
(exp(2*x) + x*log(2) + x*exp(2*x)*log(2))/log(2) + (log(2) + log(8) - 4)/( log(2)*(x + 1))
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\frac {e^{2 x} \mathrm {log}\left (2\right ) x^{2}+e^{2 x} \mathrm {log}\left (2\right ) x +e^{2 x} x +e^{2 x}+\mathrm {log}\left (2\right ) x^{2}-3 \,\mathrm {log}\left (2\right ) x +4 x}{\mathrm {log}\left (2\right ) \left (x +1\right )} \] Input:
int(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3) *log(2)+8)/(x^2+2*x+1)/log(2),x)
Output:
(e**(2*x)*log(2)*x**2 + e**(2*x)*log(2)*x + e**(2*x)*x + e**(2*x) + log(2) *x**2 - 3*log(2)*x + 4*x)/(log(2)*(x + 1))