Integrand size = 88, antiderivative size = 23 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=4+x+x^2+\frac {6+\frac {5+x}{(4+e)^2}}{4+x} \] Output:
x^2+((5+x)/(exp(1)+4)^2+6)/(4+x)+x+4
Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=\frac {\frac {97+48 e+6 e^2}{4+x}-7 (4+e)^2 (4+x)+(4+e)^2 (4+x)^2}{(4+e)^2} \] Input:
Integrate[(159 + 640*x + 272*x^2 + 32*x^3 + E^2*(10 + 40*x + 17*x^2 + 2*x^ 3) + E*(80 + 320*x + 136*x^2 + 16*x^3))/(256 + 128*x + 16*x^2 + E^2*(16 + 8*x + x^2) + E*(128 + 64*x + 8*x^2)),x]
Output:
((97 + 48*E + 6*E^2)/(4 + x) - 7*(4 + E)^2*(4 + x) + (4 + E)^2*(4 + x)^2)/ (4 + E)^2
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {2007, 2389, 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.
\(\displaystyle \int \frac {32 x^3+272 x^2+e^2 \left (2 x^3+17 x^2+40 x+10\right )+e \left (16 x^3+136 x^2+320 x+80\right )+640 x+159}{16 x^2+e^2 \left (x^2+8 x+16\right )+e \left (8 x^2+64 x+128\right )+128 x+256} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {32 x^3+272 x^2+e^2 \left (2 x^3+17 x^2+40 x+10\right )+e \left (16 x^3+136 x^2+320 x+80\right )+640 x+159}{((4+e) x+4 (4+e))^2}dx\) |
\(\Big \downarrow \) 2389 |
\(\displaystyle \int \left (2 x+\frac {-97-48 e-6 e^2}{(4+e)^2 (x+4)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2+x+\frac {97+48 e+6 e^2}{(4+e)^2 (x+4)}\) |
Input:
Int[(159 + 640*x + 272*x^2 + 32*x^3 + E^2*(10 + 40*x + 17*x^2 + 2*x^3) + E *(80 + 320*x + 136*x^2 + 16*x^3))/(256 + 128*x + 16*x^2 + E^2*(16 + 8*x + x^2) + E*(128 + 64*x + 8*x^2)),x]
Output:
x + x^2 + (97 + 48*E + 6*E^2)/((4 + E)^2*(4 + 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]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand [Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p , 0] || EqQ[n, 1])
Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(24)=48\).
Time = 0.36 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.26
method | result | size |
norman | \(\frac {\left ({\mathrm e}+4\right ) x^{3}+\left (5 \,{\mathrm e}+20\right ) x^{2}-\frac {10 \,{\mathrm e}^{2}+80 \,{\mathrm e}+159}{{\mathrm e}+4}}{\left (4+x \right ) \left ({\mathrm e}+4\right )}\) | \(52\) |
default | \(\frac {x^{2} {\mathrm e}^{2}+8 x^{2} {\mathrm e}+{\mathrm e}^{2} x +8 x \,{\mathrm e}+16 x^{2}+16 x -\frac {-6 \,{\mathrm e}^{2}-48 \,{\mathrm e}-97}{4+x}}{{\mathrm e}^{2}+8 \,{\mathrm e}+16}\) | \(62\) |
parallelrisch | \(\frac {x^{3} {\mathrm e}^{2}+5 x^{2} {\mathrm e}^{2}+8 x^{3} {\mathrm e}+40 x^{2} {\mathrm e}+16 x^{3}-10 \,{\mathrm e}^{2}+80 x^{2}-80 \,{\mathrm e}-159}{\left ({\mathrm e}^{2}+8 \,{\mathrm e}+16\right ) \left (4+x \right )}\) | \(72\) |
gosper | \(\frac {x^{3} {\mathrm e}^{2}+5 x^{2} {\mathrm e}^{2}+8 x^{3} {\mathrm e}+40 x^{2} {\mathrm e}+16 x^{3}-10 \,{\mathrm e}^{2}+80 x^{2}-80 \,{\mathrm e}-159}{{\mathrm e}^{2} x +4 \,{\mathrm e}^{2}+8 x \,{\mathrm e}+32 \,{\mathrm e}+16 x +64}\) | \(83\) |
risch | \(x^{2}+x +\frac {6 \,{\mathrm e}^{2}}{{\mathrm e}^{2} x +4 \,{\mathrm e}^{2}+8 x \,{\mathrm e}+32 \,{\mathrm e}+16 x +64}+\frac {48 \,{\mathrm e}}{{\mathrm e}^{2} x +4 \,{\mathrm e}^{2}+8 x \,{\mathrm e}+32 \,{\mathrm e}+16 x +64}+\frac {97}{{\mathrm e}^{2} x +4 \,{\mathrm e}^{2}+8 x \,{\mathrm e}+32 \,{\mathrm e}+16 x +64}\) | \(88\) |
meijerg | \(\frac {159 x}{16 \left ({\mathrm e}^{2}+8 \,{\mathrm e}+16\right ) \left (1+\frac {x}{4}\right )}+\frac {256 \left (\frac {{\mathrm e}^{2}}{8}+{\mathrm e}+2\right ) \left (-\frac {x \left (-\frac {1}{8} x^{2}+\frac {3}{2} x +12\right )}{16 \left (1+\frac {x}{4}\right )}+3 \ln \left (1+\frac {x}{4}\right )\right )}{{\mathrm e}^{2}+8 \,{\mathrm e}+16}+\frac {64 \left (\frac {17 \,{\mathrm e}^{2}}{16}+\frac {17 \,{\mathrm e}}{2}+17\right ) \left (\frac {x \left (\frac {3 x}{4}+6\right )}{3 x +12}-2 \ln \left (1+\frac {x}{4}\right )\right )}{{\mathrm e}^{2}+8 \,{\mathrm e}+16}+\frac {16 \left (\frac {5 \,{\mathrm e}^{2}}{2}+20 \,{\mathrm e}+40\right ) \left (-\frac {x}{4 \left (1+\frac {x}{4}\right )}+\ln \left (1+\frac {x}{4}\right )\right )}{{\mathrm e}^{2}+8 \,{\mathrm e}+16}+\frac {5 \,{\mathrm e}^{2} x}{8 \left ({\mathrm e}^{2}+8 \,{\mathrm e}+16\right ) \left (1+\frac {x}{4}\right )}+\frac {5 \,{\mathrm e} x}{\left ({\mathrm e}^{2}+8 \,{\mathrm e}+16\right ) \left (1+\frac {x}{4}\right )}\) | \(200\) |
Input:
int(((2*x^3+17*x^2+40*x+10)*exp(1)^2+(16*x^3+136*x^2+320*x+80)*exp(1)+32*x ^3+272*x^2+640*x+159)/((x^2+8*x+16)*exp(1)^2+(8*x^2+64*x+128)*exp(1)+16*x^ 2+128*x+256),x,method=_RETURNVERBOSE)
Output:
((exp(1)+4)*x^3+(5*exp(1)+20)*x^2-(10*exp(1)^2+80*exp(1)+159)/(exp(1)+4))/ (4+x)/(exp(1)+4)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (24) = 48\).
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.00 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=\frac {16 \, x^{3} + 80 \, x^{2} + {\left (x^{3} + 5 \, x^{2} + 4 \, x + 6\right )} e^{2} + 8 \, {\left (x^{3} + 5 \, x^{2} + 4 \, x + 6\right )} e + 64 \, x + 97}{{\left (x + 4\right )} e^{2} + 8 \, {\left (x + 4\right )} e + 16 \, x + 64} \] Input:
integrate(((2*x^3+17*x^2+40*x+10)*exp(1)^2+(16*x^3+136*x^2+320*x+80)*exp(1 )+32*x^3+272*x^2+640*x+159)/((x^2+8*x+16)*exp(1)^2+(8*x^2+64*x+128)*exp(1) +16*x^2+128*x+256),x, algorithm="fricas")
Output:
(16*x^3 + 80*x^2 + (x^3 + 5*x^2 + 4*x + 6)*e^2 + 8*(x^3 + 5*x^2 + 4*x + 6) *e + 64*x + 97)/((x + 4)*e^2 + 8*(x + 4)*e + 16*x + 64)
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=x^{2} + x + \frac {6 e^{2} + 97 + 48 e}{x \left (e^{2} + 16 + 8 e\right ) + 4 e^{2} + 64 + 32 e} \] Input:
integrate(((2*x**3+17*x**2+40*x+10)*exp(1)**2+(16*x**3+136*x**2+320*x+80)* exp(1)+32*x**3+272*x**2+640*x+159)/((x**2+8*x+16)*exp(1)**2+(8*x**2+64*x+1 28)*exp(1)+16*x**2+128*x+256),x)
Output:
x**2 + x + (6*exp(2) + 97 + 48*E)/(x*(exp(2) + 16 + 8*E) + 4*exp(2) + 64 + 32*E)
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=x^{2} + x + \frac {6 \, e^{2} + 48 \, e + 97}{x {\left (e^{2} + 8 \, e + 16\right )} + 4 \, e^{2} + 32 \, e + 64} \] Input:
integrate(((2*x^3+17*x^2+40*x+10)*exp(1)^2+(16*x^3+136*x^2+320*x+80)*exp(1 )+32*x^3+272*x^2+640*x+159)/((x^2+8*x+16)*exp(1)^2+(8*x^2+64*x+128)*exp(1) +16*x^2+128*x+256),x, algorithm="maxima")
Output:
x^2 + x + (6*e^2 + 48*e + 97)/(x*(e^2 + 8*e + 16) + 4*e^2 + 32*e + 64)
Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (24) = 48\).
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 4.39 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=\frac {x^{2} e^{4} + 16 \, x^{2} e^{3} + 96 \, x^{2} e^{2} + 256 \, x^{2} e + 256 \, x^{2} + x e^{4} + 16 \, x e^{3} + 96 \, x e^{2} + 256 \, x e + 256 \, x}{e^{4} + 16 \, e^{3} + 96 \, e^{2} + 256 \, e + 256} + \frac {6 \, e^{2} + 48 \, e + 97}{{\left (x + 4\right )} {\left (e^{2} + 8 \, e + 16\right )}} \] Input:
integrate(((2*x^3+17*x^2+40*x+10)*exp(1)^2+(16*x^3+136*x^2+320*x+80)*exp(1 )+32*x^3+272*x^2+640*x+159)/((x^2+8*x+16)*exp(1)^2+(8*x^2+64*x+128)*exp(1) +16*x^2+128*x+256),x, algorithm="giac")
Output:
(x^2*e^4 + 16*x^2*e^3 + 96*x^2*e^2 + 256*x^2*e + 256*x^2 + x*e^4 + 16*x*e^ 3 + 96*x*e^2 + 256*x*e + 256*x)/(e^4 + 16*e^3 + 96*e^2 + 256*e + 256) + (6 *e^2 + 48*e + 97)/((x + 4)*(e^2 + 8*e + 16))
Time = 2.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=x+x^2+\frac {48\,\mathrm {e}+6\,{\mathrm {e}}^2+97}{32\,\mathrm {e}+4\,{\mathrm {e}}^2+x\,\left (8\,\mathrm {e}+{\mathrm {e}}^2+16\right )+64} \] Input:
int((640*x + exp(2)*(40*x + 17*x^2 + 2*x^3 + 10) + exp(1)*(320*x + 136*x^2 + 16*x^3 + 80) + 272*x^2 + 32*x^3 + 159)/(128*x + exp(1)*(64*x + 8*x^2 + 128) + exp(2)*(8*x + x^2 + 16) + 16*x^2 + 256),x)
Output:
x + x^2 + (48*exp(1) + 6*exp(2) + 97)/(32*exp(1) + 4*exp(2) + x*(8*exp(1) + exp(2) + 16) + 64)
Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.00 \[ \int \frac {159+640 x+272 x^2+32 x^3+e^2 \left (10+40 x+17 x^2+2 x^3\right )+e \left (80+320 x+136 x^2+16 x^3\right )}{256+128 x+16 x^2+e^2 \left (16+8 x+x^2\right )+e \left (128+64 x+8 x^2\right )} \, dx=\frac {x \left (4 e^{2} x^{2}+20 e^{2} x +32 e \,x^{2}+10 e^{2}+160 e x +64 x^{2}+80 e +320 x +159\right )}{4 e^{2} x +16 e^{2}+32 e x +128 e +64 x +256} \] Input:
int(((2*x^3+17*x^2+40*x+10)*exp(1)^2+(16*x^3+136*x^2+320*x+80)*exp(1)+32*x ^3+272*x^2+640*x+159)/((x^2+8*x+16)*exp(1)^2+(8*x^2+64*x+128)*exp(1)+16*x^ 2+128*x+256),x)
Output:
(x*(4*e**2*x**2 + 20*e**2*x + 10*e**2 + 32*e*x**2 + 160*e*x + 80*e + 64*x* *2 + 320*x + 159))/(4*(e**2*x + 4*e**2 + 8*e*x + 32*e + 16*x + 64))