Integrand size = 83, antiderivative size = 28 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=-4 x+\left (x+\frac {e^5 x^2}{1+4 x^2}\right )^2-\log (3) \]
Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=2 \left (\frac {1}{2} (-4+x) x+\frac {e^5 x^3}{1+4 x^2}-\frac {e^{10} \left (1+8 x^2\right )}{32 \left (1+4 x^2\right )^2}\right ) \]
Integrate[(-4 + 2*x - 48*x^2 + 24*x^3 + 4*E^10*x^3 - 192*x^4 + 96*x^5 - 25 6*x^6 + 128*x^7 + E^5*(6*x^2 + 32*x^4 + 32*x^6))/(1 + 12*x^2 + 48*x^4 + 64 *x^6),x]
Time = 0.39 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 2070, 2345, 27, 2345, 25, 2019, 17}
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.
\(\displaystyle \int \frac {128 x^7-256 x^6+96 x^5-192 x^4+4 e^{10} x^3+24 x^3-48 x^2+e^5 \left (32 x^6+32 x^4+6 x^2\right )+2 x-4}{64 x^6+48 x^4+12 x^2+1} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {128 x^7-256 x^6+96 x^5-192 x^4+\left (24+4 e^{10}\right ) x^3-48 x^2+e^5 \left (32 x^6+32 x^4+6 x^2\right )+2 x-4}{64 x^6+48 x^4+12 x^2+1}dx\) |
\(\Big \downarrow \) 2070 |
\(\displaystyle \int \frac {128 x^7-256 x^6+96 x^5-192 x^4+\left (24+4 e^{10}\right ) x^3-48 x^2+e^5 \left (32 x^6+32 x^4+6 x^2\right )+2 x-4}{\left (4 x^2+1\right )^3}dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {e^{10}}{16 \left (4 x^2+1\right )^2}-\frac {1}{4} \int \frac {4 \left (-32 x^5+8 \left (8-e^5\right ) x^4-16 x^3+2 \left (16-3 e^5\right ) x^2-\left (2+e^{10}\right ) x+4\right )}{\left (4 x^2+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{10}}{16 \left (4 x^2+1\right )^2}-\int \frac {-32 x^5+8 \left (8-e^5\right ) x^4-16 x^3+2 \left (16-3 e^5\right ) x^2-\left (2+e^{10}\right ) x+4}{\left (4 x^2+1\right )^2}dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {1}{2} \int -\frac {-16 x^3+4 \left (8-e^5\right ) x^2-4 x-e^5+8}{4 x^2+1}dx-\frac {e^5 \left (4 x+e^5\right )}{8 \left (4 x^2+1\right )}+\frac {e^{10}}{16 \left (4 x^2+1\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {-16 x^3+4 \left (8-e^5\right ) x^2-4 x-e^5+8}{4 x^2+1}dx-\frac {e^5 \left (4 x+e^5\right )}{8 \left (4 x^2+1\right )}+\frac {e^{10}}{16 \left (4 x^2+1\right )^2}\) |
\(\Big \downarrow \) 2019 |
\(\displaystyle -\frac {1}{2} \int \left (-4 x-e^5+8\right )dx-\frac {e^5 \left (4 x+e^5\right )}{8 \left (4 x^2+1\right )}+\frac {e^{10}}{16 \left (4 x^2+1\right )^2}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {e^5 \left (4 x+e^5\right )}{8 \left (4 x^2+1\right )}+\frac {e^{10}}{16 \left (4 x^2+1\right )^2}+\frac {1}{16} \left (-4 x-e^5+8\right )^2\) |
Int[(-4 + 2*x - 48*x^2 + 24*x^3 + 4*E^10*x^3 - 192*x^4 + 96*x^5 - 256*x^6 + 128*x^7 + E^5*(6*x^2 + 32*x^4 + 32*x^6))/(1 + 12*x^2 + 48*x^4 + 64*x^6), x]
3.1.39.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
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_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px , Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x^2, 0], Expon[Px , x^2]], b = Rt[Coeff[Px, x^2, Expon[Px, x^2]], Expon[Px, x^2]]}, Int[u*(a + b*x^2)^(Expon[Px, x^2]*p), x] /; EqQ[Px, (a + b*x^2)^Expon[Px, x^2]]] /; IntegerQ[p] && PolyQ[Px, x^2] && GtQ[Expon[Px, x^2], 1] && NeQ[Coeff[Px, x^ 2, 0], 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
default | \(x^{2}+\frac {x \,{\mathrm e}^{5}}{2}-4 x +\frac {-2 x^{3} {\mathrm e}^{5}-\frac {x^{2} {\mathrm e}^{10}}{2}-\frac {x \,{\mathrm e}^{5}}{2}-\frac {{\mathrm e}^{10}}{16}}{\left (4 x^{2}+1\right )^{2}}\) | \(48\) |
risch | \(\frac {x \,{\mathrm e}^{5}}{2}+x^{2}-4 x +\frac {-\frac {x^{2} {\mathrm e}^{10}}{32}-\frac {x^{3} {\mathrm e}^{5}}{8}-\frac {{\mathrm e}^{10}}{256}-\frac {x \,{\mathrm e}^{5}}{32}}{x^{4}+\frac {1}{2} x^{2}+\frac {1}{16}}\) | \(50\) |
norman | \(\frac {\left (-64+8 \,{\mathrm e}^{5}\right ) x^{5}+\left (-32+2 \,{\mathrm e}^{5}\right ) x^{3}+x^{2}+\left ({\mathrm e}^{10}+8\right ) x^{4}-4 x +16 x^{6}}{\left (4 x^{2}+1\right )^{2}}\) | \(53\) |
gosper | \(\frac {x \left (x^{3} {\mathrm e}^{10}+8 x^{4} {\mathrm e}^{5}+16 x^{5}-64 x^{4}+2 x^{2} {\mathrm e}^{5}+8 x^{3}-32 x^{2}+x -4\right )}{16 x^{4}+8 x^{2}+1}\) | \(62\) |
parallelrisch | \(\frac {x^{4} {\mathrm e}^{10}+8 x^{5} {\mathrm e}^{5}+16 x^{6}-64 x^{5}+2 x^{3} {\mathrm e}^{5}+8 x^{4}-32 x^{3}+x^{2}-4 x}{16 x^{4}+8 x^{2}+1}\) | \(65\) |
int((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96*x^5-19 2*x^4+24*x^3-48*x^2+2*x-4)/(64*x^6+48*x^4+12*x^2+1),x,method=_RETURNVERBOS E)
x^2+1/2*x*exp(5)-4*x+8*(-1/4*x^3*exp(5)-1/16*x^2*exp(10)-1/16*x*exp(5)-1/1 28*exp(10))/(4*x^2+1)^2
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (27) = 54\).
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.46 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=\frac {256 \, x^{6} - 1024 \, x^{5} + 128 \, x^{4} - 512 \, x^{3} + 16 \, x^{2} - {\left (8 \, x^{2} + 1\right )} e^{10} + 32 \, {\left (4 \, x^{5} + x^{3}\right )} e^{5} - 64 \, x}{16 \, {\left (16 \, x^{4} + 8 \, x^{2} + 1\right )}} \]
integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96* x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64*x^6+48*x^4+12*x^2+1),x, algorithm=\
1/16*(256*x^6 - 1024*x^5 + 128*x^4 - 512*x^3 + 16*x^2 - (8*x^2 + 1)*e^10 + 32*(4*x^5 + x^3)*e^5 - 64*x)/(16*x^4 + 8*x^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=x^{2} + x \left (-4 + \frac {e^{5}}{2}\right ) + \frac {- 32 x^{3} e^{5} - 8 x^{2} e^{10} - 8 x e^{5} - e^{10}}{256 x^{4} + 128 x^{2} + 16} \]
integrate((4*x**3*exp(5)**2+(32*x**6+32*x**4+6*x**2)*exp(5)+128*x**7-256*x **6+96*x**5-192*x**4+24*x**3-48*x**2+2*x-4)/(64*x**6+48*x**4+12*x**2+1),x)
x**2 + x*(-4 + exp(5)/2) + (-32*x**3*exp(5) - 8*x**2*exp(10) - 8*x*exp(5) - exp(10))/(256*x**4 + 128*x**2 + 16)
Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=x^{2} + \frac {1}{2} \, x {\left (e^{5} - 8\right )} - \frac {32 \, x^{3} e^{5} + 8 \, x^{2} e^{10} + 8 \, x e^{5} + e^{10}}{16 \, {\left (16 \, x^{4} + 8 \, x^{2} + 1\right )}} \]
integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96* x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64*x^6+48*x^4+12*x^2+1),x, algorithm=\
x^2 + 1/2*x*(e^5 - 8) - 1/16*(32*x^3*e^5 + 8*x^2*e^10 + 8*x*e^5 + e^10)/(1 6*x^4 + 8*x^2 + 1)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=x^{2} + \frac {1}{2} \, x e^{5} - 4 \, x - \frac {32 \, x^{3} e^{5} + 8 \, x^{2} e^{10} + 8 \, x e^{5} + e^{10}}{16 \, {\left (4 \, x^{2} + 1\right )}^{2}} \]
integrate((4*x^3*exp(5)^2+(32*x^6+32*x^4+6*x^2)*exp(5)+128*x^7-256*x^6+96* x^5-192*x^4+24*x^3-48*x^2+2*x-4)/(64*x^6+48*x^4+12*x^2+1),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {-4+2 x-48 x^2+24 x^3+4 e^{10} x^3-192 x^4+96 x^5-256 x^6+128 x^7+e^5 \left (6 x^2+32 x^4+32 x^6\right )}{1+12 x^2+48 x^4+64 x^6} \, dx=x^2-\frac {4\,{\mathrm {e}}^5\,x^3+{\mathrm {e}}^{10}\,x^2+{\mathrm {e}}^5\,x+\frac {{\mathrm {e}}^{10}}{8}}{32\,x^4+16\,x^2+2}+x\,\left (\frac {{\mathrm {e}}^5}{2}-4\right ) \]
int((2*x + 4*x^3*exp(10) + exp(5)*(6*x^2 + 32*x^4 + 32*x^6) - 48*x^2 + 24* x^3 - 192*x^4 + 96*x^5 - 256*x^6 + 128*x^7 - 4)/(12*x^2 + 48*x^4 + 64*x^6 + 1),x)