Integrand size = 144, antiderivative size = 21 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=11+2 x+x^2+\frac {5}{\left (4-x^2+\log (5)\right )^2} \] Output:
2*x+5/(ln(5)+4-x^2)^2+x^2+11
Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=2 \left (x+\frac {x^2}{2}+\frac {5}{2 \left (-4+x^2-\log (5)\right )^2}\right ) \] Input:
Integrate[(128 + 148*x - 96*x^2 - 96*x^3 + 24*x^4 + 24*x^5 - 2*x^6 - 2*x^7 + (96 + 96*x - 48*x^2 - 48*x^3 + 6*x^4 + 6*x^5)*Log[5] + (24 + 24*x - 6*x ^2 - 6*x^3)*Log[5]^2 + (2 + 2*x)*Log[5]^3)/(64 - 48*x^2 + 12*x^4 - x^6 + ( 48 - 24*x^2 + 3*x^4)*Log[5] + (12 - 3*x^2)*Log[5]^2 + Log[5]^3),x]
Output:
2*(x + x^2/2 + 5/(2*(-4 + x^2 - Log[5])^2))
Time = 0.62 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2070, 2345, 27, 2019, 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.
| \(\displaystyle \int \frac {-2 x^7-2 x^6+24 x^5+24 x^4-96 x^3-96 x^2+\left (-6 x^3-6 x^2+24 x+24\right ) \log ^2(5)+\left (6 x^5+6 x^4-48 x^3-48 x^2+96 x+96\right ) \log (5)+148 x+(2 x+2) \log ^3(5)+128}{-x^6+12 x^4-48 x^2+\left (12-3 x^2\right ) \log ^2(5)+\left (3 x^4-24 x^2+48\right ) \log (5)+64+\log ^3(5)} \, dx\) |
| \(\Big \downarrow \) 2070 |
| \(\displaystyle \int \frac {-2 x^7-2 x^6+24 x^5+24 x^4-96 x^3-96 x^2+\left (-6 x^3-6 x^2+24 x+24\right ) \log ^2(5)+\left (6 x^5+6 x^4-48 x^3-48 x^2+96 x+96\right ) \log (5)+148 x+(2 x+2) \log ^3(5)+128}{\left (-x^2+4+\log (5)\right )^3}dx\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {5}{\left (-x^2+4+\log (5)\right )^2}-\frac {\int -\frac {8 \left ((4+\log (5)) x^5+(4+\log (5)) x^4-2 (4+\log (5))^2 x^3-2 (4+\log (5))^2 x^2+(4+\log (5))^3 x+(4+\log (5))^3\right )}{\left (-x^2+\log (5)+4\right )^2}dx}{4 (4+\log (5))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \int \frac {(4+\log (5)) x^5+(4+\log (5)) x^4-2 (4+\log (5))^2 x^3-2 (4+\log (5))^2 x^2+(4+\log (5))^3 x+(4+\log (5))^3}{\left (-x^2+\log (5)+4\right )^2}dx}{4+\log (5)}+\frac {5}{\left (-x^2+4+\log (5)\right )^2}\) |
| \(\Big \downarrow \) 2019 |
| \(\displaystyle \frac {2 \int \frac {(-4-\log (5)) x^3+(-4-\log (5)) x^2+\left (16+8 \log (5)+\log ^2(5)\right ) x+\log ^2(5)+8 \log (5)+16}{-x^2+\log (5)+4}dx}{4+\log (5)}+\frac {5}{\left (-x^2+4+\log (5)\right )^2}\) |
| \(\Big \downarrow \) 2019 |
| \(\displaystyle \frac {2 \int ((4+\log (5)) x+\log (5)+4)dx}{4+\log (5)}+\frac {5}{\left (-x^2+4+\log (5)\right )^2}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle \frac {5}{\left (-x^2+4+\log (5)\right )^2}+\frac {(x (4+\log (5))+4+\log (5))^2}{(4+\log (5))^2}\) |
Input:
Int[(128 + 148*x - 96*x^2 - 96*x^3 + 24*x^4 + 24*x^5 - 2*x^6 - 2*x^7 + (96 + 96*x - 48*x^2 - 48*x^3 + 6*x^4 + 6*x^5)*Log[5] + (24 + 24*x - 6*x^2 - 6 *x^3)*Log[5]^2 + (2 + 2*x)*Log[5]^3)/(64 - 48*x^2 + 12*x^4 - x^6 + (48 - 2 4*x^2 + 3*x^4)*Log[5] + (12 - 3*x^2)*Log[5]^2 + Log[5]^3),x]
Output:
5/(4 - x^2 + Log[5])^2 + (4 + Log[5] + x*(4 + Log[5]))^2/(4 + Log[5])^2
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 = 0.13 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67
| method | result | size |
| default | \(2 x +x^{2}+\frac {-20 \ln \left (5\right )-80}{\left (-4 \ln \left (5\right )-16\right ) \left (-\ln \left (5\right )-4+x^{2}\right )^{2}}\) | \(35\) |
| risch | \(x^{2}+2 x +\frac {5}{x^{4}-2 x^{2} \ln \left (5\right )+\ln \left (5\right )^{2}-8 x^{2}+8 \ln \left (5\right )+16}\) | \(37\) |
| norman | \(\frac {x^{6}+\left (-4 \ln \left (5\right )-16\right ) x^{3}+\left (2 \ln \left (5\right )^{2}+16 \ln \left (5\right )+32\right ) x +\left (-3 \ln \left (5\right )^{2}-24 \ln \left (5\right )-48\right ) x^{2}+2 x^{5}+2 \ln \left (5\right )^{3}+24 \ln \left (5\right )^{2}+96 \ln \left (5\right )+133}{\left (\ln \left (5\right )+4-x^{2}\right )^{2}}\) | \(79\) |
| gosper | \(\frac {x^{6}+2 x^{5}-3 x^{2} \ln \left (5\right )^{2}-4 x^{3} \ln \left (5\right )+2 \ln \left (5\right )^{3}+2 x \ln \left (5\right )^{2}-24 x^{2} \ln \left (5\right )-16 x^{3}+24 \ln \left (5\right )^{2}+16 x \ln \left (5\right )-48 x^{2}+96 \ln \left (5\right )+32 x +133}{x^{4}-2 x^{2} \ln \left (5\right )+\ln \left (5\right )^{2}-8 x^{2}+8 \ln \left (5\right )+16}\) | \(103\) |
| parallelrisch | \(\frac {x^{6}+2 x^{5}-3 x^{2} \ln \left (5\right )^{2}-4 x^{3} \ln \left (5\right )+2 \ln \left (5\right )^{3}+2 x \ln \left (5\right )^{2}-24 x^{2} \ln \left (5\right )-16 x^{3}+24 \ln \left (5\right )^{2}+16 x \ln \left (5\right )-48 x^{2}+96 \ln \left (5\right )+32 x +133}{x^{4}-2 x^{2} \ln \left (5\right )+\ln \left (5\right )^{2}-8 x^{2}+8 \ln \left (5\right )+16}\) | \(103\) |
Input:
int(((2+2*x)*ln(5)^3+(-6*x^3-6*x^2+24*x+24)*ln(5)^2+(6*x^5+6*x^4-48*x^3-48 *x^2+96*x+96)*ln(5)-2*x^7-2*x^6+24*x^5+24*x^4-96*x^3-96*x^2+148*x+128)/(ln (5)^3+(-3*x^2+12)*ln(5)^2+(3*x^4-24*x^2+48)*ln(5)-x^6+12*x^4-48*x^2+64),x, method=_RETURNVERBOSE)
Output:
2*x+x^2+10*(-2*ln(5)-8)/(-4*ln(5)-16)/(-ln(5)-4+x^2)^2
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (21) = 42\).
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 4.14 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=\frac {x^{6} + 2 \, x^{5} - 8 \, x^{4} - 16 \, x^{3} + {\left (x^{2} + 2 \, x\right )} \log \left (5\right )^{2} + 16 \, x^{2} - 2 \, {\left (x^{4} + 2 \, x^{3} - 4 \, x^{2} - 8 \, x\right )} \log \left (5\right ) + 32 \, x + 5}{x^{4} - 8 \, x^{2} - 2 \, {\left (x^{2} - 4\right )} \log \left (5\right ) + \log \left (5\right )^{2} + 16} \] Input:
integrate(((2+2*x)*log(5)^3+(-6*x^3-6*x^2+24*x+24)*log(5)^2+(6*x^5+6*x^4-4 8*x^3-48*x^2+96*x+96)*log(5)-2*x^7-2*x^6+24*x^5+24*x^4-96*x^3-96*x^2+148*x +128)/(log(5)^3+(-3*x^2+12)*log(5)^2+(3*x^4-24*x^2+48)*log(5)-x^6+12*x^4-4 8*x^2+64),x, algorithm="fricas")
Output:
(x^6 + 2*x^5 - 8*x^4 - 16*x^3 + (x^2 + 2*x)*log(5)^2 + 16*x^2 - 2*(x^4 + 2 *x^3 - 4*x^2 - 8*x)*log(5) + 32*x + 5)/(x^4 - 8*x^2 - 2*(x^2 - 4)*log(5) + log(5)^2 + 16)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=x^{2} + 2 x + \frac {5}{x^{4} + x^{2} \left (-8 - 2 \log {\left (5 \right )}\right ) + \log {\left (5 \right )}^{2} + 8 \log {\left (5 \right )} + 16} \] Input:
integrate(((2+2*x)*ln(5)**3+(-6*x**3-6*x**2+24*x+24)*ln(5)**2+(6*x**5+6*x* *4-48*x**3-48*x**2+96*x+96)*ln(5)-2*x**7-2*x**6+24*x**5+24*x**4-96*x**3-96 *x**2+148*x+128)/(ln(5)**3+(-3*x**2+12)*ln(5)**2+(3*x**4-24*x**2+48)*ln(5) -x**6+12*x**4-48*x**2+64),x)
Output:
x**2 + 2*x + 5/(x**4 + x**2*(-8 - 2*log(5)) + log(5)**2 + 8*log(5) + 16)
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=x^{2} + 2 \, x + \frac {5}{x^{4} - 2 \, x^{2} {\left (\log \left (5\right ) + 4\right )} + \log \left (5\right )^{2} + 8 \, \log \left (5\right ) + 16} \] Input:
integrate(((2+2*x)*log(5)^3+(-6*x^3-6*x^2+24*x+24)*log(5)^2+(6*x^5+6*x^4-4 8*x^3-48*x^2+96*x+96)*log(5)-2*x^7-2*x^6+24*x^5+24*x^4-96*x^3-96*x^2+148*x +128)/(log(5)^3+(-3*x^2+12)*log(5)^2+(3*x^4-24*x^2+48)*log(5)-x^6+12*x^4-4 8*x^2+64),x, algorithm="maxima")
Output:
x^2 + 2*x + 5/(x^4 - 2*x^2*(log(5) + 4) + log(5)^2 + 8*log(5) + 16)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=x^{2} + 2 \, x + \frac {5}{{\left (x^{2} - \log \left (5\right ) - 4\right )}^{2}} \] Input:
integrate(((2+2*x)*log(5)^3+(-6*x^3-6*x^2+24*x+24)*log(5)^2+(6*x^5+6*x^4-4 8*x^3-48*x^2+96*x+96)*log(5)-2*x^7-2*x^6+24*x^5+24*x^4-96*x^3-96*x^2+148*x +128)/(log(5)^3+(-3*x^2+12)*log(5)^2+(3*x^4-24*x^2+48)*log(5)-x^6+12*x^4-4 8*x^2+64),x, algorithm="giac")
Output:
x^2 + 2*x + 5/(x^2 - log(5) - 4)^2
Time = 3.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 5.14 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=2\,x+x^2+\left (\sum _{k=1}^3\ln \left (-{\mathrm {root}\left (1,z,k\right )}^2\,\ln \left (5\right )\,691200-\mathrm {root}\left (1,z,k\right )\,x^2\,48000-921600\,{\mathrm {root}\left (1,z,k\right )}^2-{\mathrm {root}\left (1,z,k\right )}^2\,{\ln \left (5\right )}^2\,172800-{\mathrm {root}\left (1,z,k\right )}^2\,{\ln \left (5\right )}^3\,14400+{\mathrm {root}\left (1,z,k\right )}^2\,x^2\,230400+{\mathrm {root}\left (1,z,k\right )}^2\,x^2\,\ln \left (5\right )\,115200+{\mathrm {root}\left (1,z,k\right )}^2\,x^2\,{\ln \left (5\right )}^2\,14400\right )\,\mathrm {root}\left (1,z,k\right )\right ) \] Input:
int((148*x + log(5)^3*(2*x + 2) + log(5)^2*(24*x - 6*x^2 - 6*x^3 + 24) - 9 6*x^2 - 96*x^3 + 24*x^4 + 24*x^5 - 2*x^6 - 2*x^7 + log(5)*(96*x - 48*x^2 - 48*x^3 + 6*x^4 + 6*x^5 + 96) + 128)/(log(5)*(3*x^4 - 24*x^2 + 48) - log(5 )^2*(3*x^2 - 12) + log(5)^3 - 48*x^2 + 12*x^4 - x^6 + 64),x)
Output:
2*x + x^2 + symsum(log(230400*root(1, z, k)^2*x^2 - 48000*root(1, z, k)*x^ 2 - 921600*root(1, z, k)^2 - 172800*root(1, z, k)^2*log(5)^2 - 14400*root( 1, z, k)^2*log(5)^3 - 691200*root(1, z, k)^2*log(5) + 115200*root(1, z, k) ^2*x^2*log(5) + 14400*root(1, z, k)^2*x^2*log(5)^2)*root(1, z, k), k, 1, 3 )
Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.62 \[ \int \frac {128+148 x-96 x^2-96 x^3+24 x^4+24 x^5-2 x^6-2 x^7+\left (96+96 x-48 x^2-48 x^3+6 x^4+6 x^5\right ) \log (5)+\left (24+24 x-6 x^2-6 x^3\right ) \log ^2(5)+(2+2 x) \log ^3(5)}{64-48 x^2+12 x^4-x^6+\left (48-24 x^2+3 x^4\right ) \log (5)+\left (12-3 x^2\right ) \log ^2(5)+\log ^3(5)} \, dx=\frac {\mathrm {log}\left (5\right )^{3}+4 x \mathrm {log}\left (5\right )^{2}+12 \mathrm {log}\left (5\right )^{2}-3 \,\mathrm {log}\left (5\right ) x^{4}-8 \,\mathrm {log}\left (5\right ) x^{3}+32 \,\mathrm {log}\left (5\right ) x +48 \,\mathrm {log}\left (5\right )+2 x^{6}+4 x^{5}-12 x^{4}-32 x^{3}+64 x +74}{2 \mathrm {log}\left (5\right )^{2}-4 \,\mathrm {log}\left (5\right ) x^{2}+16 \,\mathrm {log}\left (5\right )+2 x^{4}-16 x^{2}+32} \] Input:
int(((2+2*x)*log(5)^3+(-6*x^3-6*x^2+24*x+24)*log(5)^2+(6*x^5+6*x^4-48*x^3- 48*x^2+96*x+96)*log(5)-2*x^7-2*x^6+24*x^5+24*x^4-96*x^3-96*x^2+148*x+128)/ (log(5)^3+(-3*x^2+12)*log(5)^2+(3*x^4-24*x^2+48)*log(5)-x^6+12*x^4-48*x^2+ 64),x)
Output:
(log(5)**3 + 4*log(5)**2*x + 12*log(5)**2 - 3*log(5)*x**4 - 8*log(5)*x**3 + 32*log(5)*x + 48*log(5) + 2*x**6 + 4*x**5 - 12*x**4 - 32*x**3 + 64*x + 7 4)/(2*(log(5)**2 - 2*log(5)*x**2 + 8*log(5) + x**4 - 8*x**2 + 16))