Integrand size = 108, antiderivative size = 30 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\left (-9-e^x+\frac {\left (-3-2 x+x^2\right )^2}{x^2}\right ) \log ^2\left (5+x^2\right ) \] Output:
((x^2-2*x-3)^2/x^2-9-exp(x))*ln(x^2+5)^2
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.69 (sec) , antiderivative size = 572, normalized size of antiderivative = 19.07 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\frac {1}{5} \left (-128 i \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right )^2+89 \log ^2\left (i \sqrt {5}-x\right )+32 i \sqrt {5} \log ^2\left (i \sqrt {5}-x\right )+89 \log ^2\left (i \sqrt {5}+x\right )-32 i \sqrt {5} \log ^2\left (i \sqrt {5}+x\right )-256 \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right ) \log \left (\frac {10 i}{5 i-\sqrt {5} x}\right )+178 \log \left (i \sqrt {5}-x\right ) \log \left (\frac {1}{10} \left (5-i \sqrt {5} x\right )\right )+64 i \sqrt {5} \log \left (i \sqrt {5}-x\right ) \log \left (\frac {1}{10} \left (5-i \sqrt {5} x\right )\right )+178 \log \left (i \sqrt {5}+x\right ) \log \left (\frac {1}{10} \left (5+i \sqrt {5} x\right )\right )-64 i \sqrt {5} \log \left (i \sqrt {5}+x\right ) \log \left (\frac {1}{10} \left (5+i \sqrt {5} x\right )\right )-128 \sqrt {5} \arctan \left (\frac {x}{\sqrt {5}}\right ) \log \left (5+x^2\right )-178 \log \left (i \sqrt {5}-x\right ) \log \left (5+x^2\right )-64 i \sqrt {5} \log \left (i \sqrt {5}-x\right ) \log \left (5+x^2\right )-178 \log \left (i \sqrt {5}+x\right ) \log \left (5+x^2\right )+64 i \sqrt {5} \log \left (i \sqrt {5}+x\right ) \log \left (5+x^2\right )+34 \log ^2\left (5+x^2\right )-5 e^x \log ^2\left (5+x^2\right )+\frac {45 \log ^2\left (5+x^2\right )}{x^2}+\frac {60 \log ^2\left (5+x^2\right )}{x}-20 x \log ^2\left (5+x^2\right )+5 x^2 \log ^2\left (5+x^2\right )+2 \left (89-32 i \sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {i x}{2 \sqrt {5}}\right )+2 \left (89+32 i \sqrt {5}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2}+\frac {i x}{2 \sqrt {5}}\right )-128 i \sqrt {5} \operatorname {PolyLog}\left (2,\frac {5 i+\sqrt {5} x}{-5 i+\sqrt {5} x}\right )\right ) \] Input:
Integrate[((36*x^2 + 48*x^3 - 44*x^4 - 4*E^x*x^4 - 16*x^5 + 4*x^6)*Log[5 + x^2] + (-90 - 60*x - 18*x^2 - 32*x^3 + 10*x^4 - 4*x^5 + 2*x^6 + E^x*(-5*x ^3 - x^5))*Log[5 + x^2]^2)/(5*x^3 + x^5),x]
Output:
((-128*I)*Sqrt[5]*ArcTan[x/Sqrt[5]]^2 + 89*Log[I*Sqrt[5] - x]^2 + (32*I)*S qrt[5]*Log[I*Sqrt[5] - x]^2 + 89*Log[I*Sqrt[5] + x]^2 - (32*I)*Sqrt[5]*Log [I*Sqrt[5] + x]^2 - 256*Sqrt[5]*ArcTan[x/Sqrt[5]]*Log[(10*I)/(5*I - Sqrt[5 ]*x)] + 178*Log[I*Sqrt[5] - x]*Log[(5 - I*Sqrt[5]*x)/10] + (64*I)*Sqrt[5]* Log[I*Sqrt[5] - x]*Log[(5 - I*Sqrt[5]*x)/10] + 178*Log[I*Sqrt[5] + x]*Log[ (5 + I*Sqrt[5]*x)/10] - (64*I)*Sqrt[5]*Log[I*Sqrt[5] + x]*Log[(5 + I*Sqrt[ 5]*x)/10] - 128*Sqrt[5]*ArcTan[x/Sqrt[5]]*Log[5 + x^2] - 178*Log[I*Sqrt[5] - x]*Log[5 + x^2] - (64*I)*Sqrt[5]*Log[I*Sqrt[5] - x]*Log[5 + x^2] - 178* Log[I*Sqrt[5] + x]*Log[5 + x^2] + (64*I)*Sqrt[5]*Log[I*Sqrt[5] + x]*Log[5 + x^2] + 34*Log[5 + x^2]^2 - 5*E^x*Log[5 + x^2]^2 + (45*Log[5 + x^2]^2)/x^ 2 + (60*Log[5 + x^2]^2)/x - 20*x*Log[5 + x^2]^2 + 5*x^2*Log[5 + x^2]^2 + 2 *(89 - (32*I)*Sqrt[5])*PolyLog[2, 1/2 - ((I/2)*x)/Sqrt[5]] + 2*(89 + (32*I )*Sqrt[5])*PolyLog[2, 1/2 + ((I/2)*x)/Sqrt[5]] - (128*I)*Sqrt[5]*PolyLog[2 , (5*I + Sqrt[5]*x)/(-5*I + Sqrt[5]*x)])/5
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 2.64 (sec) , antiderivative size = 164, normalized size of antiderivative = 5.47, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2026, 7276, 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.
\(\displaystyle \int \frac {\left (2 x^6-4 x^5+10 x^4-32 x^3-18 x^2+e^x \left (-x^5-5 x^3\right )-60 x-90\right ) \log ^2\left (x^2+5\right )+\left (4 x^6-16 x^5-4 e^x x^4-44 x^4+48 x^3+36 x^2\right ) \log \left (x^2+5\right )}{x^5+5 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (2 x^6-4 x^5+10 x^4-32 x^3-18 x^2+e^x \left (-x^5-5 x^3\right )-60 x-90\right ) \log ^2\left (x^2+5\right )+\left (4 x^6-16 x^5-4 e^x x^4-44 x^4+48 x^3+36 x^2\right ) \log \left (x^2+5\right )}{x^3 \left (x^2+5\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (-\frac {4 x^2 \log ^2\left (x^2+5\right )}{x^2+5}+\frac {10 x \log ^2\left (x^2+5\right )}{x^2+5}-\frac {32 \log ^2\left (x^2+5\right )}{x^2+5}-\frac {18 \log ^2\left (x^2+5\right )}{\left (x^2+5\right ) x}-\frac {60 \log ^2\left (x^2+5\right )}{\left (x^2+5\right ) x^2}-\frac {16 x^2 \log \left (x^2+5\right )}{x^2+5}-\frac {44 x \log \left (x^2+5\right )}{x^2+5}+\frac {48 \log \left (x^2+5\right )}{x^2+5}-\frac {e^x \log \left (x^2+5\right ) \left (x^2 \log \left (x^2+5\right )+5 \log \left (x^2+5\right )+4 x\right )}{x^2+5}+\frac {36 \log \left (x^2+5\right )}{\left (x^2+5\right ) x}+\frac {2 x^3 \log ^2\left (x^2+5\right )}{x^2+5}-\frac {90 \log ^2\left (x^2+5\right )}{\left (x^2+5\right ) x^3}+\frac {4 x^3 \log \left (x^2+5\right )}{x^2+5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {18}{5} \operatorname {PolyLog}\left (2,-\frac {x^2}{5}\right )-\frac {18}{5} \operatorname {PolyLog}\left (2,\frac {5}{x^2+5}\right )-4 x \log ^2\left (x^2+5\right )+\frac {9 \left (x^2+5\right ) \log ^2\left (x^2+5\right )}{5 x^2}+\left (x^2+5\right ) \log ^2\left (x^2+5\right )+\frac {12 \log ^2\left (x^2+5\right )}{x}-16 \log ^2\left (x^2+5\right )-\frac {e^x \left (x^2 \log \left (x^2+5\right )+5 \log \left (x^2+5\right )\right ) \log \left (x^2+5\right )}{x^2+5}+\frac {18}{5} \log \left (1-\frac {5}{x^2+5}\right ) \log \left (x^2+5\right )-\frac {36}{5} \log (5) \log (x)\) |
Input:
Int[((36*x^2 + 48*x^3 - 44*x^4 - 4*E^x*x^4 - 16*x^5 + 4*x^6)*Log[5 + x^2] + (-90 - 60*x - 18*x^2 - 32*x^3 + 10*x^4 - 4*x^5 + 2*x^6 + E^x*(-5*x^3 - x ^5))*Log[5 + x^2]^2)/(5*x^3 + x^5),x]
Output:
(-36*Log[5]*Log[x])/5 - 16*Log[5 + x^2]^2 + (12*Log[5 + x^2]^2)/x - 4*x*Lo g[5 + x^2]^2 + (5 + x^2)*Log[5 + x^2]^2 + (9*(5 + x^2)*Log[5 + x^2]^2)/(5* x^2) - (E^x*Log[5 + x^2]*(5*Log[5 + x^2] + x^2*Log[5 + x^2]))/(5 + x^2) + (18*Log[5 + x^2]*Log[1 - 5/(5 + x^2)])/5 + (18*PolyLog[2, -1/5*x^2])/5 - ( 18*PolyLog[2, 5/(5 + x^2)])/5
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27
\[\frac {\left (x^{4}-4 x^{3}-{\mathrm e}^{x} x^{2}-11 x^{2}+12 x +9\right ) \ln \left (x^{2}+5\right )^{2}}{x^{2}}\]
Input:
int((((-x^5-5*x^3)*exp(x)+2*x^6-4*x^5+10*x^4-32*x^3-18*x^2-60*x-90)*ln(x^2 +5)^2+(-4*exp(x)*x^4+4*x^6-16*x^5-44*x^4+48*x^3+36*x^2)*ln(x^2+5))/(x^5+5* x^3),x)
Output:
(x^4-4*x^3-exp(x)*x^2-11*x^2+12*x+9)/x^2*ln(x^2+5)^2
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\frac {{\left (x^{4} - 4 \, x^{3} - x^{2} e^{x} - 11 \, x^{2} + 12 \, x + 9\right )} \log \left (x^{2} + 5\right )^{2}}{x^{2}} \] Input:
integrate((((-x^5-5*x^3)*exp(x)+2*x^6-4*x^5+10*x^4-32*x^3-18*x^2-60*x-90)* log(x^2+5)^2+(-4*exp(x)*x^4+4*x^6-16*x^5-44*x^4+48*x^3+36*x^2)*log(x^2+5)) /(x^5+5*x^3),x, algorithm="fricas")
Output:
(x^4 - 4*x^3 - x^2*e^x - 11*x^2 + 12*x + 9)*log(x^2 + 5)^2/x^2
Time = 7.40 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=- e^{x} \log {\left (x^{2} + 5 \right )}^{2} + \frac {\left (x^{4} - 4 x^{3} - 11 x^{2} + 12 x + 9\right ) \log {\left (x^{2} + 5 \right )}^{2}}{x^{2}} \] Input:
integrate((((-x**5-5*x**3)*exp(x)+2*x**6-4*x**5+10*x**4-32*x**3-18*x**2-60 *x-90)*ln(x**2+5)**2+(-4*exp(x)*x**4+4*x**6-16*x**5-44*x**4+48*x**3+36*x** 2)*ln(x**2+5))/(x**5+5*x**3),x)
Output:
-exp(x)*log(x**2 + 5)**2 + (x**4 - 4*x**3 - 11*x**2 + 12*x + 9)*log(x**2 + 5)**2/x**2
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\frac {{\left (x^{4} - 4 \, x^{3} - x^{2} e^{x} - 11 \, x^{2} + 12 \, x + 9\right )} \log \left (x^{2} + 5\right )^{2}}{x^{2}} \] Input:
integrate((((-x^5-5*x^3)*exp(x)+2*x^6-4*x^5+10*x^4-32*x^3-18*x^2-60*x-90)* log(x^2+5)^2+(-4*exp(x)*x^4+4*x^6-16*x^5-44*x^4+48*x^3+36*x^2)*log(x^2+5)) /(x^5+5*x^3),x, algorithm="maxima")
Output:
(x^4 - 4*x^3 - x^2*e^x - 11*x^2 + 12*x + 9)*log(x^2 + 5)^2/x^2
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (29) = 58\).
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.63 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\frac {x^{4} \log \left (x^{2} + 5\right )^{2} - 4 \, x^{3} \log \left (x^{2} + 5\right )^{2} - x^{2} e^{x} \log \left (x^{2} + 5\right )^{2} - 11 \, x^{2} \log \left (x^{2} + 5\right )^{2} + 12 \, x \log \left (x^{2} + 5\right )^{2} + 9 \, \log \left (x^{2} + 5\right )^{2}}{x^{2}} \] Input:
integrate((((-x^5-5*x^3)*exp(x)+2*x^6-4*x^5+10*x^4-32*x^3-18*x^2-60*x-90)* log(x^2+5)^2+(-4*exp(x)*x^4+4*x^6-16*x^5-44*x^4+48*x^3+36*x^2)*log(x^2+5)) /(x^5+5*x^3),x, algorithm="giac")
Output:
(x^4*log(x^2 + 5)^2 - 4*x^3*log(x^2 + 5)^2 - x^2*e^x*log(x^2 + 5)^2 - 11*x ^2*log(x^2 + 5)^2 + 12*x*log(x^2 + 5)^2 + 9*log(x^2 + 5)^2)/x^2
Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=-{\ln \left (x^2+5\right )}^2\,\left ({\mathrm {e}}^x+\frac {12\,x^3-6\,x^4}{x^2}-\frac {-5\,x^4+8\,x^3+12\,x+9}{x^2}+11\right ) \] Input:
int(-(log(x^2 + 5)^2*(60*x + exp(x)*(5*x^3 + x^5) + 18*x^2 + 32*x^3 - 10*x ^4 + 4*x^5 - 2*x^6 + 90) + log(x^2 + 5)*(4*x^4*exp(x) - 36*x^2 - 48*x^3 + 44*x^4 + 16*x^5 - 4*x^6))/(5*x^3 + x^5),x)
Output:
-log(x^2 + 5)^2*(exp(x) + (12*x^3 - 6*x^4)/x^2 - (12*x + 8*x^3 - 5*x^4 + 9 )/x^2 + 11)
Time = 0.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {\left (36 x^2+48 x^3-44 x^4-4 e^x x^4-16 x^5+4 x^6\right ) \log \left (5+x^2\right )+\left (-90-60 x-18 x^2-32 x^3+10 x^4-4 x^5+2 x^6+e^x \left (-5 x^3-x^5\right )\right ) \log ^2\left (5+x^2\right )}{5 x^3+x^5} \, dx=\frac {\mathrm {log}\left (x^{2}+5\right )^{2} \left (-e^{x} x^{2}+x^{4}-4 x^{3}-11 x^{2}+12 x +9\right )}{x^{2}} \] Input:
int((((-x^5-5*x^3)*exp(x)+2*x^6-4*x^5+10*x^4-32*x^3-18*x^2-60*x-90)*log(x^ 2+5)^2+(-4*exp(x)*x^4+4*x^6-16*x^5-44*x^4+48*x^3+36*x^2)*log(x^2+5))/(x^5+ 5*x^3),x)
Output:
(log(x**2 + 5)**2*( - e**x*x**2 + x**4 - 4*x**3 - 11*x**2 + 12*x + 9))/x** 2