Integrand size = 138, antiderivative size = 34 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=e^2 x \left (e^{\left (e^3-x\right )^2 x^2}+\frac {x^2 \log (x)}{5+4 x}\right ) \] Output:
exp(2)*(ln(x)*x^2/(5+4*x)+exp((-x+exp(3))^2*x^2))*x
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=e^2 \left (e^{\left (e^3-x\right )^2 x^2} x+\frac {x^3 \log (x)}{5+4 x}\right ) \] Input:
Integrate[(E^2*(5*x^2 + 4*x^3) + E^(E^6*x^2 - 2*E^3*x^3 + x^4)*(E^8*(50*x^ 2 + 80*x^3 + 32*x^4) + E^5*(-150*x^3 - 240*x^4 - 96*x^5) + E^2*(25 + 40*x + 16*x^2 + 100*x^4 + 160*x^5 + 64*x^6)) + E^2*(15*x^2 + 8*x^3)*Log[x])/(25 + 40*x + 16*x^2),x]
Output:
E^2*(E^((E^3 - x)^2*x^2)*x + (x^3*Log[x])/(5 + 4*x))
Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(34)=68\).
Time = 1.71 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2007, 7239, 27, 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.
| \(\displaystyle \int \frac {e^2 \left (4 x^3+5 x^2\right )+e^2 \left (8 x^3+15 x^2\right ) \log (x)+e^{x^4-2 e^3 x^3+e^6 x^2} \left (e^5 \left (-96 x^5-240 x^4-150 x^3\right )+e^8 \left (32 x^4+80 x^3+50 x^2\right )+e^2 \left (64 x^6+160 x^5+100 x^4+16 x^2+40 x+25\right )\right )}{16 x^2+40 x+25} \, dx\) |
| \(\Big \downarrow \) 2007 |
| \(\displaystyle \int \frac {e^2 \left (4 x^3+5 x^2\right )+e^2 \left (8 x^3+15 x^2\right ) \log (x)+e^{x^4-2 e^3 x^3+e^6 x^2} \left (e^5 \left (-96 x^5-240 x^4-150 x^3\right )+e^8 \left (32 x^4+80 x^3+50 x^2\right )+e^2 \left (64 x^6+160 x^5+100 x^4+16 x^2+40 x+25\right )\right )}{(4 x+5)^2}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^2 \left ((4 x+5) x^2+(8 x+15) x^2 \log (x)+e^{\left (e^3-x\right )^2 x^2} (4 x+5)^2 \left (4 x^4-6 e^3 x^3+2 e^6 x^2+1\right )\right )}{(4 x+5)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^2 \int \frac {(4 x+5) x^2+(8 x+15) \log (x) x^2+e^{\left (e^3-x\right )^2 x^2} (4 x+5)^2 \left (4 x^4-6 e^3 x^3+2 e^6 x^2+1\right )}{(4 x+5)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle e^2 \int \left (\frac {(8 \log (x) x+4 x+15 \log (x)+5) x^2}{(4 x+5)^2}+e^{\left (e^3-x\right )^2 x^2} \left (4 x^4-6 e^3 x^3+2 e^6 x^2+1\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^2 \left (\frac {1}{4} x^2 \log (x)+\frac {e^{\left (e^3-x\right )^2 x^2} \left (2 x^4-3 e^3 x^3+e^6 x^2\right )}{\left (e^3-x\right )^2 x-\left (e^3-x\right ) x^2}+\frac {25 x \log (x)}{16 (4 x+5)}-\frac {5}{16} x \log (x)\right )\) |
Input:
Int[(E^2*(5*x^2 + 4*x^3) + E^(E^6*x^2 - 2*E^3*x^3 + x^4)*(E^8*(50*x^2 + 80 *x^3 + 32*x^4) + E^5*(-150*x^3 - 240*x^4 - 96*x^5) + E^2*(25 + 40*x + 16*x ^2 + 100*x^4 + 160*x^5 + 64*x^6)) + E^2*(15*x^2 + 8*x^3)*Log[x])/(25 + 40* x + 16*x^2),x]
Output:
E^2*((E^((E^3 - x)^2*x^2)*(E^6*x^2 - 3*E^3*x^3 + 2*x^4))/((E^3 - x)^2*x - (E^3 - x)*x^2) - (5*x*Log[x])/16 + (x^2*Log[x])/4 + (25*x*Log[x])/(16*(5 + 4*x)))
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 119.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53
| method | result | size |
| risch | \(\frac {{\mathrm e}^{2} \left (64 x^{3}-100 x -125\right ) \ln \left (x \right )}{320+256 x}+\frac {25 \,{\mathrm e}^{2} \ln \left (x \right )}{64}+x \,{\mathrm e}^{-2 x^{3} {\mathrm e}^{3}+x^{4}+x^{2} {\mathrm e}^{6}+2}\) | \(52\) |
| parallelrisch | \(\frac {4 \ln \left (x \right ) {\mathrm e}^{2} x^{3}+16 \,{\mathrm e}^{2} x^{2} {\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}+20 x \,{\mathrm e}^{2} {\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}}{16 x +20}\) | \(72\) |
| default | \({\mathrm e}^{2} \left (\frac {x^{2}}{8}-\frac {5 x}{16}+\frac {25 \ln \left (5+4 x \right )}{64}\right )+x \,{\mathrm e}^{2} {\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}+{\mathrm e}^{2} \left (\frac {x^{2} \ln \left (x \right )}{4}-\frac {x^{2}}{8}-\frac {5 x \ln \left (x \right )}{16}+\frac {5 x}{16}-\frac {25 \ln \left (5+4 x \right )}{64}+\frac {25 \ln \left (x \right ) x}{16 \left (5+4 x \right )}\right )\) | \(90\) |
| parts | \({\mathrm e}^{2} \left (\frac {x^{2}}{8}-\frac {5 x}{16}+\frac {25 \ln \left (5+4 x \right )}{64}\right )+x \,{\mathrm e}^{2} {\mathrm e}^{x^{2} {\mathrm e}^{6}-2 x^{3} {\mathrm e}^{3}+x^{4}}+{\mathrm e}^{2} \left (\frac {x^{2} \ln \left (x \right )}{4}-\frac {x^{2}}{8}-\frac {5 x \ln \left (x \right )}{16}+\frac {5 x}{16}-\frac {25 \ln \left (5+4 x \right )}{64}+\frac {25 \ln \left (x \right ) x}{16 \left (5+4 x \right )}\right )\) | \(90\) |
Input:
int(((8*x^3+15*x^2)*exp(2)*ln(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+( -96*x^5-240*x^4-150*x^3)*exp(2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x^2+40*x +25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2))/(16* x^2+40*x+25),x,method=_RETURNVERBOSE)
Output:
1/64*exp(2)*(64*x^3-100*x-125)/(5+4*x)*ln(x)+25/64*exp(2)*ln(x)+x*exp(-2*x ^3*exp(3)+x^4+x^2*exp(6)+2)
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=\frac {x^{3} e^{2} \log \left (x\right ) + {\left (4 \, x^{2} + 5 \, x\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6} + 2\right )}}{4 \, x + 5} \] Input:
integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp (3)^2+(-96*x^5-240*x^4-150*x^3)*exp(2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x ^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2 ))/(16*x^2+40*x+25),x, algorithm="fricas")
Output:
(x^3*e^2*log(x) + (4*x^2 + 5*x)*e^(x^4 - 2*x^3*e^3 + x^2*e^6 + 2))/(4*x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (27) = 54\).
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=x e^{2} e^{x^{4} - 2 x^{3} e^{3} + x^{2} e^{6}} + \frac {25 e^{2} \log {\left (x \right )}}{64} + \frac {\left (64 x^{3} e^{2} - 100 x e^{2} - 125 e^{2}\right ) \log {\left (x \right )}}{256 x + 320} \] Input:
integrate(((8*x**3+15*x**2)*exp(2)*ln(x)+((32*x**4+80*x**3+50*x**2)*exp(2) *exp(3)**2+(-96*x**5-240*x**4-150*x**3)*exp(2)*exp(3)+(64*x**6+160*x**5+10 0*x**4+16*x**2+40*x+25)*exp(2))*exp(x**2*exp(3)**2-2*x**3*exp(3)+x**4)+(4* x**3+5*x**2)*exp(2))/(16*x**2+40*x+25),x)
Output:
x*exp(2)*exp(x**4 - 2*x**3*exp(3) + x**2*exp(6)) + 25*exp(2)*log(x)/64 + ( 64*x**3*exp(2) - 100*x*exp(2) - 125*exp(2))*log(x)/(256*x + 320)
Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (31) = 62\).
Time = 0.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 4.56 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=\frac {1}{64} \, {\left (8 \, x^{2} - 40 \, x + \frac {125}{4 \, x + 5} + 75 \, \log \left (4 \, x + 5\right )\right )} e^{2} + \frac {5}{64} \, {\left (4 \, x - \frac {25}{4 \, x + 5} - 10 \, \log \left (4 \, x + 5\right )\right )} e^{2} - \frac {25}{64} \, e^{2} \log \left (4 \, x + 5\right ) + \frac {25}{64} \, e^{2} \log \left (x\right ) - \frac {32 \, x^{3} e^{2} - 40 \, x^{2} e^{2} - 100 \, x e^{2} - 64 \, {\left (4 \, x^{2} e^{2} + 5 \, x e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )} - {\left (64 \, x^{3} e^{2} - 100 \, x e^{2} - 125 \, e^{2}\right )} \log \left (x\right )}{64 \, {\left (4 \, x + 5\right )}} \] Input:
integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp (3)^2+(-96*x^5-240*x^4-150*x^3)*exp(2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x ^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2 ))/(16*x^2+40*x+25),x, algorithm="maxima")
Output:
1/64*(8*x^2 - 40*x + 125/(4*x + 5) + 75*log(4*x + 5))*e^2 + 5/64*(4*x - 25 /(4*x + 5) - 10*log(4*x + 5))*e^2 - 25/64*e^2*log(4*x + 5) + 25/64*e^2*log (x) - 1/64*(32*x^3*e^2 - 40*x^2*e^2 - 100*x*e^2 - 64*(4*x^2*e^2 + 5*x*e^2) *e^(x^4 - 2*x^3*e^3 + x^2*e^6) - (64*x^3*e^2 - 100*x*e^2 - 125*e^2)*log(x) )/(4*x + 5)
\[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=\int { \frac {{\left (8 \, x^{3} + 15 \, x^{2}\right )} e^{2} \log \left (x\right ) + {\left (4 \, x^{3} + 5 \, x^{2}\right )} e^{2} + {\left (2 \, {\left (16 \, x^{4} + 40 \, x^{3} + 25 \, x^{2}\right )} e^{8} - 6 \, {\left (16 \, x^{5} + 40 \, x^{4} + 25 \, x^{3}\right )} e^{5} + {\left (64 \, x^{6} + 160 \, x^{5} + 100 \, x^{4} + 16 \, x^{2} + 40 \, x + 25\right )} e^{2}\right )} e^{\left (x^{4} - 2 \, x^{3} e^{3} + x^{2} e^{6}\right )}}{16 \, x^{2} + 40 \, x + 25} \,d x } \] Input:
integrate(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp (3)^2+(-96*x^5-240*x^4-150*x^3)*exp(2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x ^2+40*x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2 ))/(16*x^2+40*x+25),x, algorithm="giac")
Output:
integrate(((8*x^3 + 15*x^2)*e^2*log(x) + (4*x^3 + 5*x^2)*e^2 + (2*(16*x^4 + 40*x^3 + 25*x^2)*e^8 - 6*(16*x^5 + 40*x^4 + 25*x^3)*e^5 + (64*x^6 + 160* x^5 + 100*x^4 + 16*x^2 + 40*x + 25)*e^2)*e^(x^4 - 2*x^3*e^3 + x^2*e^6))/(1 6*x^2 + 40*x + 25), x)
Time = 1.66 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.68 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=x\,{\mathrm {e}}^{x^4-2\,{\mathrm {e}}^3\,x^3+{\mathrm {e}}^6\,x^2+2}-\frac {25\,\ln \left (x+\frac {5}{4}\right )\,{\mathrm {e}}^2}{64}+\frac {{\mathrm {e}}^2\,\left (25\,\ln \left (x+\frac {5}{4}\right )-20\,x+8\,x^2\right )}{64}+\frac {\frac {25\,x^2\,{\mathrm {e}}^2}{16}+\frac {5\,x^3\,{\mathrm {e}}^2}{8}-\frac {x^4\,{\mathrm {e}}^2}{2}+x^4\,{\mathrm {e}}^2\,\ln \left (x\right )}{4\,x^2+5\,x} \] Input:
int((exp(2)*(5*x^2 + 4*x^3) + exp(x^2*exp(6) - 2*x^3*exp(3) + x^4)*(exp(8) *(50*x^2 + 80*x^3 + 32*x^4) - exp(5)*(150*x^3 + 240*x^4 + 96*x^5) + exp(2) *(40*x + 16*x^2 + 100*x^4 + 160*x^5 + 64*x^6 + 25)) + exp(2)*log(x)*(15*x^ 2 + 8*x^3))/(40*x + 16*x^2 + 25),x)
Output:
x*exp(x^2*exp(6) - 2*x^3*exp(3) + x^4 + 2) - (25*log(x + 5/4)*exp(2))/64 + (exp(2)*(25*log(x + 5/4) - 20*x + 8*x^2))/64 + ((25*x^2*exp(2))/16 + (5*x ^3*exp(2))/8 - (x^4*exp(2))/2 + x^4*exp(2)*log(x))/(5*x + 4*x^2)
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.12 \[ \int \frac {e^2 \left (5 x^2+4 x^3\right )+e^{e^6 x^2-2 e^3 x^3+x^4} \left (e^8 \left (50 x^2+80 x^3+32 x^4\right )+e^5 \left (-150 x^3-240 x^4-96 x^5\right )+e^2 \left (25+40 x+16 x^2+100 x^4+160 x^5+64 x^6\right )\right )+e^2 \left (15 x^2+8 x^3\right ) \log (x)}{25+40 x+16 x^2} \, dx=\frac {e^{2} x \left (4 e^{e^{6} x^{2}+x^{4}} x +5 e^{e^{6} x^{2}+x^{4}}+e^{2 e^{3} x^{3}} \mathrm {log}\left (x \right ) x^{2}\right )}{e^{2 e^{3} x^{3}} \left (4 x +5\right )} \] Input:
int(((8*x^3+15*x^2)*exp(2)*log(x)+((32*x^4+80*x^3+50*x^2)*exp(2)*exp(3)^2+ (-96*x^5-240*x^4-150*x^3)*exp(2)*exp(3)+(64*x^6+160*x^5+100*x^4+16*x^2+40* x+25)*exp(2))*exp(x^2*exp(3)^2-2*x^3*exp(3)+x^4)+(4*x^3+5*x^2)*exp(2))/(16 *x^2+40*x+25),x)
Output:
(e**2*x*(4*e**(e**6*x**2 + x**4)*x + 5*e**(e**6*x**2 + x**4) + e**(2*e**3* x**3)*log(x)*x**2))/(e**(2*e**3*x**3)*(4*x + 5))