Integrand size = 73, antiderivative size = 20 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {15}{5+e+\frac {3 e^{53}}{2 x}+2 x} \] Output:
15/(2*x+exp(1)+5+3/2/x*exp(3)*exp(25)^2)
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30 x}{3 e^{53}+10 x+2 e x+4 x^2} \] Input:
Integrate[(90*E^53 - 120*x^2)/(9*E^106 + 100*x^2 + 4*E^2*x^2 + 80*x^3 + 16 *x^4 + E^53*(60*x + 12*E*x + 24*x^2) + E*(40*x^2 + 16*x^3)),x]
Output:
(30*x)/(3*E^53 + 10*x + 2*E*x + 4*x^2)
Leaf count is larger than twice the leaf count of optimal. \(91\) vs. \(2(20)=40\).
Time = 0.37 (sec) , antiderivative size = 91, normalized size of antiderivative = 4.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {6, 2459, 1380, 27, 2345, 24}
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.
\(\displaystyle \int \frac {90 e^{53}-120 x^2}{16 x^4+80 x^3+4 e^2 x^2+100 x^2+e^{53} \left (24 x^2+12 e x+60 x\right )+e \left (16 x^3+40 x^2\right )+9 e^{106}} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {90 e^{53}-120 x^2}{16 x^4+80 x^3+\left (100+4 e^2\right ) x^2+e^{53} \left (24 x^2+12 e x+60 x\right )+e \left (16 x^3+40 x^2\right )+9 e^{106}}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int \frac {-120 \left (x+\frac {1}{64} (80+16 e)\right )^2+60 (5+e) \left (x+\frac {1}{64} (80+16 e)\right )-\frac {15}{2} \left (25+10 e+e^2-12 e^{53}\right )}{16 \left (x+\frac {1}{64} (80+16 e)\right )^4-2 \left (25+10 e+e^2-12 e^{53}\right ) \left (x+\frac {1}{64} (80+16 e)\right )^2+\frac {1}{16} \left (25+10 e+e^2-12 e^{53}\right )^2}d\left (x+\frac {1}{64} (80+16 e)\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 16 \int -\frac {15 \left (16 \left (x+\frac {1}{64} (80+16 e)\right )^2-8 (5+e) \left (x+\frac {1}{64} (80+16 e)\right )-12 e^{53}+e^2+10 e+25\right )}{2 \left (-16 \left (x+\frac {1}{64} (80+16 e)\right )^2-12 e^{53}+e^2+10 e+25\right )^2}d\left (x+\frac {1}{64} (80+16 e)\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -120 \int \frac {16 \left (x+\frac {1}{64} (80+16 e)\right )^2-8 (5+e) \left (x+\frac {1}{64} (80+16 e)\right )-12 e^{53}+e^2+10 e+25}{\left (-16 \left (x+\frac {1}{64} (80+16 e)\right )^2-12 e^{53}+e^2+10 e+25\right )^2}d\left (x+\frac {1}{64} (80+16 e)\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -120 \left (-\frac {\int 0d\left (x+\frac {1}{64} (80+16 e)\right )}{2 \left (25+10 e+e^2-12 e^{53}\right )}-\frac {(5+e) \left (25+10 e+e^2-12 e^{53}\right )-4 \left (25+10 e+e^2-12 e^{53}\right ) \left (x+\frac {1}{64} (80+16 e)\right )}{4 \left (25+10 e+e^2-12 e^{53}\right ) \left (-16 \left (x+\frac {1}{64} (80+16 e)\right )^2-12 e^{53}+e^2+10 e+25\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {30 \left ((5+e) \left (25+10 e+e^2-12 e^{53}\right )-4 \left (25+10 e+e^2-12 e^{53}\right ) \left (x+\frac {1}{64} (80+16 e)\right )\right )}{\left (25+10 e+e^2-12 e^{53}\right ) \left (-16 \left (x+\frac {1}{64} (80+16 e)\right )^2-12 e^{53}+e^2+10 e+25\right )}\) |
Input:
Int[(90*E^53 - 120*x^2)/(9*E^106 + 100*x^2 + 4*E^2*x^2 + 80*x^3 + 16*x^4 + E^53*(60*x + 12*E*x + 24*x^2) + E*(40*x^2 + 16*x^3)),x]
Output:
(30*((5 + E)*(25 + 10*E + E^2 - 12*E^53) - 4*(25 + 10*E + E^2 - 12*E^53)*( (80 + 16*E)/64 + x)))/((25 + 10*E + E^2 - 12*E^53)*(25 + 10*E + E^2 - 12*E ^53 - 16*((80 + 16*E)/64 + x)^2))
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {10 x}{{\mathrm e}^{53}+\frac {2 x \,{\mathrm e}}{3}+\frac {4 x^{2}}{3}+\frac {10 x}{3}}\) | \(22\) |
gosper | \(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\) | \(28\) |
norman | \(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\) | \(28\) |
parallelrisch | \(\frac {30 x}{3 \,{\mathrm e}^{3} {\mathrm e}^{50}+2 x \,{\mathrm e}+4 x^{2}+10 x}\) | \(28\) |
default | \(\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16 \textit {\_Z}^{4}+\left (16 \,{\mathrm e}+80\right ) \textit {\_Z}^{3}+\left (24 \,{\mathrm e}^{53}+4 \,{\mathrm e}^{2}+40 \,{\mathrm e}+100\right ) \textit {\_Z}^{2}+\left (12 \,{\mathrm e}^{54}+60 \,{\mathrm e}^{53}\right ) \textit {\_Z} +9 \,{\mathrm e}^{106}\right )}{\sum }\frac {\left (3 \,{\mathrm e}^{53}-4 \textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{3 \,{\mathrm e}^{54}+12 \textit {\_R} \,{\mathrm e}^{53}+15 \,{\mathrm e}^{53}+2 \,{\mathrm e}^{2} \textit {\_R} +12 \textit {\_R}^{2} {\mathrm e}+16 \textit {\_R}^{3}+20 \textit {\_R} \,{\mathrm e}+60 \textit {\_R}^{2}+50 \textit {\_R}}\right )}{2}\) | \(119\) |
Input:
int((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*x*exp(1)+24*x^ 2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x ^3+100*x^2),x,method=_RETURNVERBOSE)
Output:
10*x/(exp(53)+2/3*x*exp(1)+4/3*x^2+10/3*x)
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30 \, x}{4 \, x^{2} + 2 \, x e + 10 \, x + 3 \, e^{53}} \] Input:
integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*exp(1)*x +24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^ 4+80*x^3+100*x^2),x, algorithm="fricas")
Output:
30*x/(4*x^2 + 2*x*e + 10*x + 3*e^53)
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30 x}{4 x^{2} + x \left (2 e + 10\right ) + 3 e^{53}} \] Input:
integrate((90*exp(3)*exp(25)**2-120*x**2)/(9*exp(3)**2*exp(25)**4+(12*exp( 1)*x+24*x**2+60*x)*exp(3)*exp(25)**2+4*x**2*exp(1)**2+(16*x**3+40*x**2)*ex p(1)+16*x**4+80*x**3+100*x**2),x)
Output:
30*x/(4*x**2 + x*(2*E + 10) + 3*exp(53))
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30 \, x}{4 \, x^{2} + 2 \, x {\left (e + 5\right )} + 3 \, e^{53}} \] Input:
integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*exp(1)*x +24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^ 4+80*x^3+100*x^2),x, algorithm="maxima")
Output:
30*x/(4*x^2 + 2*x*(e + 5) + 3*e^53)
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30 \, x}{4 \, x^{2} + 2 \, x e + 10 \, x + 3 \, e^{53}} \] Input:
integrate((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*exp(1)*x +24*x^2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^ 4+80*x^3+100*x^2),x, algorithm="giac")
Output:
30*x/(4*x^2 + 2*x*e + 10*x + 3*e^53)
Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {30\,x}{4\,x^2+\left (2\,\mathrm {e}+10\right )\,x+3\,{\mathrm {e}}^{53}} \] Input:
int((90*exp(53) - 120*x^2)/(9*exp(106) + exp(53)*(60*x + 12*x*exp(1) + 24* x^2) + exp(1)*(40*x^2 + 16*x^3) + 4*x^2*exp(2) + 100*x^2 + 80*x^3 + 16*x^4 ),x)
Output:
(30*x)/(3*exp(53) + 4*x^2 + x*(2*exp(1) + 10))
Time = 0.16 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.50 \[ \int \frac {90 e^{53}-120 x^2}{9 e^{106}+100 x^2+4 e^2 x^2+80 x^3+16 x^4+e^{53} \left (60 x+12 e x+24 x^2\right )+e \left (40 x^2+16 x^3\right )} \, dx=\frac {-45 e^{53}-60 x^{2}}{3 e^{54}+15 e^{53}+2 e^{2} x +4 e \,x^{2}+20 e x +20 x^{2}+50 x} \] Input:
int((90*exp(3)*exp(25)^2-120*x^2)/(9*exp(3)^2*exp(25)^4+(12*exp(1)*x+24*x^ 2+60*x)*exp(3)*exp(25)^2+4*x^2*exp(1)^2+(16*x^3+40*x^2)*exp(1)+16*x^4+80*x ^3+100*x^2),x)
Output:
(15*( - 3*e**53 - 4*x**2))/(3*e**54 + 15*e**53 + 2*e**2*x + 4*e*x**2 + 20* e*x + 20*x**2 + 50*x)