Integrand size = 45, antiderivative size = 23 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\left (2+\frac {3 (2+x) \left (x+e^x x \log (19)\right )}{2 x}\right )^2 \] Output:
(2+3/2/x*(x*exp(x)*ln(19)+x)*(2+x))^2
Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(23)=46\).
Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.39 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {1}{2} \left (30 x+\frac {9 x^2}{2}+3 e^x \left (20+16 x+3 x^2\right ) \log (19)+9 e^{2 x} \left (2+2 x+\frac {x^2}{2}\right ) \log ^2(19)\right ) \] Input:
Integrate[(30 + 9*x + E^x*(108 + 66*x + 9*x^2)*Log[19] + E^(2*x)*(54 + 45* x + 9*x^2)*Log[19]^2)/2,x]
Output:
(30*x + (9*x^2)/2 + 3*E^x*(20 + 16*x + 3*x^2)*Log[19] + 9*E^(2*x)*(2 + 2*x + x^2/2)*Log[19]^2)/2
Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(23)=46\).
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.43, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{2} \left (e^{2 x} \left (9 x^2+45 x+54\right ) \log ^2(19)+e^x \left (9 x^2+66 x+108\right ) \log (19)+9 x+30\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \left (9 x+9 e^{2 x} \left (x^2+5 x+6\right ) \log ^2(19)+3 e^x \left (3 x^2+22 x+36\right ) \log (19)+30\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {9 x^2}{2}+\frac {9}{2} e^{2 x} x^2 \log ^2(19)+9 e^x x^2 \log (19)+30 x+18 e^{2 x} x \log ^2(19)+18 e^{2 x} \log ^2(19)+48 e^x x \log (19)+60 e^x \log (19)\right )\) |
Input:
Int[(30 + 9*x + E^x*(108 + 66*x + 9*x^2)*Log[19] + E^(2*x)*(54 + 45*x + 9* x^2)*Log[19]^2)/2,x]
Output:
(30*x + (9*x^2)/2 + 60*E^x*Log[19] + 48*E^x*x*Log[19] + 9*E^x*x^2*Log[19] + 18*E^(2*x)*Log[19]^2 + 18*E^(2*x)*x*Log[19]^2 + (9*E^(2*x)*x^2*Log[19]^2 )/2)/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(45\) vs. \(2(20)=40\).
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00
method | result | size |
risch | \(\frac {\ln \left (19\right )^{2} \left (18+18 x +\frac {9}{2} x^{2}\right ) {\mathrm e}^{2 x}}{2}+\frac {\ln \left (19\right ) \left (9 x^{2}+48 x +60\right ) {\mathrm e}^{x}}{2}+\frac {9 x^{2}}{4}+15 x\) | \(46\) |
default | \(15 x +\frac {9 \ln \left (19\right )^{2} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}+2 x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}\right )}{2}+\frac {3 \ln \left (19\right ) \left (16 \,{\mathrm e}^{x} x +20 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{x} x^{2}\right )}{2}+\frac {9 x^{2}}{4}\) | \(60\) |
parts | \(15 x +\frac {9 \ln \left (19\right )^{2} \left (\frac {{\mathrm e}^{2 x} x^{2}}{2}+2 x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}\right )}{2}+\frac {3 \ln \left (19\right ) \left (16 \,{\mathrm e}^{x} x +20 \,{\mathrm e}^{x}+3 \,{\mathrm e}^{x} x^{2}\right )}{2}+\frac {9 x^{2}}{4}\) | \(60\) |
norman | \(9 \ln \left (19\right )^{2} {\mathrm e}^{2 x}+9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x +\frac {9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x^{2}}{4}+\frac {9 \ln \left (19\right ) {\mathrm e}^{x} x^{2}}{2}+24 x \,{\mathrm e}^{x} \ln \left (19\right )+30 \ln \left (19\right ) {\mathrm e}^{x}+\frac {9 x^{2}}{4}+15 x\) | \(66\) |
parallelrisch | \(9 \ln \left (19\right )^{2} {\mathrm e}^{2 x}+9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x +\frac {9 \ln \left (19\right )^{2} {\mathrm e}^{2 x} x^{2}}{4}+\frac {9 \ln \left (19\right ) {\mathrm e}^{x} x^{2}}{2}+24 x \,{\mathrm e}^{x} \ln \left (19\right )+30 \ln \left (19\right ) {\mathrm e}^{x}+\frac {9 x^{2}}{4}+15 x\) | \(66\) |
orering | \(\frac {\left (27 x^{6}+594 x^{5}+2367 x^{4}-20240 x^{3}-191620 x^{2}-548808 x -539784\right ) \left (\frac {\left (9 x^{2}+45 x +54\right ) \ln \left (19\right )^{2} {\mathrm e}^{2 x}}{2}+\frac {\left (9 x^{2}+66 x +108\right ) \ln \left (19\right ) {\mathrm e}^{x}}{2}+\frac {9 x}{2}+15\right )}{3 \left (3 x^{2}+16 x +24\right ) \left (6 x^{3}+71 x^{2}+280 x +360\right )}-\frac {\left (27 x^{6}+486 x^{5}+720 x^{4}-25400 x^{3}-170048 x^{2}-411216 x -350880\right ) \left (\frac {\left (18 x +45\right ) \ln \left (19\right )^{2} {\mathrm e}^{2 x}}{2}+\left (9 x^{2}+45 x +54\right ) \ln \left (19\right )^{2} {\mathrm e}^{2 x}+\frac {\left (18 x +66\right ) \ln \left (19\right ) {\mathrm e}^{x}}{2}+\frac {\left (9 x^{2}+66 x +108\right ) \ln \left (19\right ) {\mathrm e}^{x}}{2}+\frac {9}{2}\right )}{2 \left (3 x^{2}+16 x +24\right ) \left (6 x^{3}+71 x^{2}+280 x +360\right )}+\frac {\left (27 x^{6}+432 x^{5}-25400 x^{3}-144648 x^{2}-314784 x -245952\right ) \left (9 \ln \left (19\right )^{2} {\mathrm e}^{2 x}+2 \left (18 x +45\right ) \ln \left (19\right )^{2} {\mathrm e}^{2 x}+2 \left (9 x^{2}+45 x +54\right ) \ln \left (19\right )^{2} {\mathrm e}^{2 x}+9 \ln \left (19\right ) {\mathrm e}^{x}+\left (18 x +66\right ) \ln \left (19\right ) {\mathrm e}^{x}+\frac {\left (9 x^{2}+66 x +108\right ) \ln \left (19\right ) {\mathrm e}^{x}}{2}\right )}{6 \left (3 x^{2}+16 x +24\right ) \left (6 x^{3}+71 x^{2}+280 x +360\right )}\) | \(362\) |
Input:
int(1/2*(9*x^2+45*x+54)*ln(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*ln(19)*exp( x)+9/2*x+15,x,method=_RETURNVERBOSE)
Output:
1/2*ln(19)^2*(18+18*x+9/2*x^2)*exp(x)^2+1/2*ln(19)*(9*x^2+48*x+60)*exp(x)+ 9/4*x^2+15*x
Time = 0.11 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \] Input:
integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log( 19)*exp(x)+9/2*x+15,x, algorithm="fricas")
Output:
9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19 ) + 9/4*x^2 + 15*x
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.87 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {9 x^{2}}{4} + 15 x + \frac {\left (36 x^{2} \log {\left (19 \right )} + 192 x \log {\left (19 \right )} + 240 \log {\left (19 \right )}\right ) e^{x}}{8} + \frac {\left (18 x^{2} \log {\left (19 \right )}^{2} + 72 x \log {\left (19 \right )}^{2} + 72 \log {\left (19 \right )}^{2}\right ) e^{2 x}}{8} \] Input:
integrate(1/2*(9*x**2+45*x+54)*ln(19)**2*exp(x)**2+1/2*(9*x**2+66*x+108)*l n(19)*exp(x)+9/2*x+15,x)
Output:
9*x**2/4 + 15*x + (36*x**2*log(19) + 192*x*log(19) + 240*log(19))*exp(x)/8 + (18*x**2*log(19)**2 + 72*x*log(19)**2 + 72*log(19)**2)*exp(2*x)/8
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \] Input:
integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log( 19)*exp(x)+9/2*x+15,x, algorithm="maxima")
Output:
9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19 ) + 9/4*x^2 + 15*x
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {9}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{\left (2 \, x\right )} \log \left (19\right )^{2} + \frac {3}{2} \, {\left (3 \, x^{2} + 16 \, x + 20\right )} e^{x} \log \left (19\right ) + \frac {9}{4} \, x^{2} + 15 \, x \] Input:
integrate(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log( 19)*exp(x)+9/2*x+15,x, algorithm="giac")
Output:
9/4*(x^2 + 4*x + 4)*e^(2*x)*log(19)^2 + 3/2*(3*x^2 + 16*x + 20)*e^x*log(19 ) + 9/4*x^2 + 15*x
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.83 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=15\,x+9\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2+30\,{\mathrm {e}}^x\,\ln \left (19\right )+\frac {9\,x^2}{4}+24\,x\,{\mathrm {e}}^x\,\ln \left (19\right )+\frac {9\,x^2\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2}{4}+\frac {9\,x^2\,{\mathrm {e}}^x\,\ln \left (19\right )}{2}+9\,x\,{\mathrm {e}}^{2\,x}\,{\ln \left (19\right )}^2 \] Input:
int((9*x)/2 + (exp(x)*log(19)*(66*x + 9*x^2 + 108))/2 + (exp(2*x)*log(19)^ 2*(45*x + 9*x^2 + 54))/2 + 15,x)
Output:
15*x + 9*exp(2*x)*log(19)^2 + 30*exp(x)*log(19) + (9*x^2)/4 + 24*x*exp(x)* log(19) + (9*x^2*exp(2*x)*log(19)^2)/4 + (9*x^2*exp(x)*log(19))/2 + 9*x*ex p(2*x)*log(19)^2
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 3.09 \[ \int \frac {1}{2} \left (30+9 x+e^x \left (108+66 x+9 x^2\right ) \log (19)+e^{2 x} \left (54+45 x+9 x^2\right ) \log ^2(19)\right ) \, dx=\frac {9 e^{2 x} \mathrm {log}\left (19\right )^{2} x^{2}}{4}+9 e^{2 x} \mathrm {log}\left (19\right )^{2} x +9 e^{2 x} \mathrm {log}\left (19\right )^{2}+\frac {9 e^{x} \mathrm {log}\left (19\right ) x^{2}}{2}+24 e^{x} \mathrm {log}\left (19\right ) x +30 e^{x} \mathrm {log}\left (19\right )+\frac {9 x^{2}}{4}+15 x \] Input:
int(1/2*(9*x^2+45*x+54)*log(19)^2*exp(x)^2+1/2*(9*x^2+66*x+108)*log(19)*ex p(x)+9/2*x+15,x)
Output:
(3*(3*e**(2*x)*log(19)**2*x**2 + 12*e**(2*x)*log(19)**2*x + 12*e**(2*x)*lo g(19)**2 + 6*e**x*log(19)*x**2 + 32*e**x*log(19)*x + 40*e**x*log(19) + 3*x **2 + 20*x))/4