Integrand size = 81, antiderivative size = 29 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\left (2+e^{3 x}\right ) \left (\frac {1}{5} \left (-4+\frac {e^{2 x}}{x}\right )^2+x+\log (4)\right ) \] Output:
(exp(3*x)+2)*(2*ln(2)+1/5*(exp(-ln(x/exp(x))+x)-4)^2+x)
Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(29)=58\).
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {1}{5} \left (\frac {2 e^{4 x}}{x^2}+\frac {e^{7 x}}{x^2}-\frac {16 e^{2 x}}{x}-\frac {8 e^{5 x}}{x}+10 x+e^{3 x} (16+5 x+5 \log (4))\right ) \] Input:
Integrate[((E^(2*x)*(16 + E^(3*x)*(8 - 40*x) - 32*x))/x + 10*x + (E^(4*x)* (-4 + 8*x + E^(3*x)*(-2 + 7*x)))/x^2 + E^(3*x)*(53*x + 15*x^2 + 15*x*Log[4 ]))/(5*x),x]
Output:
((2*E^(4*x))/x^2 + E^(7*x)/x^2 - (16*E^(2*x))/x - (8*E^(5*x))/x + 10*x + E ^(3*x)*(16 + 5*x + 5*Log[4]))/5
Leaf count is larger than twice the leaf count of optimal. \(74\) vs. \(2(29)=58\).
Time = 0.60 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {27, 25, 2010, 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.
\(\displaystyle \int \frac {\frac {e^{4 x} \left (8 x+e^{3 x} (7 x-2)-4\right )}{x^2}+e^{3 x} \left (15 x^2+53 x+15 x \log (4)\right )+\frac {e^{2 x} \left (e^{3 x} (8-40 x)-32 x+16\right )}{x}+10 x}{5 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {\frac {e^{4 x} \left (e^{3 x} (2-7 x)-8 x+4\right )}{x^2}-10 x-e^{3 x} \left (15 x^2+15 \log (4) x+53 x\right )-\frac {8 e^{2 x} \left (e^{3 x} (1-5 x)-4 x+2\right )}{x}}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {\frac {e^{4 x} \left (e^{3 x} (2-7 x)-8 x+4\right )}{x^2}-10 x-e^{3 x} \left (15 x^2+15 \log (4) x+53 x\right )-\frac {8 e^{2 x} \left (e^{3 x} (1-5 x)-4 x+2\right )}{x}}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {16 e^{2 x} (2 x-1)}{x^2}-\frac {4 e^{4 x} (2 x-1)}{x^3}+\frac {8 e^{5 x} (5 x-1)}{x^2}-\frac {e^{7 x} (7 x-2)}{x^3}-e^{3 x} (15 x+15 \log (4)+53)-10\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {2 e^{4 x}}{x^2}+\frac {e^{7 x}}{x^2}+10 x-\frac {5 e^{3 x}}{3}-\frac {16 e^{2 x}}{x}-\frac {8 e^{5 x}}{x}+\frac {1}{3} e^{3 x} (15 x+53+15 \log (4))\right )\) |
Input:
Int[((E^(2*x)*(16 + E^(3*x)*(8 - 40*x) - 32*x))/x + 10*x + (E^(4*x)*(-4 + 8*x + E^(3*x)*(-2 + 7*x)))/x^2 + E^(3*x)*(53*x + 15*x^2 + 15*x*Log[4]))/(5 *x),x]
Output:
((-5*E^(3*x))/3 + (2*E^(4*x))/x^2 + E^(7*x)/x^2 - (16*E^(2*x))/x - (8*E^(5 *x))/x + 10*x + (E^(3*x)*(53 + 15*x + 15*Log[4]))/3)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Leaf count of result is larger than twice the leaf count of optimal. \(72\) vs. \(2(31)=62\).
Time = 4.56 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52
method | result | size |
default | \(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(73\) |
parts | \(2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{5 x}}{5 x}+\frac {{\mathrm e}^{7 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(73\) |
parallelrisch | \(\frac {{\mathrm e}^{3 x} {\mathrm e}^{4 x}}{5 x^{2}}+2 \ln \left (2\right ) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}-\frac {8 \,{\mathrm e}^{3 x} {\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}+\frac {2 \,{\mathrm e}^{4 x}}{5 x^{2}}+2 x +\frac {16 \,{\mathrm e}^{3 x}}{5}-\frac {16 \,{\mathrm e}^{-\ln \left (x \,{\mathrm e}^{-x}\right )+x}}{5}\) | \(93\) |
risch | \(2 \ln \left (2\right ) {\mathrm e}^{3 x}+x \,{\mathrm e}^{3 x}+\frac {16 \,{\mathrm e}^{3 x}}{5}+2 x +\frac {\left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{4 x}}{5 x^{2}}-\frac {8 \left ({\mathrm e}^{3 x}+2\right ) {\mathrm e}^{2 x} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} \pi }{2}} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) \pi }{2}} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \pi }{2}} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \pi }{2}}}{5 x}\) | \(143\) |
Input:
int(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-ln(x/exp(x))+x)^2+((-40*x+8)*exp(3* x)-32*x+16)*exp(-ln(x/exp(x))+x)+(30*x*ln(2)+15*x^2+53*x)*exp(3*x)+10*x)/x ,x,method=_RETURNVERBOSE)
Output:
2*x+16/5*exp(x)^3+x*exp(x)^3-8/5*exp(x)^5/x+1/5*exp(x)^7/x^2+2*ln(2)*exp(x )^3+2/5*exp(-ln(x/exp(x))+x)^2-16/5*exp(-ln(x/exp(x))+x)
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {{\left (10 \, x^{3} e^{\left (-7 \, x\right )} - 8 \, x e^{\left (-2 \, x\right )} + {\left (5 \, x^{3} + 10 \, x^{2} \log \left (2\right ) + 16 \, x^{2}\right )} e^{\left (-4 \, x\right )} - 16 \, x e^{\left (-5 \, x\right )} + 2 \, e^{\left (-3 \, x\right )} + 1\right )} e^{\left (7 \, x\right )}}{5 \, x^{2}} \] Input:
integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8) *exp(3*x)-32*x+16)*exp(-log(x/exp(x))+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x )+10*x)/x,x, algorithm="fricas")
Output:
1/5*(10*x^3*e^(-7*x) - 8*x*e^(-2*x) + (5*x^3 + 10*x^2*log(2) + 16*x^2)*e^( -4*x) - 16*x*e^(-5*x) + 2*e^(-3*x) + 1)*e^(7*x)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=2 x + \frac {- 5000 x^{5} e^{5 x} - 10000 x^{5} e^{2 x} + 625 x^{4} e^{7 x} + 1250 x^{4} e^{4 x} + \left (3125 x^{7} + 6250 x^{6} \log {\left (2 \right )} + 10000 x^{6}\right ) e^{3 x}}{3125 x^{6}} \] Input:
integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-ln(x/exp(x))+x)**2+((-40*x+8) *exp(3*x)-32*x+16)*exp(-ln(x/exp(x))+x)+(30*x*ln(2)+15*x**2+53*x)*exp(3*x) +10*x)/x,x)
Output:
2*x + (-5000*x**5*exp(5*x) - 10000*x**5*exp(2*x) + 625*x**4*exp(7*x) + 125 0*x**4*exp(4*x) + (3125*x**7 + 6250*x**6*log(2) + 10000*x**6)*exp(3*x))/(3 125*x**6)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.86 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {1}{3} \, {\left (3 \, x - 1\right )} e^{\left (3 \, x\right )} + 2 \, e^{\left (3 \, x\right )} \log \left (2\right ) + 2 \, x - 8 \, {\rm Ei}\left (5 \, x\right ) - \frac {32}{5} \, {\rm Ei}\left (2 \, x\right ) + \frac {53}{15} \, e^{\left (3 \, x\right )} + \frac {32}{5} \, \Gamma \left (-1, -2 \, x\right ) + \frac {32}{5} \, \Gamma \left (-1, -4 \, x\right ) + 8 \, \Gamma \left (-1, -5 \, x\right ) + \frac {49}{5} \, \Gamma \left (-1, -7 \, x\right ) + \frac {64}{5} \, \Gamma \left (-2, -4 \, x\right ) + \frac {98}{5} \, \Gamma \left (-2, -7 \, x\right ) \] Input:
integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8) *exp(3*x)-32*x+16)*exp(-log(x/exp(x))+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x )+10*x)/x,x, algorithm="maxima")
Output:
1/3*(3*x - 1)*e^(3*x) + 2*e^(3*x)*log(2) + 2*x - 8*Ei(5*x) - 32/5*Ei(2*x) + 53/15*e^(3*x) + 32/5*gamma(-1, -2*x) + 32/5*gamma(-1, -4*x) + 8*gamma(-1 , -5*x) + 49/5*gamma(-1, -7*x) + 64/5*gamma(-2, -4*x) + 98/5*gamma(-2, -7* x)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {5 \, x^{3} e^{\left (3 \, x\right )} + 10 \, x^{2} e^{\left (3 \, x\right )} \log \left (2\right ) + 10 \, x^{3} + 16 \, x^{2} e^{\left (3 \, x\right )} - 8 \, x e^{\left (5 \, x\right )} - 16 \, x e^{\left (2 \, x\right )} + e^{\left (7 \, x\right )} + 2 \, e^{\left (4 \, x\right )}}{5 \, x^{2}} \] Input:
integrate(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8) *exp(3*x)-32*x+16)*exp(-log(x/exp(x))+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x )+10*x)/x,x, algorithm="giac")
Output:
1/5*(5*x^3*e^(3*x) + 10*x^2*e^(3*x)*log(2) + 10*x^3 + 16*x^2*e^(3*x) - 8*x *e^(5*x) - 16*x*e^(2*x) + e^(7*x) + 2*e^(4*x))/x^2
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {\frac {2\,{\mathrm {e}}^{4\,x}}{5}+\frac {{\mathrm {e}}^{7\,x}}{5}-x\,\left (\frac {16\,{\mathrm {e}}^{2\,x}}{5}+\frac {8\,{\mathrm {e}}^{5\,x}}{5}\right )}{x^2}+x\,\left ({\mathrm {e}}^{3\,x}+2\right )+{\mathrm {e}}^{3\,x}\,\left (2\,\ln \left (2\right )+\frac {16}{5}\right ) \] Input:
int((2*x + (exp(3*x)*(53*x + 30*x*log(2) + 15*x^2))/5 - (exp(x - log(x*exp (-x)))*(32*x + exp(3*x)*(40*x - 8) - 16))/5 + (exp(2*x - 2*log(x*exp(-x))) *(8*x + exp(3*x)*(7*x - 2) - 4))/5)/x,x)
Output:
((2*exp(4*x))/5 + exp(7*x)/5 - x*((16*exp(2*x))/5 + (8*exp(5*x))/5))/x^2 + x*(exp(3*x) + 2) + exp(3*x)*(2*log(2) + 16/5)
Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.45 \[ \int \frac {\frac {e^{2 x} \left (16+e^{3 x} (8-40 x)-32 x\right )}{x}+10 x+\frac {e^{4 x} \left (-4+8 x+e^{3 x} (-2+7 x)\right )}{x^2}+e^{3 x} \left (53 x+15 x^2+15 x \log (4)\right )}{5 x} \, dx=\frac {e^{7 x}-8 e^{5 x} x +2 e^{4 x}+10 e^{3 x} \mathrm {log}\left (2\right ) x^{2}+5 e^{3 x} x^{3}+16 e^{3 x} x^{2}-16 e^{2 x} x +10 x^{3}}{5 x^{2}} \] Input:
int(1/5*(((7*x-2)*exp(3*x)+8*x-4)*exp(-log(x/exp(x))+x)^2+((-40*x+8)*exp(3 *x)-32*x+16)*exp(-log(x/exp(x))+x)+(30*x*log(2)+15*x^2+53*x)*exp(3*x)+10*x )/x,x)
Output:
(e**(7*x) - 8*e**(5*x)*x + 2*e**(4*x) + 10*e**(3*x)*log(2)*x**2 + 5*e**(3* x)*x**3 + 16*e**(3*x)*x**2 - 16*e**(2*x)*x + 10*x**3)/(5*x**2)