Integrand size = 51, antiderivative size = 22 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} e^{\frac {e^{15}}{x}} x (3-2 \log (2))^2 \]
Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} e^{\frac {e^{15}}{x}} x (-3+\log (4))^2 \]
Integrate[(E^(E^15/x)*(-36*E^15 + 36*x + (48*E^15 - 48*x)*Log[2] + (-16*E^ 15 + 16*x)*Log[2]^2))/(3*x),x]
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {27, 6, 27, 2092, 2726}
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.
\(\displaystyle \int \frac {e^{\frac {e^{15}}{x}} \left (36 x+\left (16 x-16 e^{15}\right ) \log ^2(2)+\left (48 e^{15}-48 x\right ) \log (2)-36 e^{15}\right )}{3 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {4 e^{\frac {e^{15}}{x}} \left (4 \log ^2(2) \left (e^{15}-x\right )-12 \log (2) \left (e^{15}-x\right )-9 x+9 e^{15}\right )}{x}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \frac {1}{3} \int -\frac {4 e^{\frac {e^{15}}{x}} \left (\left (-12 \log (2)+4 \log ^2(2)\right ) \left (e^{15}-x\right )-9 x+9 e^{15}\right )}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{3} \int \frac {e^{\frac {e^{15}}{x}} \left (-4 (3-\log (2)) \log (2) \left (e^{15}-x\right )-9 x+9 e^{15}\right )}{x}dx\) |
\(\Big \downarrow \) 2092 |
\(\displaystyle -\frac {4}{3} \int \frac {e^{\frac {e^{15}}{x}} \left (e^{15} (3-\log (4))^2-x (3-\log (4))^2\right )}{x}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {4}{3} e^{\frac {e^{15}}{x}} x (3-\log (4))^2\) |
Int[(E^(E^15/x)*(-36*E^15 + 36*x + (48*E^15 - 48*x)*Log[2] + (-16*E^15 + 1 6*x)*Log[2]^2))/(3*x),x]
3.19.11.3.1 Defintions of rubi rules used
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[(Px_)*(u_)^(p_.)*(z_)^(q_.), x_Symbol] :> Int[Px*ExpandToSum[z, x]^q*Ex pandToSum[u, x]^p, x] /; FreeQ[{p, q}, x] && BinomialQ[z, x] && BinomialQ[u , x] && !(BinomialMatchQ[z, x] && BinomialMatchQ[u, x])
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\left (\frac {16 \ln \left (2\right )^{2}}{3}-16 \ln \left (2\right )+12\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}\) | \(22\) |
gosper | \(\frac {4 x \left (4 \ln \left (2\right )^{2}-12 \ln \left (2\right )+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\) | \(23\) |
risch | \(\frac {4 x \left (4 \ln \left (2\right )^{2}-12 \ln \left (2\right )+9\right ) {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}}{3}\) | \(23\) |
parallelrisch | \(\frac {16 \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}} \ln \left (2\right )^{2} x}{3}-16 \ln \left (2\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}+12 x \,{\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}\) | \(38\) |
derivativedivides | \(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )+16 \ln \left (2\right ) {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}+16 \ln \left (2\right ) {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )}{3}\) | \(143\) |
default | \(-12 \,{\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-12 \,{\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )+16 \ln \left (2\right ) {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-x \,{\mathrm e}^{-15} {\mathrm e}^{\frac {{\mathrm e}^{15}}{x}}-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}+16 \ln \left (2\right ) {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )-\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )}{3}\) | \(143\) |
meijerg | \(\left (\frac {16 \ln \left (2\right )^{2}}{3}-16 \ln \left (2\right )+12\right ) {\mathrm e}^{15} \left (x \,{\mathrm e}^{-15}-14+\ln \left (x \right )-i \pi -\frac {x \,{\mathrm e}^{-15} \left (2+\frac {2 \,{\mathrm e}^{15}}{x}\right )}{2}+x \,{\mathrm e}^{-15+\frac {{\mathrm e}^{15}}{x}}+\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )+\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )+\frac {16 \ln \left (2\right )^{2} {\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )}{3}-16 \ln \left (2\right ) {\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )+12 \,{\mathrm e}^{15} \left (-\ln \left (x \right )+15+i \pi -\ln \left (-\frac {{\mathrm e}^{15}}{x}\right )-\operatorname {Ei}_{1}\left (-\frac {{\mathrm e}^{15}}{x}\right )\right )\) | \(182\) |
int(1/3*((-16*exp(15)+16*x)*ln(2)^2+(48*exp(15)-48*x)*ln(2)-36*exp(15)+36* x)*exp(exp(15)/x)/x,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} \, {\left (4 \, x \log \left (2\right )^{2} - 12 \, x \log \left (2\right ) + 9 \, x\right )} e^{\left (\frac {e^{15}}{x}\right )} \]
integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp (15)+36*x)*exp(exp(15)/x)/x,x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {\left (- 48 x \log {\left (2 \right )} + 16 x \log {\left (2 \right )}^{2} + 36 x\right ) e^{\frac {e^{15}}{x}}}{3} \]
integrate(1/3*((-16*exp(15)+16*x)*ln(2)**2+(48*exp(15)-48*x)*ln(2)-36*exp( 15)+36*x)*exp(exp(15)/x)/x,x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.86 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {16}{3} \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \left (2\right )^{2} - \frac {16}{3} \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \left (2\right )^{2} - 16 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} \log \left (2\right ) + 16 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \log \left (2\right ) + 12 \, {\rm Ei}\left (\frac {e^{15}}{x}\right ) e^{15} - 12 \, e^{15} \Gamma \left (-1, -\frac {e^{15}}{x}\right ) \]
integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp (15)+36*x)*exp(exp(15)/x)/x,x, algorithm=\
16/3*Ei(e^15/x)*e^15*log(2)^2 - 16/3*e^15*gamma(-1, -e^15/x)*log(2)^2 - 16 *Ei(e^15/x)*e^15*log(2) + 16*e^15*gamma(-1, -e^15/x)*log(2) + 12*Ei(e^15/x )*e^15 - 12*e^15*gamma(-1, -e^15/x)
Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (18) = 36\).
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4}{3} \, {\left (4 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \left (2\right )^{2} - 12 \, e^{\left (\frac {e^{15}}{x} + 45\right )} \log \left (2\right ) + 9 \, e^{\left (\frac {e^{15}}{x} + 45\right )}\right )} x e^{\left (-45\right )} \]
integrate(1/3*((-16*exp(15)+16*x)*log(2)^2+(48*exp(15)-48*x)*log(2)-36*exp (15)+36*x)*exp(exp(15)/x)/x,x, algorithm=\
Time = 10.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {e^{15}}{x}} \left (-36 e^{15}+36 x+\left (48 e^{15}-48 x\right ) \log (2)+\left (-16 e^{15}+16 x\right ) \log ^2(2)\right )}{3 x} \, dx=\frac {4\,x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{15}}{x}}\,{\left (\ln \left (4\right )-3\right )}^2}{3} \]