3.39.47 \(\int \frac {e^{\frac {5-x \log (3)}{\log (3)}} (-15-15 x-10 x^2+(2+2 x-2 x^2) \log (2)+(1+x) \log ^2(2))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \]

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 57, normalized size of antiderivative = 1.97, number of steps used = 6, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {2199, 2194, 2177, 2178} \begin {gather*} \frac {e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right ) 3^{-\frac {x}{\log (3)}}}{x}+2 e^{\frac {5}{\log (3)}} (5+\log (2)) 3^{-\frac {x}{\log (3)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((5 - x*Log[3])/Log[3])*(-15 - 15*x - 10*x^2 + (2 + 2*x - 2*x^2)*Log[2] + (1 + x)*Log[2]^2))/x^2,x]

[Out]

(2*E^(5/Log[3])*(5 + Log[2]))/3^(x/Log[3]) + (E^(5/Log[3])*(15 - Log[2]^2 - Log[4]))/(3^(x/Log[3])*x)

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-2 e^{\frac {5-x \log (3)}{\log (3)}} (5+\log (2))+\frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15+\log ^2(2)+\log (4)\right )}{x^2}+\frac {e^{\frac {5-x \log (3)}{\log (3)}} \left (-15+\log ^2(2)+\log (4)\right )}{x}\right ) \, dx\\ &=-\left ((2 (5+\log (2))) \int e^{\frac {5-x \log (3)}{\log (3)}} \, dx\right )+\left (-15+\log ^2(2)+\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x^2} \, dx+\left (-15+\log ^2(2)+\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x} \, dx\\ &=2\ 3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} (5+\log (2))+\frac {3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right )}{x}-e^{\frac {5}{\log (3)}} \text {Ei}(-x) \left (15-\log ^2(2)-\log (4)\right )+\left (15-\log ^2(2)-\log (4)\right ) \int \frac {e^{\frac {5-x \log (3)}{\log (3)}}}{x} \, dx\\ &=2\ 3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} (5+\log (2))+\frac {3^{-\frac {x}{\log (3)}} e^{\frac {5}{\log (3)}} \left (15-\log ^2(2)-\log (4)\right )}{x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{-x+\frac {5}{\log (3)}} (3+2 x-\log (2)) (5+\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5 - x*Log[3])/Log[3])*(-15 - 15*x - 10*x^2 + (2 + 2*x - 2*x^2)*Log[2] + (1 + x)*Log[2]^2))/x^2,
x]

[Out]

(E^(-x + 5/Log[3])*(3 + 2*x - Log[2])*(5 + Log[2]))/x

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 35, normalized size = 1.21 \begin {gather*} \frac {{\left (2 \, {\left (x - 1\right )} \log \relax (2) - \log \relax (2)^{2} + 10 \, x + 15\right )} e^{\left (-\frac {x \log \relax (3) - 5}{\log \relax (3)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*log(3)+5)/log(3))/x^2,x, algorithm="fr
icas")

[Out]

(2*(x - 1)*log(2) - log(2)^2 + 10*x + 15)*e^(-(x*log(3) - 5)/log(3))/x

________________________________________________________________________________________

giac [B]  time = 0.16, size = 80, normalized size = 2.76 \begin {gather*} \frac {2 \, x e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) - e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2)^{2} + 10 \, x e^{\left (-x + \frac {5}{\log \relax (3)}\right )} - 2 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) + 15 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*log(3)+5)/log(3))/x^2,x, algorithm="gi
ac")

[Out]

(2*x*e^(-x + 5/log(3))*log(2) - e^(-x + 5/log(3))*log(2)^2 + 10*x*e^(-x + 5/log(3)) - 2*e^(-x + 5/log(3))*log(
2) + 15*e^(-x + 5/log(3)))/x

________________________________________________________________________________________

maple [A]  time = 0.09, size = 30, normalized size = 1.03

method	result	size
gosper	\(-\frac {{\mathrm e}^{-\frac {x \ln \relax (3)-5}{\ln \relax (3)}} \left (\ln \relax (2)+5\right ) \left (-3-2 x +\ln \relax (2)\right )}{x}\)	\(30\)
risch	\(-\frac {\left (\ln \relax (2)^{2}-2 x \ln \relax (2)+2 \ln \relax (2)-10 x -15\right ) {\mathrm e}^{-\frac {x \ln \relax (3)-5}{\ln \relax (3)}}}{x}\)	\(37\)
norman	\(\frac {\left (-\ln \relax (2)^{2}-2 \ln \relax (2)+15\right ) {\mathrm e}^{\frac {-x \ln \relax (3)+5}{\ln \relax (3)}}+\left (2 \ln \relax (2)+10\right ) x \,{\mathrm e}^{\frac {-x \ln \relax (3)+5}{\ln \relax (3)}}}{x}\)	\(53\)
derivativedivides	\(-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}-\frac {2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}\)	\(84\)
default	\(-\frac {\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+\frac {15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+10 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}-x}-\frac {2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}}{x}+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}-x}\)	\(84\)
meijerg	\(\left (-2 \ln \relax (2)-10\right ) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (1-{\mathrm e}^{-x}\right )-\left (\ln \relax (2)^{2}+2 \ln \relax (2)-15\right ) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \expIntegralEi \left (1, x\right )+\ln \relax (2)^{2} {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )+2 \ln \relax (2) {\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )-15 \,{\mathrm e}^{\frac {5}{\ln \relax (3)}} \left (\frac {-2 x +2}{2 x}-\frac {{\mathrm e}^{-x}}{x}+\expIntegralEi \left (1, x\right )+1-\frac {1}{x}\right )\)	\(165\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+1)*ln(2)^2+(-2*x^2+2*x+2)*ln(2)-10*x^2-15*x-15)*exp((-x*ln(3)+5)/ln(3))/x^2,x,method=_RETURNVERBOSE)

[Out]

-exp(-(x*ln(3)-5)/ln(3))*(ln(2)+5)*(-3-2*x+ln(2))/x

________________________________________________________________________________________

maxima [C]  time = 0.50, size = 115, normalized size = 3.97 \begin {gather*} {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} \log \relax (2)^{2} - e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) \log \relax (2)^{2} + 2 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} \log \relax (2) - 2 \, e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) \log \relax (2) - 15 \, {\rm Ei}\left (-x\right ) e^{\frac {5}{\log \relax (3)}} + 15 \, e^{\frac {5}{\log \relax (3)}} \Gamma \left (-1, x\right ) + 2 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \log \relax (2) + 10 \, e^{\left (-x + \frac {5}{\log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*log(2)^2+(-2*x^2+2*x+2)*log(2)-10*x^2-15*x-15)*exp((-x*log(3)+5)/log(3))/x^2,x, algorithm="ma
xima")

[Out]

Ei(-x)*e^(5/log(3))*log(2)^2 - e^(5/log(3))*gamma(-1, x)*log(2)^2 + 2*Ei(-x)*e^(5/log(3))*log(2) - 2*e^(5/log(
3))*gamma(-1, x)*log(2) - 15*Ei(-x)*e^(5/log(3)) + 15*e^(5/log(3))*gamma(-1, x) + 2*e^(-x + 5/log(3))*log(2) +
10*e^(-x + 5/log(3))

________________________________________________________________________________________

mupad [B]  time = 2.23, size = 34, normalized size = 1.17 \begin {gather*} \frac {{\mathrm {e}}^{\frac {5}{\ln \relax (3)}-x}\,\left (10\,x-\ln \relax (4)+x\,\ln \relax (4)-{\ln \relax (2)}^2+15\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(x*log(3) - 5)/log(3))*(15*x - log(2)^2*(x + 1) - log(2)*(2*x - 2*x^2 + 2) + 10*x^2 + 15))/x^2,x)

[Out]

(exp(5/log(3) - x)*(10*x - log(4) + x*log(4) - log(2)^2 + 15))/x

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 34, normalized size = 1.17 \begin {gather*} \frac {\left (2 x \log {\relax (2 )} + 10 x - 2 \log {\relax (2 )} - \log {\relax (2 )}^{2} + 15\right ) e^{\frac {- x \log {\relax (3 )} + 5}{\log {\relax (3 )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+1)*ln(2)**2+(-2*x**2+2*x+2)*ln(2)-10*x**2-15*x-15)*exp((-x*ln(3)+5)/ln(3))/x**2,x)

[Out]

(2*x*log(2) + 10*x - 2*log(2) - log(2)**2 + 15)*exp((-x*log(3) + 5)/log(3))/x

________________________________________________________________________________________