\(\int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx\) [96]

Optimal result

Integrand size = 15, antiderivative size = 48 \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=-\frac {\sqrt {e^{a+b x}}}{x}+\frac {1}{2} b e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right ) \]

[Out]

-exp(b*x+a)^(1/2)/x+1/2*b*Ei(1/2*b*x)*exp(b*x+a)^(1/2)/exp(1/2*b*x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2208, 2213, 2209} \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\frac {1}{2} b e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right )-\frac {\sqrt {e^{a+b x}}}{x} \]

[In]

Int[Sqrt[E^(a + b*x)]/x^2,x]

[Out]

-(Sqrt[E^(a + b*x)]/x) + (b*Sqrt[E^(a + b*x)]*ExpIntegralEi[(b*x)/2])/(2*E^((b*x)/2))

Rule 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2213

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*F^(g*(e +
f*x)))^n/F^(g*n*(e + f*x)), Int[(c + d*x)^m*F^(g*n*(e + f*x)), x], x] /; FreeQ[{F, b, c, d, e, f, g, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {e^{a+b x}}}{x}+\frac {1}{2} b \int \frac {\sqrt {e^{a+b x}}}{x} \, dx \\ & = -\frac {\sqrt {e^{a+b x}}}{x}+\frac {1}{2} \left (b e^{\frac {1}{2} (-a-b x)} \sqrt {e^{a+b x}}\right ) \int \frac {e^{\frac {1}{2} (a+b x)}}{x} \, dx \\ & = -\frac {\sqrt {e^{a+b x}}}{x}+\frac {1}{2} b e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \text {Ei}\left (\frac {b x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\frac {e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \left (-2 e^{\frac {b x}{2}}+b x \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right )\right )}{2 x} \]

[In]

Integrate[Sqrt[E^(a + b*x)]/x^2,x]

[Out]

(Sqrt[E^(a + b*x)]*(-2*E^((b*x)/2) + b*x*ExpIntegralEi[(b*x)/2]))/(2*E^((b*x)/2)*x)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(115\) vs. \(2(37)=74\).

Time = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.42

method	result	size
meijerg	\(-\frac {\sqrt {{\mathrm e}^{b x +a}}\, {\mathrm e}^{\frac {a}{2}-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} b \left (\frac {2 \,{\mathrm e}^{-\frac {a}{2}}}{x b}+1-\ln \left (x \right )+\ln \left (2\right )-\ln \left (-b \,{\mathrm e}^{\frac {a}{2}}\right )-\frac {{\mathrm e}^{-\frac {a}{2}} \left (2+b x \,{\mathrm e}^{\frac {a}{2}}\right )}{b x}+\frac {2 \,{\mathrm e}^{-\frac {a}{2}+\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}}}{b x}+\ln \left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )+\operatorname {Ei}_{1}\left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )\right )}{2}\)	\(116\)

[In]

int(exp(b*x+a)^(1/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*exp(b*x+a)^(1/2)*exp(1/2*a-1/2*b*x*exp(1/2*a))*b*(2/x/b*exp(-1/2*a)+1-ln(x)+ln(2)-ln(-b*exp(1/2*a))-1/b/x
*exp(-1/2*a)*(2+b*x*exp(1/2*a))+2/b/x*exp(-1/2*a+1/2*b*x*exp(1/2*a))+ln(-1/2*b*x*exp(1/2*a))+Ei(1,-1/2*b*x*exp
(1/2*a)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\frac {b x {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{2 \, x} \]

[In]

integrate(exp(b*x+a)^(1/2)/x^2,x, algorithm="fricas")

[Out]

1/2*(b*x*Ei(1/2*b*x)*e^(1/2*a) - 2*e^(1/2*b*x + 1/2*a))/x

Sympy [F]

\[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\int \frac {\sqrt {e^{a} e^{b x}}}{x^{2}}\, dx \]

[In]

integrate(exp(b*x+a)**(1/2)/x**2,x)

[Out]

Integral(sqrt(exp(a)*exp(b*x))/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\frac {1}{2} \, b e^{\left (\frac {1}{2} \, a\right )} \Gamma \left (-1, -\frac {1}{2} \, b x\right ) \]

[In]

integrate(exp(b*x+a)^(1/2)/x^2,x, algorithm="maxima")

[Out]

1/2*b*e^(1/2*a)*gamma(-1, -1/2*b*x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\frac {b x {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{2 \, x} \]

[In]

integrate(exp(b*x+a)^(1/2)/x^2,x, algorithm="giac")

[Out]

1/2*(b*x*Ei(1/2*b*x)*e^(1/2*a) - 2*e^(1/2*b*x + 1/2*a))/x

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx=\int \frac {\sqrt {{\mathrm {e}}^{a+b\,x}}}{x^2} \,d x \]

[In]

int(exp(a + b*x)^(1/2)/x^2,x)

[Out]

int(exp(a + b*x)^(1/2)/x^2, x)