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

Optimal. Leaf size=48 \[ \frac {1}{2} b e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \text {Ei}\left (\frac {b x}{2}\right )-\frac {\sqrt {e^{a+b x}}}{x} \]

[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)

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Rubi [A]  time = 0.09, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2177, 2182, 2178} \[ \frac {1}{2} b e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \text {Ei}\left (\frac {b x}{2}\right )-\frac {\sqrt {e^{a+b x}}}{x} \]

Antiderivative was successfully verified.

[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 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 2182

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*} \int \frac {\sqrt {e^{a+b x}}}{x^2} \, dx &=-\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*}

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Mathematica [A]  time = 0.03, size = 47, normalized size = 0.98 \[ \frac {e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \left (b x \text {Ei}\left (\frac {b x}{2}\right )-2 e^{\frac {b x}{2}}\right )}{2 x} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.43, size = 29, normalized size = 0.60 \[ \frac {b x {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.41, size = 29, normalized size = 0.60 \[ \frac {b x {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 0.04, size = 116, normalized size = 2.42 \[ -\frac {\left (\Ei \left (1, -\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )-\ln \relax (x )-\ln \left (-b \,{\mathrm e}^{\frac {a}{2}}\right )+\ln \left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )-\frac {\left (b x \,{\mathrm e}^{\frac {a}{2}}+2\right ) {\mathrm e}^{-\frac {a}{2}}}{b x}+\frac {2 \,{\mathrm e}^{-\frac {a}{2}}}{b x}+\frac {2 \,{\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}-\frac {a}{2}}}{b x}+1+\ln \relax (2)\right ) b \,{\mathrm e}^{-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}+\frac {a}{2}} \sqrt {{\mathrm e}^{b x +a}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*exp(b*x+a)^(1/2)*exp(1/2*a-1/2*b*x*exp(1/2*a))*b*(-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))+1-ln(x)+ln(2)-ln(-b*exp(1/2*a))+2/x/b*ex
p(-1/2*a))

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maxima [A]  time = 1.24, size = 13, normalized size = 0.27 \[ \frac {1}{2} \, b e^{\left (\frac {1}{2} \, a\right )} \Gamma \left (-1, -\frac {1}{2} \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {{\mathrm {e}}^{a+b\,x}}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)^(1/2)/x^2,x)

[Out]

int(exp(a + b*x)^(1/2)/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e^{a} e^{b x}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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