3.97 \(\int \frac{\sqrt{e^{a+b x}}}{x^3} \, dx\)

Optimal. Leaf size=71 \[ \frac{1}{8} b^2 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )-\frac{\sqrt{e^{a+b x}}}{2 x^2}-\frac{b \sqrt{e^{a+b x}}}{4 x} \]

[Out]

-Sqrt[E^(a + b*x)]/(2*x^2) - (b*Sqrt[E^(a + b*x)])/(4*x) + (b^2*Sqrt[E^(a + b*x)]*ExpIntegralEi[(b*x)/2])/(8*E
^((b*x)/2))

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Rubi [A]  time = 0.119316, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2177, 2182, 2178} \[ \frac{1}{8} b^2 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )-\frac{\sqrt{e^{a+b x}}}{2 x^2}-\frac{b \sqrt{e^{a+b x}}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[E^(a + b*x)]/x^3,x]

[Out]

-Sqrt[E^(a + b*x)]/(2*x^2) - (b*Sqrt[E^(a + b*x)])/(4*x) + (b^2*Sqrt[E^(a + b*x)]*ExpIntegralEi[(b*x)/2])/(8*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 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]

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

Rubi steps

\begin{align*} \int \frac{\sqrt{e^{a+b x}}}{x^3} \, dx &=-\frac{\sqrt{e^{a+b x}}}{2 x^2}+\frac{1}{4} b \int \frac{\sqrt{e^{a+b x}}}{x^2} \, dx\\ &=-\frac{\sqrt{e^{a+b x}}}{2 x^2}-\frac{b \sqrt{e^{a+b x}}}{4 x}+\frac{1}{8} b^2 \int \frac{\sqrt{e^{a+b x}}}{x} \, dx\\ &=-\frac{\sqrt{e^{a+b x}}}{2 x^2}-\frac{b \sqrt{e^{a+b x}}}{4 x}+\frac{1}{8} \left (b^2 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}}}{2 x^2}-\frac{b \sqrt{e^{a+b x}}}{4 x}+\frac{1}{8} b^2 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0423023, size = 56, normalized size = 0.79 \[ \frac{e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \left (b^2 x^2 \text{Ei}\left (\frac{b x}{2}\right )-2 e^{\frac{b x}{2}} (b x+2)\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[E^(a + b*x)]/x^3,x]

[Out]

(Sqrt[E^(a + b*x)]*(-2*E^((b*x)/2)*(2 + b*x) + b^2*x^2*ExpIntegralEi[(b*x)/2]))/(8*E^((b*x)/2)*x^2)

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Maple [B]  time = 0.028, size = 155, normalized size = 2.2 \begin{align*}{\frac{{b}^{2}}{4}\sqrt{{{\rm e}^{bx+a}}}{{\rm e}^{a-{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}}}} \left ( -2\,{\frac{{{\rm e}^{-a}}}{{x}^{2}{b}^{2}}}-2\,{\frac{{{\rm e}^{-a/2}}}{bx}}-{\frac{3}{4}}+{\frac{\ln \left ( x \right ) }{2}}-{\frac{\ln \left ( 2 \right ) }{2}}+{\frac{1}{2}\ln \left ( -b{{\rm e}^{{\frac{a}{2}}}} \right ) }+{\frac{{{\rm e}^{-a}}}{3\,{x}^{2}{b}^{2}} \left ({\frac{9\,{b}^{2}{x}^{2}{{\rm e}^{a}}}{4}}+6\,bx{{\rm e}^{a/2}}+6 \right ) }-{\frac{2}{3\,{x}^{2}{b}^{2}}{{\rm e}^{-a+{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}}}} \left ({\frac{3\,bx}{2}{{\rm e}^{{\frac{a}{2}}}}}+3 \right ) }-{\frac{1}{2}\ln \left ( -{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}} \right ) }-{\frac{1}{2}{\it Ei} \left ( 1,-{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)^(1/2)/x^3,x)

[Out]

1/4*exp(b*x+a)^(1/2)*exp(a-1/2*b*x*exp(1/2*a))*b^2*(-2/x^2/b^2*exp(-a)-2/x/b*exp(-1/2*a)-3/4+1/2*ln(x)-1/2*ln(
2)+1/2*ln(-b*exp(1/2*a))+1/3/b^2/x^2*exp(-a)*(9/4*b^2*x^2*exp(a)+6*b*x*exp(1/2*a)+6)-2/3/b^2/x^2*exp(-a+1/2*b*
x*exp(1/2*a))*(3/2*b*x*exp(1/2*a)+3)-1/2*ln(-1/2*b*x*exp(1/2*a))-1/2*Ei(1,-1/2*b*x*exp(1/2*a)))

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Maxima [A]  time = 1.14584, size = 20, normalized size = 0.28 \begin{align*} -\frac{1}{4} \, b^{2} e^{\left (\frac{1}{2} \, a\right )} \Gamma \left (-2, -\frac{1}{2} \, b x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)^(1/2)/x^3,x, algorithm="maxima")

[Out]

-1/4*b^2*e^(1/2*a)*gamma(-2, -1/2*b*x)

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Fricas [A]  time = 1.50765, size = 101, normalized size = 1.42 \begin{align*} \frac{b^{2} x^{2}{\rm Ei}\left (\frac{1}{2} \, b x\right ) e^{\left (\frac{1}{2} \, a\right )} - 2 \,{\left (b x + 2\right )} e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

1/8*(b^2*x^2*Ei(1/2*b*x)*e^(1/2*a) - 2*(b*x + 2)*e^(1/2*b*x + 1/2*a))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e^{a} e^{b x}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)**(1/2)/x**3,x)

[Out]

Integral(sqrt(exp(a)*exp(b*x))/x**3, x)

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Giac [A]  time = 1.28556, size = 62, normalized size = 0.87 \begin{align*} \frac{b^{2} x^{2}{\rm Ei}\left (\frac{1}{2} \, b x\right ) e^{\left (\frac{1}{2} \, a\right )} - 2 \, b x e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} - 4 \, e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/8*(b^2*x^2*Ei(1/2*b*x)*e^(1/2*a) - 2*b*x*e^(1/2*b*x + 1/2*a) - 4*e^(1/2*b*x + 1/2*a))/x^2