3.98 \(\int \frac{\sqrt{e^{a+b x}}}{x^4} \, dx\)

Optimal. Leaf size=92 \[ \frac{1}{48} b^3 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )-\frac{b^2 \sqrt{e^{a+b x}}}{24 x}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}-\frac{\sqrt{e^{a+b x}}}{3 x^3} \]

[Out]

-Sqrt[E^(a + b*x)]/(3*x^3) - (b*Sqrt[E^(a + b*x)])/(12*x^2) - (b^2*Sqrt[E^(a + b*x)])/(24*x) + (b^3*Sqrt[E^(a
+ b*x)]*ExpIntegralEi[(b*x)/2])/(48*E^((b*x)/2))

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Rubi [A]  time = 0.154713, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2177, 2182, 2178} \[ \frac{1}{48} b^3 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )-\frac{b^2 \sqrt{e^{a+b x}}}{24 x}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}-\frac{\sqrt{e^{a+b x}}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[E^(a + b*x)]/x^4,x]

[Out]

-Sqrt[E^(a + b*x)]/(3*x^3) - (b*Sqrt[E^(a + b*x)])/(12*x^2) - (b^2*Sqrt[E^(a + b*x)])/(24*x) + (b^3*Sqrt[E^(a
+ b*x)]*ExpIntegralEi[(b*x)/2])/(48*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^4} \, dx &=-\frac{\sqrt{e^{a+b x}}}{3 x^3}+\frac{1}{6} b \int \frac{\sqrt{e^{a+b x}}}{x^3} \, dx\\ &=-\frac{\sqrt{e^{a+b x}}}{3 x^3}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}+\frac{1}{24} b^2 \int \frac{\sqrt{e^{a+b x}}}{x^2} \, dx\\ &=-\frac{\sqrt{e^{a+b x}}}{3 x^3}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}-\frac{b^2 \sqrt{e^{a+b x}}}{24 x}+\frac{1}{48} b^3 \int \frac{\sqrt{e^{a+b x}}}{x} \, dx\\ &=-\frac{\sqrt{e^{a+b x}}}{3 x^3}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}-\frac{b^2 \sqrt{e^{a+b x}}}{24 x}+\frac{1}{48} \left (b^3 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}}}{3 x^3}-\frac{b \sqrt{e^{a+b x}}}{12 x^2}-\frac{b^2 \sqrt{e^{a+b x}}}{24 x}+\frac{1}{48} b^3 e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \text{Ei}\left (\frac{b x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0529866, size = 64, normalized size = 0.7 \[ \frac{e^{-\frac{b x}{2}} \sqrt{e^{a+b x}} \left (b^3 x^3 \text{Ei}\left (\frac{b x}{2}\right )-2 e^{\frac{b x}{2}} \left (b^2 x^2+2 b x+8\right )\right )}{48 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[E^(a + b*x)]/x^4,x]

[Out]

(Sqrt[E^(a + b*x)]*(-2*E^((b*x)/2)*(8 + 2*b*x + b^2*x^2) + b^3*x^3*ExpIntegralEi[(b*x)/2]))/(48*E^((b*x)/2)*x^
3)

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Maple [B]  time = 0.03, size = 189, normalized size = 2.1 \begin{align*} -{\frac{{b}^{3}}{8}\sqrt{{{\rm e}^{bx+a}}}{{\rm e}^{{\frac{3\,a}{2}}-{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}}}} \left ({\frac{8}{3\,{x}^{3}{b}^{3}}{{\rm e}^{-{\frac{3\,a}{2}}}}}+2\,{\frac{{{\rm e}^{-a}}}{{x}^{2}{b}^{2}}}+{\frac{1}{bx}{{\rm e}^{-{\frac{a}{2}}}}}+{\frac{11}{36}}-{\frac{\ln \left ( x \right ) }{6}}+{\frac{\ln \left ( 2 \right ) }{6}}-{\frac{1}{6}\ln \left ( -b{{\rm e}^{{\frac{a}{2}}}} \right ) }-{\frac{1}{9\,{x}^{3}{b}^{3}}{{\rm e}^{-{\frac{3\,a}{2}}}} \left ({\frac{11\,{x}^{3}{b}^{3}}{4}{{\rm e}^{{\frac{3\,a}{2}}}}}+9\,{b}^{2}{x}^{2}{{\rm e}^{a}}+18\,bx{{\rm e}^{a/2}}+24 \right ) }+{\frac{1}{3\,{x}^{3}{b}^{3}}{{\rm e}^{-{\frac{3\,a}{2}}+{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}}}} \left ({b}^{2}{x}^{2}{{\rm e}^{a}}+2\,bx{{\rm e}^{a/2}}+8 \right ) }+{\frac{1}{6}\ln \left ( -{\frac{bx}{2}{{\rm e}^{{\frac{a}{2}}}}} \right ) }+{\frac{1}{6}{\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^4,x)

[Out]

-1/8*exp(b*x+a)^(1/2)*exp(3/2*a-1/2*b*x*exp(1/2*a))*b^3*(8/3/x^3/b^3*exp(-3/2*a)+2/x^2/b^2*exp(-a)+1/x/b*exp(-
1/2*a)+11/36-1/6*ln(x)+1/6*ln(2)-1/6*ln(-b*exp(1/2*a))-1/9/b^3/x^3*exp(-3/2*a)*(11/4*b^3*x^3*exp(3/2*a)+9*b^2*
x^2*exp(a)+18*b*x*exp(1/2*a)+24)+1/3/b^3/x^3*exp(-3/2*a+1/2*b*x*exp(1/2*a))*(b^2*x^2*exp(a)+2*b*x*exp(1/2*a)+8
)+1/6*ln(-1/2*b*x*exp(1/2*a))+1/6*Ei(1,-1/2*b*x*exp(1/2*a)))

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Maxima [A]  time = 1.13943, size = 20, normalized size = 0.22 \begin{align*} \frac{1}{8} \, b^{3} e^{\left (\frac{1}{2} \, a\right )} \Gamma \left (-3, -\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^4,x, algorithm="maxima")

[Out]

1/8*b^3*e^(1/2*a)*gamma(-3, -1/2*b*x)

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Fricas [A]  time = 1.46683, size = 119, normalized size = 1.29 \begin{align*} \frac{b^{3} x^{3}{\rm Ei}\left (\frac{1}{2} \, b x\right ) e^{\left (\frac{1}{2} \, a\right )} - 2 \,{\left (b^{2} x^{2} + 2 \, b x + 8\right )} e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{48 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

1/48*(b^3*x^3*Ei(1/2*b*x)*e^(1/2*a) - 2*(b^2*x^2 + 2*b*x + 8)*e^(1/2*b*x + 1/2*a))/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e^{a} e^{b x}}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)**(1/2)/x**4,x)

[Out]

Integral(sqrt(exp(a)*exp(b*x))/x**4, x)

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Giac [A]  time = 1.30918, size = 85, normalized size = 0.92 \begin{align*} \frac{b^{3} x^{3}{\rm Ei}\left (\frac{1}{2} \, b x\right ) e^{\left (\frac{1}{2} \, a\right )} - 2 \, b^{2} x^{2} e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} - 4 \, b x e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} - 16 \, e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{48 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

1/48*(b^3*x^3*Ei(1/2*b*x)*e^(1/2*a) - 2*b^2*x^2*e^(1/2*b*x + 1/2*a) - 4*b*x*e^(1/2*b*x + 1/2*a) - 16*e^(1/2*b*
x + 1/2*a))/x^3