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

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

[Out]

-1/3*exp(b*x+a)^(1/2)/x^3-1/12*b*exp(b*x+a)^(1/2)/x^2-1/24*b^2*exp(b*x+a)^(1/2)/x+1/48*b^3*Ei(1/2*b*x)*exp(b*x
+a)^(1/2)/exp(1/2*b*x)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2177, 2182, 2178} \[ \frac {1}{48} b^3 e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \text {Ei}\left (\frac {b x}{2}\right )-\frac {b^2 \sqrt {e^{a+b x}}}{24 x}-\frac {b \sqrt {e^{a+b x}}}{12 x^2}-\frac {\sqrt {e^{a+b x}}}{3 x^3} \]

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 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^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 = 0.06, size = 64, normalized size = 0.70 \[ \frac {e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \left (b^3 x^3 \text {Ei}\left (\frac {b x}{2}\right )-2 e^{\frac {b x}{2}} \left (b^2 x^2+2 b x+8\right )\right )}{48 x^3} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 46, normalized size = 0.50 \[ \frac {b^{3} x^{3} {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, {\left (b^{2} x^{2} + 2 \, b x + 8\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{48 \, x^{3}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.36, size = 63, normalized size = 0.68 \[ \frac {b^{3} x^{3} {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, b^{2} x^{2} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - 4 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - 16 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{48 \, x^{3}} \]

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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 189, normalized size = 2.05 \[ -\frac {\left (\frac {\Ei \left (1, -\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )}{6}-\frac {\ln \relax (x )}{6}-\frac {\ln \left (-b \,{\mathrm e}^{\frac {a}{2}}\right )}{6}+\frac {\ln \left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )}{6}+\frac {{\mathrm e}^{-\frac {a}{2}}}{b x}+\frac {2 \,{\mathrm e}^{-a}}{b^{2} x^{2}}-\frac {\left (\frac {11 b^{3} x^{3} {\mathrm e}^{\frac {3 a}{2}}}{4}+9 b^{2} x^{2} {\mathrm e}^{a}+18 b x \,{\mathrm e}^{\frac {a}{2}}+24\right ) {\mathrm e}^{-\frac {3 a}{2}}}{9 b^{3} x^{3}}+\frac {8 \,{\mathrm e}^{-\frac {3 a}{2}}}{3 b^{3} x^{3}}+\frac {\left (b^{2} x^{2} {\mathrm e}^{a}+2 b x \,{\mathrm e}^{\frac {a}{2}}+8\right ) {\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}-\frac {3 a}{2}}}{3 b^{3} x^{3}}+\frac {11}{36}+\frac {\ln \relax (2)}{6}\right ) b^{3} {\mathrm e}^{-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}+\frac {3 a}{2}} \sqrt {{\mathrm e}^{b x +a}}}{8} \]

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*(-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))+11/36-1/6*ln(x)+1/6*ln(2)-1/6*ln(-b*exp(1/2*a))
+8/3/x^3/b^3*exp(-3/2*a)+2/b^2/x^2*exp(-a)+1/b/x*exp(-1/2*a))

________________________________________________________________________________________

maxima [A]  time = 1.25, size = 15, normalized size = 0.16 \[ \frac {1}{8} \, b^{3} e^{\left (\frac {1}{2} \, a\right )} \Gamma \left (-3, -\frac {1}{2} \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^3*e^(1/2*a)*gamma(-3, -1/2*b*x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {{\mathrm {e}}^{a+b\,x}}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)^(1/2)/x^4,x)

[Out]

int(exp(a + b*x)^(1/2)/x^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e^{a} e^{b x}}}{x^{4}}\, dx \]

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)

________________________________________________________________________________________