Integrand size = 15, antiderivative size = 92 \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=-\frac {\sqrt {e^{a+b x}}}{3 x^3}-\frac {b \sqrt {e^{a+b x}}}{12 x^2}-\frac {b^2 \sqrt {e^{a+b x}}}{24 x}+\frac {1}{48} b^3 e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right ) \] Output:
-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*exp(b*x+a)^(1/2)*Ei(1/2*b*x)/exp(1/2*b*x)
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\frac {e^{-\frac {b x}{2}} \sqrt {e^{a+b x}} \left (-2 e^{\frac {b x}{2}} \left (8+2 b x+b^2 x^2\right )+b^3 x^3 \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right )\right )}{48 x^3} \] Input:
Integrate[Sqrt[E^(a + b*x)]/x^4,x]
Output:
(Sqrt[E^(a + b*x)]*(-2*E^((b*x)/2)*(8 + 2*b*x + b^2*x^2) + b^3*x^3*ExpInte gralEi[(b*x)/2]))/(48*E^((b*x)/2)*x^3)
Time = 0.64 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2608, 2608, 2608, 2613, 2609}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {1}{6} b \int \frac {\sqrt {e^{a+b x}}}{x^3}dx-\frac {\sqrt {e^{a+b x}}}{3 x^3}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{4} b \int \frac {\sqrt {e^{a+b x}}}{x^2}dx-\frac {\sqrt {e^{a+b x}}}{2 x^2}\right )-\frac {\sqrt {e^{a+b x}}}{3 x^3}\) |
\(\Big \downarrow \) 2608 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{4} b \left (\frac {1}{2} b \int \frac {\sqrt {e^{a+b x}}}{x}dx-\frac {\sqrt {e^{a+b x}}}{x}\right )-\frac {\sqrt {e^{a+b x}}}{2 x^2}\right )-\frac {\sqrt {e^{a+b x}}}{3 x^3}\) |
\(\Big \downarrow \) 2613 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{4} b \left (\frac {1}{2} b e^{\frac {1}{2} (-a-b x)} \sqrt {e^{a+b x}} \int \frac {e^{\frac {1}{2} (a+b x)}}{x}dx-\frac {\sqrt {e^{a+b x}}}{x}\right )-\frac {\sqrt {e^{a+b x}}}{2 x^2}\right )-\frac {\sqrt {e^{a+b x}}}{3 x^3}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{4} b \left (\frac {1}{2} b e^{\frac {1}{2} (-a-b x)+\frac {a}{2}} \sqrt {e^{a+b x}} \operatorname {ExpIntegralEi}\left (\frac {b x}{2}\right )-\frac {\sqrt {e^{a+b x}}}{x}\right )-\frac {\sqrt {e^{a+b x}}}{2 x^2}\right )-\frac {\sqrt {e^{a+b x}}}{3 x^3}\) |
Input:
Int[Sqrt[E^(a + b*x)]/x^4,x]
Output:
-1/3*Sqrt[E^(a + b*x)]/x^3 + (b*(-1/2*Sqrt[E^(a + b*x)]/x^2 + (b*(-(Sqrt[E ^(a + b*x)]/x) + (b*E^(a/2 + (-a - b*x)/2)*Sqrt[E^(a + b*x)]*ExpIntegralEi [(b*x)/2])/2))/4))/6
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] - Simp[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] && In tegerQ[2*m] && !TrueQ[$UseGamma]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(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]
Leaf count of result is larger than twice the leaf count of optimal. \(188\) vs. \(2(69)=138\).
Time = 0.06 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.05
method | result | size |
meijerg | \(-\frac {\sqrt {{\mathrm e}^{b x +a}}\, {\mathrm e}^{\frac {3 a}{2}-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} b^{3} \left (\frac {8 \,{\mathrm e}^{-\frac {3 a}{2}}}{3 x^{3} b^{3}}+\frac {2 \,{\mathrm e}^{-a}}{x^{2} b^{2}}+\frac {{\mathrm e}^{-\frac {a}{2}}}{x b}+\frac {11}{36}-\frac {\ln \left (x \right )}{6}+\frac {\ln \left (2\right )}{6}-\frac {\ln \left (-b \,{\mathrm e}^{\frac {a}{2}}\right )}{6}-\frac {{\mathrm e}^{-\frac {3 a}{2}} \left (\frac {11 b^{3} x^{3} {\mathrm e}^{\frac {3 a}{2}}}{4}+9 b^{2} x^{2} {\mathrm e}^{a}+18 b x \,{\mathrm e}^{\frac {a}{2}}+24\right )}{9 b^{3} x^{3}}+\frac {{\mathrm e}^{-\frac {3 a}{2}+\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} \left (b^{2} x^{2} {\mathrm e}^{a}+2 b x \,{\mathrm e}^{\frac {a}{2}}+8\right )}{3 b^{3} x^{3}}+\frac {\ln \left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )}{6}+\frac {\operatorname {expIntegral}_{1}\left (-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}\right )}{6}\right )}{8}\) | \(189\) |
Input:
int(exp(b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)
Output:
-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*l n(-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)))
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\frac {b^{3} x^{3} {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, {\left (b^{2} x^{2} + 2 \, b x + 8\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{48 \, x^{3}} \] Input:
integrate(exp(b*x+a)^(1/2)/x^4,x, algorithm="fricas")
Output:
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
\[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\int \frac {\sqrt {e^{a} e^{b x}}}{x^{4}}\, dx \] Input:
integrate(exp(b*x+a)**(1/2)/x**4,x)
Output:
Integral(sqrt(exp(a)*exp(b*x))/x**4, x)
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\frac {1}{8} \, b^{3} e^{\left (\frac {1}{2} \, a\right )} \Gamma \left (-3, -\frac {1}{2} \, b x\right ) \] Input:
integrate(exp(b*x+a)^(1/2)/x^4,x, algorithm="maxima")
Output:
1/8*b^3*e^(1/2*a)*gamma(-3, -1/2*b*x)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\frac {b^{3} x^{3} {\rm Ei}\left (\frac {1}{2} \, b x\right ) e^{\left (\frac {1}{2} \, a\right )} - 2 \, b^{2} x^{2} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - 4 \, b x e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )} - 16 \, e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{48 \, x^{3}} \] Input:
integrate(exp(b*x+a)^(1/2)/x^4,x, algorithm="giac")
Output:
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
Timed out. \[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\int \frac {\sqrt {{\mathrm {e}}^{a+b\,x}}}{x^4} \,d x \] Input:
int(exp(a + b*x)^(1/2)/x^4,x)
Output:
int(exp(a + b*x)^(1/2)/x^4, x)
\[ \int \frac {\sqrt {e^{a+b x}}}{x^4} \, dx=\frac {-2 e^{\frac {b x}{2}+\frac {a}{2}} b^{2} x^{2}-4 e^{\frac {b x}{2}+\frac {a}{2}} b x -16 e^{\frac {b x}{2}+\frac {a}{2}}+\left (\int \frac {e^{\frac {b x}{2}+\frac {a}{2}}}{x}d x \right ) b^{3} x^{3}}{48 x^{3}} \] Input:
int(exp(b*x+a)^(1/2)/x^4,x)
Output:
( - 2*e**((a + b*x)/2)*b**2*x**2 - 4*e**((a + b*x)/2)*b*x - 16*e**((a + b* x)/2) + int(e**((a + b*x)/2)/x,x)*b**3*x**3)/(48*x**3)