\(\int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx\) [1784]

Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=-\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}}+\frac {3 a \sqrt {a+\frac {b}{x}}}{4 b^2 \sqrt {x}}-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{5/2}} \]

[Out]

-3/4*a^2*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))/b^(5/2)-1/2*(a+b/x)^(1/2)/b/x^(3/2)+3/4*a*(a+b/x)^(1/2)/b^2/x^
(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {344, 327, 223, 212} \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{4 b^{5/2}}+\frac {3 a \sqrt {a+\frac {b}{x}}}{4 b^2 \sqrt {x}}-\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}} \]

[In]

Int[1/(Sqrt[a + b/x]*x^(7/2)),x]

[Out]

-1/2*Sqrt[a + b/x]/(b*x^(3/2)) + (3*a*Sqrt[a + b/x])/(4*b^2*Sqrt[x]) - (3*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*S
qrt[x])])/(4*b^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 344

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[-k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {x^4}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = -\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}}+\frac {(3 a) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{2 b} \\ & = -\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}}+\frac {3 a \sqrt {a+\frac {b}{x}}}{4 b^2 \sqrt {x}}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{4 b^2} \\ & = -\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}}+\frac {3 a \sqrt {a+\frac {b}{x}}}{4 b^2 \sqrt {x}}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^2} \\ & = -\frac {\sqrt {a+\frac {b}{x}}}{2 b x^{3/2}}+\frac {3 a \sqrt {a+\frac {b}{x}}}{4 b^2 \sqrt {x}}-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\frac {\sqrt {b} \left (-2 b^2+a b x+3 a^2 x^2\right )-3 a^{5/2} \sqrt {1+\frac {b}{a x}} x^{5/2} \text {arcsinh}\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{4 b^{5/2} \sqrt {a+\frac {b}{x}} x^{5/2}} \]

[In]

Integrate[1/(Sqrt[a + b/x]*x^(7/2)),x]

[Out]

(Sqrt[b]*(-2*b^2 + a*b*x + 3*a^2*x^2) - 3*a^(5/2)*Sqrt[1 + b/(a*x)]*x^(5/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])]
)/(4*b^(5/2)*Sqrt[a + b/x]*x^(5/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{2} x^{2}-3 a x \sqrt {a x +b}\, \sqrt {b}+2 b^{\frac {3}{2}} \sqrt {a x +b}\right )}{4 x^{\frac {3}{2}} b^{\frac {5}{2}} \sqrt {a x +b}}\)	\(74\)
risch	\(\frac {\left (a x +b \right ) \left (3 a x -2 b \right )}{4 b^{2} x^{\frac {5}{2}} \sqrt {\frac {a x +b}{x}}}-\frac {3 a^{2} \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {a x +b}}{4 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}\)	\(75\)

[In]

int(1/(a+b/x)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*((a*x+b)/x)^(1/2)/x^(3/2)/b^(5/2)*(3*arctanh((a*x+b)^(1/2)/b^(1/2))*a^2*x^2-3*a*x*(a*x+b)^(1/2)*b^(1/2)+2
*b^(3/2)*(a*x+b)^(1/2))/(a*x+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} x^{2} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + 2 \, {\left (3 \, a b x - 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{8 \, b^{3} x^{2}}, \frac {3 \, a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (3 \, a b x - 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{4 \, b^{3} x^{2}}\right ] \]

[In]

integrate(1/(a+b/x)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*sqrt(b)*x^2*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) + 2*(3*a*b*x - 2*b^2)*sqrt(x)
*sqrt((a*x + b)/x))/(b^3*x^2), 1/4*(3*a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (3*a*b*x
- 2*b^2)*sqrt(x)*sqrt((a*x + b)/x))/(b^3*x^2)]

Sympy [A] (verification not implemented)

Time = 10.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\frac {3 a^{\frac {3}{2}}}{4 b^{2} \sqrt {x} \sqrt {1 + \frac {b}{a x}}} + \frac {\sqrt {a}}{4 b x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} - \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{4 b^{\frac {5}{2}}} - \frac {1}{2 \sqrt {a} x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} \]

[In]

integrate(1/(a+b/x)**(1/2)/x**(7/2),x)

[Out]

3*a**(3/2)/(4*b**2*sqrt(x)*sqrt(1 + b/(a*x))) + sqrt(a)/(4*b*x**(3/2)*sqrt(1 + b/(a*x))) - 3*a**2*asinh(sqrt(b
)/(sqrt(a)*sqrt(x)))/(4*b**(5/2)) - 1/(2*sqrt(a)*x**(5/2)*sqrt(1 + b/(a*x)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\frac {3 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} x^{\frac {3}{2}} - 5 \, \sqrt {a + \frac {b}{x}} a^{2} b \sqrt {x}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} b^{2} x^{2} - 2 \, {\left (a + \frac {b}{x}\right )} b^{3} x + b^{4}\right )}} \]

[In]

integrate(1/(a+b/x)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

3/8*a^2*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/b^(5/2) + 1/4*(3*(a + b/x)^(3
/2)*a^2*x^(3/2) - 5*sqrt(a + b/x)*a^2*b*sqrt(x))/((a + b/x)^2*b^2*x^2 - 2*(a + b/x)*b^3*x + b^4)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\frac {\frac {3 \, a^{3} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{3} - 5 \, \sqrt {a x + b} a^{3} b}{a^{2} b^{2} x^{2}}}{4 \, a \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/(a+b/x)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

1/4*(3*a^3*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(a*x + b)^(3/2)*a^3 - 5*sqrt(a*x + b)*a^3*b)/(a^
2*b^2*x^2))/(a*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{7/2}} \, dx=\int \frac {1}{x^{7/2}\,\sqrt {a+\frac {b}{x}}} \,d x \]

[In]

int(1/(x^(7/2)*(a + b/x)^(1/2)),x)

[Out]

int(1/(x^(7/2)*(a + b/x)^(1/2)), x)