\(\int \frac {(a+\frac {b}{x^2})^{5/2}}{x} \, dx\) [1911]

Optimal result

Integrand size = 15, antiderivative size = 72 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=-a^2 \sqrt {a+\frac {b}{x^2}}-\frac {1}{3} a \left (a+\frac {b}{x^2}\right )^{3/2}-\frac {1}{5} \left (a+\frac {b}{x^2}\right )^{5/2}+a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 52, 65, 214} \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )-a^2 \sqrt {a+\frac {b}{x^2}}-\frac {1}{3} a \left (a+\frac {b}{x^2}\right )^{3/2}-\frac {1}{5} \left (a+\frac {b}{x^2}\right )^{5/2} \]

[In]

Int[(a + b/x^2)^(5/2)/x,x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (-3 b^2-11 a b x^2-23 a^2 x^4+\frac {30 a^{5/2} x^5 \text {arctanh}\left (\frac {\sqrt {a} x}{-\sqrt {b}+\sqrt {b+a x^2}}\right )}{\sqrt {b+a x^2}}\right )}{15 x^4} \]

[In]

Integrate[(a + b/x^2)^(5/2)/x,x]

[Out]

(Sqrt[a + b/x^2]*(-3*b^2 - 11*a*b*x^2 - 23*a^2*x^4 + (30*a^(5/2)*x^5*ArcTanh[(Sqrt[a]*x)/(-Sqrt[b] + Sqrt[b +
a*x^2])])/Sqrt[b + a*x^2]))/(15*x^4)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17

method	result	size
risch	\(-\frac {\left (23 a^{2} x^{4}+11 a b \,x^{2}+3 b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{15 x^{4}}+\frac {a^{\frac {5}{2}} \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{\sqrt {a \,x^{2}+b}}\)	\(84\)
default	\(\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} \left (8 a^{\frac {7}{2}} \left (a \,x^{2}+b \right )^{\frac {5}{2}} x^{6}+10 a^{\frac {7}{2}} \left (a \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{6}+15 a^{\frac {7}{2}} \sqrt {a \,x^{2}+b}\, b^{2} x^{6}-8 a^{\frac {5}{2}} \left (a \,x^{2}+b \right )^{\frac {7}{2}} x^{4}-2 a^{\frac {3}{2}} \left (a \,x^{2}+b \right )^{\frac {7}{2}} b \,x^{2}+15 \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right ) a^{3} b^{3} x^{5}-3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b^{2} \sqrt {a}\right )}{15 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{3} \sqrt {a}}\)	\(166\)

[In]

int((a+b/x^2)^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

-1/15*(23*a^2*x^4+11*a*b*x^2+3*b^2)/x^4*((a*x^2+b)/x^2)^(1/2)+a^(5/2)*ln(a^(1/2)*x+(a*x^2+b)^(1/2))*((a*x^2+b)
/x^2)^(1/2)*x/(a*x^2+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.35 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=\left [\frac {15 \, a^{\frac {5}{2}} x^{4} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) - 2 \, {\left (23 \, a^{2} x^{4} + 11 \, a b x^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{30 \, x^{4}}, -\frac {15 \, \sqrt {-a} a^{2} x^{4} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (23 \, a^{2} x^{4} + 11 \, a b x^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{15 \, x^{4}}\right ] \]

[In]

integrate((a+b/x^2)^(5/2)/x,x, algorithm="fricas")

[Out]

[1/30*(15*a^(5/2)*x^4*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) - 2*(23*a^2*x^4 + 11*a*b*x^2 + 3
*b^2)*sqrt((a*x^2 + b)/x^2))/x^4, -1/15*(15*sqrt(-a)*a^2*x^4*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2
+ b)) + (23*a^2*x^4 + 11*a*b*x^2 + 3*b^2)*sqrt((a*x^2 + b)/x^2))/x^4]

Sympy [A] (verification not implemented)

Time = 2.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=- \frac {23 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{2}}}}{15} - \frac {a^{\frac {5}{2}} \log {\left (\frac {b}{a x^{2}} \right )}}{2} + a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b}{a x^{2}}} + 1 \right )} - \frac {11 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{2}}}}{15 x^{2}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{2}}}}{5 x^{4}} \]

[In]

integrate((a+b/x**2)**(5/2)/x,x)

[Out]

-23*a**(5/2)*sqrt(1 + b/(a*x**2))/15 - a**(5/2)*log(b/(a*x**2))/2 + a**(5/2)*log(sqrt(1 + b/(a*x**2)) + 1) - 1
1*a**(3/2)*b*sqrt(1 + b/(a*x**2))/(15*x**2) - sqrt(a)*b**2*sqrt(1 + b/(a*x**2))/(5*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=-\frac {1}{2} \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right ) - \frac {1}{5} \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} - \frac {1}{3} \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a - \sqrt {a + \frac {b}{x^{2}}} a^{2} \]

[In]

integrate((a+b/x^2)^(5/2)/x,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (56) = 112\).

Time = 0.79 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.50 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=-\frac {1}{2} \, a^{\frac {5}{2}} \log \left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{8} a^{\frac {5}{2}} b \mathrm {sgn}\left (x\right ) - 90 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{6} a^{\frac {5}{2}} b^{2} \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{4} a^{\frac {5}{2}} b^{3} \mathrm {sgn}\left (x\right ) - 70 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} a^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (x\right ) + 23 \, a^{\frac {5}{2}} b^{5} \mathrm {sgn}\left (x\right )\right )}}{15 \, {\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b\right )}^{5}} \]

[In]

integrate((a+b/x^2)^(5/2)/x,x, algorithm="giac")

[Out]

-1/2*a^(5/2)*log((sqrt(a)*x - sqrt(a*x^2 + b))^2)*sgn(x) + 2/15*(45*(sqrt(a)*x - sqrt(a*x^2 + b))^8*a^(5/2)*b*
sgn(x) - 90*(sqrt(a)*x - sqrt(a*x^2 + b))^6*a^(5/2)*b^2*sgn(x) + 140*(sqrt(a)*x - sqrt(a*x^2 + b))^4*a^(5/2)*b
^3*sgn(x) - 70*(sqrt(a)*x - sqrt(a*x^2 + b))^2*a^(5/2)*b^4*sgn(x) + 23*a^(5/2)*b^5*sgn(x))/((sqrt(a)*x - sqrt(
a*x^2 + b))^2 - b)^5

Mupad [B] (verification not implemented)

Time = 6.17 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x} \, dx=-\frac {a\,{\left (a+\frac {b}{x^2}\right )}^{3/2}}{3}-\frac {{\left (a+\frac {b}{x^2}\right )}^{5/2}}{5}-a^2\,\sqrt {a+\frac {b}{x^2}}-a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x^2}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \]

[In]

int((a + b/x^2)^(5/2)/x,x)

[Out]

- a^(5/2)*atan(((a + b/x^2)^(1/2)*1i)/a^(1/2))*1i - (a*(a + b/x^2)^(3/2))/3 - (a + b/x^2)^(5/2)/5 - a^2*(a + b
/x^2)^(1/2)