\(\int (a+\frac {b}{x^2})^{5/2} \, dx\) [1910]

Optimal result

Integrand size = 11, antiderivative size = 86 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=-\frac {15 a b \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}+\left (a+\frac {b}{x^2}\right )^{5/2} x-\frac {15}{8} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right ) \]

[Out]

-5/4*b*(a+b/x^2)^(3/2)/x+(a+b/x^2)^(5/2)*x-15/8*a^2*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))*b^(1/2)-15/8*a*b*(a+b/x
^2)^(1/2)/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {248, 283, 201, 223, 212} \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=-\frac {15}{8} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )+x \left (a+\frac {b}{x^2}\right )^{5/2}-\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}-\frac {15 a b \sqrt {a+\frac {b}{x^2}}}{8 x} \]

[In]

Int[(a + b/x^2)^(5/2),x]

[Out]

(-15*a*b*Sqrt[a + b/x^2])/(8*x) - (5*b*(a + b/x^2)^(3/2))/(4*x) + (a + b/x^2)^(5/2)*x - (15*a^2*Sqrt[b]*ArcTan
h[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/8

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 283

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

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (a+b x^2\right )^{5/2}}{x^2} \, dx,x,\frac {1}{x}\right ) \\ & = \left (a+\frac {b}{x^2}\right )^{5/2} x-(5 b) \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}+\left (a+\frac {b}{x^2}\right )^{5/2} x-\frac {1}{4} (15 a b) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {15 a b \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}+\left (a+\frac {b}{x^2}\right )^{5/2} x-\frac {1}{8} \left (15 a^2 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {15 a b \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}+\left (a+\frac {b}{x^2}\right )^{5/2} x-\frac {1}{8} \left (15 a^2 b\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right ) \\ & = -\frac {15 a b \sqrt {a+\frac {b}{x^2}}}{8 x}-\frac {5 b \left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}+\left (a+\frac {b}{x^2}\right )^{5/2} x-\frac {15}{8} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=-\frac {\sqrt {a+\frac {b}{x^2}} \left (\sqrt {b+a x^2} \left (2 b^2+9 a b x^2-8 a^2 x^4\right )+15 a^2 \sqrt {b} x^4 \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )\right )}{8 x^3 \sqrt {b+a x^2}} \]

[In]

Integrate[(a + b/x^2)^(5/2),x]

[Out]

-1/8*(Sqrt[a + b/x^2]*(Sqrt[b + a*x^2]*(2*b^2 + 9*a*b*x^2 - 8*a^2*x^4) + 15*a^2*Sqrt[b]*x^4*ArcTanh[Sqrt[b + a
*x^2]/Sqrt[b]]))/(x^3*Sqrt[b + a*x^2])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16

method	result	size
risch	\(-\frac {b \left (9 a \,x^{2}+2 b \right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{8 x^{3}}+\frac {\left (-\frac {15 \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a^{2}}{8}+\sqrt {a \,x^{2}+b}\, a^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{\sqrt {a \,x^{2}+b}}\)	\(100\)
default	\(-\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x \left (-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{2} x^{4}+15 b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) a^{2} x^{4}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a \,x^{2}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{2} b \,x^{4}-15 \sqrt {a \,x^{2}+b}\, a^{2} b^{2} x^{4}+2 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b \right )}{8 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{2}}\)	\(144\)

[In]

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

[Out]

-1/8*b*(9*a*x^2+2*b)/x^3*((a*x^2+b)/x^2)^(1/2)+(-15/8*b^(1/2)*ln((2*b+2*b^(1/2)*(a*x^2+b)^(1/2))/x)*a^2+(a*x^2
+b)^(1/2)*a^2)*((a*x^2+b)/x^2)^(1/2)*x/(a*x^2+b)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.00 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x^{3} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, x^{3}}, \frac {15 \, a^{2} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, x^{3}}\right ] \]

[In]

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

[Out]

[1/16*(15*a^2*sqrt(b)*x^3*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) + 2*(8*a^2*x^4 - 9*a*b*x
^2 - 2*b^2)*sqrt((a*x^2 + b)/x^2))/x^3, 1/8*(15*a^2*sqrt(-b)*x^3*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^
2 + b)) + (8*a^2*x^4 - 9*a*b*x^2 - 2*b^2)*sqrt((a*x^2 + b)/x^2))/x^3]

Sympy [A] (verification not implemented)

Time = 2.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {a^{\frac {5}{2}} x}{\sqrt {1 + \frac {b}{a x^{2}}}} - \frac {a^{\frac {3}{2}} b}{8 x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {11 \sqrt {a} b^{2}}{8 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8} - \frac {b^{3}}{4 \sqrt {a} x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} \]

[In]

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

[Out]

a**(5/2)*x/sqrt(1 + b/(a*x**2)) - a**(3/2)*b/(8*x*sqrt(1 + b/(a*x**2))) - 11*sqrt(a)*b**2/(8*x**3*sqrt(1 + b/(
a*x**2))) - 15*a**2*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x))/8 - b**3/(4*sqrt(a)*x**5*sqrt(1 + b/(a*x**2)))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.51 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\sqrt {a + \frac {b}{x^{2}}} a^{2} x + \frac {15}{16} \, a^{2} \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right ) - \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} b x^{3} - 7 \, \sqrt {a + \frac {b}{x^{2}}} a^{2} b^{2} x}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} b x^{2} + b^{2}\right )}} \]

[In]

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

[Out]

sqrt(a + b/x^2)*a^2*x + 15/16*a^2*sqrt(b)*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b))) - 1
/8*(9*(a + b/x^2)^(3/2)*a^2*b*x^3 - 7*sqrt(a + b/x^2)*a^2*b^2*x)/((a + b/x^2)^2*x^4 - 2*(a + b/x^2)*b*x^2 + b^
2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {\frac {15 \, a^{3} b \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + 8 \, \sqrt {a x^{2} + b} a^{3} \mathrm {sgn}\left (x\right ) - \frac {9 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{3} b \mathrm {sgn}\left (x\right ) - 7 \, \sqrt {a x^{2} + b} a^{3} b^{2} \mathrm {sgn}\left (x\right )}{a^{2} x^{4}}}{8 \, a} \]

[In]

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

[Out]

1/8*(15*a^3*b*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) + 8*sqrt(a*x^2 + b)*a^3*sgn(x) - (9*(a*x^2 + b)
^(3/2)*a^3*b*sgn(x) - 7*sqrt(a*x^2 + b)*a^3*b^2*sgn(x))/(a^2*x^4))/a

Mupad [B] (verification not implemented)

Time = 6.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.42 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {x\,{\left (a\,x^2+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b}{a\,x^2}\right )}{{\left (\frac {b}{a}+x^2\right )}^{5/2}} \]

[In]

int((a + b/x^2)^(5/2),x)

[Out]

(x*(b + a*x^2)^(5/2)*hypergeom([-5/2, -1/2], 1/2, -b/(a*x^2)))/(b/a + x^2)^(5/2)