Integrand size = 11, antiderivative size = 91 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=-\frac {b^2 \sqrt {a+\frac {b}{x^2}}}{4 x^3}-\frac {9 a b \sqrt {a+\frac {b}{x^2}}}{8 x}+a^2 \sqrt {a+\frac {b}{x^2}} x-\frac {15}{8} a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right ) \] Output:
-1/4*b^2*(a+b/x^2)^(1/2)/x^3-9/8*a*b*(a+b/x^2)^(1/2)/x+a^2*(a+b/x^2)^(1/2) *x-15/8*a^2*b^(1/2)*arctanh(b^(1/2)/(a+b/x^2)^(1/2)/x)
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=-\frac {\sqrt {a+\frac {b}{x^2}} \left (\sqrt {b+a x^2} \left (2 b^2+9 a b x^2-8 a^2 x^4\right )+15 a^2 \sqrt {b} x^4 \text {arctanh}\left (\frac {\sqrt {b+a x^2}}{\sqrt {b}}\right )\right )}{8 x^3 \sqrt {b+a x^2}} \] Input:
Integrate[(a + b/x^2)^(5/2),x]
Output:
-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) + 1 5*a^2*Sqrt[b]*x^4*ArcTanh[Sqrt[b + a*x^2]/Sqrt[b]]))/(x^3*Sqrt[b + a*x^2])
Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {773, 247, 211, 211, 224, 219}
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.
\(\displaystyle \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 773 |
\(\displaystyle -\int \left (a+\frac {b}{x^2}\right )^{5/2} x^2d\frac {1}{x}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \int \left (a+\frac {b}{x^2}\right )^{3/2}d\frac {1}{x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \int \sqrt {a+\frac {b}{x^2}}d\frac {1}{x}+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a+\frac {b}{x^2}}}d\frac {1}{x}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b}{x^2}}d\frac {1}{\sqrt {a+\frac {b}{x^2}} x}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle x \left (a+\frac {b}{x^2}\right )^{5/2}-5 b \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{2 \sqrt {b}}+\frac {\sqrt {a+\frac {b}{x^2}}}{2 x}\right )+\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{4 x}\right )\) |
Input:
Int[(a + b/x^2)^(5/2),x]
Output:
(a + b/x^2)^(5/2)*x - 5*b*((a + b/x^2)^(3/2)/(4*x) + (3*a*(Sqrt[a + b/x^2] /(2*x) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*Sqrt[b])))/4)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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] && !IntegerQ[p]
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10
method | result | size |
risch | \(-\frac {b \left (9 a \,x^{2}+2 b \right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}{8 x^{3}}+\frac {\left (-\frac {15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) \sqrt {b}\, a^{2}}{8}+\sqrt {a \,x^{2}+b}\, a^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}{\sqrt {a \,x^{2}+b}}\) | \(100\) |
default | \(-\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} x \left (-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{2} x^{4}+15 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,x^{2}+b}}{x}\right ) b^{\frac {5}{2}} a^{2} x^{4}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a \,x^{2}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{2} b \,x^{4}-15 \sqrt {a \,x^{2}+b}\, a^{2} b^{2} x^{4}+2 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b \right )}{8 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{2}}\) | \(144\) |
Input:
int((a+b/x^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/8*b*(9*a*x^2+2*b)/x^3*((a*x^2+b)/x^2)^(1/2)+(-15/8*ln((2*b+2*b^(1/2)*(a *x^2+b)^(1/2))/x)*b^(1/2)*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)
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.82 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\left [\frac {15 \, a^{2} \sqrt {b} x^{3} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{16 \, x^{3}}, \frac {15 \, a^{2} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{b}\right ) + {\left (8 \, a^{2} x^{4} - 9 \, a b x^{2} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{8 \, x^{3}}\right ] \] Input:
integrate((a+b/x^2)^(5/2),x, algorithm="fricas")
Output:
[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)/b) + (8 *a^2*x^4 - 9*a*b*x^2 - 2*b^2)*sqrt((a*x^2 + b)/x^2))/x^3]
Time = 2.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {a^{\frac {5}{2}} x}{\sqrt {1 + \frac {b}{a x^{2}}}} - \frac {a^{\frac {3}{2}} b}{8 x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {11 \sqrt {a} b^{2}}{8 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{8} - \frac {b^{3}}{4 \sqrt {a} x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} \] Input:
integrate((a+b/x**2)**(5/2),x)
Output:
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)))
Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.43 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\sqrt {a + \frac {b}{x^{2}}} a^{2} x + \frac {15}{16} \, a^{2} \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right ) - \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{2} b x^{3} - 7 \, \sqrt {a + \frac {b}{x^{2}}} a^{2} b^{2} x}{8 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (a + \frac {b}{x^{2}}\right )} b x^{2} + b^{2}\right )}} \] Input:
integrate((a+b/x^2)^(5/2),x, algorithm="maxima")
Output:
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)
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.05 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {\frac {15 \, a^{3} b \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + 8 \, \sqrt {a x^{2} + b} a^{3} \mathrm {sgn}\left (x\right ) - \frac {9 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{3} b \mathrm {sgn}\left (x\right ) - 7 \, \sqrt {a x^{2} + b} a^{3} b^{2} \mathrm {sgn}\left (x\right )}{a^{2} x^{4}}}{8 \, a} \] Input:
integrate((a+b/x^2)^(5/2),x, algorithm="giac")
Output:
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
Time = 0.74 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.40 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {x\,{\left (a\,x^2+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b}{a\,x^2}\right )}{{\left (\frac {b}{a}+x^2\right )}^{5/2}} \] Input:
int((a + b/x^2)^(5/2),x)
Output:
(x*(b + a*x^2)^(5/2)*hypergeom([-5/2, -1/2], 1/2, -b/(a*x^2)))/(b/a + x^2) ^(5/2)
Time = 0.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.25 \[ \int \left (a+\frac {b}{x^2}\right )^{5/2} \, dx=\frac {8 \sqrt {a \,x^{2}+b}\, a^{2} x^{4}-9 \sqrt {a \,x^{2}+b}\, a b \,x^{2}-2 \sqrt {a \,x^{2}+b}\, b^{2}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x -\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,x^{2}+b}+\sqrt {a}\, x +\sqrt {b}}{\sqrt {b}}\right ) a^{2} x^{4}}{8 x^{4}} \] Input:
int((a+b/x^2)^(5/2),x)
Output:
(8*sqrt(a*x**2 + b)*a**2*x**4 - 9*sqrt(a*x**2 + b)*a*b*x**2 - 2*sqrt(a*x** 2 + b)*b**2 + 15*sqrt(b)*log((sqrt(a*x**2 + b) + sqrt(a)*x - sqrt(b))/sqrt (b))*a**2*x**4 - 15*sqrt(b)*log((sqrt(a*x**2 + b) + sqrt(a)*x + sqrt(b))/s qrt(b))*a**2*x**4)/(8*x**4)