3.1912 \(\int \frac {(a+\frac {b}{x^2})^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=92 \[ -\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{16 \sqrt {b}}-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x} \]

[Out]

-5/24*a*(a+b/x^2)^(3/2)/x-1/6*(a+b/x^2)^(5/2)/x-5/16*a^3*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(1/2)-5/16*a^2*(
a+b/x^2)^(1/2)/x

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Rubi [A]  time = 0.04, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {335, 195, 217, 206} \[ -\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{16 \sqrt {b}}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2)/x^2,x]

[Out]

(-5*a^2*Sqrt[a + b/x^2])/(16*x) - (5*a*(a + b/x^2)^(3/2))/(24*x) - (a + b/x^2)^(5/2)/(6*x) - (5*a^3*ArcTanh[Sq
rt[b]/(Sqrt[a + b/x^2]*x)])/(16*Sqrt[b])

Rule 195

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 206

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

Rule 217

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 335

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

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{x^2} \, dx &=-\operatorname {Subst}\left (\int \left (a+b x^2\right )^{5/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {1}{6} (5 a) \operatorname {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {1}{8} \left (5 a^2\right ) \operatorname {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {1}{16} \left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {1}{16} \left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{16 x}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{24 x}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{6 x}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{16 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 96, normalized size = 1.04 \[ -\frac {\sqrt {a+\frac {b}{x^2}} \left (15 a^3 x^6 \sqrt {\frac {a x^2}{b}+1} \tanh ^{-1}\left (\sqrt {\frac {a x^2}{b}+1}\right )+33 a^3 x^6+59 a^2 b x^4+34 a b^2 x^2+8 b^3\right )}{48 x^5 \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2)/x^2,x]

[Out]

-1/48*(Sqrt[a + b/x^2]*(8*b^3 + 34*a*b^2*x^2 + 59*a^2*b*x^4 + 33*a^3*x^6 + 15*a^3*x^6*Sqrt[1 + (a*x^2)/b]*ArcT
anh[Sqrt[1 + (a*x^2)/b]]))/(x^5*(b + a*x^2))

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fricas [A]  time = 0.81, size = 185, normalized size = 2.01 \[ \left [\frac {15 \, a^{3} \sqrt {b} x^{5} \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{96 \, b x^{5}}, \frac {15 \, a^{3} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - {\left (33 \, a^{2} b x^{4} + 26 \, a b^{2} x^{2} + 8 \, b^{3}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{48 \, b x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(15*a^3*sqrt(b)*x^5*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(33*a^2*b*x^4 + 26*a
*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b*x^5), 1/48*(15*a^3*sqrt(-b)*x^5*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)
/x^2)/(a*x^2 + b)) - (33*a^2*b*x^4 + 26*a*b^2*x^2 + 8*b^3)*sqrt((a*x^2 + b)/x^2))/(b*x^5)]

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giac [A]  time = 0.23, size = 95, normalized size = 1.03 \[ \frac {\frac {15 \, a^{4} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} - \frac {33 \, {\left (a x^{2} + b\right )}^{\frac {5}{2}} a^{4} \mathrm {sgn}\relax (x) - 40 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{4} b \mathrm {sgn}\relax (x) + 15 \, \sqrt {a x^{2} + b} a^{4} b^{2} \mathrm {sgn}\relax (x)}{a^{3} x^{6}}}{48 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(15*a^4*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - (33*(a*x^2 + b)^(5/2)*a^4*sgn(x) - 40*(a*x^2 +
b)^(3/2)*a^4*b*sgn(x) + 15*sqrt(a*x^2 + b)*a^4*b^2*sgn(x))/(a^3*x^6))/a

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maple [B]  time = 0.01, size = 166, normalized size = 1.80 \[ -\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} \left (15 a^{3} b^{\frac {5}{2}} x^{6} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {a \,x^{2}+b}\, a^{3} b^{2} x^{6}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{3} b \,x^{6}-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{3} x^{6}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a^{2} x^{4}+2 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a b \,x^{2}+8 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b^{2}\right )}{48 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((a*x^2+b)/x^2)^(5/2)/x*(-3*(a*x^2+b)^(5/2)*x^6*a^3+15*b^(5/2)*ln(2*(b+(a*x^2+b)^(1/2)*b^(1/2))/x)*x^6*a
^3+3*(a*x^2+b)^(7/2)*x^4*a^2-5*(a*x^2+b)^(3/2)*x^6*a^3*b-15*(a*x^2+b)^(1/2)*x^6*a^3*b^2+2*(a*x^2+b)^(7/2)*x^2*
a*b+8*(a*x^2+b)^(7/2)*b^2)/(a*x^2+b)^(5/2)/b^3

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maxima [B]  time = 1.99, size = 152, normalized size = 1.65 \[ \frac {5 \, a^{3} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{32 \, \sqrt {b}} - \frac {33 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{3} x^{5} - 40 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{3} b x^{3} + 15 \, \sqrt {a + \frac {b}{x^{2}}} a^{3} b^{2} x}{48 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{3} x^{6} - 3 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b x^{4} + 3 \, {\left (a + \frac {b}{x^{2}}\right )} b^{2} x^{2} - b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/32*a^3*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/sqrt(b) - 1/48*(33*(a + b/x^2)^(5/2)
*a^3*x^5 - 40*(a + b/x^2)^(3/2)*a^3*b*x^3 + 15*sqrt(a + b/x^2)*a^3*b^2*x)/((a + b/x^2)^3*x^6 - 3*(a + b/x^2)^2
*b*x^4 + 3*(a + b/x^2)*b^2*x^2 - b^3)

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mupad [B]  time = 1.79, size = 39, normalized size = 0.42 \[ -\frac {{\left (a\,x^2+b\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b}{a\,x^2}\right )}{x\,{\left (\frac {b}{a}+x^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(5/2)/x^2,x)

[Out]

-((b + a*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -b/(a*x^2)))/(x*(b/a + x^2)^(5/2))

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sympy [A]  time = 4.82, size = 99, normalized size = 1.08 \[ - \frac {11 a^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x^{2}}}}{16 x} - \frac {13 a^{\frac {3}{2}} b \sqrt {1 + \frac {b}{a x^{2}}}}{24 x^{3}} - \frac {\sqrt {a} b^{2} \sqrt {1 + \frac {b}{a x^{2}}}}{6 x^{5}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{16 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11*a**(5/2)*sqrt(1 + b/(a*x**2))/(16*x) - 13*a**(3/2)*b*sqrt(1 + b/(a*x**2))/(24*x**3) - sqrt(a)*b**2*sqrt(1
+ b/(a*x**2))/(6*x**5) - 5*a**3*asinh(sqrt(b)/(sqrt(a)*x))/(16*sqrt(b))

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