3.1914 \(\int \frac {(a+\frac {b}{x^2})^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=116 \[ \frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{128 b^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x^2}}}{128 b x}-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3} \]

[Out]

-5/48*a*(a+b/x^2)^(3/2)/x^3-1/8*(a+b/x^2)^(5/2)/x^3+5/128*a^4*arctanh(b^(1/2)/x/(a+b/x^2)^(1/2))/b^(3/2)-5/64*
a^2*(a+b/x^2)^(1/2)/x^3-5/128*a^3*(a+b/x^2)^(1/2)/b/x

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Rubi [A]  time = 0.06, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 279, 321, 217, 206} \[ \frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )}{128 b^{3/2}}-\frac {5 a^3 \sqrt {a+\frac {b}{x^2}}}{128 b x}-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2)/x^4,x]

[Out]

(-5*a^2*Sqrt[a + b/x^2])/(64*x^3) - (5*a*(a + b/x^2)^(3/2))/(48*x^3) - (a + b/x^2)^(5/2)/(8*x^3) - (5*a^3*Sqrt
[a + b/x^2])/(128*b*x) + (5*a^4*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(128*b^(3/2))

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 279

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

Rule 321

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 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^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \left (a+b x^2\right )^{5/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {1}{8} (5 a) \operatorname {Subst}\left (\int x^2 \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}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {1}{16} \left (5 a^2\right ) \operatorname {Subst}\left (\int x^2 \sqrt {a+b x^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {1}{64} \left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {5 a^3 \sqrt {a+\frac {b}{x^2}}}{128 b x}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{x}\right )}{128 b}\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {5 a^3 \sqrt {a+\frac {b}{x^2}}}{128 b x}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x^2}} x}\right )}{128 b}\\ &=-\frac {5 a^2 \sqrt {a+\frac {b}{x^2}}}{64 x^3}-\frac {5 a \left (a+\frac {b}{x^2}\right )^{3/2}}{48 x^3}-\frac {\left (a+\frac {b}{x^2}\right )^{5/2}}{8 x^3}-\frac {5 a^3 \sqrt {a+\frac {b}{x^2}}}{128 b x}+\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x^2}} x}\right )}{128 b^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.42 \[ -\frac {a^4 x^5 \left (a+\frac {b}{x^2}\right )^{5/2} \left (a x^2+b\right ) \, _2F_1\left (\frac {7}{2},5;\frac {9}{2};\frac {a x^2}{b}+1\right )}{7 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2)/x^4,x]

[Out]

-1/7*(a^4*(a + b/x^2)^(5/2)*x^5*(b + a*x^2)*Hypergeometric2F1[7/2, 5, 9/2, 1 + (a*x^2)/b])/b^5

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fricas [A]  time = 0.98, size = 206, normalized size = 1.78 \[ \left [\frac {15 \, a^{4} \sqrt {b} x^{7} \log \left (-\frac {a x^{2} + 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, {\left (15 \, a^{3} b x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a b^{3} x^{2} + 48 \, b^{4}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{768 \, b^{2} x^{7}}, -\frac {15 \, a^{4} \sqrt {-b} x^{7} \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (15 \, a^{3} b x^{6} + 118 \, a^{2} b^{2} x^{4} + 136 \, a b^{3} x^{2} + 48 \, b^{4}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{384 \, b^{2} x^{7}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/768*(15*a^4*sqrt(b)*x^7*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*(15*a^3*b*x^6 + 118
*a^2*b^2*x^4 + 136*a*b^3*x^2 + 48*b^4)*sqrt((a*x^2 + b)/x^2))/(b^2*x^7), -1/384*(15*a^4*sqrt(-b)*x^7*arctan(sq
rt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (15*a^3*b*x^6 + 118*a^2*b^2*x^4 + 136*a*b^3*x^2 + 48*b^4)*sqrt((
a*x^2 + b)/x^2))/(b^2*x^7)]

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giac [A]  time = 0.25, size = 119, normalized size = 1.03 \[ -\frac {\frac {15 \, a^{5} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b} b} + \frac {15 \, {\left (a x^{2} + b\right )}^{\frac {7}{2}} a^{5} \mathrm {sgn}\relax (x) + 73 \, {\left (a x^{2} + b\right )}^{\frac {5}{2}} a^{5} b \mathrm {sgn}\relax (x) - 55 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{5} b^{2} \mathrm {sgn}\relax (x) + 15 \, \sqrt {a x^{2} + b} a^{5} b^{3} \mathrm {sgn}\relax (x)}{a^{4} b x^{8}}}{384 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^(5/2)/x^4,x, algorithm="giac")

[Out]

-1/384*(15*a^5*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + (15*(a*x^2 + b)^(7/2)*a^5*sgn(x) + 73*(a
*x^2 + b)^(5/2)*a^5*b*sgn(x) - 55*(a*x^2 + b)^(3/2)*a^5*b^2*sgn(x) + 15*sqrt(a*x^2 + b)*a^5*b^3*sgn(x))/(a^4*b
*x^8))/a

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maple [B]  time = 0.02, size = 186, normalized size = 1.60 \[ \frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} \left (15 a^{4} b^{\frac {5}{2}} x^{8} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {a \,x^{2}+b}\, a^{4} b^{2} x^{8}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a^{4} b \,x^{8}-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a^{4} x^{8}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a^{3} x^{6}+2 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a^{2} b \,x^{4}+8 \left (a \,x^{2}+b \right )^{\frac {7}{2}} a \,b^{2} x^{2}-48 \left (a \,x^{2}+b \right )^{\frac {7}{2}} b^{3}\right )}{384 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)^(5/2)/x^4,x)

[Out]

1/384*((a*x^2+b)/x^2)^(5/2)/x^3*(-3*(a*x^2+b)^(5/2)*x^8*a^4+3*(a*x^2+b)^(7/2)*x^6*a^3-5*(a*x^2+b)^(3/2)*x^8*a^
4*b+15*b^(5/2)*ln(2*(b+(a*x^2+b)^(1/2)*b^(1/2))/x)*x^8*a^4-15*(a*x^2+b)^(1/2)*x^8*a^4*b^2+2*(a*x^2+b)^(7/2)*x^
4*a^2*b+8*(a*x^2+b)^(7/2)*x^2*a*b^2-48*(a*x^2+b)^(7/2)*b^3)/(a*x^2+b)^(5/2)/b^4

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maxima [B]  time = 1.97, size = 190, normalized size = 1.64 \[ -\frac {5 \, a^{4} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right )}{256 \, b^{\frac {3}{2}}} - \frac {15 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {7}{2}} a^{4} x^{7} + 73 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} a^{4} b x^{5} - 55 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{4} b^{2} x^{3} + 15 \, \sqrt {a + \frac {b}{x^{2}}} a^{4} b^{3} x}{384 \, {\left ({\left (a + \frac {b}{x^{2}}\right )}^{4} b x^{8} - 4 \, {\left (a + \frac {b}{x^{2}}\right )}^{3} b^{2} x^{6} + 6 \, {\left (a + \frac {b}{x^{2}}\right )}^{2} b^{3} x^{4} - 4 \, {\left (a + \frac {b}{x^{2}}\right )} b^{4} x^{2} + b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^(5/2)/x^4,x, algorithm="maxima")

[Out]

-5/256*a^4*log((sqrt(a + b/x^2)*x - sqrt(b))/(sqrt(a + b/x^2)*x + sqrt(b)))/b^(3/2) - 1/384*(15*(a + b/x^2)^(7
/2)*a^4*x^7 + 73*(a + b/x^2)^(5/2)*a^4*b*x^5 - 55*(a + b/x^2)^(3/2)*a^4*b^2*x^3 + 15*sqrt(a + b/x^2)*a^4*b^3*x
)/((a + b/x^2)^4*b*x^8 - 4*(a + b/x^2)^3*b^2*x^6 + 6*(a + b/x^2)^2*b^3*x^4 - 4*(a + b/x^2)*b^4*x^2 + b^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{x^2}\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(5/2)/x^4,x)

[Out]

int((a + b/x^2)^(5/2)/x^4, x)

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sympy [A]  time = 8.19, size = 150, normalized size = 1.29 \[ - \frac {5 a^{\frac {7}{2}}}{128 b x \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {133 a^{\frac {5}{2}}}{384 x^{3} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {127 a^{\frac {3}{2}} b}{192 x^{5} \sqrt {1 + \frac {b}{a x^{2}}}} - \frac {23 \sqrt {a} b^{2}}{48 x^{7} \sqrt {1 + \frac {b}{a x^{2}}}} + \frac {5 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x} \right )}}{128 b^{\frac {3}{2}}} - \frac {b^{3}}{8 \sqrt {a} x^{9} \sqrt {1 + \frac {b}{a x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)**(5/2)/x**4,x)

[Out]

-5*a**(7/2)/(128*b*x*sqrt(1 + b/(a*x**2))) - 133*a**(5/2)/(384*x**3*sqrt(1 + b/(a*x**2))) - 127*a**(3/2)*b/(19
2*x**5*sqrt(1 + b/(a*x**2))) - 23*sqrt(a)*b**2/(48*x**7*sqrt(1 + b/(a*x**2))) + 5*a**4*asinh(sqrt(b)/(sqrt(a)*
x))/(128*b**(3/2)) - b**3/(8*sqrt(a)*x**9*sqrt(1 + b/(a*x**2)))

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