3.1908 \(\int (a+\frac {b}{x^2})^{5/2} x^2 \, dx\)

Optimal. Leaf size=88 \[ -\frac {5}{2} a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )-\frac {5 b^2 \sqrt {a+\frac {b}{x^2}}}{2 x}+\frac {5}{3} b x \left (a+\frac {b}{x^2}\right )^{3/2}+\frac {1}{3} x^3 \left (a+\frac {b}{x^2}\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {335, 277, 195, 217, 206} \[ -\frac {5 b^2 \sqrt {a+\frac {b}{x^2}}}{2 x}-\frac {5}{2} a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b}}{x \sqrt {a+\frac {b}{x^2}}}\right )+\frac {1}{3} x^3 \left (a+\frac {b}{x^2}\right )^{5/2}+\frac {5}{3} b x \left (a+\frac {b}{x^2}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2)*x^2,x]

[Out]

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

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 277

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]

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

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.53 \[ \frac {a x^5 \left (a+\frac {b}{x^2}\right )^{5/2} \left (a x^2+b\right ) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {a x^2}{b}+1\right )}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2)*x^2,x]

[Out]

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

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fricas [A]  time = 0.92, size = 165, normalized size = 1.88 \[ \left [\frac {15 \, a b^{\frac {3}{2}} x \log \left (-\frac {a x^{2} - 2 \, \sqrt {b} x \sqrt {\frac {a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) + 2 \, {\left (2 \, a^{2} x^{4} + 14 \, a b x^{2} - 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{12 \, x}, \frac {15 \, a \sqrt {-b} b x \arctan \left (\frac {\sqrt {-b} x \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + {\left (2 \, a^{2} x^{4} + 14 \, a b x^{2} - 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{6 \, x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(15*a*b^(3/2)*x*log(-(a*x^2 - 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) + 2*(2*a^2*x^4 + 14*a*b*x^2
- 3*b^2)*sqrt((a*x^2 + b)/x^2))/x, 1/6*(15*a*sqrt(-b)*b*x*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b))
+ (2*a^2*x^4 + 14*a*b*x^2 - 3*b^2)*sqrt((a*x^2 + b)/x^2))/x]

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giac [A]  time = 0.22, size = 90, normalized size = 1.02 \[ \frac {\frac {15 \, a^{2} b^{2} \arctan \left (\frac {\sqrt {a x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} + 2 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{2} \mathrm {sgn}\relax (x) + 12 \, \sqrt {a x^{2} + b} a^{2} b \mathrm {sgn}\relax (x) - \frac {3 \, \sqrt {a x^{2} + b} a b^{2} \mathrm {sgn}\relax (x)}{x^{2}}}{6 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a^2*b^2*arctan(sqrt(a*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) + 2*(a*x^2 + b)^(3/2)*a^2*sgn(x) + 12*sqrt(a*
x^2 + b)*a^2*b*sgn(x) - 3*sqrt(a*x^2 + b)*a*b^2*sgn(x)/x^2)/a

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maple [A]  time = 0.01, size = 122, normalized size = 1.39 \[ -\frac {\left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} \left (15 a \,b^{\frac {5}{2}} x^{2} \ln \left (\frac {2 b +2 \sqrt {a \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {a \,x^{2}+b}\, a \,b^{2} x^{2}-5 \left (a \,x^{2}+b \right )^{\frac {3}{2}} a b \,x^{2}-3 \left (a \,x^{2}+b \right )^{\frac {5}{2}} a \,x^{2}+3 \left (a \,x^{2}+b \right )^{\frac {7}{2}}\right ) x^{3}}{6 \left (a \,x^{2}+b \right )^{\frac {5}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.80, size = 105, normalized size = 1.19 \[ \frac {1}{3} \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a x^{3} + 2 \, \sqrt {a + \frac {b}{x^{2}}} a b x - \frac {\sqrt {a + \frac {b}{x^{2}}} a b^{2} x}{2 \, {\left ({\left (a + \frac {b}{x^{2}}\right )} x^{2} - b\right )}} + \frac {5}{4} \, a b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} x - \sqrt {b}}{\sqrt {a + \frac {b}{x^{2}}} x + \sqrt {b}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/x^2)^(5/2),x)

[Out]

int(x^2*(a + b/x^2)^(5/2), x)

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sympy [A]  time = 4.00, size = 112, normalized size = 1.27 \[ \frac {a^{2} \sqrt {b} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3} + \frac {7 a b^{\frac {3}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3} + \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a x^{2}}{b} \right )}}{4} - \frac {5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a x^{2}}{b} + 1} + 1 \right )}}{2} - \frac {b^{\frac {5}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*sqrt(b)*x**2*sqrt(a*x**2/b + 1)/3 + 7*a*b**(3/2)*sqrt(a*x**2/b + 1)/3 + 5*a*b**(3/2)*log(a*x**2/b)/4 - 5*
a*b**(3/2)*log(sqrt(a*x**2/b + 1) + 1)/2 - b**(5/2)*sqrt(a*x**2/b + 1)/(2*x**2)

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