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

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

[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

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Rubi [A]  time = 0.0431205, antiderivative size = 88, normalized size of antiderivative = 1., 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 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]

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 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 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 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])

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*}

Mathematica [C]  time = 0.0151627, 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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Maple [A]  time = 0.006, size = 122, normalized size = 1.4 \begin{align*} -{\frac{{x}^{3}}{6\,b} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{5}{2}}} \left ( -3\, \left ( a{x}^{2}+b \right ) ^{5/2}{x}^{2}a+3\, \left ( a{x}^{2}+b \right ) ^{7/2}-5\, \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{2}ab+15\,{b}^{5/2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{2}a-15\,\sqrt{a{x}^{2}+b}{x}^{2}a{b}^{2} \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58441, size = 385, normalized size = 4.38 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 4.18992, size = 112, normalized size = 1.27 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.22412, size = 101, normalized size = 1.15 \begin{align*} \frac{1}{6} \,{\left (\frac{15 \, b^{2} \arctan \left (\frac{\sqrt{a x^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 2 \,{\left (a x^{2} + b\right )}^{\frac{3}{2}} + 12 \, \sqrt{a x^{2} + b} b - \frac{3 \, \sqrt{a x^{2} + b} b^{2}}{a x^{2}}\right )} a \mathrm{sgn}\left (x\right ) \end{align*}

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