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

Optimal. Leaf size=83 \[ \frac{15}{8} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{5}{4} b x \left (a+b x^2\right )^{3/2}+\frac{15}{8} a b x \sqrt{a+b x^2} \]

[Out]

(15*a*b*x*Sqrt[a + b*x^2])/8 + (5*b*x*(a + b*x^2)^(3/2))/4 - (a + b*x^2)^(5/2)/x + (15*a^2*Sqrt[b]*ArcTanh[(Sq
rt[b]*x)/Sqrt[a + b*x^2]])/8

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Rubi [A]  time = 0.0250642, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {277, 195, 217, 206} \[ \frac{15}{8} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{5}{4} b x \left (a+b x^2\right )^{3/2}+\frac{15}{8} a b x \sqrt{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^2,x]

[Out]

(15*a*b*x*Sqrt[a + b*x^2])/8 + (5*b*x*(a + b*x^2)^(3/2))/4 - (a + b*x^2)^(5/2)/x + (15*a^2*Sqrt[b]*ArcTanh[(Sq
rt[b]*x)/Sqrt[a + b*x^2]])/8

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 \frac{\left (a+b x^2\right )^{5/2}}{x^2} \, dx &=-\frac{\left (a+b x^2\right )^{5/2}}{x}+(5 b) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac{5}{4} b x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{1}{4} (15 a b) \int \sqrt{a+b x^2} \, dx\\ &=\frac{15}{8} a b x \sqrt{a+b x^2}+\frac{5}{4} b x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{1}{8} \left (15 a^2 b\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{15}{8} a b x \sqrt{a+b x^2}+\frac{5}{4} b x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{1}{8} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{15}{8} a b x \sqrt{a+b x^2}+\frac{5}{4} b x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{x}+\frac{15}{8} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0075302, size = 52, normalized size = 0.63 \[ -\frac{a^2 \sqrt{a+b x^2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^2,x]

[Out]

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

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Maple [A]  time = 0.004, size = 85, normalized size = 1. \begin{align*} -{\frac{1}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{bx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,bx}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{15\,abx}{8}\sqrt{b{x}^{2}+a}}+{\frac{15\,{a}^{2}}{8}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57637, size = 329, normalized size = 3.96 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \, b^{2} x^{4} + 9 \, a b x^{2} - 8 \, a^{2}\right )} \sqrt{b x^{2} + a}}{16 \, x}, -\frac{15 \, a^{2} \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, b^{2} x^{4} + 9 \, a b x^{2} - 8 \, a^{2}\right )} \sqrt{b x^{2} + a}}{8 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*a^2*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*x^4 + 9*a*b*x^2 - 8*a^2)*sq
rt(b*x^2 + a))/x, -1/8*(15*a^2*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*x^4 + 9*a*b*x^2 - 8*a^2)
*sqrt(b*x^2 + a))/x]

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Sympy [A]  time = 3.67323, size = 117, normalized size = 1.41 \begin{align*} - \frac{a^{\frac{5}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{a^{\frac{3}{2}} b x}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 \sqrt{a} b^{2} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8} + \frac{b^{3} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)/(x*sqrt(1 + b*x**2/a)) + a**(3/2)*b*x/(8*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b**2*x**3/(8*sqrt(1 + b*x*
*2/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/8 + b**3*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 2.86719, size = 117, normalized size = 1.41 \begin{align*} -\frac{15}{16} \, a^{2} \sqrt{b} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \, a^{3} \sqrt{b}}{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a} + \frac{1}{8} \,{\left (2 \, b^{2} x^{2} + 9 \, a b\right )} \sqrt{b x^{2} + a} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/16*a^2*sqrt(b)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2*a^3*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)
+ 1/8*(2*b^2*x^2 + 9*a*b)*sqrt(b*x^2 + a)*x