3.257 \(\int \frac{(c \sqrt{a+b x^2})^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=115 \[ -\frac{\left (c \sqrt{a+b x^2}\right )^{3/2}}{x}+\frac{3 b x \left (c \sqrt{a+b x^2}\right )^{3/2}}{a+b x^2}-\frac{3 \sqrt{b} \left (c \sqrt{a+b x^2}\right )^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a} \left (\frac{b x^2}{a}+1\right )^{3/4}} \]

[Out]

-((c*Sqrt[a + b*x^2])^(3/2)/x) + (3*b*x*(c*Sqrt[a + b*x^2])^(3/2))/(a + b*x^2) - (3*Sqrt[b]*(c*Sqrt[a + b*x^2]
)^(3/2)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(Sqrt[a]*(1 + (b*x^2)/a)^(3/4))

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Rubi [A]  time = 0.137411, antiderivative size = 142, normalized size of antiderivative = 1.23, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6720, 277, 229, 227, 196} \[ -\frac{c \sqrt{a+b x^2} \sqrt{c \sqrt{a+b x^2}}}{x}+\frac{3 b c x \sqrt{c \sqrt{a+b x^2}}}{\sqrt{a+b x^2}}-\frac{3 \sqrt{a} \sqrt{b} c \sqrt [4]{\frac{b x^2}{a}+1} \sqrt{c \sqrt{a+b x^2}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2)/x^2,x]

[Out]

(3*b*c*x*Sqrt[c*Sqrt[a + b*x^2]])/Sqrt[a + b*x^2] - (c*Sqrt[c*Sqrt[a + b*x^2]]*Sqrt[a + b*x^2])/x - (3*Sqrt[a]
*Sqrt[b]*c*Sqrt[c*Sqrt[a + b*x^2]]*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[a +
b*x^2]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{\left (c \sqrt{a+b x^2}\right )^{3/2}}{x^2} \, dx &=\frac{\left (c \sqrt{c \sqrt{a+b x^2}}\right ) \int \frac{\left (a+b x^2\right )^{3/4}}{x^2} \, dx}{\sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{x}+\frac{\left (3 b c \sqrt{c \sqrt{a+b x^2}}\right ) \int \frac{1}{\sqrt [4]{a+b x^2}} \, dx}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{x}+\frac{\left (3 b c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1+\frac{b x^2}{a}}} \, dx}{2 \sqrt{a+b x^2}}\\ &=\frac{3 b c x \sqrt{c \sqrt{a+b x^2}}}{\sqrt{a+b x^2}}-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{x}-\frac{\left (3 b c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{2 \sqrt{a+b x^2}}\\ &=\frac{3 b c x \sqrt{c \sqrt{a+b x^2}}}{\sqrt{a+b x^2}}-\frac{c \sqrt{c \sqrt{a+b x^2}} \sqrt{a+b x^2}}{x}-\frac{3 \sqrt{a} \sqrt{b} c \sqrt{c \sqrt{a+b x^2}} \sqrt [4]{1+\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{\sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0101591, size = 55, normalized size = 0.48 \[ -\frac{\left (c \sqrt{a+b x^2}\right )^{3/2} \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \left (\frac{b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sqrt[a + b*x^2])^(3/2)/x^2,x]

[Out]

-(((c*Sqrt[a + b*x^2])^(3/2)*Hypergeometric2F1[-3/4, -1/2, 1/2, -((b*x^2)/a)])/(x*(1 + (b*x^2)/a)^(3/4)))

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Maple [F]  time = 0.009, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ( c\sqrt{b{x}^{2}+a} \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(b*x^2+a)^(1/2))^(3/2)/x^2,x)

[Out]

int((c*(b*x^2+a)^(1/2))^(3/2)/x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\sqrt{b x^{2} + a} c\right )^{\frac{3}{2}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^2,x, algorithm="maxima")

[Out]

integrate((sqrt(b*x^2 + a)*c)^(3/2)/x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{\sqrt{b x^{2} + a} c} c}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^2,x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)*c/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \sqrt{a + b x^{2}}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x**2+a)**(1/2))**(3/2)/x**2,x)

[Out]

Integral((c*sqrt(a + b*x**2))**(3/2)/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^2,x, algorithm="giac")

[Out]

Exception raised: TypeError