3.209 \(\int \frac{1}{(a-b x^2)^{3/2} \sqrt{a^2-b^2 x^4}} \, dx\)

Optimal. Leaf size=124 \[ \frac{x \left (a+b x^2\right )}{4 a^2 \sqrt{a-b x^2} \sqrt{a^2-b^2 x^4}}+\frac{3 \sqrt{a-b x^2} \sqrt{a+b x^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a+b x^2}}\right )}{4 \sqrt{2} a^2 \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

[Out]

(x*(a + b*x^2))/(4*a^2*Sqrt[a - b*x^2]*Sqrt[a^2 - b^2*x^4]) + (3*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTanh[(Sqrt
[2]*Sqrt[b]*x)/Sqrt[a + b*x^2]])/(4*Sqrt[2]*a^2*Sqrt[b]*Sqrt[a^2 - b^2*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.0538451, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1152, 382, 377, 208} \[ \frac{x \left (a+b x^2\right )}{4 a^2 \sqrt{a-b x^2} \sqrt{a^2-b^2 x^4}}+\frac{3 \sqrt{a-b x^2} \sqrt{a+b x^2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a+b x^2}}\right )}{4 \sqrt{2} a^2 \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^2))/(4*a^2*Sqrt[a - b*x^2]*Sqrt[a^2 - b^2*x^4]) + (3*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTanh[(Sqrt
[2]*Sqrt[b]*x)/Sqrt[a + b*x^2]])/(4*Sqrt[2]*a^2*Sqrt[b]*Sqrt[a^2 - b^2*x^4])

Rule 1152

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^FracPart[p]/((d + e*x
^2)^FracPart[p]*(a/d + (c*x^2)/e)^FracPart[p]), Int[(d + e*x^2)^(p + q)*(a/d + (c*x^2)/e)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.083406, size = 110, normalized size = 0.89 \[ \frac{\sqrt{a^2-b^2 x^4} \left (2 \sqrt{b} x \sqrt{a+b x^2}+3 \sqrt{2} \left (a-b x^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a+b x^2}}\right )\right )}{8 a^2 \sqrt{b} \left (a-b x^2\right )^{3/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.041, size = 510, normalized size = 4.1 \begin{align*} -{\frac{1}{4\,b{x}^{2}-4\,a}\sqrt{-b{x}^{2}+a}\sqrt{-{b}^{2}{x}^{4}+{a}^{2}}{b}^{{\frac{5}{2}}} \left ( 3\,\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{b{x}^{2}+a}+\sqrt{ab}x+a \right ) }{bx-\sqrt{ab}}} \right ) \sqrt{2}{x}^{2}{b}^{3/2}\sqrt{a}-3\,\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{b{x}^{2}+a}-\sqrt{ab}x+a \right ) }{bx+\sqrt{ab}}} \right ) \sqrt{2}{x}^{2}{b}^{3/2}\sqrt{a}-3\,\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{b{x}^{2}+a}+\sqrt{ab}x+a \right ) }{bx-\sqrt{ab}}} \right ) \sqrt{2}{a}^{3/2}\sqrt{b}+3\,\ln \left ( 2\,{\frac{b \left ( \sqrt{2}\sqrt{a}\sqrt{b{x}^{2}+a}-\sqrt{ab}x+a \right ) }{bx+\sqrt{ab}}} \right ) \sqrt{2}{a}^{3/2}\sqrt{b}+4\,\ln \left ({\frac{1}{\sqrt{b}} \left ( \sqrt{b}\sqrt{-{\frac{ \left ( bx+\sqrt{-ab} \right ) \left ( -bx+\sqrt{-ab} \right ) }{b}}}+bx \right ) } \right ){x}^{2}b\sqrt{ab}-4\,\ln \left ({\frac{\sqrt{b}\sqrt{b{x}^{2}+a}+bx}{\sqrt{b}}} \right ){x}^{2}b\sqrt{ab}-4\,\sqrt{b}\sqrt{ab}\sqrt{b{x}^{2}+a}x-4\,\ln \left ({\frac{1}{\sqrt{b}} \left ( \sqrt{b}\sqrt{-{\frac{ \left ( bx+\sqrt{-ab} \right ) \left ( -bx+\sqrt{-ab} \right ) }{b}}}+bx \right ) } \right ) a\sqrt{ab}+4\,\ln \left ({\frac{\sqrt{b}\sqrt{b{x}^{2}+a}+bx}{\sqrt{b}}} \right ) a\sqrt{ab} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}} \left ( \sqrt{-ab}+\sqrt{ab} \right ) ^{-2} \left ( \sqrt{-ab}-\sqrt{ab} \right ) ^{-2} \left ( bx+\sqrt{ab} \right ) ^{-1} \left ( bx-\sqrt{ab} \right ) ^{-1}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}}{\left (-b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-b^2*x^4 + a^2)*(-b*x^2 + a)^(3/2)), x)

________________________________________________________________________________________

Fricas [A]  time = 2.35277, size = 660, normalized size = 5.32 \begin{align*} \left [\frac{4 \, \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{-b x^{2} + a} b x + 3 \, \sqrt{2}{\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \sqrt{b} \log \left (-\frac{3 \, b^{2} x^{4} - 2 \, a b x^{2} - 2 \, \sqrt{2} \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{-b x^{2} + a} \sqrt{b} x - a^{2}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}\right )}{16 \,{\left (a^{2} b^{3} x^{4} - 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}, \frac{2 \, \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{-b x^{2} + a} b x + 3 \, \sqrt{2}{\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{2} \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{-b x^{2} + a} \sqrt{-b}}{2 \,{\left (b^{2} x^{3} - a b x\right )}}\right )}{8 \,{\left (a^{2} b^{3} x^{4} - 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*b*x + 3*sqrt(2)*(b^2*x^4 - 2*a*b*x^2 + a^2)*sqrt(b)*log(-(3*b^2
*x^4 - 2*a*b*x^2 - 2*sqrt(2)*sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*sqrt(b)*x - a^2)/(b^2*x^4 - 2*a*b*x^2 + a^2
)))/(a^2*b^3*x^4 - 2*a^3*b^2*x^2 + a^4*b), 1/8*(2*sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*b*x + 3*sqrt(2)*(b^2*x
^4 - 2*a*b*x^2 + a^2)*sqrt(-b)*arctan(1/2*sqrt(2)*sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*sqrt(-b)/(b^2*x^3 - a*
b*x)))/(a^2*b^3*x^4 - 2*a^3*b^2*x^2 + a^4*b)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x**2+a)**(3/2)/(-b**2*x**4+a**2)**(1/2),x)

[Out]

Integral(1/(sqrt(-(-a + b*x**2)*(a + b*x**2))*(a - b*x**2)**(3/2)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}}{\left (-b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-b*x^2+a)^(3/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-b^2*x^4 + a^2)*(-b*x^2 + a)^(3/2)), x)