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]

1/4*x*(b*x^2+a)/a^2/(-b*x^2+a)^(1/2)/(-b^2*x^4+a^2)^(1/2)+3/8*arctanh(x*2^(1/2)*b^(1/2)/(b*x^2+a)^(1/2))*(-b*x
^2+a)^(1/2)*(b*x^2+a)^(1/2)/a^2*2^(1/2)/b^(1/2)/(-b^2*x^4+a^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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])

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fricas [A]  time = 0.65, size = 302, normalized size = 2.44 \[ \left [\frac {4 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} b x + 3 \, \sqrt {2} {\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \sqrt {b} \log \left (-\frac {3 \, b^{2} x^{4} - 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {b} x - a^{2}}{b^{2} x^{4} - 2 \, a b x^{2} + a^{2}}\right )}{16 \, {\left (a^{2} b^{3} x^{4} - 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}, \frac {2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} b x + 3 \, \sqrt {2} {\left (b^{2} x^{4} - 2 \, a b x^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {-b}}{2 \, {\left (b^{2} x^{3} - a b x\right )}}\right )}{8 \, {\left (a^{2} b^{3} x^{4} - 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}\right ] \]

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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (-b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.04, size = 510, normalized size = 4.11 \[ \frac {\sqrt {-b \,x^{2}+a}\, \sqrt {-b^{2} x^{4}+a^{2}}\, \left (3 \sqrt {2}\, \sqrt {a}\, b^{\frac {3}{2}} x^{2} \ln \left (\frac {2 \left (a -\sqrt {a b}\, x +\sqrt {2}\, \sqrt {b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x +\sqrt {a b}}\right )-3 \sqrt {2}\, \sqrt {a}\, b^{\frac {3}{2}} x^{2} \ln \left (\frac {2 \left (a +\sqrt {a b}\, x +\sqrt {2}\, \sqrt {b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x -\sqrt {a b}}\right )-4 \sqrt {a b}\, b \,x^{2} \ln \left (\frac {b x +\sqrt {-\frac {\left (b x +\sqrt {-a b}\right ) \left (-b x +\sqrt {-a b}\right )}{b}}\, \sqrt {b}}{\sqrt {b}}\right )+4 \sqrt {a b}\, b \,x^{2} \ln \left (\frac {b x +\sqrt {b \,x^{2}+a}\, \sqrt {b}}{\sqrt {b}}\right )-3 \sqrt {2}\, a^{\frac {3}{2}} \sqrt {b}\, \ln \left (\frac {2 \left (a -\sqrt {a b}\, x +\sqrt {2}\, \sqrt {b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x +\sqrt {a b}}\right )+3 \sqrt {2}\, a^{\frac {3}{2}} \sqrt {b}\, \ln \left (\frac {2 \left (a +\sqrt {a b}\, x +\sqrt {2}\, \sqrt {b \,x^{2}+a}\, \sqrt {a}\right ) b}{b x -\sqrt {a b}}\right )+4 \sqrt {a b}\, a \ln \left (\frac {b x +\sqrt {-\frac {\left (b x +\sqrt {-a b}\right ) \left (-b x +\sqrt {-a b}\right )}{b}}\, \sqrt {b}}{\sqrt {b}}\right )-4 \sqrt {a b}\, a \ln \left (\frac {b x +\sqrt {b \,x^{2}+a}\, \sqrt {b}}{\sqrt {b}}\right )+4 \sqrt {a b}\, \sqrt {b \,x^{2}+a}\, \sqrt {b}\, x \right ) b^{\frac {5}{2}}}{4 \left (b \,x^{2}-a \right ) \sqrt {b \,x^{2}+a}\, \left (\sqrt {-a b}+\sqrt {a b}\right )^{2} \left (\sqrt {-a b}-\sqrt {a b}\right )^{2} \sqrt {a b}\, \left (b x -\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (-b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a^2-b^2\,x^4}\,{\left (a-b\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )} \left (a - b x^{2}\right )^{\frac {3}{2}}}\, dx \]

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)

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