\(\int \frac {1}{(a+b x^2)^{3/2} \sqrt {a^2-b^2 x^4}} \, dx\) [203]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {x \left (a-b x^2\right )}{4 a^2 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}}+\frac {3 \sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{4 \sqrt {2} a^2 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[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*arctan(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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1166, 390, 385, 211} \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {3 \sqrt {a+b x^2} \sqrt {a-b x^2} \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )}{4 \sqrt {2} a^2 \sqrt {b} \sqrt {a^2-b^2 x^4}}+\frac {x \left (a-b x^2\right )}{4 a^2 \sqrt {a+b x^2} \sqrt {a^2-b^2 x^4}} \]

[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]*ArcTan[(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 211

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

Rule 385

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 390

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] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1166

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/e)*x^2)^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*} \text {integral}& = \frac {\left (\sqrt {a-b x^2} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {a-b x^2} \left (a+b x^2\right )^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}{\sqrt {a-b x^2} \left (a+b x^2\right )} \, 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 ) \text {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} \tan ^{-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] (verified)

Time = 2.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\frac {\sqrt {a^2-b^2 x^4} \left (2 \sqrt {b} x \sqrt {a-b x^2}+3 \sqrt {2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {a-b x^2}}\right )\right )}{8 a^2 \sqrt {b} \sqrt {a-b x^2} \left (a+b x^2\right )^{3/2}} \]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(487\) vs. \(2(101)=202\).

Time = 0.23 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.90

method	result	size
default	\(-\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, b^{\frac {5}{2}} \left (3 \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}-\sqrt {-a b}\, x +a \right )}{b x -\sqrt {-a b}}\right ) \sqrt {2}\, b^{\frac {3}{2}} x^{2} \sqrt {a}-3 \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}+\sqrt {-a b}\, x +a \right )}{b x +\sqrt {-a b}}\right ) \sqrt {2}\, b^{\frac {3}{2}} x^{2} \sqrt {a}+3 \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}-\sqrt {-a b}\, x +a \right )}{b x -\sqrt {-a b}}\right ) \sqrt {2}\, a^{\frac {3}{2}} \sqrt {b}-3 \ln \left (\frac {2 b \left (\sqrt {2}\, \sqrt {a}\, \sqrt {-b \,x^{2}+a}+\sqrt {-a b}\, x +a \right )}{b x +\sqrt {-a b}}\right ) \sqrt {2}\, a^{\frac {3}{2}} \sqrt {b}+4 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) b \,x^{2} \sqrt {-a b}-4 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right ) b \,x^{2} \sqrt {-a b}-4 \sqrt {-b \,x^{2}+a}\, \sqrt {b}\, \sqrt {-a b}\, x +4 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) a \sqrt {-a b}-4 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {\frac {\left (-b x +\sqrt {a b}\right ) \left (b x +\sqrt {a b}\right )}{b}}}\right ) a \sqrt {-a b}\right )}{4 \sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \left (-\sqrt {-a b}+\sqrt {a b}\right )^{2} \left (\sqrt {-a b}+\sqrt {a b}\right )^{2} \left (b x +\sqrt {-a b}\right ) \left (b x -\sqrt {-a b}\right ) \sqrt {-a b}}\)	\(488\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\left [\frac {4 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} b x - 3 \, \sqrt {2} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-b} \log \left (-\frac {3 \, b^{2} x^{4} + 2 \, a b x^{2} - 2 \, \sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{16 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}, \frac {2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} b x - 3 \, \sqrt {2} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{2 \, {\left (b^{2} x^{3} + a b x\right )}}\right )}{8 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}}\right ] \]

[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]

\[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\int \frac {1}{\sqrt {- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )} \left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\int { \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\int { \frac {1}{\sqrt {-b^{2} x^{4} + a^{2}} {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{3/2} \sqrt {a^2-b^2 x^4}} \, dx=\int \frac {1}{\sqrt {a^2-b^2\,x^4}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

[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)