\(\int \frac {(a-b x^2)^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx\) [206]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {\left (a-b x^2\right )^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx=-\frac {x \sqrt {a-b x^2} \left (a+b x^2\right )}{2 \sqrt {a^2-b^2 x^4}}+\frac {3 a \sqrt {a-b x^2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1166, 396, 223, 212} \[ \int \frac {\left (a-b x^2\right )^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {3 a \sqrt {a-b x^2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b} \sqrt {a^2-b^2 x^4}}-\frac {x \sqrt {a-b x^2} \left (a+b x^2\right )}{2 \sqrt {a^2-b^2 x^4}} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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

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

Mathematica [A] (verified)

Time = 2.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a-b x^2\right )^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {1}{2} \left (-\frac {x \sqrt {a^2-b^2 x^4}}{\sqrt {a-b x^2}}-\frac {3 a \log \left (-a+b x^2\right )}{\sqrt {b}}+\frac {3 a \log \left (a b x-b^2 x^3+\sqrt {b} \sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}\right )}{\sqrt {b}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (-x \sqrt {b}\, \sqrt {b \,x^{2}+a}+3 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a \right )}{2 \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}\, \sqrt {b}}\)	\(74\)
risch	\(\frac {x \sqrt {b \,x^{2}+a}\, \sqrt {\frac {\left (-b \,x^{2}+a \right ) \left (-b^{2} x^{4}+a^{2}\right )}{\left (b \,x^{2}-a \right )^{2}}}\, \left (b \,x^{2}-a \right )}{2 \sqrt {-b \,x^{2}+a}\, \sqrt {-b^{2} x^{4}+a^{2}}}-\frac {3 a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {\frac {\left (-b \,x^{2}+a \right ) \left (-b^{2} x^{4}+a^{2}\right )}{\left (b \,x^{2}-a \right )^{2}}}\, \left (b \,x^{2}-a \right )}{2 \sqrt {b}\, \sqrt {-b \,x^{2}+a}\, \sqrt {-b^{2} x^{4}+a^{2}}}\)	\(170\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.17 \[ \int \frac {\left (a-b x^2\right )^{3/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\left [\frac {2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} b x + 3 \, {\left (a b x^{2} - a^{2}\right )} \sqrt {b} \log \left (\frac {2 \, b^{2} x^{4} - a b x^{2} - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {b} x - a^{2}}{b x^{2} - a}\right )}{4 \, {\left (b^{2} x^{2} - a b\right )}}, \frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} b x + 3 \, {\left (a b x^{2} - a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {-b}}{b^{2} x^{3} - a b x}\right )}{2 \, {\left (b^{2} x^{2} - a b\right )}}\right ] \]

[In]

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

[Out]

[1/4*(2*sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*b*x + 3*(a*b*x^2 - a^2)*sqrt(b)*log((2*b^2*x^4 - a*b*x^2 - 2*sqr
t(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*sqrt(b)*x - a^2)/(b*x^2 - a)))/(b^2*x^2 - a*b), 1/2*(sqrt(-b^2*x^4 + a^2)*s
qrt(-b*x^2 + a)*b*x + 3*(a*b*x^2 - a^2)*sqrt(-b)*arctan(sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)*sqrt(-b)/(b^2*x^
3 - a*b*x)))/(b^2*x^2 - a*b)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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