\(\int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx\) [291]

Optimal result

Integrand size = 20, antiderivative size = 60 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=-\frac {\sqrt {a x^2-b x^4}}{2 b}+\frac {a \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a x^2-b x^4}}\right )}{2 b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2043, 654, 634, 209} \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=\frac {a \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a x^2-b x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {a x^2-b x^4}}{2 b} \]

[In]

Int[x^3/Sqrt[a*x^2 - b*x^4],x]

[Out]

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

Rule 209

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

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=\frac {x \left (\sqrt {b} x \left (-a+b x^2\right )+2 a \sqrt {a-b x^2} \arctan \left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a-b x^2}}\right )\right )}{2 b^{3/2} \sqrt {x^2 \left (a-b x^2\right )}} \]

[In]

Integrate[x^3/Sqrt[a*x^2 - b*x^4],x]

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(-\frac {\arctan \left (\frac {-2 b \,x^{2}+a}{2 \sqrt {b}\, \sqrt {x^{2} \left (-b \,x^{2}+a \right )}}\right ) a +2 \sqrt {b}\, \sqrt {x^{2} \left (-b \,x^{2}+a \right )}}{4 b^{\frac {3}{2}}}\)	\(56\)
default	\(-\frac {x \sqrt {-b \,x^{2}+a}\, \left (x \sqrt {-b \,x^{2}+a}\, b^{\frac {3}{2}}-a \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) b \right )}{2 \sqrt {-b \,x^{4}+a \,x^{2}}\, b^{\frac {5}{2}}}\)	\(67\)
risch	\(-\frac {x^{2} \left (-b \,x^{2}+a \right )}{2 b \sqrt {x^{2} \left (-b \,x^{2}+a \right )}}+\frac {a \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) x \sqrt {-b \,x^{2}+a}}{2 b^{\frac {3}{2}} \sqrt {x^{2} \left (-b \,x^{2}+a \right )}}\)	\(79\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.00 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=\left [-\frac {a \sqrt {-b} \log \left (2 \, b x^{2} - a - 2 \, \sqrt {-b x^{4} + a x^{2}} \sqrt {-b}\right ) + 2 \, \sqrt {-b x^{4} + a x^{2}} b}{4 \, b^{2}}, -\frac {a \sqrt {b} \arctan \left (\frac {\sqrt {-b x^{4} + a x^{2}} \sqrt {b}}{b x^{2} - a}\right ) + \sqrt {-b x^{4} + a x^{2}} b}{2 \, b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=-\frac {a \arcsin \left (-\frac {2 \, b x^{2} - a}{a}\right )}{4 \, b^{\frac {3}{2}}} - \frac {\sqrt {-b x^{4} + a x^{2}}}{2 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=\frac {a \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, \sqrt {-b} b} - \frac {\sqrt {-b x^{2} + a} x}{2 \, b \mathrm {sgn}\left (x\right )} - \frac {a \log \left ({\left | -\sqrt {-b} x + \sqrt {-b x^{2} + a} \right |}\right )}{2 \, \sqrt {-b} b \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/4*a*log(abs(a))*sgn(x)/(sqrt(-b)*b) - 1/2*sqrt(-b*x^2 + a)*x/(b*sgn(x)) - 1/2*a*log(abs(-sqrt(-b)*x + sqrt(-
b*x^2 + a)))/(sqrt(-b)*b*sgn(x))

Mupad [B] (verification not implemented)

Time = 13.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt {a x^2-b x^4}} \, dx=-\frac {\sqrt {a\,x^2-b\,x^4}}{2\,b}-\frac {a\,\ln \left (\frac {\frac {a}{2}-b\,x^2}{\sqrt {-b}}+\sqrt {a\,x^2-b\,x^4}\right )}{4\,{\left (-b\right )}^{3/2}} \]

[In]

int(x^3/(a*x^2 - b*x^4)^(1/2),x)

[Out]

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