3.970 \(\int \frac {x^3}{\sqrt {a+b x^2-c x^4}} \, dx\)

Optimal. Leaf size=70 \[ -\frac {b \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{4 c^{3/2}}-\frac {\sqrt {a+b x^2-c x^4}}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1114, 640, 621, 204} \[ -\frac {b \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{4 c^{3/2}}-\frac {\sqrt {a+b x^2-c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

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

Rule 204

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

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 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 70, normalized size = 1.00 \[ -\frac {b \tan ^{-1}\left (\frac {b-2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\right )}{4 c^{3/2}}-\frac {\sqrt {a+b x^2-c x^4}}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + b*x^2 - c*x^4],x]

[Out]

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

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fricas [A]  time = 0.83, size = 169, normalized size = 2.41 \[ \left [-\frac {b \sqrt {-c} \log \left (8 \, c^{2} x^{4} - 8 \, b c x^{2} + b^{2} - 4 \, \sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {-c} - 4 \, a c\right ) + 4 \, \sqrt {-c x^{4} + b x^{2} + a} c}{8 \, c^{2}}, -\frac {b \sqrt {c} \arctan \left (\frac {\sqrt {-c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} - b\right )} \sqrt {c}}{2 \, {\left (c^{2} x^{4} - b c x^{2} - a c\right )}}\right ) + 2 \, \sqrt {-c x^{4} + b x^{2} + a} c}{4 \, c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 70, normalized size = 1.00 \[ -\frac {b \log \left ({\left | 2 \, {\left (\sqrt {-c} x^{2} - \sqrt {-c x^{4} + b x^{2} + a}\right )} \sqrt {-c} + b \right |}\right )}{4 \, \sqrt {-c} c} - \frac {\sqrt {-c x^{4} + b x^{2} + a}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 58, normalized size = 0.83 \[ \frac {b \arctan \left (\frac {\left (x^{2}-\frac {b}{2 c}\right ) \sqrt {c}}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.47, size = 50, normalized size = 0.71 \[ -\frac {b \arcsin \left (-\frac {2 \, c x^{2} - b}{\sqrt {b^{2} + 4 \, a c}}\right )}{4 \, c^{\frac {3}{2}}} - \frac {\sqrt {-c x^{4} + b x^{2} + a}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.59, size = 62, normalized size = 0.89 \[ -\frac {\sqrt {-c\,x^4+b\,x^2+a}}{2\,c}-\frac {b\,\ln \left (\frac {\frac {b}{2}-c\,x^2}{\sqrt {-c}}+\sqrt {-c\,x^4+b\,x^2+a}\right )}{4\,{\left (-c\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*x^2 - c*x^4)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {a + b x^{2} - c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/sqrt(a + b*x**2 - c*x**4), x)

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