3.10.70 \(\int \frac {x^3}{\sqrt {a+b x^2-c x^4}} \, dx\) [970]

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

[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 = 0.04, 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 = {1128, 654, 635, 210} \begin {gather*} -\frac {b \text {ArcTan}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[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))

Rule 210

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

Rule 635

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 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 1128

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} \text {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 \text {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 \text {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 [B] Leaf count is larger than twice the leaf count of optimal. \(394\) vs. \(2(70)=140\).
time = 2.12, size = 394, normalized size = 5.63 \begin {gather*} \frac {1}{8} \left (\frac {16 a^2 \left (b^4 \sqrt {-c^2}+8 a b^2 c \sqrt {-c^2}-16 a^2 \left (-c^2\right )^{3/2}+8 b^3 \sqrt {-c} c^{3/2} x^2-32 a b \left (-c^2\right )^{3/2} x^2\right )}{b^3 \sqrt {c} \left (b^2+4 a c\right ) \left (b^2+4 a c+8 b c x^2\right )}-\frac {4 \sqrt {a+b x^2-c x^4} \left (16 a^2 c^{7/2}+4 a b c \left (b \left (c^{3/2}-\sqrt {-c} \sqrt {-c^2}\right )+8 c^{5/2} x^2\right )-b^3 \left (b \sqrt {-c} \sqrt {-c^2}-4 c \left (c^{3/2}-\sqrt {-c} \sqrt {-c^2}\right ) x^2\right )\right )}{c^{5/2} \left (b^2+4 a c\right ) \left (b^2+4 a c+8 b c x^2\right )}+\frac {2 b \tan ^{-1}\left (\frac {2 \sqrt {c} \left (-\sqrt {-c} x^2+\sqrt {a+b x^2-c x^4}\right )}{b}\right )}{c^{3/2}}+\frac {b \log \left (b^2+4 b c x^2+4 c \left (a-2 \left (c x^4+\sqrt {-c} x^2 \sqrt {a+b x^2-c x^4}\right )\right )\right )}{(-c)^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*a^2*(b^4*Sqrt[-c^2] + 8*a*b^2*c*Sqrt[-c^2] - 16*a^2*(-c^2)^(3/2) + 8*b^3*Sqrt[-c]*c^(3/2)*x^2 - 32*a*b*(-
c^2)^(3/2)*x^2))/(b^3*Sqrt[c]*(b^2 + 4*a*c)*(b^2 + 4*a*c + 8*b*c*x^2)) - (4*Sqrt[a + b*x^2 - c*x^4]*(16*a^2*c^
(7/2) + 4*a*b*c*(b*(c^(3/2) - Sqrt[-c]*Sqrt[-c^2]) + 8*c^(5/2)*x^2) - b^3*(b*Sqrt[-c]*Sqrt[-c^2] - 4*c*(c^(3/2
) - Sqrt[-c]*Sqrt[-c^2])*x^2)))/(c^(5/2)*(b^2 + 4*a*c)*(b^2 + 4*a*c + 8*b*c*x^2)) + (2*b*ArcTan[(2*Sqrt[c]*(-(
Sqrt[-c]*x^2) + Sqrt[a + b*x^2 - c*x^4]))/b])/c^(3/2) + (b*Log[b^2 + 4*b*c*x^2 + 4*c*(a - 2*(c*x^4 + Sqrt[-c]*
x^2*Sqrt[a + b*x^2 - c*x^4]))])/(-c)^(3/2))/8

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Maple [A]
time = 0.04, size = 58, normalized size = 0.83

method	result	size
default	\(-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}}{2 c}+\frac {b \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}\)	\(58\)
risch	\(-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}}{2 c}+\frac {b \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}\)	\(58\)
elliptic	\(-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}}{2 c}+\frac {b \arctan \left (\frac {\sqrt {c}\, \left (x^{2}-\frac {b}{2 c}\right )}{\sqrt {-c \,x^{4}+b \,x^{2}+a}}\right )}{4 c^{\frac {3}{2}}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 50, normalized size = 0.71 \begin {gather*} -\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} \end {gather*}

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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Fricas [A]
time = 0.35, size = 169, normalized size = 2.41 \begin {gather*} \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 ] \end {gather*}

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

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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Giac [A]
time = 3.74, size = 70, normalized size = 1.00 \begin {gather*} -\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} \end {gather*}

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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Mupad [B]
time = 4.59, size = 62, normalized size = 0.89 \begin {gather*} -\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}} \end {gather*}

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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