\(\int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\) [996]

Optimal result

Integrand size = 28, antiderivative size = 22 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {b x^2+c x^4}}{c x} \]

[Out]

(c*x^4+b*x^2)^(1/2)/c/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3, 1602} \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {b x^2+c x^4}}{c x} \]

[In]

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

[Out]

Sqrt[b*x^2 + c*x^4]/(c*x)

Rule 3

Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /;
FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {\sqrt {b x^2+c x^4}}{c x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )}}{c x} \]

[In]

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

[Out]

Sqrt[x^2*(b + c*x^2)]/(c*x)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
trager	\(\frac {\sqrt {c \,x^{4}+b \,x^{2}}}{c x}\)	\(21\)
gosper	\(\frac {\left (c \,x^{2}+b \right ) x}{c \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(26\)
default	\(\frac {\left (c \,x^{2}+b \right ) x}{c \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(26\)
risch	\(\frac {x \left (c \,x^{2}+b \right )}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c}\)	\(26\)

[In]

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

[Out]

(c*x^4+b*x^2)^(1/2)/c/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}}}{c x} \]

[In]

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

[Out]

sqrt(c*x^4 + b*x^2)/(c*x)

Sympy [F]

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

[In]

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

[Out]

Integral(x**2/sqrt(x**2*(b + c*x**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{2} + b}}{c} \]

[In]

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

[Out]

sqrt(c*x^2 + b)/c

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {\sqrt {b} \mathrm {sgn}\left (x\right )}{c} + \frac {\sqrt {c x^{2} + b}}{c \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-sqrt(b)*sgn(x)/c + sqrt(c*x^2 + b)/(c*sgn(x))

Mupad [B] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}}{c\,x} \]

[In]

int(x^2/(b*x^2 + c*x^4)^(1/2),x)

[Out]

(b*x^2 + c*x^4)^(1/2)/(c*x)