\(\int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\) [994]

Optimal result

Integrand size = 28, antiderivative size = 50 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}+\frac {x \sqrt {b x^2+c x^4}}{3 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3, 2041, 1602} \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x} \]

[In]

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

[Out]

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

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]

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^4}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {x \sqrt {b x^2+c x^4}}{3 c}-\frac {(2 b) \int \frac {x^2}{\sqrt {b x^2+c x^4}} \, dx}{3 c} \\ & = -\frac {2 b \sqrt {b x^2+c x^4}}{3 c^2 x}+\frac {x \sqrt {b x^2+c x^4}}{3 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\left (-2 b+c x^2\right ) \sqrt {x^2 \left (b+c x^2\right )}}{3 c^2 x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.64

method	result	size
trager	\(-\frac {\left (-c \,x^{2}+2 b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3 c^{2} x}\)	\(32\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-c \,x^{2}+2 b \right ) x}{3 c^{2} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(37\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-c \,x^{2}+2 b \right ) x}{3 c^{2} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(37\)
risch	\(-\frac {x \left (c \,x^{2}+b \right ) \left (-c \,x^{2}+2 b \right )}{3 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, c^{2}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.60 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}} {\left (c x^{2} - 2 \, b\right )}}{3 \, c^{2} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {c^{2} x^{4} - b c x^{2} - 2 \, b^{2}}{3 \, \sqrt {c x^{2} + b} c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=\frac {2 \, b^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{3 \, c^{2}} + \frac {{\left (c x^{2} + b\right )}^{\frac {3}{2}}}{3 \, c^{2} \mathrm {sgn}\left (x\right )} - \frac {\sqrt {c x^{2} + b} b}{c^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

2/3*b^(3/2)*sgn(x)/c^2 + 1/3*(c*x^2 + b)^(3/2)/(c^2*sgn(x)) - sqrt(c*x^2 + b)*b/(c^2*sgn(x))

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.66 \[ \int \frac {x^4}{\sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {2\,b}{3\,c^2}-\frac {x^2}{3\,c}\right )}{x} \]

[In]

int(x^4/(b*x^2 + c*x^4)^(1/2),x)

[Out]

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