\(\int x^2 \sqrt {b x^2+c x^4} \, dx\) [232]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=-\frac {2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3}+\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x} \]

[Out]

-2/15*b*(c*x^4+b*x^2)^(3/2)/c^2/x^3+1/5*(c*x^4+b*x^2)^(3/2)/c/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2025} \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3} \]

[In]

Int[x^2*Sqrt[b*x^2 + c*x^4],x]

[Out]

(-2*b*(b*x^2 + c*x^4)^(3/2))/(15*c^2*x^3) + (b*x^2 + c*x^4)^(3/2)/(5*c*x)

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

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}& = \frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x}-\frac {(2 b) \int \sqrt {b x^2+c x^4} \, dx}{5 c} \\ & = -\frac {2 b \left (b x^2+c x^4\right )^{3/2}}{15 c^2 x^3}+\frac {\left (b x^2+c x^4\right )^{3/2}}{5 c x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-2 b^2+b c x^2+3 c^2 x^4\right )}{15 c^2 x} \]

[In]

Integrate[x^2*Sqrt[b*x^2 + c*x^4],x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-2*b^2 + b*c*x^2 + 3*c^2*x^4))/(15*c^2*x)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.75

method	result	size
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-3 c \,x^{2}+2 b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 c^{2} x}\)	\(39\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-3 c \,x^{2}+2 b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 c^{2} x}\)	\(39\)
trager	\(-\frac {\left (-3 c^{2} x^{4}-b c \,x^{2}+2 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15 c^{2} x}\)	\(43\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-3 c^{2} x^{4}-b c \,x^{2}+2 b^{2}\right )}{15 x \,c^{2}}\)	\(43\)

[In]

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

[Out]

-1/15*(c*x^2+b)*(-3*c*x^2+2*b)*(c*x^4+b*x^2)^(1/2)/c^2/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.79 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, c^{2} x} \]

[In]

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

[Out]

1/15*(3*c^2*x^4 + b*c*x^2 - 2*b^2)*sqrt(c*x^4 + b*x^2)/(c^2*x)

Sympy [F]

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

[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 = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.65 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {{\left (3 \, c^{2} x^{4} + b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{2} + b}}{15 \, c^{2}} \]

[In]

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

[Out]

1/15*(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 = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {2 \, b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{15 \, c^{2}} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b \mathrm {sgn}\left (x\right )}{15 \, c^{2}} \]

[In]

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

[Out]

2/15*b^(5/2)*sgn(x)/c^2 + 1/15*(3*(c*x^2 + b)^(5/2)*sgn(x) - 5*(c*x^2 + b)^(3/2)*b*sgn(x))/c^2

Mupad [B] (verification not implemented)

Time = 12.88 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.79 \[ \int x^2 \sqrt {b x^2+c x^4} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-2\,b^2+b\,c\,x^2+3\,c^2\,x^4\right )}{15\,c^2\,x} \]

[In]

int(x^2*(b*x^2 + c*x^4)^(1/2),x)

[Out]

((b*x^2 + c*x^4)^(1/2)*(3*c^2*x^4 - 2*b^2 + b*c*x^2))/(15*c^2*x)