\(\int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [542]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2} \]

[Out]

1/6*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/b^2-1/4*a*(b*x^2+a)*((b*x^2+a)^2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 654, 623} \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}-\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2} \]

[In]

Int[x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-1/4*(a*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/b^2 + (a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/(6*b^2)

Rule 623

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

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 1125

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] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \sqrt {a^2+2 a b x+b^2 x^2} \, dx,x,x^2\right ) \\ & = \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2}-\frac {a \text {Subst}\left (\int \sqrt {a^2+2 a b x+b^2 x^2} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{6 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a x^4+2 b x^6\right )}{12 \left (a+b x^2\right )} \]

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(3*a*x^4 + 2*b*x^6))/(12*(a + b*x^2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.36

method	result	size
pseudoelliptic	\(\frac {x^{4} \left (2 b \,x^{2}+3 a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{12}\)	\(24\)
gosper	\(\frac {x^{4} \left (2 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 b \,x^{2}+12 a}\)	\(36\)
default	\(\frac {x^{4} \left (2 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 b \,x^{2}+12 a}\)	\(36\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,x^{6}}{6 b \,x^{2}+6 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \,x^{4}}{4 b \,x^{2}+4 a}\)	\(54\)

[In]

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

[Out]

1/12*x^4*(2*b*x^2+3*a)*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.19 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{6} \, b x^{6} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/6*b*x^6 + 1/4*a*x^4

Sympy [F]

\[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\int x^{3} \sqrt {\left (a + b x^{2}\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(x**3*sqrt((a + b*x**2)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.19 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{6} \, b x^{6} + \frac {1}{4} \, a x^{4} \]

[In]

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

[Out]

1/6*b*x^6 + 1/4*a*x^4

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.34 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{12} \, {\left (2 \, b x^{6} + 3 \, a x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

1/12*(2*b*x^6 + 3*a*x^4)*sgn(b*x^2 + a)

Mupad [B] (verification not implemented)

Time = 13.50 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int x^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (8\,b^2\,\left (a^2+b^2\,x^4\right )-12\,a^2\,b^2+4\,a\,b^3\,x^2\right )}{48\,b^4} \]

[In]

int(x^3*((a + b*x^2)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*(8*b^2*(a^2 + b^2*x^4) - 12*a^2*b^2 + 4*a*b^3*x^2))/(48*b^4)