\(\int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [543]

Optimal result

Integrand size = 24, antiderivative size = 36 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1121, 623} \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 b} \]

[In]

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

[Out]

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

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 1121

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(36)=72\).

Time = 0.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.44 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {x^2 \left (2 a+b x^2\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{-4 a^2-4 a b x^2+4 \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {x^{2} \left (b \,x^{2}+2 a \right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{4}\)	\(23\)
default	\(\frac {\left (b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b}\)	\(24\)
risch	\(\frac {\left (b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b}\)	\(24\)
gosper	\(\frac {x^{2} \left (b \,x^{2}+2 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b \,x^{2}+4 a}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.36 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{2} \, a x^{2} \]

[In]

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

[Out]

1/4*b*x^4 + 1/2*a*x^2

Sympy [F]

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

[In]

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

[Out]

Integral(x*sqrt((a + b*x**2)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.39 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {{\left (b x^{2} + a\right )}^{2}}{4 \, b} \]

[In]

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

[Out]

1/4*(b*x^2 + a)^2/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.61 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{4} \, {\left (b x^{4} + 2 \, a x^{2}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

1/4*(b*x^4 + 2*a*x^2)*sgn(b*x^2 + a)

Mupad [B] (verification not implemented)

Time = 13.72 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int x \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\left (\frac {a}{4\,b}+\frac {x^2}{4}\right )\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4} \]

[In]

int(x*((a + b*x^2)^2)^(1/2),x)

[Out]

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