\(\int (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [576]

Optimal result

Integrand size = 22, antiderivative size = 159 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac {a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac {3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac {b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3} \]

[Out]

a^3*x*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(b*x^2+a)^3+a^2*b*x^3*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(b*x^2+a)^3+3/5*a*b^2*
x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(b*x^2+a)^3+1/7*b^3*x^7*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(b*x^2+a)^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1102, 200} \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac {a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac {b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}+\frac {a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3} \]

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(a^3*x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b*x^2)^3 + (a^2*b*x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(a + b
*x^2)^3 + (3*a*b^2*x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/(5*(a + b*x^2)^3) + (b^3*x^7*(a^2 + 2*a*b*x^2 + b^2*
x^4)^(3/2))/(7*(a + b*x^2)^3)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 1102

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \int \left (2 a b+2 b^2 x^2\right )^3 \, dx}{\left (2 a b+2 b^2 x^2\right )^3} \\ & = \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \int \left (8 a^3 b^3+24 a^2 b^4 x^2+24 a b^5 x^4+8 b^6 x^6\right ) \, dx}{\left (2 a b+2 b^2 x^2\right )^3} \\ & = \frac {a^3 x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac {a^2 b x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac {3 a b^2 x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac {b^3 x^7 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.37 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (35 a^3 x+35 a^2 b x^3+21 a b^2 x^5+5 b^3 x^7\right )}{35 \left (a+b x^2\right )} \]

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(Sqrt[(a + b*x^2)^2]*(35*a^3*x + 35*a^2*b*x^3 + 21*a*b^2*x^5 + 5*b^3*x^7))/(35*(a + b*x^2))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.35

method	result	size
gosper	\(\frac {x \left (5 b^{3} x^{6}+21 b^{2} x^{4} a +35 a^{2} b \,x^{2}+35 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{35 \left (b \,x^{2}+a \right )^{3}}\)	\(56\)
default	\(\frac {x \left (5 b^{3} x^{6}+21 b^{2} x^{4} a +35 a^{2} b \,x^{2}+35 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{35 \left (b \,x^{2}+a \right )^{3}}\)	\(56\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} x^{7}}{7 b \,x^{2}+7 a}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{2} x^{5} a}{5 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b \,x^{3}}{b \,x^{2}+a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} x}{b \,x^{2}+a}\)	\(112\)

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/35*x*(5*b^3*x^6+21*a*b^2*x^4+35*a^2*b*x^2+35*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.19 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{7} \, b^{3} x^{7} + \frac {3}{5} \, a b^{2} x^{5} + a^{2} b x^{3} + a^{3} x \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/7*b^3*x^7 + 3/5*a*b^2*x^5 + a^2*b*x^3 + a^3*x

Sympy [F]

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

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.19 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{7} \, b^{3} x^{7} + \frac {3}{5} \, a b^{2} x^{5} + a^{2} b x^{3} + a^{3} x \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/7*b^3*x^7 + 3/5*a*b^2*x^5 + a^2*b*x^3 + a^3*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.40 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{7} \, b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{5} \, a b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{2} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/7*b^3*x^7*sgn(b*x^2 + a) + 3/5*a*b^2*x^5*sgn(b*x^2 + a) + a^2*b*x^3*sgn(b*x^2 + a) + a^3*x*sgn(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \]

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)