3.7.41 \(\int \frac {x^2}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [641]

Optimal. Leaf size=129 \[ \frac {x}{8 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1126, 294, 205, 211} \begin {gather*} \frac {x}{8 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(8*a*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - x/(4*b*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((a + b*x^2)
*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 294

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

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {x^2}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {x}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a b+b^2 x^2\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^2} \, dx}{4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {x}{8 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a b+b^2 x^2\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {x}{8 a b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{4 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 81, normalized size = 0.63 \begin {gather*} \frac {\sqrt {a} \sqrt {b} x \left (-a+b x^2\right )+\left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{3/2} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[b]*x*(-a + b*x^2) + (a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(3/2)*(a + b*x^2)*Sq
rt[(a + b*x^2)^2])

________________________________________________________________________________________

Maple [A]
time = 0.05, size = 99, normalized size = 0.77

method	result	size
default	\(-\frac {\left (-\arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{2} x^{4}-\sqrt {a b}\, b \,x^{3}-2 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a b \,x^{2}+\sqrt {a b}\, a x -a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, b a \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}\)	\(99\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {x^{3}}{8 a}-\frac {x}{8 b}\right )}{\left (b \,x^{2}+a \right )^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b a}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 62, normalized size = 0.48 \begin {gather*} \frac {b x^{3} - a x}{8 \, {\left (a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^3 - a*x)/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b) + 1/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b)

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 190, normalized size = 1.47 \begin {gather*} \left [\frac {2 \, a b^{2} x^{3} - 2 \, a^{2} b x - {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}, \frac {a b^{2} x^{3} - a^{2} b x + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*a*b^2*x^3 - 2*a^2*b*x - (b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^
2 + a)))/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2), 1/8*(a*b^2*x^3 - a^2*b*x + (b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(
a*b)*arctan(sqrt(a*b)*x/a))/(a^2*b^4*x^4 + 2*a^3*b^3*x^2 + a^4*b^2)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 4.49, size = 70, normalized size = 0.54 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {b x^{3} - a x}{8 \, {\left (b x^{2} + a\right )}^{2} a b \mathrm {sgn}\left (b x^{2} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b*sgn(b*x^2 + a)) + 1/8*(b*x^3 - a*x)/((b*x^2 + a)^2*a*b*sgn(b*x^2 + a)
)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________