3.655 \(\int \frac {x^2}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=213 \[ \frac {5 x}{192 a^2 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{7/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 x}{128 a^3 b \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

5/128*x/a^3/b/((b*x^2+a)^2)^(1/2)-1/8*x/b/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)+1/48*x/a/b/(b*x^2+a)^2/((b*x^2+a)^2)
^(1/2)+5/192*x/a^2/b/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+5/128*(b*x^2+a)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/b^(3/2)/(
(b*x^2+a)^2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1112, 288, 199, 205} \[ \frac {5 x}{128 a^3 b \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 x}{192 a^2 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{7/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(5*x)/(128*a^3*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - x/(8*b*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + x/
(48*a*b*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (5*x)/(192*a^2*b*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4]) + (5*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(7/2)*b^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rule 288

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 1112

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 )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac {x^2}{\left (a b+b^2 x^2\right )^5} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^4} \, dx}{8 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^3} \, dx}{48 a \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 x}{192 a^2 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^2} \, dx}{64 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 x}{128 a^3 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 x}{192 a^2 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{128 a^3 b \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5 x}{128 a^3 b \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{48 a b \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 x}{192 a^2 b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{7/2} b^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 105, normalized size = 0.49 \[ \frac {\sqrt {a} \sqrt {b} x \left (-15 a^3+73 a^2 b x^2+55 a b^2 x^4+15 b^3 x^6\right )+15 \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{384 a^{7/2} b^{3/2} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(Sqrt[a]*Sqrt[b]*x*(-15*a^3 + 73*a^2*b*x^2 + 55*a*b^2*x^4 + 15*b^3*x^6) + 15*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/
Sqrt[a]])/(384*a^(7/2)*b^(3/2)*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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fricas [A]  time = 1.05, size = 324, normalized size = 1.52 \[ \left [\frac {30 \, a b^{4} x^{7} + 110 \, a^{2} b^{3} x^{5} + 146 \, a^{3} b^{2} x^{3} - 30 \, a^{4} b x - 15 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{768 \, {\left (a^{4} b^{6} x^{8} + 4 \, a^{5} b^{5} x^{6} + 6 \, a^{6} b^{4} x^{4} + 4 \, a^{7} b^{3} x^{2} + a^{8} b^{2}\right )}}, \frac {15 \, a b^{4} x^{7} + 55 \, a^{2} b^{3} x^{5} + 73 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x + 15 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{384 \, {\left (a^{4} b^{6} x^{8} + 4 \, a^{5} b^{5} x^{6} + 6 \, a^{6} b^{4} x^{4} + 4 \, a^{7} b^{3} x^{2} + a^{8} b^{2}\right )}}\right ] \]

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)^(5/2),x, algorithm="fricas")

[Out]

[1/768*(30*a*b^4*x^7 + 110*a^2*b^3*x^5 + 146*a^3*b^2*x^3 - 30*a^4*b*x - 15*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*
x^4 + 4*a^3*b*x^2 + a^4)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^4*b^6*x^8 + 4*a^5*b^5*x^
6 + 6*a^6*b^4*x^4 + 4*a^7*b^3*x^2 + a^8*b^2), 1/384*(15*a*b^4*x^7 + 55*a^2*b^3*x^5 + 73*a^3*b^2*x^3 - 15*a^4*b
*x + 15*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^4*b^6*
x^8 + 4*a^5*b^5*x^6 + 6*a^6*b^4*x^4 + 4*a^7*b^3*x^2 + a^8*b^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.02, size = 172, normalized size = 0.81 \[ \frac {\left (15 b^{4} x^{8} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+60 a \,b^{3} x^{6} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+15 \sqrt {a b}\, b^{3} x^{7}+90 a^{2} b^{2} x^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+55 \sqrt {a b}\, a \,b^{2} x^{5}+60 a^{3} b \,x^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+73 \sqrt {a b}\, a^{2} b \,x^{3}+15 a^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-15 \sqrt {a b}\, a^{3} x \right ) \left (b \,x^{2}+a \right )}{384 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} a^{3} b} \]

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)^(5/2),x)

[Out]

1/384*(15*b^4*x^8*arctan(1/(a*b)^(1/2)*b*x)+15*(a*b)^(1/2)*b^3*x^7+60*a*b^3*x^6*arctan(1/(a*b)^(1/2)*b*x)+55*(
a*b)^(1/2)*a*b^2*x^5+90*a^2*b^2*x^4*arctan(1/(a*b)^(1/2)*b*x)+73*(a*b)^(1/2)*a^2*b*x^3+60*a^3*b*x^2*arctan(1/(
a*b)^(1/2)*b*x)-15*(a*b)^(1/2)*a^3*x+15*a^4*arctan(1/(a*b)^(1/2)*b*x))*(b*x^2+a)/(a*b)^(1/2)/b/a^3/((b*x^2+a)^
2)^(5/2)

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maxima [A]  time = 2.90, size = 109, normalized size = 0.51 \[ \frac {15 \, b^{3} x^{7} + 55 \, a b^{2} x^{5} + 73 \, a^{2} b x^{3} - 15 \, a^{3} x}{384 \, {\left (a^{3} b^{5} x^{8} + 4 \, a^{4} b^{4} x^{6} + 6 \, a^{5} b^{3} x^{4} + 4 \, a^{6} b^{2} x^{2} + a^{7} b\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{3} b} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/384*(15*b^3*x^7 + 55*a*b^2*x^5 + 73*a^2*b*x^3 - 15*a^3*x)/(a^3*b^5*x^8 + 4*a^4*b^4*x^6 + 6*a^5*b^3*x^4 + 4*a
^6*b^2*x^2 + a^7*b) + 5/128*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

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)^(5/2),x)

[Out]

int(x^2/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

Integral(x**2/((a + b*x**2)**2)**(5/2), x)

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