3.81 \(\int \frac {x^2 (d+e x^2)}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac {x (b d-5 a e)}{8 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x (b d-a e)}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) (3 a e+b d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

1/8*(-5*a*e+b*d)*x/a/b^2/((b*x^2+a)^2)^(1/2)-1/4*(-a*e+b*d)*x/b^2/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+1/8*(3*a*e+b*d
)*(b*x^2+a)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^(5/2)/((b*x^2+a)^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1250, 455, 385, 205} \[ \frac {x (b d-5 a e)}{8 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x (b d-a e)}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) (3 a e+b d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - 5*a*e)*x)/(8*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - ((b*d - a*e)*x)/(4*b^2*(a + b*x^2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]) + ((b*d + 3*a*e)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(5/2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4])

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^2 \left (d+e x^2\right )}{\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 (d+e x^2\right )}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac {(b d-a e) x}{4 b^2 \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 {-b (b d-a e)-4 b^2 e x^2}{\left (a b+b^2 x^2\right )^2} \, dx}{4 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {(b d-5 a e) x}{8 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left ((b d+3 a e) \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {(b d-5 a e) x}{8 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(b d-a e) x}{4 b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(b d+3 a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 108, normalized size = 0.71 \[ \frac {\left (a+b x^2\right )^2 (3 a e+b d) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-\sqrt {a} \sqrt {b} x \left (3 a^2 e+a b \left (d+5 e x^2\right )-b^2 d x^2\right )}{8 a^{3/2} b^{5/2} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[a]*Sqrt[b]*x*(3*a^2*e - b^2*d*x^2 + a*b*(d + 5*e*x^2))) + (b*d + 3*a*e)*(a + b*x^2)^2*ArcTan[(Sqrt[b]*
x)/Sqrt[a]])/(8*a^(3/2)*b^(5/2)*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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fricas [A]  time = 0.75, size = 300, normalized size = 1.96 \[ \left [\frac {2 \, {\left (a b^{3} d - 5 \, a^{2} b^{2} e\right )} x^{3} - {\left ({\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} + a^{2} b d + 3 \, a^{3} e + 2 \, {\left (a b^{2} d + 3 \, a^{2} b e\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (a^{2} b^{2} d + 3 \, a^{3} b e\right )} x}{16 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, \frac {{\left (a b^{3} d - 5 \, a^{2} b^{2} e\right )} x^{3} + {\left ({\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} + a^{2} b d + 3 \, a^{3} e + 2 \, {\left (a b^{2} d + 3 \, a^{2} b e\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (a^{2} b^{2} d + 3 \, a^{3} b e\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(2*(a*b^3*d - 5*a^2*b^2*e)*x^3 - ((b^3*d + 3*a*b^2*e)*x^4 + a^2*b*d + 3*a^3*e + 2*(a*b^2*d + 3*a^2*b*e)*
x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(a^2*b^2*d + 3*a^3*b*e)*x)/(a^2*b^5*x^4 + 2*
a^3*b^4*x^2 + a^4*b^3), 1/8*((a*b^3*d - 5*a^2*b^2*e)*x^3 + ((b^3*d + 3*a*b^2*e)*x^4 + a^2*b*d + 3*a^3*e + 2*(a
*b^2*d + 3*a^2*b*e)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (a^2*b^2*d + 3*a^3*b*e)*x)/(a^2*b^5*x^4 + 2*a^3*b^4
*x^2 + a^4*b^3)]

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

[Out]

sage0*x

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maple [A]  time = 0.02, size = 188, normalized size = 1.23 \[ -\frac {\left (-3 a \,b^{2} e \,x^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-b^{3} d \,x^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-6 a^{2} b e \,x^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-2 a \,b^{2} d \,x^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+5 \sqrt {a b}\, a b e \,x^{3}-\sqrt {a b}\, b^{2} d \,x^{3}-3 a^{3} e \arctan \left (\frac {b x}{\sqrt {a b}}\right )-a^{2} b d \arctan \left (\frac {b x}{\sqrt {a b}}\right )+3 \sqrt {a b}\, a^{2} e x +\sqrt {a b}\, a b d x \right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-3*arctan(1/(a*b)^(1/2)*b*x)*x^4*a*b^2*e-arctan(1/(a*b)^(1/2)*b*x)*x^4*b^3*d+5*(a*b)^(1/2)*x^3*a*b*e-(a*
b)^(1/2)*x^3*b^2*d-6*arctan(1/(a*b)^(1/2)*b*x)*x^2*a^2*b*e-2*arctan(1/(a*b)^(1/2)*b*x)*x^2*a*b^2*d+3*(a*b)^(1/
2)*x*a^2*e+(a*b)^(1/2)*x*a*b*d-3*arctan(1/(a*b)^(1/2)*b*x)*a^3*e-arctan(1/(a*b)^(1/2)*b*x)*a^2*b*d)*(b*x^2+a)/
(a*b)^(1/2)/a/b^2/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 1.49, size = 125, normalized size = 0.82 \[ -\frac {1}{8} \, e {\left (\frac {5 \, b x^{3} + 3 \, a x}{b^{4} x^{4} + 2 \, a b^{3} x^{2} + a^{2} b^{2}} - \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}}\right )} + \frac {1}{8} \, d {\left (\frac {b x^{3} - a x}{a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{2} + a^{3} b} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*e*((5*b*x^3 + 3*a*x)/(b^4*x^4 + 2*a*b^3*x^2 + a^2*b^2) - 3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^2)) + 1/8*d
*((b*x^3 - a*x)/(a*b^3*x^4 + 2*a^2*b^2*x^2 + a^3*b) + arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (d + e x^{2}\right )}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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