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

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

[Out]

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

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Rubi [A]  time = 0.0526389, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1112, 288, 205} \[ -\frac{x^3}{4 b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 x}{8 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0311322, size = 84, normalized size = 0.66 \[ \frac{3 \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )-\sqrt{a} \sqrt{b} x \left (3 a+5 b x^2\right )}{8 \sqrt{a} b^{5/2} \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.224, size = 97, normalized size = 0.8 \begin{align*} -{\frac{b{x}^{2}+a}{8\,{b}^{2}} \left ( -3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{4}{b}^{2}+5\,\sqrt{ab}{x}^{3}b-6\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}ab+3\,\sqrt{ab}xa-3\,{a}^{2}\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.31183, size = 404, normalized size = 3.16 \begin{align*} \left [-\frac{10 \, a b^{2} x^{3} + 6 \, a^{2} b x + 3 \,{\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 b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}, -\frac{5 \, a b^{2} x^{3} + 3 \, a^{2} b x - 3 \,{\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 b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(10*a*b^2*x^3 + 6*a^2*b*x + 3*(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*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3), -1/8*(5*a*b^2*x^3 + 3*a^2*b*x - 3*(b^2*x^4 + 2*a*b*x^2 + a
^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a*b^5*x^4 + 2*a^2*b^4*x^2 + a^3*b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x