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

Optimal. Leaf size=64 \[ -\frac{3 x}{8 b^2 \left (a+b x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{5/2}}-\frac{x^3}{4 b \left (a+b x^2\right )^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0197249, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {288, 205} \[ -\frac{3 x}{8 b^2 \left (a+b x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{5/2}}-\frac{x^3}{4 b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0384839, size = 55, normalized size = 0.86 \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 \sqrt{a} b^{5/2}}-\frac{3 a x+5 b x^3}{8 b^2 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2)^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 47, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( -{\frac{5\,{x}^{3}}{8\,b}}-{\frac{3\,ax}{8\,{b}^{2}}} \right ) }+{\frac{3}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.28876, size = 404, normalized size = 6.31 \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*x^2+a)^3,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)]

________________________________________________________________________________________

Sympy [A]  time = 0.48288, size = 109, normalized size = 1.7 \begin{align*} - \frac{3 \sqrt{- \frac{1}{a b^{5}}} \log{\left (- a b^{2} \sqrt{- \frac{1}{a b^{5}}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{a b^{5}}} \log{\left (a b^{2} \sqrt{- \frac{1}{a b^{5}}} + x \right )}}{16} - \frac{3 a x + 5 b x^{3}}{8 a^{2} b^{2} + 16 a b^{3} x^{2} + 8 b^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a*b**5))*log(-a*b**2*sqrt(-1/(a*b**5)) + x)/16 + 3*sqrt(-1/(a*b**5))*log(a*b**2*sqrt(-1/(a*b**5))
+ x)/16 - (3*a*x + 5*b*x**3)/(8*a**2*b**2 + 16*a*b**3*x**2 + 8*b**4*x**4)

________________________________________________________________________________________

Giac [A]  time = 1.36908, size = 61, normalized size = 0.95 \begin{align*} \frac{3 \, \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{2}} - \frac{5 \, b x^{3} + 3 \, a x}{8 \,{\left (b x^{2} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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