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

Optimal. Leaf size=129 \[ \frac{x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{a x \left (a+b x^2\right )}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0460382, antiderivative size = 129, 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, 302, 205} \[ \frac{x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{a x \left (a+b x^2\right )}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{a^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-((a*x*(a + b*x^2))/(b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])) + (x^3*(a + b*x^2))/(3*b*Sqrt[a^2 + 2*a*b*x^2 + b^2
*x^4]) + (a^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 0.0250801, size = 66, normalized size = 0.51 \[ \frac{\left (a+b x^2\right ) \left (3 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )+\sqrt{b} x \left (b x^2-3 a\right )\right )}{3 b^{5/2} \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.217, size = 63, normalized size = 0.5 \begin{align*}{\frac{b{x}^{2}+a}{3\,{b}^{2}} \left ( \sqrt{ab}{x}^{3}b-3\,\sqrt{ab}xa+3\,{a}^{2}\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) \right ){\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((b*x^2+a)^2)^(1/2),x)

[Out]

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

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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*x^2+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34042, size = 217, normalized size = 1.68 \begin{align*} \left [\frac{2 \, b x^{3} + 3 \, a \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right ) - 6 \, a x}{6 \, b^{2}}, \frac{b x^{3} + 3 \, a \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) - 3 \, a x}{3 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x^2+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(2*b*x^3 + 3*a*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) - 6*a*x)/b^2, 1/3*(b*x^3 + 3*a*
sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - 3*a*x)/b^2]

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Sympy [A]  time = 0.351204, size = 80, normalized size = 0.62 \begin{align*} - \frac{a x}{b^{2}} - \frac{\sqrt{- \frac{a^{3}}{b^{5}}} \log{\left (x - \frac{b^{2} \sqrt{- \frac{a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac{\sqrt{- \frac{a^{3}}{b^{5}}} \log{\left (x + \frac{b^{2} \sqrt{- \frac{a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac{x^{3}}{3 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/((b*x**2+a)**2)**(1/2),x)

[Out]

-a*x/b**2 - sqrt(-a**3/b**5)*log(x - b**2*sqrt(-a**3/b**5)/a)/2 + sqrt(-a**3/b**5)*log(x + b**2*sqrt(-a**3/b**
5)/a)/2 + x**3/(3*b)

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Giac [A]  time = 1.16783, size = 86, normalized size = 0.67 \begin{align*} \frac{a^{2} \arctan \left (\frac{b x}{\sqrt{a b}}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{a b} b^{2}} + \frac{b^{2} x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) - 3 \, a b x \mathrm{sgn}\left (b x^{2} + a\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x^2+a)^2)^(1/2),x, algorithm="giac")

[Out]

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