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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 129, normalized size of antiderivative = 1.00, 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 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 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 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^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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.71, size = 99, normalized size = 0.77 \[ \left [\frac {2 \, b x^{3} + 3 \, a \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) - 6 \, a x}{6 \, b^{2}}, \frac {b x^{3} + 3 \, a \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) - 3 \, a x}{3 \, b^{2}}\right ] \]

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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giac [A]  time = 0.16, size = 64, normalized size = 0.50 \[ \frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {a b} b^{2}} + \frac {b^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 3 \, a b x \mathrm {sgn}\left (b x^{2} + a\right )}{3 \, b^{3}} \]

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

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maple [A]  time = 0.01, size = 63, normalized size = 0.49 \[ \frac {\left (b \,x^{2}+a \right ) \left (\sqrt {a b}\, b \,x^{3}+3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )-3 \sqrt {a b}\, a x \right )}{3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {a b}\, b^{2}} \]

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(1/(a*b)^(1/2)*b*x))/((b*x^2+a)^2)^(1/2)/b^2/(a
*b)^(1/2)

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maxima [A]  time = 2.93, size = 37, normalized size = 0.29 \[ \frac {a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{2}} + \frac {b x^{3} - 3 \, a x}{3 \, b^{2}} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((a + b*x^2)^2)^(1/2),x)

[Out]

int(x^4/((a + b*x^2)^2)^(1/2), x)

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sympy [A]  time = 0.21, size = 80, normalized size = 0.62 \[ - \frac {a x}{b^{2}} - \frac {\sqrt {- \frac {a^{3}}{b^{5}}} \log {\left (x - \frac {b^{2} \sqrt {- \frac {a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac {\sqrt {- \frac {a^{3}}{b^{5}}} \log {\left (x + \frac {b^{2} \sqrt {- \frac {a^{3}}{b^{5}}}}{a} \right )}}{2} + \frac {x^{3}}{3 b} \]

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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