3.654 \(\int \frac{x^4}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=212 \[ -\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{64 a b^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 x}{128 a^2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{5/2} b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

(3*x)/(128*a^2*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - x^3/(8*b*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
- x/(16*b^2*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + x/(64*a*b^2*(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]])/(128*a^(5/2)*b^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0878467, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1112, 288, 199, 205} \[ -\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{64 a b^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 x}{128 a^2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{3 \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{5/2} b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(3*x)/(128*a^2*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - x^3/(8*b*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
- x/(16*b^2*(a + b*x^2)^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + x/(64*a*b^2*(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]])/(128*a^(5/2)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^2\right )\right ) \int \frac{x^4}{\left (a b+b^2 x^2\right )^5} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (3 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{x^2}{\left (a b+b^2 x^2\right )^4} \, dx}{8 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a b+b^2 x^2\right ) \int \frac{1}{\left (a b+b^2 x^2\right )^3} \, dx}{16 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{64 a b^2 \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}{\left (a b+b^2 x^2\right )^2} \, dx}{64 a b \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 x}{128 a^2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{64 a b^2 \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}{128 a^2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{3 x}{128 a^2 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x^3}{8 b \left (a+b x^2\right )^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x}{16 b^2 \left (a+b x^2\right )^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{64 a b^2 \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 )}{128 a^{5/2} b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.04082, size = 105, normalized size = 0.5 \[ \frac{\sqrt{a} \sqrt{b} x \left (-11 a^2 b x^2-3 a^3+11 a b^2 x^4+3 b^3 x^6\right )+3 \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{128 a^{5/2} b^{5/2} \left (a+b x^2\right )^3 \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]

[Out]

(Sqrt[a]*Sqrt[b]*x*(-3*a^3 - 11*a^2*b*x^2 + 11*a*b^2*x^4 + 3*b^3*x^6) + 3*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqr
t[a]])/(128*a^(5/2)*b^(5/2)*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.228, size = 172, normalized size = 0.8 \begin{align*}{\frac{b{x}^{2}+a}{128\,{b}^{2}{a}^{2}} \left ( 3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{8}{b}^{4}+3\,\sqrt{ab}{x}^{7}{b}^{3}+12\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{6}a{b}^{3}+11\,\sqrt{ab}{x}^{5}a{b}^{2}+18\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{4}{a}^{2}{b}^{2}-11\,\sqrt{ab}{x}^{3}{a}^{2}b+12\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}{a}^{3}b-3\,\sqrt{ab}x{a}^{3}+3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){a}^{4} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

1/128*(3*arctan(b*x/(a*b)^(1/2))*x^8*b^4+3*(a*b)^(1/2)*x^7*b^3+12*arctan(b*x/(a*b)^(1/2))*x^6*a*b^3+11*(a*b)^(
1/2)*x^5*a*b^2+18*arctan(b*x/(a*b)^(1/2))*x^4*a^2*b^2-11*(a*b)^(1/2)*x^3*a^2*b+12*arctan(b*x/(a*b)^(1/2))*x^2*
a^3*b-3*(a*b)^(1/2)*x*a^3+3*arctan(b*x/(a*b)^(1/2))*a^4)*(b*x^2+a)/(a*b)^(1/2)/a^2/b^2/((b*x^2+a)^2)^(5/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52245, size = 674, normalized size = 3.18 \begin{align*} \left [\frac{6 \, a b^{4} x^{7} + 22 \, a^{2} b^{3} x^{5} - 22 \, a^{3} b^{2} x^{3} - 6 \, a^{4} b x - 3 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{256 \,{\left (a^{3} b^{7} x^{8} + 4 \, a^{4} b^{6} x^{6} + 6 \, a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{2} + a^{7} b^{3}\right )}}, \frac{3 \, a b^{4} x^{7} + 11 \, a^{2} b^{3} x^{5} - 11 \, a^{3} b^{2} x^{3} - 3 \, a^{4} b x + 3 \,{\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{128 \,{\left (a^{3} b^{7} x^{8} + 4 \, a^{4} b^{6} x^{6} + 6 \, a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{2} + a^{7} 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)^(5/2),x, algorithm="fricas")

[Out]

[1/256*(6*a*b^4*x^7 + 22*a^2*b^3*x^5 - 22*a^3*b^2*x^3 - 6*a^4*b*x - 3*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 +
4*a^3*b*x^2 + a^4)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^3*b^7*x^8 + 4*a^4*b^6*x^6 + 6
*a^5*b^5*x^4 + 4*a^6*b^4*x^2 + a^7*b^3), 1/128*(3*a*b^4*x^7 + 11*a^2*b^3*x^5 - 11*a^3*b^2*x^3 - 3*a^4*b*x + 3*
(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^3*b^7*x^8 + 4*
a^4*b^6*x^6 + 6*a^5*b^5*x^4 + 4*a^6*b^4*x^2 + a^7*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{5}{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)**(5/2),x)

[Out]

Integral(x**4/((a + b*x**2)**2)**(5/2), 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)^(5/2),x, algorithm="giac")

[Out]

sage0*x