3.7 \(\int \frac{1}{(a^2+2 a b x^2+b^2 x^4)^{9/4}} \, dx\)

Optimal. Leaf size=135 \[ \frac{16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{8 x \left (a+b x^2\right )}{35 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac{6 x}{35 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac{x \left (a+b x^2\right )}{7 a \left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \]

[Out]

(x*(a + b*x^2))/(7*a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(9/4)) + (6*x)/(35*a^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/4)) + (
8*x*(a + b*x^2))/(35*a^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/4)) + (16*x)/(35*a^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)
)

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Rubi [A]  time = 0.0432201, antiderivative size = 148, normalized size of antiderivative = 1.1, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1089, 192, 191} \[ \frac{8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-9/4),x]

[Out]

(16*x)/(35*a^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)) + x/(7*a*(a + b*x^2)^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)) +
(6*x)/(35*a^2*(a + b*x^2)^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(1/4)) + (8*x)/(35*a^3*(a + b*x^2)*(a^2 + 2*a*b*x^2 +
b^2*x^4)^(1/4))

Rule 1089

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 + c*x^4)^FracPart[p]
)/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2
- 4*a*c, 0] &&  !IntegerQ[2*p]

Rule 192

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, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{9/4}} \, dx &=\frac{\sqrt{1+\frac{b x^2}{a}} \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{9/2}} \, dx}{a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (6 \sqrt{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{7/2}} \, dx}{7 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (24 \sqrt{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/2}} \, dx}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (16 \sqrt{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/2}} \, dx}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{16 x}{35 a^4 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{x}{7 a \left (a+b x^2\right )^3 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{6 x}{35 a^2 \left (a+b x^2\right )^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac{8 x}{35 a^3 \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0236541, size = 62, normalized size = 0.46 \[ \frac{x \left (70 a^2 b x^2+35 a^3+56 a b^2 x^4+16 b^3 x^6\right )}{35 a^4 \left (a+b x^2\right )^3 \sqrt [4]{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-9/4),x]

[Out]

(x*(35*a^3 + 70*a^2*b*x^2 + 56*a*b^2*x^4 + 16*b^3*x^6))/(35*a^4*(a + b*x^2)^3*((a + b*x^2)^2)^(1/4))

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Maple [A]  time = 0.045, size = 66, normalized size = 0.5 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) x \left ( 16\,{b}^{3}{x}^{6}+56\,{b}^{2}{x}^{4}a+70\,{a}^{2}b{x}^{2}+35\,{a}^{3} \right ) }{35\,{a}^{4}} \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{-{\frac{9}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x)

[Out]

1/35*(b*x^2+a)*x*(16*b^3*x^6+56*a*b^2*x^4+70*a^2*b*x^2+35*a^3)/a^4/(b^2*x^4+2*a*b*x^2+a^2)^(9/4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-9/4), x)

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Fricas [A]  time = 1.37772, size = 216, normalized size = 1.6 \begin{align*} \frac{{\left (16 \, b^{3} x^{7} + 56 \, a b^{2} x^{5} + 70 \, a^{2} b x^{3} + 35 \, a^{3} x\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}}}{35 \,{\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="fricas")

[Out]

1/35*(16*b^3*x^7 + 56*a*b^2*x^5 + 70*a^2*b*x^3 + 35*a^3*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)/(a^4*b^4*x^8 + 4*
a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 + a^8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac{9}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(9/4),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-9/4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{9}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(9/4),x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-9/4), x)