3.6 \(\int \frac{1}{(a^2+2 a b x^2+b^2 x^4)^{7/4}} \, dx\)

Optimal. Leaf size=105 \[ \frac{8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{x \left (a+b x^2\right )}{5 a \left (a^2+2 a b x^2+b^2 x^4\right )^{7/4}} \]

[Out]

(x*(a + b*x^2))/(5*a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(7/4)) + (4*x)/(15*a^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)) + (
8*x*(a + b*x^2))/(15*a^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0236083, antiderivative size = 107, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1089, 192, 191} \[ \frac{8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-7/4),x]

[Out]

(4*x)/(15*a^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)) + x/(5*a*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4)) + (8*
x*(a + b*x^2))/(15*a^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/4))

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 )^{7/4}} \, dx &=\frac{\left (1+\frac{b x^2}{a}\right )^{3/2} \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{7/2}} \, dx}{a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac{x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{\left (4 \left (1+\frac{b x^2}{a}\right )^{3/2}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/2}} \, dx}{5 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac{4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{\left (8 \left (1+\frac{b x^2}{a}\right )^{3/2}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{3/2}} \, dx}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ &=\frac{4 x}{15 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{x}{5 a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}+\frac{8 x \left (a+b x^2\right )}{15 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0150957, size = 51, normalized size = 0.49 \[ \frac{x \left (15 a^2+20 a b x^2+8 b^2 x^4\right )}{15 a^3 \left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-7/4),x]

[Out]

(x*(15*a^2 + 20*a*b*x^2 + 8*b^2*x^4))/(15*a^3*(a + b*x^2)*((a + b*x^2)^2)^(3/4))

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 55, normalized size = 0.5 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) x \left ( 8\,{b}^{2}{x}^{4}+20\,ab{x}^{2}+15\,{a}^{2} \right ) }{15\,{a}^{3}} \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{-{\frac{7}{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)^(7/4),x)

[Out]

1/15*(b*x^2+a)*x*(8*b^2*x^4+20*a*b*x^2+15*a^2)/a^3/(b^2*x^4+2*a*b*x^2+a^2)^(7/4)

________________________________________________________________________________________

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{7}{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)^(7/4),x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-7/4), x)

________________________________________________________________________________________

Fricas [A]  time = 1.42588, size = 170, normalized size = 1.62 \begin{align*} \frac{{\left (8 \, b^{2} x^{5} + 20 \, a b x^{3} + 15 \, a^{2} x\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac{1}{4}}}{15 \,{\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\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)^(7/4),x, algorithm="fricas")

[Out]

1/15*(8*b^2*x^5 + 20*a*b*x^3 + 15*a^2*x)*(b^2*x^4 + 2*a*b*x^2 + a^2)^(1/4)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^
5*b*x^2 + a^6)

________________________________________________________________________________________

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{7}{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)**(7/4),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-7/4), x)

________________________________________________________________________________________

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{7}{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)^(7/4),x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-7/4), x)