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

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

[Out]

1/5*x*(b*x^2+a)/a/(b^2*x^4+2*a*b*x^2+a^2)^(7/4)+4/15*x/a^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/4)+8/15*x*(b*x^2+a)/a^3/
(b^2*x^4+2*a*b*x^2+a^2)^(3/4)

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Rubi [A]  time = 0.02, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.02, 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))

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fricas [A]  time = 0.56, size = 80, normalized size = 0.76 \[ \frac {{\left (8 \, b^{2} x^{5} + 20 \, a b x^{3} + 15 \, a^{2} x\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}}}{15 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {7}{4}}}\,{d x} \]

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)

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maple [A]  time = 0.00, size = 55, normalized size = 0.52 \[ \frac {\left (b \,x^{2}+a \right ) \left (8 b^{2} x^{4}+20 a b \,x^{2}+15 a^{2}\right ) x}{15 \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {7}{4}} a^{3}} \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {7}{4}}}\,{d x} \]

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)

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mupad [B]  time = 4.21, size = 56, normalized size = 0.53 \[ \frac {x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{1/4}\,\left (15\,a^2+20\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,{\left (b\,x^2+a\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(7/4),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {7}{4}}}\, dx \]

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)

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