3.1.5 \(\int \frac {1}{(a^2+2 a b x^2+b^2 x^4)^{5/4}} \, dx\) [5]

Optimal. Leaf size=68 \[ \frac {x \left (a+b x^2\right )}{3 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/4}}+\frac {2 x}{3 a^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 70, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1103, 198, 197} \begin {gather*} \frac {x}{3 a \left (a+b x^2\right ) \sqrt [4]{a^2+2 a b x^2+b^2 x^4}}+\frac {2 x}{3 a^2 \sqrt [4]{a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

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 198

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 1103

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

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Mathematica [A]
time = 0.05, size = 38, normalized size = 0.56 \begin {gather*} \frac {\left (a+b x^2\right ) \left (3 a x+2 b x^3\right )}{3 a^2 \left (\left (a+b x^2\right )^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 44, normalized size = 0.65

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right ) x \left (2 b \,x^{2}+3 a \right )}{3 a^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{4}}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/4),x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-5/4), x)

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Fricas [A]
time = 0.43, size = 58, normalized size = 0.85 \begin {gather*} \frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{\frac {1}{4}} {\left (2 \, b x^{3} + 3 \, a x\right )}}{3 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(5/4),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-5/4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/4),x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^(-5/4), x)

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Mupad [B]
time = 4.20, size = 45, normalized size = 0.66 \begin {gather*} \frac {x\,\left (2\,b\,x^2+3\,a\right )\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/4}}{3\,a^2\,{\left (b\,x^2+a\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/4),x)

[Out]

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

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