3.9.57 \(\int \frac {1}{(a+b x^2)^{7/4}} \, dx\) [857]

Optimal. Leaf size=78 \[ \frac {2 x}{3 a \left (a+b x^2\right )^{3/4}}+\frac {2 \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \sqrt {b} \left (a+b x^2\right )^{3/4}} \]

[Out]

2/3*x/a/(b*x^2+a)^(3/4)+2/3*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^
(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^2+a)^(3/4)/a^(1/2)/b^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {205, 239, 237} \begin {gather*} \frac {2 \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {2 x}{3 a \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(-7/4),x]

[Out]

(2*x)/(3*a*(a + b*x^2)^(3/4)) + (2*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*Sqrt[
a]*Sqrt[b]*(a + b*x^2)^(3/4))

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + b*(x^2
/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^2\right )^{7/4}} \, dx &=\frac {2 x}{3 a \left (a+b x^2\right )^{3/4}}+\frac {\int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{3 a}\\ &=\frac {2 x}{3 a \left (a+b x^2\right )^{3/4}}+\frac {\left (1+\frac {b x^2}{a}\right )^{3/4} \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{3 a \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 x}{3 a \left (a+b x^2\right )^{3/4}}+\frac {2 \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 \sqrt {a} \sqrt {b} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.94, size = 55, normalized size = 0.71 \begin {gather*} \frac {x \left (2+\left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};-\frac {b x^2}{a}\right )\right )}{3 a \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(-7/4),x]

[Out]

(x*(2 + (1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b*x^2)/a)]))/(3*a*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(-7/4), x)

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Fricas [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*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.52, size = 24, normalized size = 0.31 \begin {gather*} \frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {7}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*hyper((1/2, 7/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(7/4)

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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*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(-7/4), x)

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Mupad [B]
time = 4.88, size = 37, normalized size = 0.47 \begin {gather*} \frac {x\,{\left (\frac {b\,x^2}{a}+1\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{7/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^2)^(7/4),x)

[Out]

(x*((b*x^2)/a + 1)^(7/4)*hypergeom([1/2, 7/4], 3/2, -(b*x^2)/a))/(a + b*x^2)^(7/4)

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