3.9.73 \(\int \frac {1}{(a+b x^4)^{5/2}} \, dx\) [873]

Optimal. Leaf size=127 \[ \frac {x}{6 a \left (a+b x^4\right )^{3/2}}+\frac {5 x}{12 a^2 \sqrt {a+b x^4}}+\frac {5 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt {a+b x^4}} \]

[Out]

1/6*x/a/(b*x^4+a)^(3/2)+5/12*x/a^2/(b*x^4+a)^(1/2)+5/24*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arcta
n(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)
/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(9/4)/b^(1/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {205, 226} \begin {gather*} \frac {5 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {5 x}{12 a^2 \sqrt {a+b x^4}}+\frac {x}{6 a \left (a+b x^4\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(-5/2),x]

[Out]

x/(6*a*(a + b*x^4)^(3/2)) + (5*x)/(12*a^2*Sqrt[a + b*x^4]) + (5*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt
[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(24*a^(9/4)*b^(1/4)*Sqrt[a + b*x^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 226

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x^4\right )^{5/2}} \, dx &=\frac {x}{6 a \left (a+b x^4\right )^{3/2}}+\frac {5 \int \frac {1}{\left (a+b x^4\right )^{3/2}} \, dx}{6 a}\\ &=\frac {x}{6 a \left (a+b x^4\right )^{3/2}}+\frac {5 x}{12 a^2 \sqrt {a+b x^4}}+\frac {5 \int \frac {1}{\sqrt {a+b x^4}} \, dx}{12 a^2}\\ &=\frac {x}{6 a \left (a+b x^4\right )^{3/2}}+\frac {5 x}{12 a^2 \sqrt {a+b x^4}}+\frac {5 \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt {a+b x^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 = 10.03, size = 72, normalized size = 0.57 \begin {gather*} \frac {7 a x+5 b x^5+5 x \left (a+b x^4\right ) \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )}{12 a^2 \left (a+b x^4\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(-5/2),x]

[Out]

(7*a*x + 5*b*x^5 + 5*x*(a + b*x^4)*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(12*a^2
*(a + b*x^4)^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 0.15, size = 123, normalized size = 0.97

method	result	size
default	\(\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(123\)
elliptic	\(\frac {x \sqrt {b \,x^{4}+a}}{6 a \,b^{2} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {5 x}{12 a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {5 \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{12 a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a*x/b^2*(b*x^4+a)^(1/2)/(x^4+a/b)^2+5/12/a^2*x/((x^4+a/b)*b)^(1/2)+5/12/a^2/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I
/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*b^(1/2))^(1
/2),I)

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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^4+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(-5/2), x)

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Fricas [A]
time = 0.07, size = 101, normalized size = 0.80 \begin {gather*} -\frac {5 \, {\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, b^{2} x^{5} + 7 \, a b x\right )} \sqrt {b x^{4} + a}}{12 \, {\left (a^{2} b^{3} x^{8} + 2 \, a^{3} b^{2} x^{4} + a^{4} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(5*(b^2*x^8 + 2*a*b*x^4 + a^2)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin(x*(-b/a)^(1/4)), -1) - (5*b^2*x^5
+ 7*a*b*x)*sqrt(b*x^4 + a))/(a^2*b^3*x^8 + 2*a^3*b^2*x^4 + a^4*b)

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Sympy [C] Result contains complex when optimal does not.
time = 0.44, size = 36, normalized size = 0.28 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/2)*gamma(5/4))

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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^4+a)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(-5/2), x)

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Mupad [B]
time = 1.16, size = 37, normalized size = 0.29 \begin {gather*} \frac {x\,{\left (\frac {b\,x^4}{a}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {5}{2};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (b\,x^4+a\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^4)^(5/2),x)

[Out]

(x*((b*x^4)/a + 1)^(5/2)*hypergeom([1/4, 5/2], 5/4, -(b*x^4)/a))/(a + b*x^4)^(5/2)

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