3.222 \(\int \frac {1}{(a+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=108 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {x}{2 a \sqrt {a+c x^4}} \]

[Out]

1/2*x/a/(c*x^4+a)^(1/2)+1/4*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*Ellipt
icF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1
/2)/a^(5/4)/c^(1/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {199, 220} \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {x}{2 a \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(-3/2),x]

[Out]

x/(2*a*Sqrt[a + c*x^4]) + ((Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*Arc
Tan[(c^(1/4)*x)/a^(1/4)], 1/2])/(4*a^(5/4)*c^(1/4)*Sqrt[a + c*x^4])

Rule 199

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

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 55, normalized size = 0.51 \[ \frac {x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )+x}{2 a \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(-3/2),x]

[Out]

(x + x*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^4)/a)])/(2*a*Sqrt[a + c*x^4])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + a}}{c^{2} x^{8} + 2 \, a c x^{4} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + a)/(c^2*x^8 + 2*a*c*x^4 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + a)^(-3/2), x)

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maple [C]  time = 0.00, size = 94, normalized size = 0.87 \[ \frac {x}{2 \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}\, a}+\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{2 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^4+a)^(3/2),x)

[Out]

1/2/((x^4+a/c)*c)^(1/2)/a*x+1/2/a/(I/a^(1/2)*c^(1/2))^(1/2)*(-I/a^(1/2)*c^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*c^(1/2
)*x^2+1)^(1/2)/(c*x^4+a)^(1/2)*EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + a)^(-3/2), x)

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mupad [B]  time = 2.66, size = 37, normalized size = 0.34 \[ \frac {x\,{\left (\frac {c\,x^4}{a}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{2};\ \frac {5}{4};\ -\frac {c\,x^4}{a}\right )}{{\left (c\,x^4+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + c*x^4)^(3/2),x)

[Out]

(x*((c*x^4)/a + 1)^(3/2)*hypergeom([1/4, 3/2], 5/4, -(c*x^4)/a))/(a + c*x^4)^(3/2)

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sympy [C]  time = 0.81, size = 36, normalized size = 0.33 \[ \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**4+a)**(3/2),x)

[Out]

x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(5/4))

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