3.209 \(\int \sqrt {a+c x^4} \, dx\)

Optimal. Leaf size=105 \[ \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {1}{3} x \sqrt {a+c x^4} \]

[Out]

1/3*x*(c*x^4+a)^(1/2)+1/3*a^(3/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*
EllipticF(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)/c^(1/4)/(c*x^4+a)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 105, 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 = {195, 220} \[ \frac {a^{3/4} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {1}{3} x \sqrt {a+c x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c*x^4],x]

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[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 \sqrt {a+c x^4} \, dx &=\frac {1}{3} x \sqrt {a+c x^4}+\frac {1}{3} (2 a) \int \frac {1}{\sqrt {a+c x^4}} \, dx\\ &=\frac {1}{3} x \sqrt {a+c x^4}+\frac {a^{3/4} \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 )}{3 \sqrt [4]{c} \sqrt {a+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 89, normalized size = 0.85 \[ \frac {x \left (a+c x^4\right )-\frac {2 i a \sqrt {\frac {c x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}} x\right )\right |-1\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}}}{3 \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c*x^4],x]

[Out]

(x*(a + c*x^4) - ((2*I)*a*Sqrt[1 + (c*x^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[a]]*x], -1])/Sqrt[(I*S
qrt[c])/Sqrt[a]])/(3*Sqrt[a + c*x^4])

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{4} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + a), x)

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maple [C]  time = 0.00, size = 85, normalized size = 0.81 \[ \frac {2 \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {\sqrt {c \,x^{4}+a}\, x}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(c*x^4+a)^(1/2)+2/3*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 \[ \int \sqrt {c x^{4} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*x^4 + a), x)

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mupad [B]  time = 2.64, size = 37, normalized size = 0.35 \[ \frac {x\,\sqrt {c\,x^4+a}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {c\,x^4}{a}\right )}{\sqrt {\frac {c\,x^4}{a}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(1/2),x)

[Out]

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

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sympy [C]  time = 0.79, size = 37, normalized size = 0.35 \[ \frac {\sqrt {a} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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