3.1.4 \(\int \frac {A+B x^2}{\sqrt {a+c x^4}} \, dx\) [4]

Optimal. Leaf size=226 \[ \frac {B x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} B \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} \left (B+\frac {A \sqrt {c}}{\sqrt {a}}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}} \]

[Out]

B*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-a^(1/4)*B*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*a
rctan(c^(1/4)*x/a^(1/4)))*EllipticE(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^(3/4)/(c*x^4+a)^(1/2)+1/2*a^(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)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*c^(
1/2))*(B+A*c^(1/2)/a^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^4+a)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1212, 226, 1210} \begin {gather*} \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {A \sqrt {c}}{\sqrt {a}}+B\right ) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} B \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {B x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/Sqrt[a + c*x^4],x]

[Out]

(B*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*B*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/
(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c^(3/4)*Sqrt[a + c*x^4]) + (a^(1/4)
*(B + (A*Sqrt[c])/Sqrt[a])*(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])/(2*c^(3/4)*Sqrt[a + c*x^4])

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]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1212

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{\sqrt {a+c x^4}} \, dx &=\left (A+\frac {\sqrt {a} B}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx-\frac {\left (\sqrt {a} B\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{\sqrt {c}}\\ &=\frac {B x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} B \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {\left (\sqrt {a} B+A \sqrt {c}\right ) \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 )}{2 \sqrt [4]{a} c^{3/4} \sqrt {a+c 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 = 10.04, size = 77, normalized size = 0.34 \begin {gather*} \frac {\sqrt {1+\frac {c x^4}{a}} \left (3 A x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )+B x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )\right )}{3 \sqrt {a+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/Sqrt[a + c*x^4],x]

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 0.12, size = 169, normalized size = 0.75

method	result	size
default	\(\frac {i B \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {A \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(169\)
elliptic	\(\frac {i B \sqrt {a}\, \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\frac {A \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

I*B*a^(1/2)/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a
)^(1/2)/c^(1/2)*(EllipticF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I))+A/(I/a^(1/
2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*
(I/a^(1/2)*c^(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((B*x^2+A)/(c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/sqrt(c*x^4 + a), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [C] Result contains complex when optimal does not.
time = 0.93, size = 78, normalized size = 0.35 \begin {gather*} \frac {A x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {B x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + B*x**3*gamma(3/4)*
hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(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((B*x^2+A)/(c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/sqrt(c*x^4 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {B\,x^2+A}{\sqrt {c\,x^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(a + c*x^4)^(1/2),x)

[Out]

int((A + B*x^2)/(a + c*x^4)^(1/2), x)

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