3.153 \(\int \frac {d+e x^2}{\sqrt {a+c x^4}} \, dx\)

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

[Out]

e*x*(c*x^4+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-a^(1/4)*e*(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))*(e+d*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 = 0.07, 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 = {1198, 220, 1196} \[ \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {e x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/Sqrt[a + c*x^4],x]

[Out]

(e*x*Sqrt[a + c*x^4])/(Sqrt[c]*(Sqrt[a] + Sqrt[c]*x^2)) - (a^(1/4)*e*(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)
*((Sqrt[c]*d)/Sqrt[a] + e)*(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 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]

Rule 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1198

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 {d+e x^2}{\sqrt {a+c x^4}} \, dx &=-\frac {\left (\sqrt {a} e\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{\sqrt {c}}+\left (d+\frac {\sqrt {a} e}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx\\ &=\frac {e x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} e \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 {c} d+\sqrt {a} e\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]  time = 0.03, size = 77, normalized size = 0.34 \[ \frac {\sqrt {\frac {c x^4}{a}+1} \left (3 d x \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right )+e x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right )\right )}{3 \sqrt {a+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x^{2} + d}{\sqrt {c x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{\sqrt {c x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)/sqrt(c*x^4 + a), x)

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maple [C]  time = 0.00, size = 169, normalized size = 0.75 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, d \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\frac {i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}\, e}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*e*a^(1/2)/(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)/c^(1/2)*(EllipticF((I/a^(1/2)*c^(1/2))^(1/2)*x,I)-EllipticE((I/a^(1/2)*c^(1/2))^(1/2)*x,I))+d/(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 \[ \int \frac {e x^{2} + d}{\sqrt {c x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)/sqrt(c*x^4 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {e\,x^2+d}{\sqrt {c\,x^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)/(a + c*x^4)^(1/2),x)

[Out]

int((d + e*x^2)/(a + c*x^4)^(1/2), x)

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sympy [C]  time = 2.06, size = 78, normalized size = 0.35 \[ \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*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)) + e*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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