3.776 \(\int \frac{1}{\sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \, dx\)

Optimal. Leaf size=227 \[ \frac{\sqrt [4]{4 a d^2+c^3} \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{2 \sqrt [4]{c} d \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.173098, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {1106, 1103} \[ \frac{\sqrt [4]{4 a d^2+c^3} \sqrt{\frac{d^2 \left (4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4\right )}{\left (4 a d^2+c^3\right ) \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right )^2}} \left (\frac{d^2 \left (\frac{c}{d}+x\right )^2}{\sqrt{4 a d^2+c^3}}+\sqrt{c}\right ) F\left (2 \tan ^{-1}\left (\frac{c+d x}{\sqrt [4]{c} \sqrt [4]{c^3+4 a d^2}}\right )|\frac{1}{2} \left (\frac{c^{3/2}}{\sqrt{c^3+4 a d^2}}+1\right )\right )}{2 \sqrt [4]{c} d \sqrt{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4],x]

[Out]

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

Rule 1106

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - (b*d)/(8*e) + (c - (3*d^2)/(8*
e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 2.04817, size = 822, normalized size = 3.62 \[ \frac{2 \left (-c-d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right ) \left (c+d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right ) \sqrt{-\frac{\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d} \left (c+d x-\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right )}{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}+\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (-c-d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}} \sqrt{-\frac{\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d} \left (c+d x+\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right )}{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}-\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (-c-d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}} F\left (\sin ^{-1}\left (\sqrt{\frac{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}-\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (c+d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}+\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (-c-d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}}\right )|\frac{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}+\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right )^2}{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}-\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right )^2}\right )}{d \sqrt{c^2-2 i \sqrt{a} \sqrt{c} d} \sqrt{\frac{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}-\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (c+d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}{\left (\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}+\sqrt{c^2+2 i \sqrt{a} d \sqrt{c}}\right ) \left (-c-d x+\sqrt{c^2-2 i \sqrt{a} \sqrt{c} d}\right )}} \sqrt{x^2 (2 c+d x)^2+4 a c}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4],x]

[Out]

(2*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x)*(c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x)*Sqrt[-((Sq
rt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*(c - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*
Sqrt[c]*d] + Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x)))]*Sqrt[-((
Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*(c + Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a
]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x)))]*Ellipt
icF[ArcSin[Sqrt[((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(c + Sqrt[c^2 - (
2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])*(
-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x))]], (Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + Sqrt[c^2 + (2*I)*Sq
rt[a]*Sqrt[c]*d])^2/(Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])^2])/(d*Sqrt[c^
2 - (2*I)*Sqrt[a]*Sqrt[c]*d]*Sqrt[((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - Sqrt[c^2 + (2*I)*Sqrt[a]*Sqrt[c]*d])
*(c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + d*x))/((Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] + Sqrt[c^2 + (2*I)*Sqr
t[a]*Sqrt[c]*d])*(-c + Sqrt[c^2 - (2*I)*Sqrt[a]*Sqrt[c]*d] - d*x))]*Sqrt[4*a*c + x^2*(2*c + d*x)^2])

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Maple [B]  time = 0.026, size = 1056, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/sqrt(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 a c + 4 c^{2} x^{2} + 4 c d x^{3} + d^{2} x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(4*a*c + 4*c**2*x**2 + 4*c*d*x**3 + d**2*x**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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