3.1003 \(\int \frac{1}{\sqrt{a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx\)

Optimal. Leaf size=131 \[ \frac{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right ) \sqrt{\frac{a+b d^4 \left (\frac{c}{d}+x\right )^4}{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d \sqrt{a+b d^4 \left (\frac{c}{d}+x\right )^4}} \]

[Out]

((Sqrt[a] + Sqrt[b]*d^2*(c/d + x)^2)*Sqrt[(a + b*d^4*(c/d + x)^4)/(Sqrt[a] + Sqrt[b]*d^2*(c/d + x)^2)^2]*Ellip
ticF[2*ArcTan[(b^(1/4)*(c + d*x))/a^(1/4)], 1/2])/(2*a^(1/4)*b^(1/4)*d*Sqrt[a + b*d^4*(c/d + x)^4])

________________________________________________________________________________________

Rubi [A]  time = 0.100435, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.041, Rules used = {1106, 220} \[ \frac{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right ) \sqrt{\frac{a+b d^4 \left (\frac{c}{d}+x\right )^4}{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d \sqrt{a+b d^4 \left (\frac{c}{d}+x\right )^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*c^4 + 4*b*c^3*d*x + 6*b*c^2*d^2*x^2 + 4*b*c*d^3*x^3 + b*d^4*x^4],x]

[Out]

((Sqrt[a] + Sqrt[b]*d^2*(c/d + x)^2)*Sqrt[(a + b*d^4*(c/d + x)^4)/(Sqrt[a] + Sqrt[b]*d^2*(c/d + x)^2)^2]*Ellip
ticF[2*ArcTan[(b^(1/4)*(c + d*x))/a^(1/4)], 1/2])/(2*a^(1/4)*b^(1/4)*d*Sqrt[a + b*d^4*(c/d + 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 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}{\sqrt{a+b c^4+4 b c^3 d x+6 b c^2 d^2 x^2+4 b c d^3 x^3+b d^4 x^4}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b d^4 x^4}} \, dx,x,\frac{c}{d}+x\right )\\ &=\frac{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right ) \sqrt{\frac{a+b d^4 \left (\frac{c}{d}+x\right )^4}{\left (\sqrt{a}+\sqrt{b} d^2 \left (\frac{c}{d}+x\right )^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{b} d \sqrt{a+b d^4 \left (\frac{c}{d}+x\right )^4}}\\ \end{align*}

Mathematica [C]  time = 0.0573804, size = 90, normalized size = 0.69 \[ -\frac{i \sqrt{\frac{a+b (c+d x)^4}{a}} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} (c+d x)\right )\right |-1\right )}{d \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b (c+d x)^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*c^4 + 4*b*c^3*d*x + 6*b*c^2*d^2*x^2 + 4*b*c*d^3*x^3 + b*d^4*x^4],x]

[Out]

((-I)*Sqrt[(a + b*(c + d*x)^4)/a]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*(c + d*x)], -1])/(Sqrt[(I*Sqrt
[b])/Sqrt[a]]*d*Sqrt[a + b*(c + d*x)^4])

________________________________________________________________________________________

Maple [C]  time = 0.019, size = 1036, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x)

[Out]

2*((1/b*(-a*b^3)^(1/4)-c)/d-(-I/b*(-a*b^3)^(1/4)-c)/d)*(((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)*(
x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))
^(1/2)*(x-(I/b*(-a*b^3)^(1/4)-c)/d)^2*(((I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-1/b*(-a*b^3)^(
1/4)-c)/d)/((-1/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*(((I/b*(-a
*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b
^3)^(1/4)-c)/d)/(x-(I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)/((-I/b*(-a*b^3)^(1/4)-c)/d-(I/b*(-a*b^3)^(1/4)-c)/d)/((I/b
*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(b*d^4*(x-(1/b*(-a*b^3)^(1/4)-c)/d)*(x-(I/b*(-a*b^3)^(1/4)-c)/d
)*(x-(-1/b*(-a*b^3)^(1/4)-c)/d)*(x-(-I/b*(-a*b^3)^(1/4)-c)/d))^(1/2)*EllipticF((((-I/b*(-a*b^3)^(1/4)-c)/d-(I/
b*(-a*b^3)^(1/4)-c)/d)*(x-(1/b*(-a*b^3)^(1/4)-c)/d)/((-I/b*(-a*b^3)^(1/4)-c)/d-(1/b*(-a*b^3)^(1/4)-c)/d)/(x-(I
/b*(-a*b^3)^(1/4)-c)/d))^(1/2),(((I/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)*((1/b*(-a*b^3)^(1/4)-c)/d
-(-I/b*(-a*b^3)^(1/4)-c)/d)/((1/b*(-a*b^3)^(1/4)-c)/d-(-1/b*(-a*b^3)^(1/4)-c)/d)/((I/b*(-a*b^3)^(1/4)-c)/d-(-I
/b*(-a*b^3)^(1/4)-c)/d))^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*d**4*x**4+4*b*c*d**3*x**3+6*b*c**2*d**2*x**2+4*b*c**3*d*x+b*c**4+a)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*c**4 + 4*b*c**3*d*x + 6*b*c**2*d**2*x**2 + 4*b*c*d**3*x**3 + b*d**4*x**4), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*d^4*x^4+4*b*c*d^3*x^3+6*b*c^2*d^2*x^2+4*b*c^3*d*x+b*c^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + a), x)