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]

1/2*(cos(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(d*x+c)/a^(1/4)))*EllipticF(sin(2*ar
ctan(b^(1/4)*(d*x+c)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+d^2*(c/d+x)^2*b^(1/2))*((a+b*d^4*(c/d+x)^4)/(a^(1/2)+d^2*
(c/d+x)^2*b^(1/2))^2)^(1/2)/a^(1/4)/b^(1/4)/d/(a+b*d^4*(c/d+x)^4)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [C]  time = 0.06, 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])

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}\,{d x} \]

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)

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maple [C]  time = 0.06, size = 1036, normalized size = 7.91 \[ \text {result too large to display} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4} + a}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {b\,c^4+4\,b\,c^3\,d\,x+6\,b\,c^2\,d^2\,x^2+4\,b\,c\,d^3\,x^3+b\,d^4\,x^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b c^{4} + 4 b c^{3} d x + 6 b c^{2} d^{2} x^{2} + 4 b c d^{3} x^{3} + b d^{4} x^{4}}}\, dx \]

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)

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