3.1.2 \(\int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx\)

Optimal. Leaf size=145 \[ \frac {\sqrt {2 \sqrt {a} \sqrt {c}+b} \tanh ^{-1}\left (\frac {x \sqrt {2 \sqrt {a} \sqrt {c}+b}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}-\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} \tanh ^{-1}\left (\frac {x \sqrt {b-2 \sqrt {a} \sqrt {c}}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \]

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2071, 1093, 208} \begin {gather*} \frac {\sqrt {2 \sqrt {a} \sqrt {c}+b} \tanh ^{-1}\left (\frac {x \sqrt {2 \sqrt {a} \sqrt {c}+b}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}-\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} \tanh ^{-1}\left (\frac {x \sqrt {b-2 \sqrt {a} \sqrt {c}}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/(a*d - c*d*x^4),x]

[Out]

-(Sqrt[b - 2*Sqrt[a]*Sqrt[c]]*ArcTanh[(Sqrt[b - 2*Sqrt[a]*Sqrt[c]]*x)/Sqrt[a + b*x^2 + c*x^4]])/(4*Sqrt[a]*Sqr
t[c]*d) + (Sqrt[b + 2*Sqrt[a]*Sqrt[c]]*ArcTanh[(Sqrt[b + 2*Sqrt[a]*Sqrt[c]]*x)/Sqrt[a + b*x^2 + c*x^4]])/(4*Sq
rt[a]*Sqrt[c]*d)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 1093

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

Rule 2071

Int[Sqrt[v_]/((d_) + (e_.)*(x_)^4), x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coeff[v, x,
4]}, Dist[a/d, Subst[Int[1/(1 - 2*b*x^2 + (b^2 - 4*a*c)*x^4), x], x, x/Sqrt[v]], x] /; EqQ[c*d + a*e, 0] && P
osQ[a*c]] /; FreeQ[{d, e}, x] && PolyQ[v, x^2, 2]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-2 b x^2+\left (b^2-4 a c\right ) x^4} \, dx,x,\frac {x}{\sqrt {a+b x^2+c x^4}}\right )}{d}\\ &=\frac {\left (b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{-b-2 \sqrt {a} \sqrt {c}+\left (b^2-4 a c\right ) x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}-\frac {\left (b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{-b+2 \sqrt {a} \sqrt {c}+\left (b^2-4 a c\right ) x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}\\ &=-\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}+\frac {\sqrt {b+2 \sqrt {a} \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {b+2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.61, size = 441, normalized size = 3.04 \begin {gather*} \frac {i \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \left (2 \sqrt {a} \sqrt {c} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\left (2 \sqrt {a} \sqrt {c}+b\right ) \Pi \left (\frac {-b-\sqrt {b^2-4 a c}}{2 \sqrt {a} \sqrt {c}};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+\left (b-2 \sqrt {a} \sqrt {c}\right ) \Pi \left (\frac {b+\sqrt {b^2-4 a c}}{2 \sqrt {a} \sqrt {c}};i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 \sqrt {2} \sqrt {a} \sqrt {c} d \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*
c])]*(2*Sqrt[a]*Sqrt[c]*EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c]
)/(b - Sqrt[b^2 - 4*a*c])] - (b + 2*Sqrt[a]*Sqrt[c])*EllipticPi[(-b - Sqrt[b^2 - 4*a*c])/(2*Sqrt[a]*Sqrt[c]),
I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] + (b -
2*Sqrt[a]*Sqrt[c])*EllipticPi[(b + Sqrt[b^2 - 4*a*c])/(2*Sqrt[a]*Sqrt[c]), I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[
b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])]))/(Sqrt[2]*Sqrt[a]*Sqrt[c]*Sqrt[c/(b + Sqr
t[b^2 - 4*a*c])]*d*Sqrt[a + b*x^2 + c*x^4])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.76, size = 153, normalized size = 1.06 \begin {gather*} \frac {\sqrt {2 \sqrt {a} \sqrt {c}-b} \tan ^{-1}\left (\frac {x \sqrt {2 \sqrt {a} \sqrt {c}-b}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}-\frac {\sqrt {-2 \sqrt {a} \sqrt {c}-b} \tan ^{-1}\left (\frac {x \sqrt {-2 \sqrt {a} \sqrt {c}-b}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x^2 + c*x^4]/(a*d - c*d*x^4),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 9.59, size = 603, normalized size = 4.16 \begin {gather*} \frac {1}{8} \, \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}} \log \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} + x^{2}\right )} + {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} + a d x\right )} \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}}}{c x^{4} - a}\right ) - \frac {1}{8} \, \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}} \log \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} + x^{2}\right )} - {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} + a d x\right )} \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}}}{c x^{4} - a}\right ) + \frac {1}{8} \, \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}} \log \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} - x^{2}\right )} + {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} - a d x\right )} \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}}}{c x^{4} - a}\right ) - \frac {1}{8} \, \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}} \log \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} - x^{2}\right )} - {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} - a d x\right )} \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}}}{c x^{4} - a}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt((2*a*c*d^2*sqrt(1/(a*c*d^4)) + b)/(a*c*d^2))*log((sqrt(c*x^4 + b*x^2 + a)*(a*d^2*sqrt(1/(a*c*d^4)) +
x^2) + (a*c*d^3*x^3*sqrt(1/(a*c*d^4)) + a*d*x)*sqrt((2*a*c*d^2*sqrt(1/(a*c*d^4)) + b)/(a*c*d^2)))/(c*x^4 - a))
- 1/8*sqrt((2*a*c*d^2*sqrt(1/(a*c*d^4)) + b)/(a*c*d^2))*log((sqrt(c*x^4 + b*x^2 + a)*(a*d^2*sqrt(1/(a*c*d^4))
+ x^2) - (a*c*d^3*x^3*sqrt(1/(a*c*d^4)) + a*d*x)*sqrt((2*a*c*d^2*sqrt(1/(a*c*d^4)) + b)/(a*c*d^2)))/(c*x^4 -
a)) + 1/8*sqrt(-(2*a*c*d^2*sqrt(1/(a*c*d^4)) - b)/(a*c*d^2))*log(-(sqrt(c*x^4 + b*x^2 + a)*(a*d^2*sqrt(1/(a*c*
d^4)) - x^2) + (a*c*d^3*x^3*sqrt(1/(a*c*d^4)) - a*d*x)*sqrt(-(2*a*c*d^2*sqrt(1/(a*c*d^4)) - b)/(a*c*d^2)))/(c*
x^4 - a)) - 1/8*sqrt(-(2*a*c*d^2*sqrt(1/(a*c*d^4)) - b)/(a*c*d^2))*log(-(sqrt(c*x^4 + b*x^2 + a)*(a*d^2*sqrt(1
/(a*c*d^4)) - x^2) - (a*c*d^3*x^3*sqrt(1/(a*c*d^4)) - a*d*x)*sqrt(-(2*a*c*d^2*sqrt(1/(a*c*d^4)) - b)/(a*c*d^2)
))/(c*x^4 - a))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {\sqrt {c x^{4} + b x^{2} + a}}{c d x^{4} - a d}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.04, size = 238, normalized size = 1.64 \begin {gather*} -\frac {\sqrt {2}\, b \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{\sqrt {-2 b -4 \sqrt {a c}}\, x}\right )}{4 \sqrt {a c}\, \sqrt {-2 b -4 \sqrt {a c}}\, d}+\frac {\sqrt {2}\, b \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{\sqrt {-2 b +4 \sqrt {a c}}\, x}\right )}{4 \sqrt {a c}\, \sqrt {-2 b +4 \sqrt {a c}}\, d}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{\sqrt {-2 b -4 \sqrt {a c}}\, x}\right )}{2 \sqrt {-2 b -4 \sqrt {a c}}\, d}-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{\sqrt {-2 b +4 \sqrt {a c}}\, x}\right )}{2 \sqrt {-2 b +4 \sqrt {a c}}\, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*2^(1/2)/(-4*(a*c)^(1/2)-2*b)^(1/2)*arctan((c*x^4+b*x^2+a)^(1/2)*2^(1/2)/x/(-4*(a*c)^(1/2)-2*b)^(1/2))-1
/4/d*2^(1/2)/(a*c)^(1/2)/(-4*(a*c)^(1/2)-2*b)^(1/2)*arctan((c*x^4+b*x^2+a)^(1/2)*2^(1/2)/x/(-4*(a*c)^(1/2)-2*b
)^(1/2))*b-1/2/d*2^(1/2)/(4*(a*c)^(1/2)-2*b)^(1/2)*arctan((c*x^4+b*x^2+a)^(1/2)*2^(1/2)/x/(4*(a*c)^(1/2)-2*b)^
(1/2))+1/4/d*2^(1/2)/(a*c)^(1/2)/(4*(a*c)^(1/2)-2*b)^(1/2)*arctan((c*x^4+b*x^2+a)^(1/2)*2^(1/2)/x/(4*(a*c)^(1/
2)-2*b)^(1/2))*b

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {\sqrt {c x^{4} + b x^{2} + a}}{c d x^{4} - a d}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt {a + b x^{2} + c x^{4}}}{- a + c x^{4}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(sqrt(a + b*x**2 + c*x**4)/(-a + c*x**4), x)/d

________________________________________________________________________________________