\(\int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx\) [35]

Optimal result

Integrand size = 30, antiderivative size = 145 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=-\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} \text {arctanh}\left (\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}+\frac {\sqrt {b+2 \sqrt {a} \sqrt {c}} \text {arctanh}\left (\frac {\sqrt {b+2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2096, 1107, 214} \[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\frac {\sqrt {2 \sqrt {a} \sqrt {c}+b} \text {arctanh}\left (\frac {x \sqrt {2 \sqrt {a} \sqrt {c}+b}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d}-\frac {\sqrt {b-2 \sqrt {a} \sqrt {c}} \text {arctanh}\left (\frac {x \sqrt {b-2 \sqrt {a} \sqrt {c}}}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \]

[In]

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

Rule 214

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

Rule 1107

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 2096

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*} \text {integral}& = \frac {\text {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 ) \text {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 ) \text {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 [A] (verified)

Time = 0.91 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\frac {-\sqrt {-b-2 \sqrt {a} \sqrt {c}} \arctan \left (\frac {\sqrt {-b-2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )+\sqrt {-b+2 \sqrt {a} \sqrt {c}} \arctan \left (\frac {\sqrt {-b+2 \sqrt {a} \sqrt {c}} x}{\sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a} \sqrt {c} d} \]

[In]

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

[Out]

(-(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[-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)

Maple [A] (verified)

Time = 1.67 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(-\frac {\left (2 \sqrt {a c}+b \right ) \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {-2 \sqrt {a c}-b}}\right )+\sqrt {2 \sqrt {a c}-b}\, \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{x \sqrt {2 \sqrt {a c}-b}}\right ) \sqrt {-2 \sqrt {a c}-b}}{4 \sqrt {-2 \sqrt {a c}-b}\, \sqrt {a c}\, d}\)	\(126\)
default	\(\frac {\left (-\frac {\left (2 \sqrt {a c}+b \right ) \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{x \sqrt {-4 \sqrt {a c}-2 b}}\right )}{2 \sqrt {a c}\, \sqrt {-4 \sqrt {a c}-2 b}}-\frac {\left (2 \sqrt {a c}-b \right ) \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{x \sqrt {4 \sqrt {a c}-2 b}}\right )}{2 \sqrt {a c}\, \sqrt {4 \sqrt {a c}-2 b}}\right ) \sqrt {2}}{2 d}\)	\(140\)
elliptic	\(\frac {2 \left (-\frac {\left (2 \sqrt {a c}+b \right ) \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{x \sqrt {-4 \sqrt {a c}-2 b}}\right )}{8 \sqrt {a c}\, \sqrt {-4 \sqrt {a c}-2 b}}-\frac {\left (2 \sqrt {a c}-b \right ) \arctan \left (\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {2}}{x \sqrt {4 \sqrt {a c}-2 b}}\right )}{8 \sqrt {a c}\, \sqrt {4 \sqrt {a c}-2 b}}\right ) \sqrt {2}}{d}\)	\(140\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 603 vs. \(2 (105) = 210\).

Time = 1.78 (sec) , antiderivative size = 603, normalized size of antiderivative = 4.16 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\frac {1}{8} \, \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}} \log \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} + x^{2}\right )} + {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} + a d x\right )} \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}}}{c x^{4} - a}\right ) - \frac {1}{8} \, \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}} \log \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} + x^{2}\right )} - {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} + a d x\right )} \sqrt {\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} + b}{a c d^{2}}}}{c x^{4} - a}\right ) + \frac {1}{8} \, \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}} \log \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} - x^{2}\right )} + {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} - a d x\right )} \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}}}{c x^{4} - a}\right ) - \frac {1}{8} \, \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}} \log \left (-\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (a d^{2} \sqrt {\frac {1}{a c d^{4}}} - x^{2}\right )} - {\left (a c d^{3} x^{3} \sqrt {\frac {1}{a c d^{4}}} - a d x\right )} \sqrt {-\frac {2 \, a c d^{2} \sqrt {\frac {1}{a c d^{4}}} - b}{a c d^{2}}}}{c x^{4} - a}\right ) \]

[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))

Sympy [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=- \frac {\int \frac {\sqrt {a + b x^{2} + c x^{4}}}{- a + c x^{4}}\, dx}{d} \]

[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

Maxima [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\int { -\frac {\sqrt {c x^{4} + b x^{2} + a}}{c d x^{4} - a d} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\int { -\frac {\sqrt {c x^{4} + b x^{2} + a}}{c d x^{4} - a d} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2+c x^4}}{a d-c d x^4} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^2+a}}{a\,d-c\,d\,x^4} \,d x \]

[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)