3.142 \(\int \frac {A+B x^2}{\sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=55 \[ \frac {B \sqrt {b x^2+c x^4}}{c x}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {b}} \]

[Out]

-A*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(1/2)+B*(c*x^4+b*x^2)^(1/2)/c/x

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1145, 2008, 206} \[ \frac {B \sqrt {b x^2+c x^4}}{c x}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(B*Sqrt[b*x^2 + c*x^4])/(c*x) - (A*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/Sqrt[b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1145

Int[((d_) + (e_.)*(x_)^2)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*(b*x^2 + c*x^4)^(p + 1))/(c
*(4*p + 3)*x), x] - Dist[(b*e*(2*p + 1) - c*d*(4*p + 3))/(c*(4*p + 3)), Int[(b*x^2 + c*x^4)^p, x], x] /; FreeQ
[{b, c, d, e, p}, x] &&  !IntegerQ[p] && NeQ[4*p + 3, 0] && NeQ[b*e*(2*p + 1) - c*d*(4*p + 3), 0]

Rule 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 73, normalized size = 1.33 \[ \frac {x \left (\sqrt {b} B \left (b+c x^2\right )-A c \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{\sqrt {b} c \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(Sqrt[b]*B*(b + c*x^2) - A*c*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(Sqrt[b]*c*Sqrt[x^2*(b + c*
x^2)])

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 138, normalized size = 2.51 \[ \left [\frac {A \sqrt {b} c x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} B b}{2 \, b c x}, \frac {A \sqrt {-b} c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} B b}{b c x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(A*sqrt(b)*c*x*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*sqrt(c*x^4 + b*x^2)*B*b)/(b*
c*x), (A*sqrt(-b)*c*x*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2)*B*b)/(b*c*x)]

________________________________________________________________________________________

giac [A]  time = 0.33, size = 60, normalized size = 1.09 \[ \frac {A \log \left ({\left (\sqrt {c + \frac {b}{x^{2}}} - \frac {\sqrt {b}}{x}\right )}^{2}\right )}{2 \, \sqrt {b}} - \frac {2 \, B \sqrt {b}}{{\left (\sqrt {c + \frac {b}{x^{2}}} - \frac {\sqrt {b}}{x}\right )}^{2} - c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/2*A*log((sqrt(c + b/x^2) - sqrt(b)/x)^2)/sqrt(b) - 2*B*sqrt(b)/((sqrt(c + b/x^2) - sqrt(b)/x)^2 - c)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 72, normalized size = 1.31 \[ -\frac {\sqrt {c \,x^{2}+b}\, \left (A c \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-\sqrt {c \,x^{2}+b}\, B \sqrt {b}\right ) x}{\sqrt {c \,x^{4}+b \,x^{2}}\, \sqrt {b}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)

[Out]

-x*(c*x^2+b)^(1/2)*(A*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*c-B*(c*x^2+b)^(1/2)*b^(1/2))/(c*x^4+b*x^2)^(1/2)/c/b
^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/sqrt(c*x^4 + b*x^2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {B\,x^2+A}{\sqrt {c\,x^4+b\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int((A + B*x^2)/(b*x^2 + c*x^4)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x^{2}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral((A + B*x**2)/sqrt(x**2*(b + c*x**2)), x)

________________________________________________________________________________________