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

Optimal. Leaf size=57 \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\frac {A \sqrt {b x^2+c x^4}}{b x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 792, 620, 206} \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\frac {A \sqrt {b x^2+c x^4}}{b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 74, normalized size = 1.30 \[ \frac {b B x \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )-A \sqrt {c} \left (b+c x^2\right )}{b \sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 136, normalized size = 2.39 \[ \left [\frac {B b \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} A c}{2 \, b c x^{2}}, -\frac {B b \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} A c}{b c x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 66, normalized size = 1.16 \[ -\frac {B \log \left ({\left | 2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} + b \right |}\right )}{2 \, \sqrt {c}} + \frac {A}{\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*B*log(abs(2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2))*sqrt(c) + b))/sqrt(c) + A/(sqrt(c)*x^2 - sqrt(c*x^4 + b*x
^2))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 67, normalized size = 1.18 \[ -\frac {\sqrt {c \,x^{2}+b}\, \left (-B b x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+\sqrt {c \,x^{2}+b}\, A \sqrt {c}\right )}{\sqrt {c \,x^{4}+b \,x^{2}}\, b \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.41, size = 56, normalized size = 0.98 \[ \frac {B \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{2 \, \sqrt {c}} - \frac {\sqrt {c x^{4} + b x^{2}} A}{b x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.51, size = 57, normalized size = 1.00 \[ \frac {B\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,\sqrt {c}}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{b\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________