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

Optimal. Leaf size=68 \[ -\frac {(2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3} \]

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Rubi [A]  time = 0.12, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2038, 2008, 206} \begin {gather*} -\frac {(2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(2*b*x^3) - ((2*b*B - A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))

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 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]

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^2}{x^2 \sqrt {b x^2+c x^4}} \, dx &=-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}-\frac {(-2 b B+A c) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{2 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}+\frac {(-2 b B+A c) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{2 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3}-\frac {(2 b B-A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 87, normalized size = 1.28 \begin {gather*} \frac {x \sqrt {b+c x^2} \left (-\frac {2 \left (b B-\frac {A c}{2}\right ) \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {A \sqrt {b+c x^2}}{b x^2}\right )}{2 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[b + c*x^2]*(-((A*Sqrt[b + c*x^2])/(b*x^2)) - (2*(b*B - (A*c)/2)*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]])/b^(3
/2)))/(2*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.16, size = 67, normalized size = 0.99 \begin {gather*} \frac {(A c-2 b B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {A \sqrt {b x^2+c x^4}}{2 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*Sqrt[b*x^2 + c*x^4])/(b*x^3) + ((-2*b*B + A*c)*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2))

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fricas [A]  time = 0.44, size = 152, normalized size = 2.24 \begin {gather*} \left [-\frac {{\left (2 \, B b - A c\right )} \sqrt {b} x^{3} \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}} A b}{4 \, b^{2} x^{3}}, \frac {{\left (2 \, B b - A c\right )} \sqrt {-b} x^{3} \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}} A b}{2 \, b^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-2,[0,1,2]%%%},0,%%%{1,[0,2,4]%%%}] at parameters values [64.3995612673,65,-85]Warning, choosing
root of [1,0,%%%{-4,[1,0,0]%%%}+%%%{-2,[0,1,2]%%%},0,%%%{1,[0,2,4]%%%}] at parameters values [66.1769613782,9
3,91]-4*A/8/b/x*sqrt(b*(1/x)^2+c)+2*(-A*c+2*B*b)/4/b/sqrt(b)*ln(abs(sqrt(b*(1/x)^2+c)-sqrt(b)/x))

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maple [A]  time = 0.06, size = 105, normalized size = 1.54 \begin {gather*} -\frac {\sqrt {c \,x^{2}+b}\, \left (-A b c \,x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+2 B \,b^{2} x^{2} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+\sqrt {c \,x^{2}+b}\, A \,b^{\frac {3}{2}}\right )}{2 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {5}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x*(c*x^2+b)^(1/2)*(2*B*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^2*b^2-A*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*
x^2*b*c+A*b^(3/2)*(c*x^2+b)^(1/2))/(c*x^4+b*x^2)^(1/2)/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^2/(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^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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