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

Optimal. Leaf size=103 \[ \frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}}-\frac {\sqrt {b x^2+c x^4} (4 b B-3 A c)}{8 b^2 x^3}-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5} \]

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Rubi [A]  time = 0.17, antiderivative size = 103, 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 = {2038, 2025, 2008, 206} \begin {gather*} -\frac {\sqrt {b x^2+c x^4} (4 b B-3 A c)}{8 b^2 x^3}+\frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}}-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[b*x^2 + c*x^4])/(4*b*x^5) - ((4*b*B - 3*A*c)*Sqrt[b*x^2 + c*x^4])/(8*b^2*x^3) + (c*(4*b*B - 3*A*c)*Ar
cTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(8*b^(5/2))

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 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

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^4 \sqrt {b x^2+c x^4}} \, dx &=-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5}-\frac {(-4 b B+3 A c) \int \frac {1}{x^2 \sqrt {b x^2+c x^4}} \, dx}{4 b}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5}-\frac {(4 b B-3 A c) \sqrt {b x^2+c x^4}}{8 b^2 x^3}-\frac {(c (4 b B-3 A c)) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{8 b^2}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5}-\frac {(4 b B-3 A c) \sqrt {b x^2+c x^4}}{8 b^2 x^3}+\frac {(c (4 b B-3 A c)) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{8 b^2}\\ &=-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5}-\frac {(4 b B-3 A c) \sqrt {b x^2+c x^4}}{8 b^2 x^3}+\frac {c (4 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 104, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (b \sqrt {\frac {c x^2}{b}+1} \left (2 A b-3 A c x^2+4 b B x^2\right )+c x^4 (3 A c-4 b B) \tanh ^{-1}\left (\sqrt {\frac {c x^2}{b}+1}\right )\right )}{8 b^3 x^5 \sqrt {\frac {c x^2}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[x^2*(b + c*x^2)]*(b*(2*A*b + 4*b*B*x^2 - 3*A*c*x^2)*Sqrt[1 + (c*x^2)/b] + c*(-4*b*B + 3*A*c)*x^4*Ar
cTanh[Sqrt[1 + (c*x^2)/b]]))/(b^3*x^5*Sqrt[1 + (c*x^2)/b])

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IntegrateAlgebraic [A]  time = 0.22, size = 89, normalized size = 0.86 \begin {gather*} \frac {\left (4 b B c-3 A c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}}+\frac {\sqrt {b x^2+c x^4} \left (-2 A b+3 A c x^2-4 b B x^2\right )}{8 b^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*A*b - 4*b*B*x^2 + 3*A*c*x^2)*Sqrt[b*x^2 + c*x^4])/(8*b^2*x^5) + ((4*b*B*c - 3*A*c^2)*ArcTanh[(Sqrt[b]*x)/
Sqrt[b*x^2 + c*x^4]])/(8*b^(5/2))

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fricas [A]  time = 0.44, size = 199, normalized size = 1.93 \begin {gather*} \left [-\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {b} x^{5} \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}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{16 \, b^{3} x^{5}}, -\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {-b} x^{5} \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}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{8 \, b^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((4*B*b*c - 3*A*c^2)*sqrt(b)*x^5*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)*(2*A*b^2 + (4*B*b^2 - 3*A*b*c)*x^2))/(b^3*x^5), -1/8*((4*B*b*c - 3*A*c^2)*sqrt(-b)*x^5*arctan(sqrt
(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + sqrt(c*x^4 + b*x^2)*(2*A*b^2 + (4*B*b^2 - 3*A*b*c)*x^2))/(b^3*x^5)]

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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^4/(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]2*(-4*b^2*A/32/b^3/x/x-(8*b^2*B-6*b*A*c)/32/b^3)/x*sqrt(b*(1/x)^2+c)+2*(3*A*c^2-4*B*b*c)/16/b^2/sqrt(b)*l
n(abs(sqrt(b*(1/x)^2+c)-sqrt(b)/x))

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maple [A]  time = 0.06, size = 146, normalized size = 1.42 \begin {gather*} -\frac {\sqrt {c \,x^{2}+b}\, \left (3 A b \,c^{2} x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-4 B \,b^{2} c \,x^{4} \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-3 \sqrt {c \,x^{2}+b}\, A \,b^{\frac {3}{2}} c \,x^{2}+4 \sqrt {c \,x^{2}+b}\, B \,b^{\frac {5}{2}} x^{2}+2 \sqrt {c \,x^{2}+b}\, A \,b^{\frac {5}{2}}\right )}{8 \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {7}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*x^2+b)^(1/2)*(3*A*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*b*c^2+4*B*(c*x^2+b)^(1/2)*b^(5/2)*x^2-4*B*ln
(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*x^4*b^2*c-3*A*(c*x^2+b)^(1/2)*b^(3/2)*x^2*c+2*A*(c*x^2+b)^(1/2)*b^(5/2))/x^3
/(c*x^4+b*x^2)^(1/2)/b^(7/2)

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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^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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