3.2.56 \(\int \frac {x^2 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=64 \[ -\frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}}-\frac {x (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

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Rubi [A]  time = 0.14, antiderivative size = 64, 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 = {2037, 2008, 206} \begin {gather*} -\frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}}-\frac {x (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x)/(b*c*Sqrt[b*x^2 + c*x^4])) - (A*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/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 2037

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 73, normalized size = 1.14 \begin {gather*} -\frac {x \left (\sqrt {b} (b B-A c)+A c \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{b^{3/2} c \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.80, size = 74, normalized size = 1.16 \begin {gather*} \frac {\sqrt {b x^2+c x^4} (A c-b B)}{b c x \left (b+c x^2\right )}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 199, normalized size = 3.11 \begin {gather*} \left [\frac {{\left (A c^{2} x^{3} + A b c x\right )} \sqrt {b} \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 (B b^{2} - A b c\right )}}{2 \, {\left (b^{2} c^{2} x^{3} + b^{3} c x\right )}}, \frac {{\left (A c^{2} x^{3} + A b c x\right )} \sqrt {-b} \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 (B b^{2} - A b c\right )}}{b^{2} c^{2} x^{3} + b^{3} c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

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maple [A]  time = 0.06, size = 79, normalized size = 1.23 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (-\sqrt {c \,x^{2}+b}\, A b c \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+A \,b^{\frac {3}{2}} c -B \,b^{\frac {5}{2}}\right ) x^{3}}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {5}{2}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (B x^{2} + A\right )} x^{2}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^2/(c*x^4 + b*x^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**2*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)

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