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

Optimal. Leaf size=51 \[ \frac {x}{b \sqrt {b x^2+c x^4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}} \]

[Out]

-arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(3/2)+x/b/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2023, 2008, 206} \[ \frac {x}{b \sqrt {b x^2+c x^4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(b*x^2 + c*x^4)^(3/2),x]

[Out]

x/(b*Sqrt[b*x^2 + c*x^4]) - 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 2023

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*(n - j)*(p + 1)), x] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] &
& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {x}{b \sqrt {b x^2+c x^4}}+\frac {\int \frac {1}{\sqrt {b x^2+c x^4}} \, dx}{b}\\ &=\frac {x}{b \sqrt {b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )}{b}\\ &=\frac {x}{b \sqrt {b x^2+c x^4}}-\frac {\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 [C]  time = 0.01, size = 38, normalized size = 0.75 \[ \frac {x \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c x^2}{b}+1\right )}{b \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (c*x^2)/b])/(b*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.59, size = 162, normalized size = 3.18 \[ \left [\frac {{\left (c x^{3} + b x\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} b}{2 \, {\left (b^{2} c x^{3} + b^{3} x\right )}}, \frac {{\left (c x^{3} + b x\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + \sqrt {c x^{4} + b x^{2}} b}{b^{2} c x^{3} + b^{3} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(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]1/b/x*sqrt(b*(1/x)^2+c)/(b*(1/x)^2+c)+1/b/sqrt(b)*ln(abs(sqrt(b*(1/x)^2+c)-sqrt(b)/x))

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maple [A]  time = 0.01, size = 65, normalized size = 1.27 \[ \frac {\left (c \,x^{2}+b \right ) \left (-\sqrt {c \,x^{2}+b}\, b \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )+b^{\frac {3}{2}}\right ) x^{3}}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(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 \[ \int \frac {x^2}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(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 \[ \int \frac {x^{2}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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