3.3.83 \(\int \frac {x^2}{(b x^2+c x^4)^{3/2}} \, dx\) [283]

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 = 0.04, 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 = {2048, 2033, 212} \begin {gather*} \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 {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

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 2048

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 {\text {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 [A]
time = 0.04, size = 59, normalized size = 1.16 \begin {gather*} \frac {x \left (\sqrt {b}-\sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{b^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 65, normalized size = 1.27

method	result	size
default	\(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (b^{\frac {3}{2}}-\ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \sqrt {c \,x^{2}+b}\right )}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {5}{2}}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*(c*x^2+b)*(b^(3/2)-ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.37, size = 162, normalized size = 3.18 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\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/(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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Giac [A]
time = 4.76, size = 79, normalized size = 1.55 \begin {gather*} -\frac {{\left (\sqrt {b} \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + \sqrt {-b}\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-b} b^{\frac {3}{2}}} + \frac {\arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b \mathrm {sgn}\left (x\right )} + \frac {1}{\sqrt {c x^{2} + b} b \mathrm {sgn}\left (x\right )} \end {gather*}

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]

-(sqrt(b)*arctan(sqrt(b)/sqrt(-b)) + sqrt(-b))*sgn(x)/(sqrt(-b)*b^(3/2)) + arctan(sqrt(c*x^2 + b)/sqrt(-b))/(s
qrt(-b)*b*sgn(x)) + 1/(sqrt(c*x^2 + b)*b*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\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/(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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