3.1002 \(\int \frac{1}{x^4 \sqrt{2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=87 \[ -\frac{3 c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{5/2}}+\frac{3 c \sqrt{b x^2+c x^4}}{8 b^2 x^3}-\frac{\sqrt{b x^2+c x^4}}{4 b x^5} \]

[Out]

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

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Rubi [A]  time = 0.0995023, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3, 2025, 2008, 206} \[ -\frac{3 c^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{b x^2+c x^4}}\right )}{8 b^{5/2}}+\frac{3 c \sqrt{b x^2+c x^4}}{8 b^2 x^3}-\frac{\sqrt{b x^2+c x^4}}{4 b x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]

[Out]

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

Rule 3

Int[(u_.)*((a_) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n + c*x^(2*n))^p, x] /;
FreeQ[{a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[a, 0]

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 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 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])

Rubi steps

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

Mathematica [C]  time = 0.0119407, size = 44, normalized size = 0.51 \[ -\frac{c^2 \sqrt{x^2 \left (b+c x^2\right )} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{c x^2}{b}+1\right )}{b^3 x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[2 + 2*a - 2*(1 + a) + b*x^2 + c*x^4]),x]

[Out]

-((c^2*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[1/2, 3, 3/2, 1 + (c*x^2)/b])/(b^3*x))

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Maple [A]  time = 0.047, size = 94, normalized size = 1.1 \begin{align*} -{\frac{1}{8\,{x}^{3}}\sqrt{c{x}^{2}+b} \left ( 3\,\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{c{x}^{2}+b}+b}{x}} \right ){x}^{4}b{c}^{2}-3\,\sqrt{c{x}^{2}+b}{b}^{3/2}{x}^{2}c+2\,\sqrt{c{x}^{2}+b}{b}^{5/2} \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*x^2+b)^(1/2)*(3*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*x^4*b*c^2-3*(c*x^2+b)^(1/2)*b^(3/2)*x^2*c+2*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{2}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^4 + b*x^2)*x^4), x)

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Fricas [A]  time = 1.57034, size = 366, normalized size = 4.21 \begin{align*} \left [\frac{3 \, \sqrt{b} c^{2} 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 (3 \, b c x^{2} - 2 \, b^{2}\right )}}{16 \, b^{3} x^{5}}, \frac{3 \, \sqrt{-b} c^{2} 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 (3 \, b c x^{2} - 2 \, b^{2}\right )}}{8 \, b^{3} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \sqrt{x^{2} \left (b + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**4*sqrt(x**2*(b + c*x**2))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError