3.11.2 \(\int \frac {1}{x^4 \sqrt {2+2 a-2 (1+a)+b x^2+c x^4}} \, dx\) [1002]

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

[Out]

-3/8*c^2*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2))/b^(5/2)-1/4*(c*x^4+b*x^2)^(1/2)/b/x^5+3/8*c*(c*x^4+b*x^2)^(1/2
)/b^2/x^3

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Rubi [A]
time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, 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, 2050, 2033, 212} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Sqrt[b*x^2 + c*x^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 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 2050

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]

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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 91, normalized size = 1.05 \begin {gather*} \frac {\sqrt {b} \left (-2 b^2+b c x^2+3 c^2 x^4\right )-3 c^2 x^4 \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{8 b^{5/2} x^3 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 94, normalized size = 1.08

method	result	size
default	\(-\frac {\sqrt {c \,x^{2}+b}\, \left (3 \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,c^{2} x^{4}-3 b^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, c \,x^{2}+2 \sqrt {c \,x^{2}+b}\, b^{\frac {5}{2}}\right )}{8 x^{3} \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {7}{2}}}\)	\(94\)
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (-3 c \,x^{2}+2 b \right )}{8 b^{2} x^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {3 c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{8 b^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/x^3*(c*x^2+b)^(1/2)*(3*ln(2*(b^(1/2)*(c*x^2+b)^(1/2)+b)/x)*b*c^2*x^4-3*b^(3/2)*(c*x^2+b)^(1/2)*c*x^2+2*(c
*x^2+b)^(1/2)*b^(5/2))/(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*} \text {Failed to integrate} \end {gather*}

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 = 0.37, size = 163, normalized size = 1.87 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{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(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 [A]
time = 4.51, size = 79, normalized size = 0.91 \begin {gather*} \frac {\frac {3 \, c^{3} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} c^{3} - 5 \, \sqrt {c x^{2} + b} b c^{3}}{b^{2} c^{2} x^{4}}}{8 \, c \mathrm {sgn}\left (x\right )} \end {gather*}

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]

1/8*(3*c^3*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x^2 + b)^(3/2)*c^3 - 5*sqrt(c*x^2 + b)*b*c^
3)/(b^2*c^2*x^4))/(c*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{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(1/(x^4*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

int(1/(x^4*(b*x^2 + c*x^4)^(1/2)), x)

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