3.972 \(\int \frac{a+\frac{b}{x^2}}{\sqrt{c+\frac{d}{x^2}} x^4} \, dx\)

Optimal. Leaf size=93 \[ \frac{\sqrt{c+\frac{d}{x^2}} (3 b c-4 a d)}{8 d^2 x}-\frac{c (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{5/2}}-\frac{b \sqrt{c+\frac{d}{x^2}}}{4 d x^3} \]

[Out]

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

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Rubi [A]  time = 0.0502702, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 335, 321, 217, 206} \[ \frac{\sqrt{c+\frac{d}{x^2}} (3 b c-4 a d)}{8 d^2 x}-\frac{c (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{5/2}}-\frac{b \sqrt{c+\frac{d}{x^2}}}{4 d x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A]  time = 0.121172, size = 107, normalized size = 1.15 \[ -\frac{\left (c x^2+d\right ) \left (d \sqrt{\frac{c x^2}{d}+1} \left (4 a d x^2-3 b c x^2+2 b d\right )+c x^4 (3 b c-4 a d) \tanh ^{-1}\left (\sqrt{\frac{c x^2}{d}+1}\right )\right )}{8 d^3 x^5 \sqrt{c+\frac{d}{x^2}} \sqrt{\frac{c x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + c*x^2)*(d*Sqrt[1 + (c*x^2)/d]*(2*b*d - 3*b*c*x^2 + 4*a*d*x^2) + c*(3*b*c - 4*a*d)*x^4*ArcTanh[Sqrt[1 +
(c*x^2)/d]]))/(8*d^3*Sqrt[c + d/x^2]*x^5*Sqrt[1 + (c*x^2)/d])

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Maple [A]  time = 0.01, size = 146, normalized size = 1.6 \begin{align*} -{\frac{1}{8\,{x}^{5}}\sqrt{c{x}^{2}+d} \left ( -4\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}ac{d}^{2}+3\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}b{c}^{2}d+4\,{d}^{5/2}\sqrt{c{x}^{2}+d}{x}^{2}a-3\,{d}^{3/2}\sqrt{c{x}^{2}+d}{x}^{2}bc+2\,{d}^{5/2}\sqrt{c{x}^{2}+d}b \right ){\frac{1}{\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}}}}{d}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*x^2+d)^(1/2)*(-4*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*x^4*a*c*d^2+3*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)
*x^4*b*c^2*d+4*d^(5/2)*(c*x^2+d)^(1/2)*x^2*a-3*d^(3/2)*(c*x^2+d)^(1/2)*x^2*b*c+2*d^(5/2)*(c*x^2+d)^(1/2)*b)/((
c*x^2+d)/x^2)^(1/2)/x^5/d^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72066, size = 459, normalized size = 4.94 \begin{align*} \left [-\frac{{\left (3 \, b c^{2} - 4 \, a c d\right )} \sqrt{d} x^{3} \log \left (-\frac{c x^{2} + 2 \, \sqrt{d} x \sqrt{\frac{c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \,{\left (2 \, b d^{2} -{\left (3 \, b c d - 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \, d^{3} x^{3}}, \frac{{\left (3 \, b c^{2} - 4 \, a c d\right )} \sqrt{-d} x^{3} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) -{\left (2 \, b d^{2} -{\left (3 \, b c d - 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \, d^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((3*b*c^2 - 4*a*c*d)*sqrt(d)*x^3*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) + 2*(2*b*d
^2 - (3*b*c*d - 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^3), 1/8*((3*b*c^2 - 4*a*c*d)*sqrt(-d)*x^3*arctan(s
qrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (2*b*d^2 - (3*b*c*d - 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3
*x^3)]

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Sympy [A]  time = 8.33416, size = 150, normalized size = 1.61 \begin{align*} - \frac{a \sqrt{c} \sqrt{1 + \frac{d}{c x^{2}}}}{2 d x} + \frac{a c \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2 d^{\frac{3}{2}}} + \frac{3 b c^{\frac{3}{2}}}{8 d^{2} x \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{b \sqrt{c}}{8 d x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 b c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 d^{\frac{5}{2}}} - \frac{b}{4 \sqrt{c} x^{5} \sqrt{1 + \frac{d}{c x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*d*x) + a*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*d**(3/2)) + 3*b*c**(3/2)/(8*d**2*x
*sqrt(1 + d/(c*x**2))) + b*sqrt(c)/(8*d*x**3*sqrt(1 + d/(c*x**2))) - 3*b*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d*
*(5/2)) - b/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2)))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + \frac{b}{x^{2}}}{\sqrt{c + \frac{d}{x^{2}}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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