3.975 \(\int \frac{a+\frac{b}{x^2}}{(c+\frac{d}{x^2})^{3/2} x} \, dx\)

Optimal. Leaf size=52 \[ \frac{b c-a d}{c d \sqrt{c+\frac{d}{x^2}}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0400453, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {446, 78, 63, 208} \[ \frac{b c-a d}{c d \sqrt{c+\frac{d}{x^2}}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x),x]

[Out]

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

Rule 446

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0623979, size = 73, normalized size = 1.4 \[ \frac{\sqrt{c} x (b c-a d)+a d^{3/2} \sqrt{\frac{c x^2}{d}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{c^{3/2} d x \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(b*c - a*d)*x + a*d^(3/2)*Sqrt[1 + (c*x^2)/d]*ArcSinh[(Sqrt[c]*x)/Sqrt[d]])/(c^(3/2)*d*Sqrt[c + d/x^2
]*x)

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Maple [A]  time = 0.008, size = 75, normalized size = 1.4 \begin{align*}{\frac{c{x}^{2}+d}{d{x}^{3}} \left ({c}^{{\frac{5}{2}}}xb-{c}^{{\frac{3}{2}}}xad+\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) \sqrt{c{x}^{2}+d}acd \right ) \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)/(c+d/x^2)^(3/2)/x,x)

[Out]

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

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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)/(c+d/x^2)^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60452, size = 427, normalized size = 8.21 \begin{align*} \left [\frac{2 \,{\left (b c^{2} - a c d\right )} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} +{\left (a c d x^{2} + a d^{2}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} - d\right )}{2 \,{\left (c^{3} d x^{2} + c^{2} d^{2}\right )}}, \frac{{\left (b c^{2} - a c d\right )} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} -{\left (a c d x^{2} + a d^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right )}{c^{3} d x^{2} + c^{2} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 10.2146, size = 49, normalized size = 0.94 \begin{align*} - \frac{a \operatorname{atan}{\left (\frac{\sqrt{c + \frac{d}{x^{2}}}}{\sqrt{- c}} \right )}}{c \sqrt{- c}} - \frac{a d - b c}{c d \sqrt{c + \frac{d}{x^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*atan(sqrt(c + d/x**2)/sqrt(-c))/(c*sqrt(-c)) - (a*d - b*c)/(c*d*sqrt(c + d/x**2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + \frac{b}{x^{2}}}{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x,x, algorithm="giac")

[Out]

integrate((a + b/x^2)/((c + d/x^2)^(3/2)*x), x)