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

Optimal. Leaf size=59 \[ -a \sqrt{c+\frac{d}{x^2}}+a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0431102, antiderivative size = 59, 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 = {446, 80, 50, 63, 208} \[ -a \sqrt{c+\frac{d}{x^2}}+a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.116312, size = 82, normalized size = 1.39 \[ \frac{\sqrt{c+\frac{d}{x^2}} \left (\frac{3 a \sqrt{c} \sqrt{d} x^3 \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{\sqrt{\frac{c x^2}{d}+1}}-3 a d x^2-b \left (c x^2+d\right )\right )}{3 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 110, normalized size = 1.9 \begin{align*}{\frac{1}{3\,{x}^{2}d}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 3\,{c}^{3/2}\sqrt{c{x}^{2}+d}{x}^{4}a-3\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}a+3\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ){x}^{3}acd-\sqrt{c} \left ( c{x}^{2}+d \right ) ^{{\frac{3}{2}}}b \right ){\frac{1}{\sqrt{c{x}^{2}+d}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(c+d/x^2)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3588, size = 385, normalized size = 6.53 \begin{align*} \left [\frac{3 \, a \sqrt{c} d x^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} - d\right ) - 2 \,{\left ({\left (b c + 3 \, a d\right )} x^{2} + b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{6 \, d x^{2}}, -\frac{3 \, a \sqrt{-c} d x^{2} \arctan \left (\frac{\sqrt{-c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left ({\left (b c + 3 \, a d\right )} x^{2} + b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{3 \, d x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*a*sqrt(c)*d*x^2*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*((b*c + 3*a*d)*x^2 + b*d)*
sqrt((c*x^2 + d)/x^2))/(d*x^2), -1/3*(3*a*sqrt(-c)*d*x^2*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)
) + ((b*c + 3*a*d)*x^2 + b*d)*sqrt((c*x^2 + d)/x^2))/(d*x^2)]

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Sympy [A]  time = 17.0405, size = 75, normalized size = 1.27 \begin{align*} \frac{a \left (- \frac{2 c \operatorname{atan}{\left (\frac{\sqrt{c + \frac{d}{x^{2}}}}{\sqrt{- c}} \right )}}{\sqrt{- c}} - 2 \sqrt{c + \frac{d}{x^{2}}}\right )}{2} + \frac{b \left (\begin{cases} - \frac{\sqrt{c}}{x^{2}} & \text{for}\: d = 0 \\- \frac{2 \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3 d} & \text{otherwise} \end{cases}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-2*c*atan(sqrt(c + d/x**2)/sqrt(-c))/sqrt(-c) - 2*sqrt(c + d/x**2))/2 + b*Piecewise((-sqrt(c)/x**2, Eq(d, 0
)), (-2*(c + d/x**2)**(3/2)/(3*d), True))/2

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Giac [B]  time = 1.39762, size = 220, normalized size = 3.73 \begin{align*} -\frac{1}{2} \, a \sqrt{c} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \,{\left (3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} a \sqrt{c} d \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} a \sqrt{c} d^{2} \mathrm{sgn}\left (x\right ) + b c^{\frac{3}{2}} d^{2} \mathrm{sgn}\left (x\right ) + 3 \, a \sqrt{c} d^{3} \mathrm{sgn}\left (x\right )\right )}}{3 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*sqrt(c)*log((sqrt(c)*x - sqrt(c*x^2 + d))^2)*sgn(x) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(3/2)*
sgn(x) + 3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*sqrt(c)*d*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*sqrt(c)*d^
2*sgn(x) + b*c^(3/2)*d^2*sgn(x) + 3*a*sqrt(c)*d^3*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^3