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

Optimal. Leaf size=84 \[ -\frac{\sqrt{c+\frac{d}{x^2}} (a d+2 b c)}{2 c}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{2 \sqrt{c}}+\frac{a x^2 \left (c+\frac{d}{x^2}\right )^{3/2}}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0579218, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 78, 50, 63, 208} \[ -\frac{\sqrt{c+\frac{d}{x^2}} (a d+2 b c)}{2 c}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{2 \sqrt{c}}+\frac{a x^2 \left (c+\frac{d}{x^2}\right )^{3/2}}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2]*x,x]

[Out]

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

Mathematica [A]  time = 0.160214, size = 71, normalized size = 0.85 \[ \frac{1}{2} \sqrt{c+\frac{d}{x^2}} \left (\frac{x (a d+2 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{\sqrt{c} \sqrt{d} \sqrt{\frac{c x^2}{d}+1}}+a x^2-2 b\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 127, normalized size = 1.5 \begin{align*}{\frac{1}{2\,d}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 2\,{c}^{3/2}\sqrt{c{x}^{2}+d}{x}^{2}b+\sqrt{c}\sqrt{c{x}^{2}+d}{x}^{2}ad-2\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{3/2}b+\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) xa{d}^{2}+2\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) xbcd \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)*x*(c+d/x^2)^(1/2),x)

[Out]

1/2*((c*x^2+d)/x^2)^(1/2)*(2*c^(3/2)*(c*x^2+d)^(1/2)*x^2*b+c^(1/2)*(c*x^2+d)^(1/2)*x^2*a*d-2*c^(1/2)*(c*x^2+d)
^(3/2)*b+ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x*a*d^2+2*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x*b*c*d)/(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)*x*(c+d/x^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32003, size = 358, normalized size = 4.26 \begin{align*} \left [\frac{{\left (2 \, b c + a d\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 (a c x^{2} - 2 \, b c\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{4 \, c}, -\frac{{\left (2 \, b c + a d\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) -{\left (a c x^{2} - 2 \, b c\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{2 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 32.2153, size = 107, normalized size = 1.27 \begin{align*} \frac{a \sqrt{d} x \sqrt{\frac{c x^{2}}{d} + 1}}{2} + \frac{a d \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )}}{2 \sqrt{c}} + b \sqrt{c} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )} - \frac{b c x}{\sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}} - \frac{b \sqrt{d}}{x \sqrt{\frac{c x^{2}}{d} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sqrt(d)*x*sqrt(c*x**2/d + 1)/2 + a*d*asinh(sqrt(c)*x/sqrt(d))/(2*sqrt(c)) + b*sqrt(c)*asinh(sqrt(c)*x/sqrt(d
)) - b*c*x/(sqrt(d)*sqrt(c*x**2/d + 1)) - b*sqrt(d)/(x*sqrt(c*x**2/d + 1))

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Giac [A]  time = 1.25489, size = 124, normalized size = 1.48 \begin{align*} \frac{1}{2} \, \sqrt{c x^{2} + d} a x \mathrm{sgn}\left (x\right ) + \frac{2 \, b \sqrt{c} d \mathrm{sgn}\left (x\right )}{{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d} - \frac{{\left (2 \, b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + a \sqrt{c} d \mathrm{sgn}\left (x\right )\right )} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2}\right )}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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