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

Optimal. Leaf size=76 \[ a c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )-\frac{1}{3} a \left (c+\frac{d}{x^2}\right )^{3/2}-a c \sqrt{c+\frac{d}{x^2}}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0541351, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ a c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )-\frac{1}{3} a \left (c+\frac{d}{x^2}\right )^{3/2}-a c \sqrt{c+\frac{d}{x^2}}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.046791, size = 90, normalized size = 1.18 \[ -\frac{\sqrt{c+\frac{d}{x^2}} \left (5 a d^2 x^2 \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x^2}{d}\right )+3 b \left (c x^2+d\right )^2 \sqrt{\frac{c x^2}{d}+1}\right )}{15 d x^4 \sqrt{\frac{c x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d/x^2]*(3*b*(d + c*x^2)^2*Sqrt[1 + (c*x^2)/d] + 5*a*d^2*x^2*Hypergeometric2F1[-3/2, -3/2, -1/2, -((
c*x^2)/d)]))/(15*d*x^4*Sqrt[1 + (c*x^2)/d])

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Maple [B]  time = 0.014, size = 153, normalized size = 2. \begin{align*}{\frac{1}{15\,{x}^{2}{d}^{2}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 10\,{c}^{5/2} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{6}a+15\,{c}^{5/2}\sqrt{c{x}^{2}+d}{x}^{6}ad-10\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{4}a+15\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ){x}^{5}a{c}^{2}{d}^{2}-5\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}ad-3\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{5/2}bd \right ) \left ( c{x}^{2}+d \right ) ^{-{\frac{3}{2}}}{\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)^(3/2)/x,x)

[Out]

1/15*((c*x^2+d)/x^2)^(3/2)*(10*c^(5/2)*(c*x^2+d)^(3/2)*x^6*a+15*c^(5/2)*(c*x^2+d)^(1/2)*x^6*a*d-10*c^(3/2)*(c*
x^2+d)^(5/2)*x^4*a+15*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x^5*a*c^2*d^2-5*c^(1/2)*(c*x^2+d)^(5/2)*x^2*a*d-3*c^(1/2)*
(c*x^2+d)^(5/2)*b*d)/x^2/(c*x^2+d)^(3/2)/d^2/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)^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34711, size = 493, normalized size = 6.49 \begin{align*} \left [\frac{15 \, a c^{\frac{3}{2}} d x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} - d\right ) - 2 \,{\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} +{\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{30 \, d x^{4}}, -\frac{15 \, a \sqrt{-c} c d x^{4} \arctan \left (\frac{\sqrt{-c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left ({\left (3 \, b c^{2} + 20 \, a c d\right )} x^{4} + 3 \, b d^{2} +{\left (6 \, b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{15 \, d x^{4}}\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]

[1/30*(15*a*c^(3/2)*d*x^4*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*((3*b*c^2 + 20*a*c*d)*x^
4 + 3*b*d^2 + (6*b*c*d + 5*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d*x^4), -1/15*(15*a*sqrt(-c)*c*d*x^4*arctan(sqr
t(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + ((3*b*c^2 + 20*a*c*d)*x^4 + 3*b*d^2 + (6*b*c*d + 5*a*d^2)*x^2)*
sqrt((c*x^2 + d)/x^2))/(d*x^4)]

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Sympy [A]  time = 37.2231, size = 73, normalized size = 0.96 \begin{align*} - \frac{a c^{2} \operatorname{atan}{\left (\frac{\sqrt{c + \frac{d}{x^{2}}}}{\sqrt{- c}} \right )}}{\sqrt{- c}} - a c \sqrt{c + \frac{d}{x^{2}}} - \frac{a \left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3} - \frac{b \left (c + \frac{d}{x^{2}}\right )^{\frac{5}{2}}}{5 d} \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*c**2*atan(sqrt(c + d/x**2)/sqrt(-c))/sqrt(-c) - a*c*sqrt(c + d/x**2) - a*(c + d/x**2)**(3/2)/3 - b*(c + d/x
**2)**(5/2)/(5*d)

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Giac [B]  time = 1.80624, size = 343, normalized size = 4.51 \begin{align*} -\frac{1}{2} \, a c^{\frac{3}{2}} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \,{\left (15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{8} b c^{\frac{5}{2}} \mathrm{sgn}\left (x\right ) + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{8} a c^{\frac{3}{2}} d \mathrm{sgn}\left (x\right ) - 90 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{6} a c^{\frac{3}{2}} d^{2} \mathrm{sgn}\left (x\right ) + 30 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} b c^{\frac{5}{2}} d^{2} \mathrm{sgn}\left (x\right ) + 110 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} a c^{\frac{3}{2}} d^{3} \mathrm{sgn}\left (x\right ) - 70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} a c^{\frac{3}{2}} d^{4} \mathrm{sgn}\left (x\right ) + 3 \, b c^{\frac{5}{2}} d^{4} \mathrm{sgn}\left (x\right ) + 20 \, a c^{\frac{3}{2}} d^{5} \mathrm{sgn}\left (x\right )\right )}}{15 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d\right )}^{5}} \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]

-1/2*a*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + d))^2)*sgn(x) + 2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(5/2
)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(3/2)*d*sgn(x) - 90*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(3/2
)*d^2*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(5/2)*d^2*sgn(x) + 110*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a
*c^(3/2)*d^3*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(3/2)*d^4*sgn(x) + 3*b*c^(5/2)*d^4*sgn(x) + 20*a*
c^(3/2)*d^5*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5