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

Optimal. Leaf size=110 \[ -\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (3 a d+2 b c)}{6 c}-\frac{1}{2} \sqrt{c+\frac{d}{x^2}} (3 a d+2 b c)+\frac{1}{2} \sqrt{c} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )+\frac{a x^2 \left (c+\frac{d}{x^2}\right )^{5/2}}{2 c} \]

[Out]

-((2*b*c + 3*a*d)*Sqrt[c + d/x^2])/2 - ((2*b*c + 3*a*d)*(c + d/x^2)^(3/2))/(6*c) + (a*(c + d/x^2)^(5/2)*x^2)/(
2*c) + (Sqrt[c]*(2*b*c + 3*a*d)*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/2

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Rubi [A]  time = 0.0727332, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (3 a d+2 b c)}{6 c}-\frac{1}{2} \sqrt{c+\frac{d}{x^2}} (3 a d+2 b c)+\frac{1}{2} \sqrt{c} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )+\frac{a x^2 \left (c+\frac{d}{x^2}\right )^{5/2}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*b*c + 3*a*d)*Sqrt[c + d/x^2])/2 - ((2*b*c + 3*a*d)*(c + d/x^2)^(3/2))/(6*c) + (a*(c + d/x^2)^(5/2)*x^2)/(
2*c) + (Sqrt[c]*(2*b*c + 3*a*d)*ArcTanh[Sqrt[c + d/x^2]/Sqrt[c]])/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 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 ) \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^2} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^2}{2 c}-\frac{\left (b c+\frac{3 a d}{2}\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x} \, dx,x,\frac{1}{x^2}\right )}{2 c}\\ &=-\frac{(2 b c+3 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{6 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^2}{2 c}-\frac{1}{4} (2 b c+3 a d) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{1}{2} (2 b c+3 a d) \sqrt{c+\frac{d}{x^2}}-\frac{(2 b c+3 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{6 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^2}{2 c}-\frac{1}{4} (c (2 b c+3 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{1}{2} (2 b c+3 a d) \sqrt{c+\frac{d}{x^2}}-\frac{(2 b c+3 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{6 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^2}{2 c}-\frac{(c (2 b c+3 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{1}{2} (2 b c+3 a d) \sqrt{c+\frac{d}{x^2}}-\frac{(2 b c+3 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{6 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^2}{2 c}+\frac{1}{2} \sqrt{c} (2 b c+3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0827385, size = 78, normalized size = 0.71 \[ \frac{1}{3} \sqrt{c+\frac{d}{x^2}} \left (-\frac{(3 a d+2 b c) \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{c x^2}{d}\right )}{\sqrt{\frac{c x^2}{d}+1}}-\frac{b \left (c x^2+d\right )^2}{d x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 216, normalized size = 2. \begin{align*}{\frac{1}{6\,{d}^{2}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 6\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{4}ad+4\,{c}^{5/2} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{4}b-6\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}ad-4\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}b+9\,{c}^{3/2}\sqrt{c{x}^{2}+d}{x}^{4}a{d}^{2}+6\,{c}^{5/2}\sqrt{c{x}^{2}+d}{x}^{4}bd+9\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ){x}^{3}ac{d}^{3}+6\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ){x}^{3}b{c}^{2}{d}^{2}-2\,\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]

1/6*((c*x^2+d)/x^2)^(3/2)*(6*c^(3/2)*(c*x^2+d)^(3/2)*x^4*a*d+4*c^(5/2)*(c*x^2+d)^(3/2)*x^4*b-6*c^(1/2)*(c*x^2+
d)^(5/2)*x^2*a*d-4*c^(3/2)*(c*x^2+d)^(5/2)*x^2*b+9*c^(3/2)*(c*x^2+d)^(1/2)*x^4*a*d^2+6*c^(5/2)*(c*x^2+d)^(1/2)
*x^4*b*d+9*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x^3*a*c*d^3+6*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*x^3*b*c^2*d^2-2*c^(1/2)*(
c*x^2+d)^(5/2)*b*d)/(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 = 1.36788, size = 456, normalized size = 4.15 \begin{align*} \left [\frac{3 \,{\left (2 \, b c + 3 \, a d\right )} \sqrt{c} 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 (3 \, a c x^{4} - 2 \,{\left (4 \, b c + 3 \, a d\right )} x^{2} - 2 \, b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{12 \, x^{2}}, -\frac{3 \,{\left (2 \, b c + 3 \, a d\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c} x^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) -{\left (3 \, a c x^{4} - 2 \,{\left (4 \, b c + 3 \, a d\right )} x^{2} - 2 \, b d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{6 \, 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)^(3/2)*x,x, algorithm="fricas")

[Out]

[1/12*(3*(2*b*c + 3*a*d)*sqrt(c)*x^2*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) + 2*(3*a*c*x^4 -
2*(4*b*c + 3*a*d)*x^2 - 2*b*d)*sqrt((c*x^2 + d)/x^2))/x^2, -1/6*(3*(2*b*c + 3*a*d)*sqrt(-c)*x^2*arctan(sqrt(-c
)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (3*a*c*x^4 - 2*(4*b*c + 3*a*d)*x^2 - 2*b*d)*sqrt((c*x^2 + d)/x^2))/
x^2]

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Sympy [A]  time = 33.146, size = 187, normalized size = 1.7 \begin{align*} \frac{3 a \sqrt{c} d \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )}}{2} + \frac{a c \sqrt{d} x \sqrt{\frac{c x^{2}}{d} + 1}}{2} - \frac{a c \sqrt{d} x}{\sqrt{\frac{c x^{2}}{d} + 1}} - \frac{a d^{\frac{3}{2}}}{x \sqrt{\frac{c x^{2}}{d} + 1}} + b c^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )} - \frac{b c^{2} x}{\sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}} - \frac{b c \sqrt{d}}{x \sqrt{\frac{c x^{2}}{d} + 1}} + b d \left (\begin{cases} - \frac{\sqrt{c}}{2 x^{2}} & \text{for}\: d = 0 \\- \frac{\left (c + \frac{d}{x^{2}}\right )^{\frac{3}{2}}}{3 d} & \text{otherwise} \end{cases}\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)

[Out]

3*a*sqrt(c)*d*asinh(sqrt(c)*x/sqrt(d))/2 + a*c*sqrt(d)*x*sqrt(c*x**2/d + 1)/2 - a*c*sqrt(d)*x/sqrt(c*x**2/d +
1) - a*d**(3/2)/(x*sqrt(c*x**2/d + 1)) + b*c**(3/2)*asinh(sqrt(c)*x/sqrt(d)) - b*c**2*x/(sqrt(d)*sqrt(c*x**2/d
+ 1)) - b*c*sqrt(d)/(x*sqrt(c*x**2/d + 1)) + b*d*Piecewise((-sqrt(c)/(2*x**2), Eq(d, 0)), (-(c + d/x**2)**(3/
2)/(3*d), True))

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Giac [B]  time = 1.71049, size = 304, normalized size = 2.76 \begin{align*} \frac{1}{2} \, \sqrt{c x^{2} + d} a c x \mathrm{sgn}\left (x\right ) - \frac{1}{4} \,{\left (2 \, b c^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 3 \, a \sqrt{c} d \mathrm{sgn}\left (x\right )\right )} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2}\right ) + \frac{2 \,{\left (6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} b c^{\frac{3}{2}} d \mathrm{sgn}\left (x\right ) + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{4} a \sqrt{c} d^{2} \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} b c^{\frac{3}{2}} d^{2} \mathrm{sgn}\left (x\right ) - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} a \sqrt{c} d^{3} \mathrm{sgn}\left (x\right ) + 4 \, b c^{\frac{3}{2}} d^{3} \mathrm{sgn}\left (x\right ) + 3 \, a \sqrt{c} d^{4} \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)^(3/2)*x,x, algorithm="giac")

[Out]

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