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

Optimal. Leaf size=112 \[ -\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (4 a d+b c)}{4 c x}-\frac{3 \sqrt{c+\frac{d}{x^2}} (4 a d+b c)}{8 x}-\frac{3 c (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 \sqrt{d}}+\frac{a x \left (c+\frac{d}{x^2}\right )^{5/2}}{c} \]

[Out]

(-3*(b*c + 4*a*d)*Sqrt[c + d/x^2])/(8*x) - ((b*c + 4*a*d)*(c + d/x^2)^(3/2))/(4*c*x) + (a*(c + d/x^2)^(5/2)*x)
/c - (3*c*(b*c + 4*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(8*Sqrt[d])

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Rubi [A]  time = 0.0623911, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {375, 453, 195, 217, 206} \[ -\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (4 a d+b c)}{4 c x}-\frac{3 \sqrt{c+\frac{d}{x^2}} (4 a d+b c)}{8 x}-\frac{3 c (4 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 \sqrt{d}}+\frac{a x \left (c+\frac{d}{x^2}\right )^{5/2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*(c + d/x^2)^(3/2),x]

[Out]

(-3*(b*c + 4*a*d)*Sqrt[c + d/x^2])/(8*x) - ((b*c + 4*a*d)*(c + d/x^2)^(3/2))/(4*c*x) + (a*(c + d/x^2)^(5/2)*x)
/c - (3*c*(b*c + 4*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(8*Sqrt[d])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 453

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x^2}\right ) \left (c+\frac{d}{x^2}\right )^{3/2} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x}{c}+\frac{(-b c-4 a d) \operatorname{Subst}\left (\int \left (c+d x^2\right )^{3/2} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{(b c+4 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{4 c x}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x}{c}-\frac{1}{4} (3 (b c+4 a d)) \operatorname{Subst}\left (\int \sqrt{c+d x^2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{3 (b c+4 a d) \sqrt{c+\frac{d}{x^2}}}{8 x}-\frac{(b c+4 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{4 c x}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x}{c}-\frac{1}{8} (3 c (b c+4 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{3 (b c+4 a d) \sqrt{c+\frac{d}{x^2}}}{8 x}-\frac{(b c+4 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{4 c x}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x}{c}-\frac{1}{8} (3 c (b c+4 a d)) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{1}{\sqrt{c+\frac{d}{x^2}} x}\right )\\ &=-\frac{3 (b c+4 a d) \sqrt{c+\frac{d}{x^2}}}{8 x}-\frac{(b c+4 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{4 c x}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x}{c}-\frac{3 c (b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{8 \sqrt{d}}\\ \end{align*}

Mathematica [C]  time = 0.0321634, size = 68, normalized size = 0.61 \[ \frac{\sqrt{c+\frac{d}{x^2}} \left (c x^2+d\right )^2 \left (c x^4 (4 a d+b c) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{c x^2}{d}+1\right )-5 b d^2\right )}{20 d^3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*(d + c*x^2)^2*(-5*b*d^2 + c*(b*c + 4*a*d)*x^4*Hypergeometric2F1[2, 5/2, 7/2, 1 + (c*x^2)/d]))
/(20*d^3*x^3)

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Maple [B]  time = 0.012, size = 213, normalized size = 1.9 \begin{align*} -{\frac{1}{8\,x{d}^{2}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 12\,{d}^{5/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}ac+3\,{d}^{3/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{4}b{c}^{2}-4\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{4}acd- \left ( c{x}^{2}+d \right ) ^{{\frac{3}{2}}}{x}^{4}b{c}^{2}+4\, \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}ad+ \left ( c{x}^{2}+d \right ) ^{{\frac{5}{2}}}{x}^{2}bc-12\,\sqrt{c{x}^{2}+d}{x}^{4}ac{d}^{2}-3\,\sqrt{c{x}^{2}+d}{x}^{4}b{c}^{2}d+2\, \left ( c{x}^{2}+d \right ) ^{5/2}bd \right ) \left ( c{x}^{2}+d \right ) ^{-{\frac{3}{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)

[Out]

-1/8*((c*x^2+d)/x^2)^(3/2)/x*(12*d^(5/2)*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*x^4*a*c+3*d^(3/2)*ln(2*(d^(1/2)*(
c*x^2+d)^(1/2)+d)/x)*x^4*b*c^2-4*(c*x^2+d)^(3/2)*x^4*a*c*d-(c*x^2+d)^(3/2)*x^4*b*c^2+4*(c*x^2+d)^(5/2)*x^2*a*d
+(c*x^2+d)^(5/2)*x^2*b*c-12*(c*x^2+d)^(1/2)*x^4*a*c*d^2-3*(c*x^2+d)^(1/2)*x^4*b*c^2*d+2*(c*x^2+d)^(5/2)*b*d)/(
c*x^2+d)^(3/2)/d^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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.40256, size = 490, normalized size = 4.38 \begin{align*} \left [\frac{3 \,{\left (b c^{2} + 4 \, a c d\right )} \sqrt{d} x^{3} \log \left (-\frac{c x^{2} - 2 \, \sqrt{d} x \sqrt{\frac{c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \,{\left (8 \, a c d x^{4} - 2 \, b d^{2} -{\left (5 \, b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \, d x^{3}}, \frac{3 \,{\left (b c^{2} + 4 \, a c d\right )} \sqrt{-d} x^{3} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left (8 \, a c d x^{4} - 2 \, b d^{2} -{\left (5 \, b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \, d x^{3}}\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, algorithm="fricas")

[Out]

[1/16*(3*(b*c^2 + 4*a*c*d)*sqrt(d)*x^3*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) + 2*(8*a*c*
d*x^4 - 2*b*d^2 - (5*b*c*d + 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d*x^3), 1/8*(3*(b*c^2 + 4*a*c*d)*sqrt(-d)*x
^3*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (8*a*c*d*x^4 - 2*b*d^2 - (5*b*c*d + 4*a*d^2)*x^2)*sq
rt((c*x^2 + d)/x^2))/(d*x^3)]

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Sympy [B]  time = 11.1277, size = 216, normalized size = 1.93 \begin{align*} \frac{a c^{\frac{3}{2}} x}{\sqrt{1 + \frac{d}{c x^{2}}}} - \frac{a \sqrt{c} d \sqrt{1 + \frac{d}{c x^{2}}}}{2 x} + \frac{a \sqrt{c} d}{x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 a c \sqrt{d} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2} - \frac{b c^{\frac{3}{2}} \sqrt{1 + \frac{d}{c x^{2}}}}{2 x} - \frac{b c^{\frac{3}{2}}}{8 x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 b \sqrt{c} d}{8 x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 b c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 \sqrt{d}} - \frac{b d^{2}}{4 \sqrt{c} x^{5} \sqrt{1 + \frac{d}{c 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)

[Out]

a*c**(3/2)*x/sqrt(1 + d/(c*x**2)) - a*sqrt(c)*d*sqrt(1 + d/(c*x**2))/(2*x) + a*sqrt(c)*d/(x*sqrt(1 + d/(c*x**2
))) - 3*a*c*sqrt(d)*asinh(sqrt(d)/(sqrt(c)*x))/2 - b*c**(3/2)*sqrt(1 + d/(c*x**2))/(2*x) - b*c**(3/2)/(8*x*sqr
t(1 + d/(c*x**2))) - 3*b*sqrt(c)*d/(8*x**3*sqrt(1 + d/(c*x**2))) - 3*b*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*sqrt
(d)) - b*d**2/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2)))

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Giac [A]  time = 1.16603, size = 196, normalized size = 1.75 \begin{align*} \frac{8 \, \sqrt{c x^{2} + d} a c^{2} \mathrm{sgn}\left (x\right ) + \frac{3 \,{\left (b c^{3} \mathrm{sgn}\left (x\right ) + 4 \, a c^{2} d \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + d}}{\sqrt{-d}}\right )}{\sqrt{-d}} - \frac{5 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} b c^{3} \mathrm{sgn}\left (x\right ) + 4 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} a c^{2} d \mathrm{sgn}\left (x\right ) - 3 \, \sqrt{c x^{2} + d} b c^{3} d \mathrm{sgn}\left (x\right ) - 4 \, \sqrt{c x^{2} + d} a c^{2} d^{2} \mathrm{sgn}\left (x\right )}{c^{2} x^{4}}}{8 \, c} \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, algorithm="giac")

[Out]

1/8*(8*sqrt(c*x^2 + d)*a*c^2*sgn(x) + 3*(b*c^3*sgn(x) + 4*a*c^2*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/sqr
t(-d) - (5*(c*x^2 + d)^(3/2)*b*c^3*sgn(x) + 4*(c*x^2 + d)^(3/2)*a*c^2*d*sgn(x) - 3*sqrt(c*x^2 + d)*b*c^3*d*sgn
(x) - 4*sqrt(c*x^2 + d)*a*c^2*d^2*sgn(x))/(c^2*x^4))/c