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

Optimal. Leaf size=123 \[ \frac{c^2 (b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{16 d^{3/2}}+\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (b c-6 a d)}{24 d x}+\frac{c \sqrt{c+\frac{d}{x^2}} (b c-6 a d)}{16 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x} \]

[Out]

(c*(b*c - 6*a*d)*Sqrt[c + d/x^2])/(16*d*x) + ((b*c - 6*a*d)*(c + d/x^2)^(3/2))/(24*d*x) - (b*(c + d/x^2)^(5/2)
)/(6*d*x) + (c^2*(b*c - 6*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*d^(3/2))

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Rubi [A]  time = 0.0638926, antiderivative size = 123, 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 = {459, 335, 195, 217, 206} \[ \frac{c^2 (b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{16 d^{3/2}}+\frac{\left (c+\frac{d}{x^2}\right )^{3/2} (b c-6 a d)}{24 d x}+\frac{c \sqrt{c+\frac{d}{x^2}} (b c-6 a d)}{16 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(b*c - 6*a*d)*Sqrt[c + d/x^2])/(16*d*x) + ((b*c - 6*a*d)*(c + d/x^2)^(3/2))/(24*d*x) - (b*(c + d/x^2)^(5/2)
)/(6*d*x) + (c^2*(b*c - 6*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*d^(3/2))

Rule 459

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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 \frac{\left (a+\frac{b}{x^2}\right ) \left (c+\frac{d}{x^2}\right )^{3/2}}{x^2} \, dx &=-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}+\frac{(-b c+6 a d) \int \frac{\left (c+\frac{d}{x^2}\right )^{3/2}}{x^2} \, dx}{6 d}\\ &=-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}-\frac{(-b c+6 a d) \operatorname{Subst}\left (\int \left (c+d x^2\right )^{3/2} \, dx,x,\frac{1}{x}\right )}{6 d}\\ &=\frac{(b c-6 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{24 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}+\frac{(c (b c-6 a d)) \operatorname{Subst}\left (\int \sqrt{c+d x^2} \, dx,x,\frac{1}{x}\right )}{8 d}\\ &=\frac{c (b c-6 a d) \sqrt{c+\frac{d}{x^2}}}{16 d x}+\frac{(b c-6 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{24 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}+\frac{\left (c^2 (b c-6 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{16 d}\\ &=\frac{c (b c-6 a d) \sqrt{c+\frac{d}{x^2}}}{16 d x}+\frac{(b c-6 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{24 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}+\frac{\left (c^2 (b c-6 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{1}{\sqrt{c+\frac{d}{x^2}} x}\right )}{16 d}\\ &=\frac{c (b c-6 a d) \sqrt{c+\frac{d}{x^2}}}{16 d x}+\frac{(b c-6 a d) \left (c+\frac{d}{x^2}\right )^{3/2}}{24 d x}-\frac{b \left (c+\frac{d}{x^2}\right )^{5/2}}{6 d x}+\frac{c^2 (b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{16 d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.099036, size = 126, normalized size = 1.02 \[ -\frac{\sqrt{c+\frac{d}{x^2}} \left (\left (c x^2+d\right ) \left (6 a d x^2 \left (5 c x^2+2 d\right )+b \left (3 c^2 x^4+14 c d x^2+8 d^2\right )\right )+3 c^2 x^6 \sqrt{\frac{c x^2}{d}+1} (6 a d-b c) \tanh ^{-1}\left (\sqrt{\frac{c x^2}{d}+1}\right )\right )}{48 d x^5 \left (c x^2+d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c + d/x^2]*((d + c*x^2)*(6*a*d*x^2*(2*d + 5*c*x^2) + b*(8*d^2 + 14*c*d*x^2 + 3*c^2*x^4)) + 3*c^2*(-(b*c
) + 6*a*d)*x^6*Sqrt[1 + (c*x^2)/d]*ArcTanh[Sqrt[1 + (c*x^2)/d]]))/(48*d*x^5*(d + c*x^2))

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Maple [B]  time = 0.013, size = 259, normalized size = 2.1 \begin{align*} -{\frac{1}{48\,{x}^{3}{d}^{3}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 18\,{d}^{5/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{6}a{c}^{2}-3\,{d}^{3/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{6}b{c}^{3}-6\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{6}a{c}^{2}d+ \left ( c{x}^{2}+d \right ) ^{{\frac{3}{2}}}{x}^{6}b{c}^{3}+6\, \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{4}acd- \left ( c{x}^{2}+d \right ) ^{{\frac{5}{2}}}{x}^{4}b{c}^{2}-18\,\sqrt{c{x}^{2}+d}{x}^{6}a{c}^{2}{d}^{2}+3\,\sqrt{c{x}^{2}+d}{x}^{6}b{c}^{3}d+12\, \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}a{d}^{2}-2\, \left ( c{x}^{2}+d \right ) ^{5/2}{x}^{2}bcd+8\, \left ( c{x}^{2}+d \right ) ^{5/2}b{d}^{2} \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^2,x)

[Out]

-1/48*((c*x^2+d)/x^2)^(3/2)/x^3*(18*d^(5/2)*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*x^6*a*c^2-3*d^(3/2)*ln(2*(d^(1
/2)*(c*x^2+d)^(1/2)+d)/x)*x^6*b*c^3-6*(c*x^2+d)^(3/2)*x^6*a*c^2*d+(c*x^2+d)^(3/2)*x^6*b*c^3+6*(c*x^2+d)^(5/2)*
x^4*a*c*d-(c*x^2+d)^(5/2)*x^4*b*c^2-18*(c*x^2+d)^(1/2)*x^6*a*c^2*d^2+3*(c*x^2+d)^(1/2)*x^6*b*c^3*d+12*(c*x^2+d
)^(5/2)*x^2*a*d^2-2*(c*x^2+d)^(5/2)*x^2*b*c*d+8*(c*x^2+d)^(5/2)*b*d^2)/(c*x^2+d)^(3/2)/d^3

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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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45102, size = 562, normalized size = 4.57 \begin{align*} \left [-\frac{3 \,{\left (b c^{3} - 6 \, a c^{2} d\right )} \sqrt{d} x^{5} \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 (3 \,{\left (b c^{2} d + 10 \, a c d^{2}\right )} x^{4} + 8 \, b d^{3} + 2 \,{\left (7 \, b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{96 \, d^{2} x^{5}}, -\frac{3 \,{\left (b c^{3} - 6 \, a c^{2} d\right )} \sqrt{-d} x^{5} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left (3 \,{\left (b c^{2} d + 10 \, a c d^{2}\right )} x^{4} + 8 \, b d^{3} + 2 \,{\left (7 \, b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{48 \, d^{2} x^{5}}\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^2,x, algorithm="fricas")

[Out]

[-1/96*(3*(b*c^3 - 6*a*c^2*d)*sqrt(d)*x^5*log(-(c*x^2 - 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) + 2*(3*(
b*c^2*d + 10*a*c*d^2)*x^4 + 8*b*d^3 + 2*(7*b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5), -1/48*(3*
(b*c^3 - 6*a*c^2*d)*sqrt(-d)*x^5*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (3*(b*c^2*d + 10*a*c*d
^2)*x^4 + 8*b*d^3 + 2*(7*b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5)]

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Sympy [B]  time = 17.6721, size = 253, normalized size = 2.06 \begin{align*} - \frac{a c^{\frac{3}{2}} \sqrt{1 + \frac{d}{c x^{2}}}}{2 x} - \frac{a c^{\frac{3}{2}}}{8 x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 a \sqrt{c} d}{8 x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 a c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 \sqrt{d}} - \frac{a d^{2}}{4 \sqrt{c} x^{5} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{b c^{\frac{5}{2}}}{16 d x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{17 b c^{\frac{3}{2}}}{48 x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{11 b \sqrt{c} d}{24 x^{5} \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{b c^{3} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{16 d^{\frac{3}{2}}} - \frac{b d^{2}}{6 \sqrt{c} x^{7} \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**2,x)

[Out]

-a*c**(3/2)*sqrt(1 + d/(c*x**2))/(2*x) - a*c**(3/2)/(8*x*sqrt(1 + d/(c*x**2))) - 3*a*sqrt(c)*d/(8*x**3*sqrt(1
+ d/(c*x**2))) - 3*a*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*sqrt(d)) - a*d**2/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2))
) - b*c**(5/2)/(16*d*x*sqrt(1 + d/(c*x**2))) - 17*b*c**(3/2)/(48*x**3*sqrt(1 + d/(c*x**2))) - 11*b*sqrt(c)*d/(
24*x**5*sqrt(1 + d/(c*x**2))) + b*c**3*asinh(sqrt(d)/(sqrt(c)*x))/(16*d**(3/2)) - b*d**2/(6*sqrt(c)*x**7*sqrt(
1 + d/(c*x**2)))

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Giac [A]  time = 1.19783, size = 234, normalized size = 1.9 \begin{align*} -\frac{\frac{3 \,{\left (b c^{4} \mathrm{sgn}\left (x\right ) - 6 \, a c^{3} d \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + d}}{\sqrt{-d}}\right )}{\sqrt{-d} d} + \frac{3 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} b c^{4} \mathrm{sgn}\left (x\right ) + 30 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} a c^{3} d \mathrm{sgn}\left (x\right ) + 8 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} b c^{4} d \mathrm{sgn}\left (x\right ) - 48 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} a c^{3} d^{2} \mathrm{sgn}\left (x\right ) - 3 \, \sqrt{c x^{2} + d} b c^{4} d^{2} \mathrm{sgn}\left (x\right ) + 18 \, \sqrt{c x^{2} + d} a c^{3} d^{3} \mathrm{sgn}\left (x\right )}{c^{3} d x^{6}}}{48 \, 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^2,x, algorithm="giac")

[Out]

-1/48*(3*(b*c^4*sgn(x) - 6*a*c^3*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d) + (3*(c*x^2 + d)^(5/2
)*b*c^4*sgn(x) + 30*(c*x^2 + d)^(5/2)*a*c^3*d*sgn(x) + 8*(c*x^2 + d)^(3/2)*b*c^4*d*sgn(x) - 48*(c*x^2 + d)^(3/
2)*a*c^3*d^2*sgn(x) - 3*sqrt(c*x^2 + d)*b*c^4*d^2*sgn(x) + 18*sqrt(c*x^2 + d)*a*c^3*d^3*sgn(x))/(c^3*d*x^6))/c