3.944 \(\int \frac{(a+\frac{b}{x^2}) \sqrt{c+\frac{d}{x^2}}}{x^4} \, dx\)

Optimal. Leaf size=123 \[ -\frac{c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{16 d^{5/2}}+\frac{c \sqrt{c+\frac{d}{x^2}} (b c-2 a d)}{16 d^2 x}+\frac{\sqrt{c+\frac{d}{x^2}} (b c-2 a d)}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3} \]

[Out]

((b*c - 2*a*d)*Sqrt[c + d/x^2])/(8*d*x^3) - (b*(c + d/x^2)^(3/2))/(6*d*x^3) + (c*(b*c - 2*a*d)*Sqrt[c + d/x^2]
)/(16*d^2*x) - (c^2*(b*c - 2*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*d^(5/2))

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Rubi [A]  time = 0.0805208, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {459, 335, 279, 321, 217, 206} \[ -\frac{c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{16 d^{5/2}}+\frac{c \sqrt{c+\frac{d}{x^2}} (b c-2 a d)}{16 d^2 x}+\frac{\sqrt{c+\frac{d}{x^2}} (b c-2 a d)}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*Sqrt[c + d/x^2])/x^4,x]

[Out]

((b*c - 2*a*d)*Sqrt[c + d/x^2])/(8*d*x^3) - (b*(c + d/x^2)^(3/2))/(6*d*x^3) + (c*(b*c - 2*a*d)*Sqrt[c + d/x^2]
)/(16*d^2*x) - (c^2*(b*c - 2*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*d^(5/2))

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 279

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

Rule 321

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

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 ) \sqrt{c+\frac{d}{x^2}}}{x^4} \, dx &=-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}+\frac{(-3 b c+6 a d) \int \frac{\sqrt{c+\frac{d}{x^2}}}{x^4} \, dx}{6 d}\\ &=-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}-\frac{(-3 b c+6 a d) \operatorname{Subst}\left (\int x^2 \sqrt{c+d x^2} \, dx,x,\frac{1}{x}\right )}{6 d}\\ &=\frac{(b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}+\frac{(c (b c-2 a d)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{8 d}\\ &=\frac{(b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}+\frac{c (b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{16 d^2 x}-\frac{\left (c^2 (b c-2 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{16 d^2}\\ &=\frac{(b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}+\frac{c (b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{16 d^2 x}-\frac{\left (c^2 (b c-2 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^2}\\ &=\frac{(b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{8 d x^3}-\frac{b \left (c+\frac{d}{x^2}\right )^{3/2}}{6 d x^3}+\frac{c (b c-2 a d) \sqrt{c+\frac{d}{x^2}}}{16 d^2 x}-\frac{c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{16 d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.025872, size = 68, normalized size = 0.55 \[ \frac{\sqrt{c+\frac{d}{x^2}} \left (c x^2+d\right ) \left (c^2 x^6 (b c-2 a d) \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x^2}{d}+1\right )-b d^3\right )}{6 d^4 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x^2)*Sqrt[c + d/x^2])/x^4,x]

[Out]

(Sqrt[c + d/x^2]*(d + c*x^2)*(-(b*d^3) + c^2*(b*c - 2*a*d)*x^6*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x^2)/d]))
/(6*d^4*x^5)

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Maple [B]  time = 0.014, size = 220, normalized size = 1.8 \begin{align*}{\frac{1}{48\,{x}^{5}{d}^{3}}\sqrt{{\frac{c{x}^{2}+d}{{x}^{2}}}} \left ( 6\,{d}^{3/2}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{6}a{c}^{2}-3\,\sqrt{d}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ){x}^{6}b{c}^{3}-6\,\sqrt{c{x}^{2}+d}{x}^{6}a{c}^{2}d+3\,\sqrt{c{x}^{2}+d}{x}^{6}b{c}^{3}+6\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{4}acd-3\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{4}b{c}^{2}-12\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}a{d}^{2}+6\, \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}bcd-8\, \left ( c{x}^{2}+d \right ) ^{3/2}b{d}^{2} \right ){\frac{1}{\sqrt{c{x}^{2}+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*(c+d/x^2)^(1/2)/x^4,x)

[Out]

1/48*((c*x^2+d)/x^2)^(1/2)/x^5*(6*d^(3/2)*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*x^6*a*c^2-3*d^(1/2)*ln(2*(d^(1/2
)*(c*x^2+d)^(1/2)+d)/x)*x^6*b*c^3-6*(c*x^2+d)^(1/2)*x^6*a*c^2*d+3*(c*x^2+d)^(1/2)*x^6*b*c^3+6*(c*x^2+d)^(3/2)*
x^4*a*c*d-3*(c*x^2+d)^(3/2)*x^4*b*c^2-12*(c*x^2+d)^(3/2)*x^2*a*d^2+6*(c*x^2+d)^(3/2)*x^2*b*c*d-8*(c*x^2+d)^(3/
2)*b*d^2)/(c*x^2+d)^(1/2)/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)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38809, size = 552, normalized size = 4.49 \begin{align*} \left [-\frac{3 \,{\left (b c^{3} - 2 \, 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 - 2 \, a c d^{2}\right )} x^{4} - 8 \, b d^{3} - 2 \,{\left (b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{96 \, d^{3} x^{5}}, \frac{3 \,{\left (b c^{3} - 2 \, 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 - 2 \, a c d^{2}\right )} x^{4} - 8 \, b d^{3} - 2 \,{\left (b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{48 \, d^{3} 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)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 10.8581, size = 226, normalized size = 1.84 \begin{align*} - \frac{a c^{\frac{3}{2}}}{8 d x \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{3 a \sqrt{c}}{8 x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{a c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 d^{\frac{3}{2}}} - \frac{a d}{4 \sqrt{c} x^{5} \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{b c^{\frac{5}{2}}}{16 d^{2} x \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{b c^{\frac{3}{2}}}{48 d x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{5 b \sqrt{c}}{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{5}{2}}} - \frac{b d}{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)**(1/2)/x**4,x)

[Out]

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

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Giac [A]  time = 1.24678, size = 207, normalized size = 1.68 \begin{align*} \frac{\frac{3 \,{\left (b c^{4} \mathrm{sgn}\left (x\right ) - 2 \, a c^{3} d \mathrm{sgn}\left (x\right )\right )} \arctan \left (\frac{\sqrt{c x^{2} + d}}{\sqrt{-d}}\right )}{\sqrt{-d} d^{2}} + \frac{3 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} b c^{4} \mathrm{sgn}\left (x\right ) - 6 \,{\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 ) - 3 \, \sqrt{c x^{2} + d} b c^{4} d^{2} \mathrm{sgn}\left (x\right ) + 6 \, \sqrt{c x^{2} + d} a c^{3} d^{3} \mathrm{sgn}\left (x\right )}{c^{3} d^{2} 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)^(1/2)/x^4,x, algorithm="giac")

[Out]

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