3.984 \(\int \frac{a+\frac{b}{x^2}}{(c+\frac{d}{x^2})^{3/2} x^4} \, dx\)

Optimal. Leaf size=92 \[ -\frac{3 b c-2 a d}{2 d^2 x \sqrt{c+\frac{d}{x^2}}}+\frac{(3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 d^{5/2}}-\frac{b}{2 d x^3 \sqrt{c+\frac{d}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.050077, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 335, 288, 217, 206} \[ -\frac{3 b c-2 a d}{2 d^2 x \sqrt{c+\frac{d}{x^2}}}+\frac{(3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{2 d^{5/2}}-\frac{b}{2 d x^3 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x^4),x]

[Out]

-b/(2*d*Sqrt[c + d/x^2]*x^3) - (3*b*c - 2*a*d)/(2*d^2*Sqrt[c + d/x^2]*x) + ((3*b*c - 2*a*d)*ArcTanh[Sqrt[d]/(S
qrt[c + d/x^2]*x)])/(2*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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{a+\frac{b}{x^2}}{\left (c+\frac{d}{x^2}\right )^{3/2} x^4} \, dx &=-\frac{b}{2 d \sqrt{c+\frac{d}{x^2}} x^3}+\frac{(-3 b c+2 a d) \int \frac{1}{\left (c+\frac{d}{x^2}\right )^{3/2} x^4} \, dx}{2 d}\\ &=-\frac{b}{2 d \sqrt{c+\frac{d}{x^2}} x^3}-\frac{(-3 b c+2 a d) \operatorname{Subst}\left (\int \frac{x^2}{\left (c+d x^2\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{2 d}\\ &=-\frac{b}{2 d \sqrt{c+\frac{d}{x^2}} x^3}-\frac{3 b c-2 a d}{2 d^2 \sqrt{c+\frac{d}{x^2}} x}+\frac{(3 b c-2 a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{2 d^2}\\ &=-\frac{b}{2 d \sqrt{c+\frac{d}{x^2}} x^3}-\frac{3 b c-2 a d}{2 d^2 \sqrt{c+\frac{d}{x^2}} x}+\frac{(3 b c-2 a d) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{1}{\sqrt{c+\frac{d}{x^2}} x}\right )}{2 d^2}\\ &=-\frac{b}{2 d \sqrt{c+\frac{d}{x^2}} x^3}-\frac{3 b c-2 a d}{2 d^2 \sqrt{c+\frac{d}{x^2}} x}+\frac{(3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{2 d^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0244457, size = 57, normalized size = 0.62 \[ \frac{x^2 (2 a d-3 b c) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{c x^2}{d}+1\right )-b d}{2 d^2 x^3 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x^4),x]

[Out]

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

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Maple [A]  time = 0.01, size = 132, normalized size = 1.4 \begin{align*}{\frac{c{x}^{2}+d}{2\,{x}^{5}} \left ( 2\,{d}^{5/2}{x}^{2}a-3\,{d}^{3/2}{x}^{2}bc-2\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ) \sqrt{c{x}^{2}+d}{x}^{2}a{d}^{2}+3\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ) \sqrt{c{x}^{2}+d}{x}^{2}bcd-{d}^{{\frac{5}{2}}}b \right ) \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{d}^{-{\frac{7}{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^4,x)

[Out]

1/2*(c*x^2+d)*(2*d^(5/2)*x^2*a-3*d^(3/2)*x^2*b*c-2*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*(c*x^2+d)^(1/2)*x^2*a*d
^2+3*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*(c*x^2+d)^(1/2)*x^2*b*c*d-d^(5/2)*b)/((c*x^2+d)/x^2)^(3/2)/x^5/d^(7/2
)

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.08669, size = 551, normalized size = 5.99 \begin{align*} \left [-\frac{{\left ({\left (3 \, b c^{2} - 2 \, a c d\right )} x^{3} +{\left (3 \, b c d - 2 \, a d^{2}\right )} x\right )} \sqrt{d} \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 (b d^{2} +{\left (3 \, b c d - 2 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{4 \,{\left (c d^{3} x^{3} + d^{4} x\right )}}, -\frac{{\left ({\left (3 \, b c^{2} - 2 \, a c d\right )} x^{3} +{\left (3 \, b c d - 2 \, a d^{2}\right )} x\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left (b d^{2} +{\left (3 \, b c d - 2 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{2 \,{\left (c d^{3} x^{3} + d^{4} x\right )}}\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^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 19.5526, size = 262, normalized size = 2.85 \begin{align*} a \left (\frac{c d^{2} x^{2} \log{\left (\frac{c x^{2}}{d} \right )}}{2 c d^{\frac{7}{2}} x^{2} + 2 d^{\frac{9}{2}}} - \frac{2 c d^{2} x^{2} \log{\left (\sqrt{\frac{c x^{2}}{d} + 1} + 1 \right )}}{2 c d^{\frac{7}{2}} x^{2} + 2 d^{\frac{9}{2}}} + \frac{2 d^{3} \sqrt{\frac{c x^{2}}{d} + 1}}{2 c d^{\frac{7}{2}} x^{2} + 2 d^{\frac{9}{2}}} + \frac{d^{3} \log{\left (\frac{c x^{2}}{d} \right )}}{2 c d^{\frac{7}{2}} x^{2} + 2 d^{\frac{9}{2}}} - \frac{2 d^{3} \log{\left (\sqrt{\frac{c x^{2}}{d} + 1} + 1 \right )}}{2 c d^{\frac{7}{2}} x^{2} + 2 d^{\frac{9}{2}}}\right ) + b \left (- \frac{3 \sqrt{c}}{2 d^{2} x \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{3 c \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{2 d^{\frac{5}{2}}} - \frac{1}{2 \sqrt{c} d x^{3} \sqrt{1 + \frac{d}{c 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**4,x)

[Out]

a*(c*d**2*x**2*log(c*x**2/d)/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) - 2*c*d**2*x**2*log(sqrt(c*x**2/d + 1) + 1)/(2*c
*d**(7/2)*x**2 + 2*d**(9/2)) + 2*d**3*sqrt(c*x**2/d + 1)/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) + d**3*log(c*x**2/d)
/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) - 2*d**3*log(sqrt(c*x**2/d + 1) + 1)/(2*c*d**(7/2)*x**2 + 2*d**(9/2))) + b*(
-3*sqrt(c)/(2*d**2*x*sqrt(1 + d/(c*x**2))) + 3*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*d**(5/2)) - 1/(2*sqrt(c)*d*x**3
*sqrt(1 + d/(c*x**2))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + \frac{b}{x^{2}}}{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} x^{4}}\,{d x} \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^4,x, algorithm="giac")

[Out]

integrate((a + b/x^2)/((c + d/x^2)^(3/2)*x^4), x)