3.985 \(\int \frac{a+\frac{b}{x^2}}{(c+\frac{d}{x^2})^{3/2} x^6} \, dx\)

Optimal. Leaf size=123 \[ \frac{3 \sqrt{c+\frac{d}{x^2}} (5 b c-4 a d)}{8 d^3 x}-\frac{5 b c-4 a d}{4 d^2 x^3 \sqrt{c+\frac{d}{x^2}}}-\frac{3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{7/2}}-\frac{b}{4 d x^5 \sqrt{c+\frac{d}{x^2}}} \]

[Out]

-b/(4*d*Sqrt[c + d/x^2]*x^5) - (5*b*c - 4*a*d)/(4*d^2*Sqrt[c + d/x^2]*x^3) + (3*(5*b*c - 4*a*d)*Sqrt[c + d/x^2
])/(8*d^3*x) - (3*c*(5*b*c - 4*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(8*d^(7/2))

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Rubi [A]  time = 0.0689806, 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, 288, 321, 217, 206} \[ \frac{3 \sqrt{c+\frac{d}{x^2}} (5 b c-4 a d)}{8 d^3 x}-\frac{5 b c-4 a d}{4 d^2 x^3 \sqrt{c+\frac{d}{x^2}}}-\frac{3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{x \sqrt{c+\frac{d}{x^2}}}\right )}{8 d^{7/2}}-\frac{b}{4 d x^5 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x^6),x]

[Out]

-b/(4*d*Sqrt[c + d/x^2]*x^5) - (5*b*c - 4*a*d)/(4*d^2*Sqrt[c + d/x^2]*x^3) + (3*(5*b*c - 4*a*d)*Sqrt[c + d/x^2
])/(8*d^3*x) - (3*c*(5*b*c - 4*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(8*d^(7/2))

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 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{a+\frac{b}{x^2}}{\left (c+\frac{d}{x^2}\right )^{3/2} x^6} \, dx &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}+\frac{(-5 b c+4 a d) \int \frac{1}{\left (c+\frac{d}{x^2}\right )^{3/2} x^6} \, dx}{4 d}\\ &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}-\frac{(-5 b c+4 a d) \operatorname{Subst}\left (\int \frac{x^4}{\left (c+d x^2\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{4 d}\\ &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}-\frac{5 b c-4 a d}{4 d^2 \sqrt{c+\frac{d}{x^2}} x^3}+\frac{(3 (5 b c-4 a d)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{4 d^2}\\ &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}-\frac{5 b c-4 a d}{4 d^2 \sqrt{c+\frac{d}{x^2}} x^3}+\frac{3 (5 b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d^3 x}-\frac{(3 c (5 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x^2}} \, dx,x,\frac{1}{x}\right )}{8 d^3}\\ &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}-\frac{5 b c-4 a d}{4 d^2 \sqrt{c+\frac{d}{x^2}} x^3}+\frac{3 (5 b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d^3 x}-\frac{(3 c (5 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 )}{8 d^3}\\ &=-\frac{b}{4 d \sqrt{c+\frac{d}{x^2}} x^5}-\frac{5 b c-4 a d}{4 d^2 \sqrt{c+\frac{d}{x^2}} x^3}+\frac{3 (5 b c-4 a d) \sqrt{c+\frac{d}{x^2}}}{8 d^3 x}-\frac{3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d}}{\sqrt{c+\frac{d}{x^2}} x}\right )}{8 d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0272797, size = 60, normalized size = 0.49 \[ \frac{c x^4 (5 b c-4 a d) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{c x^2}{d}+1\right )-b d^2}{4 d^3 x^5 \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x^6),x]

[Out]

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

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Maple [A]  time = 0.011, size = 157, normalized size = 1.3 \begin{align*} -{\frac{c{x}^{2}+d}{8\,{x}^{7}} \left ( 12\,{d}^{5/2}{x}^{4}ac-15\,{d}^{3/2}{x}^{4}b{c}^{2}-12\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ) \sqrt{c{x}^{2}+d}{x}^{4}ac{d}^{2}+15\,\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{c{x}^{2}+d}+d}{x}} \right ) \sqrt{c{x}^{2}+d}{x}^{4}b{c}^{2}d+4\,{d}^{7/2}{x}^{2}a-5\,{d}^{5/2}{x}^{2}bc+2\,{d}^{7/2}b \right ) \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{d}^{-{\frac{9}{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^6,x)

[Out]

-1/8*(c*x^2+d)*(12*d^(5/2)*x^4*a*c-15*d^(3/2)*x^4*b*c^2-12*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*(c*x^2+d)^(1/2)
*x^4*a*c*d^2+15*ln(2*(d^(1/2)*(c*x^2+d)^(1/2)+d)/x)*(c*x^2+d)^(1/2)*x^4*b*c^2*d+4*d^(7/2)*x^2*a-5*d^(5/2)*x^2*
b*c+2*d^(7/2)*b)/((c*x^2+d)/x^2)^(3/2)/x^7/d^(9/2)

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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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.13007, size = 680, normalized size = 5.53 \begin{align*} \left [-\frac{3 \,{\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} +{\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\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 (3 \,{\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} +{\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \,{\left (c d^{4} x^{5} + d^{5} x^{3}\right )}}, \frac{3 \,{\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} +{\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x \sqrt{\frac{c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) +{\left (3 \,{\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} +{\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \,{\left (c d^{4} x^{5} + d^{5} x^{3}\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^6,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 34.5534, size = 180, normalized size = 1.46 \begin{align*} a \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 ) + b \left (\frac{15 c^{\frac{3}{2}}}{8 d^{3} x \sqrt{1 + \frac{d}{c x^{2}}}} + \frac{5 \sqrt{c}}{8 d^{2} x^{3} \sqrt{1 + \frac{d}{c x^{2}}}} - \frac{15 c^{2} \operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{c} x} \right )}}{8 d^{\frac{7}{2}}} - \frac{1}{4 \sqrt{c} d x^{5} \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**6,x)

[Out]

a*(-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)))) + b*(15*c**(3/2)/(8*d**3*x*sqrt(1 + d/(c*x**2))) + 5*sqrt(c)/(8*d**2*x**3*sqrt(1 +
d/(c*x**2))) - 15*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d**(7/2)) - 1/(4*sqrt(c)*d*x**5*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^{6}}\,{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^6,x, algorithm="giac")

[Out]

integrate((a + b/x^2)/((c + d/x^2)^(3/2)*x^6), x)