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

Optimal. Leaf size=79 \[ \frac{2 x \sqrt{c+\frac{d}{x^2}} (3 b c-4 a d)}{3 c^3}-\frac{x (3 b c-4 a d)}{3 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^3}{3 c \sqrt{c+\frac{d}{x^2}}} \]

[Out]

-((3*b*c - 4*a*d)*x)/(3*c^2*Sqrt[c + d/x^2]) + (2*(3*b*c - 4*a*d)*Sqrt[c + d/x^2]*x)/(3*c^3) + (a*x^3)/(3*c*Sq
rt[c + d/x^2])

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Rubi [A]  time = 0.0301135, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {453, 192, 191} \[ \frac{2 x \sqrt{c+\frac{d}{x^2}} (3 b c-4 a d)}{3 c^3}-\frac{x (3 b c-4 a d)}{3 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^3}{3 c \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*c - 4*a*d)*x)/(3*c^2*Sqrt[c + d/x^2]) + (2*(3*b*c - 4*a*d)*Sqrt[c + d/x^2]*x)/(3*c^3) + (a*x^3)/(3*c*Sq
rt[c + d/x^2])

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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0376584, size = 57, normalized size = 0.72 \[ \frac{a \left (c^2 x^4-4 c d x^2-8 d^2\right )+3 b c \left (c x^2+2 d\right )}{3 c^3 x \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 66, normalized size = 0.8 \begin{align*}{\frac{ \left ( a{x}^{4}{c}^{2}-4\,acd{x}^{2}+3\,b{c}^{2}{x}^{2}-8\,a{d}^{2}+6\,bcd \right ) \left ( c{x}^{2}+d \right ) }{3\,{x}^{3}{c}^{3}} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*c^2*x^4-4*a*c*d*x^2+3*b*c^2*x^2-8*a*d^2+6*b*c*d)*(c*x^2+d)/((c*x^2+d)/x^2)^(3/2)/x^3/c^3

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Maxima [A]  time = 0.942641, size = 122, normalized size = 1.54 \begin{align*} b{\left (\frac{\sqrt{c + \frac{d}{x^{2}}} x}{c^{2}} + \frac{d}{\sqrt{c + \frac{d}{x^{2}}} c^{2} x}\right )} + \frac{1}{3} \, a{\left (\frac{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} x^{3} - 6 \, \sqrt{c + \frac{d}{x^{2}}} d x}{c^{3}} - \frac{3 \, d^{2}}{\sqrt{c + \frac{d}{x^{2}}} c^{3} x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^2/(c+d/x^2)^(3/2),x, algorithm="maxima")

[Out]

b*(sqrt(c + d/x^2)*x/c^2 + d/(sqrt(c + d/x^2)*c^2*x)) + 1/3*a*(((c + d/x^2)^(3/2)*x^3 - 6*sqrt(c + d/x^2)*d*x)
/c^3 - 3*d^2/(sqrt(c + d/x^2)*c^3*x))

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Fricas [A]  time = 1.31819, size = 147, normalized size = 1.86 \begin{align*} \frac{{\left (a c^{2} x^{5} +{\left (3 \, b c^{2} - 4 \, a c d\right )} x^{3} + 2 \,{\left (3 \, b c d - 4 \, a d^{2}\right )} x\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{3 \,{\left (c^{4} x^{2} + c^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^2/(c+d/x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 7.56112, size = 267, normalized size = 3.38 \begin{align*} a \left (\frac{c^{3} d^{\frac{9}{2}} x^{6} \sqrt{\frac{c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac{3 c^{2} d^{\frac{11}{2}} x^{4} \sqrt{\frac{c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac{12 c d^{\frac{13}{2}} x^{2} \sqrt{\frac{c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac{8 d^{\frac{15}{2}} \sqrt{\frac{c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}}\right ) + b \left (\frac{x^{2}}{c \sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}} + \frac{2 \sqrt{d}}{c^{2} \sqrt{\frac{c x^{2}}{d} + 1}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(c**3*d**(9/2)*x**6*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 3*c**2*d**(11/2
)*x**4*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 12*c*d**(13/2)*x**2*sqrt(c*x**
2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 8*d**(15/2)*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**
4 + 6*c**4*d**5*x**2 + 3*c**3*d**6)) + b*(x**2/(c*sqrt(d)*sqrt(c*x**2/d + 1)) + 2*sqrt(d)/(c**2*sqrt(c*x**2/d
+ 1)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^2/(c+d/x^2)^(3/2),x, algorithm="giac")

[Out]

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