3.955 \(\int (a+\frac{b}{x^2}) (c+\frac{d}{x^2})^{3/2} x^8 \, dx\)

Optimal. Leaf size=84 \[ \frac{x^7 \left (c+\frac{d}{x^2}\right )^{5/2} (9 b c-4 a d)}{63 c^2}-\frac{2 d x^5 \left (c+\frac{d}{x^2}\right )^{5/2} (9 b c-4 a d)}{315 c^3}+\frac{a x^9 \left (c+\frac{d}{x^2}\right )^{5/2}}{9 c} \]

[Out]

(-2*d*(9*b*c - 4*a*d)*(c + d/x^2)^(5/2)*x^5)/(315*c^3) + ((9*b*c - 4*a*d)*(c + d/x^2)^(5/2)*x^7)/(63*c^2) + (a
*(c + d/x^2)^(5/2)*x^9)/(9*c)

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Rubi [A]  time = 0.0415588, antiderivative size = 84, 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, 271, 264} \[ \frac{x^7 \left (c+\frac{d}{x^2}\right )^{5/2} (9 b c-4 a d)}{63 c^2}-\frac{2 d x^5 \left (c+\frac{d}{x^2}\right )^{5/2} (9 b c-4 a d)}{315 c^3}+\frac{a x^9 \left (c+\frac{d}{x^2}\right )^{5/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*(c + d/x^2)^(3/2)*x^8,x]

[Out]

(-2*d*(9*b*c - 4*a*d)*(c + d/x^2)^(5/2)*x^5)/(315*c^3) + ((9*b*c - 4*a*d)*(c + d/x^2)^(5/2)*x^7)/(63*c^2) + (a
*(c + d/x^2)^(5/2)*x^9)/(9*c)

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 271

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

Rule 264

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

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x^2}\right ) \left (c+\frac{d}{x^2}\right )^{3/2} x^8 \, dx &=\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^9}{9 c}+\frac{(9 b c-4 a d) \int \left (c+\frac{d}{x^2}\right )^{3/2} x^6 \, dx}{9 c}\\ &=\frac{(9 b c-4 a d) \left (c+\frac{d}{x^2}\right )^{5/2} x^7}{63 c^2}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^9}{9 c}-\frac{(2 d (9 b c-4 a d)) \int \left (c+\frac{d}{x^2}\right )^{3/2} x^4 \, dx}{63 c^2}\\ &=-\frac{2 d (9 b c-4 a d) \left (c+\frac{d}{x^2}\right )^{5/2} x^5}{315 c^3}+\frac{(9 b c-4 a d) \left (c+\frac{d}{x^2}\right )^{5/2} x^7}{63 c^2}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^9}{9 c}\\ \end{align*}

Mathematica [A]  time = 0.0413909, size = 66, normalized size = 0.79 \[ \frac{x \sqrt{c+\frac{d}{x^2}} \left (c x^2+d\right )^2 \left (a \left (35 c^2 x^4-20 c d x^2+8 d^2\right )+9 b c \left (5 c x^2-2 d\right )\right )}{315 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)*(c + d/x^2)^(3/2)*x^8,x]

[Out]

(Sqrt[c + d/x^2]*x*(d + c*x^2)^2*(9*b*c*(-2*d + 5*c*x^2) + a*(8*d^2 - 20*c*d*x^2 + 35*c^2*x^4)))/(315*c^3)

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Maple [A]  time = 0.006, size = 67, normalized size = 0.8 \begin{align*}{\frac{{x}^{3} \left ( 35\,a{x}^{4}{c}^{2}-20\,acd{x}^{2}+45\,b{c}^{2}{x}^{2}+8\,a{d}^{2}-18\,bcd \right ) \left ( c{x}^{2}+d \right ) }{315\,{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)*(c+d/x^2)^(3/2)*x^8,x)

[Out]

1/315*((c*x^2+d)/x^2)^(3/2)*x^3*(35*a*c^2*x^4-20*a*c*d*x^2+45*b*c^2*x^2+8*a*d^2-18*b*c*d)*(c*x^2+d)/c^3

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Maxima [A]  time = 0.960015, size = 122, normalized size = 1.45 \begin{align*} \frac{{\left (5 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{7}{2}} x^{7} - 7 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}} d x^{5}\right )} b}{35 \, c^{2}} + \frac{{\left (35 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{9}{2}} x^{9} - 90 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{7}{2}} d x^{7} + 63 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}} d^{2} x^{5}\right )} a}{315 \, c^{3}} \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^8,x, algorithm="maxima")

[Out]

1/35*(5*(c + d/x^2)^(7/2)*x^7 - 7*(c + d/x^2)^(5/2)*d*x^5)*b/c^2 + 1/315*(35*(c + d/x^2)^(9/2)*x^9 - 90*(c + d
/x^2)^(7/2)*d*x^7 + 63*(c + d/x^2)^(5/2)*d^2*x^5)*a/c^3

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Fricas [A]  time = 1.33022, size = 232, normalized size = 2.76 \begin{align*} \frac{{\left (35 \, a c^{4} x^{9} + 5 \,{\left (9 \, b c^{4} + 10 \, a c^{3} d\right )} x^{7} + 3 \,{\left (24 \, b c^{3} d + a c^{2} d^{2}\right )} x^{5} +{\left (9 \, b c^{2} d^{2} - 4 \, a c d^{3}\right )} x^{3} - 2 \,{\left (9 \, b c d^{3} - 4 \, a d^{4}\right )} x\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{315 \, c^{3}} \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^8,x, algorithm="fricas")

[Out]

1/315*(35*a*c^4*x^9 + 5*(9*b*c^4 + 10*a*c^3*d)*x^7 + 3*(24*b*c^3*d + a*c^2*d^2)*x^5 + (9*b*c^2*d^2 - 4*a*c*d^3
)*x^3 - 2*(9*b*c*d^3 - 4*a*d^4)*x)*sqrt((c*x^2 + d)/x^2)/c^3

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Sympy [B]  time = 10.1723, size = 1340, normalized size = 15.95 \begin{align*} \text{result too large to display} \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**8,x)

[Out]

35*a*c**8*d**(19/2)*x**14*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 +
315*c**4*d**12) + 110*a*c**7*d**(21/2)*x**12*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 9
45*c**5*d**11*x**2 + 315*c**4*d**12) + 114*a*c**6*d**(23/2)*x**10*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945
*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 40*a*c**5*d**(25/2)*x**8*sqrt(c*x**2/d + 1)/(315*c*
*7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 15*a*c**5*d**(11/2)*x**10*sqrt(c*
x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) - 5*a*c**4*d**(27/2)*x**6*sqrt(c*x**2/d
+ 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 33*a*c**4*d**(13/2)*x
**8*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) - 30*a*c**3*d**(29/2)*x**4*sq
rt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 17*a*c**3
*d**(15/2)*x**6*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) - 40*a*c**2*d**(3
1/2)*x**2*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12)
+ 3*a*c**2*d**(17/2)*x**4*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) - 16*a
*c*d**(33/2)*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**
12) + 12*a*c*d**(19/2)*x**2*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 8*a
*d**(21/2)*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 15*b*c**6*d**(9/2)*x
**10*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 33*b*c**5*d**(11/2)*x**8*s
qrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 17*b*c**4*d**(13/2)*x**6*sqrt(c*
x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 3*b*c**3*d**(15/2)*x**4*sqrt(c*x**2/d
+ 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 12*b*c**2*d**(17/2)*x**2*sqrt(c*x**2/d + 1)/(
105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 8*b*c*d**(19/2)*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x
**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + b*d**(3/2)*x**4*sqrt(c*x**2/d + 1)/5 + b*d**(5/2)*x**2*sqrt(c*x**2
/d + 1)/(15*c) - 2*b*d**(7/2)*sqrt(c*x**2/d + 1)/(15*c**2)

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Giac [B]  time = 1.10413, size = 288, normalized size = 3.43 \begin{align*} \frac{\frac{21 \,{\left (3 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} d\right )} b d \mathrm{sgn}\left (x\right )}{c} + \frac{3 \,{\left (15 \,{\left (c x^{2} + d\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} d + 35 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} d^{2}\right )} b \mathrm{sgn}\left (x\right )}{c} + \frac{3 \,{\left (15 \,{\left (c x^{2} + d\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} d + 35 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} d^{2}\right )} a d \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{{\left (35 \,{\left (c x^{2} + d\right )}^{\frac{9}{2}} - 135 \,{\left (c x^{2} + d\right )}^{\frac{7}{2}} d + 189 \,{\left (c x^{2} + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (c x^{2} + d\right )}^{\frac{3}{2}} d^{3}\right )} a \mathrm{sgn}\left (x\right )}{c^{2}}}{315 \, c} + \frac{2 \,{\left (9 \, b c d^{\frac{7}{2}} - 4 \, a d^{\frac{9}{2}}\right )} \mathrm{sgn}\left (x\right )}{315 \, c^{3}} \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^8,x, algorithm="giac")

[Out]

1/315*(21*(3*(c*x^2 + d)^(5/2) - 5*(c*x^2 + d)^(3/2)*d)*b*d*sgn(x)/c + 3*(15*(c*x^2 + d)^(7/2) - 42*(c*x^2 + d
)^(5/2)*d + 35*(c*x^2 + d)^(3/2)*d^2)*b*sgn(x)/c + 3*(15*(c*x^2 + d)^(7/2) - 42*(c*x^2 + d)^(5/2)*d + 35*(c*x^
2 + d)^(3/2)*d^2)*a*d*sgn(x)/c^2 + (35*(c*x^2 + d)^(9/2) - 135*(c*x^2 + d)^(7/2)*d + 189*(c*x^2 + d)^(5/2)*d^2
- 105*(c*x^2 + d)^(3/2)*d^3)*a*sgn(x)/c^2)/c + 2/315*(9*b*c*d^(7/2) - 4*a*d^(9/2))*sgn(x)/c^3