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

Optimal. Leaf size=111 \[ \frac{x^3 (5 b c-6 a d)}{15 c^2 \sqrt{c+\frac{d}{x^2}}}-\frac{8 d x \sqrt{c+\frac{d}{x^2}} (5 b c-6 a d)}{15 c^4}+\frac{4 d x (5 b c-6 a d)}{15 c^3 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}} \]

[Out]

(4*d*(5*b*c - 6*a*d)*x)/(15*c^3*Sqrt[c + d/x^2]) - (8*d*(5*b*c - 6*a*d)*Sqrt[c + d/x^2]*x)/(15*c^4) + ((5*b*c
- 6*a*d)*x^3)/(15*c^2*Sqrt[c + d/x^2]) + (a*x^5)/(5*c*Sqrt[c + d/x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0488855, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {453, 271, 192, 191} \[ \frac{x^3 (5 b c-6 a d)}{15 c^2 \sqrt{c+\frac{d}{x^2}}}-\frac{8 d x \sqrt{c+\frac{d}{x^2}} (5 b c-6 a d)}{15 c^4}+\frac{4 d x (5 b c-6 a d)}{15 c^3 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*d*(5*b*c - 6*a*d)*x)/(15*c^3*Sqrt[c + d/x^2]) - (8*d*(5*b*c - 6*a*d)*Sqrt[c + d/x^2]*x)/(15*c^4) + ((5*b*c
- 6*a*d)*x^3)/(15*c^2*Sqrt[c + d/x^2]) + (a*x^5)/(5*c*Sqrt[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 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 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^4}{\left (c+\frac{d}{x^2}\right )^{3/2}} \, dx &=\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}}+\frac{(5 b c-6 a d) \int \frac{x^2}{\left (c+\frac{d}{x^2}\right )^{3/2}} \, dx}{5 c}\\ &=\frac{(5 b c-6 a d) x^3}{15 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}}-\frac{(4 d (5 b c-6 a d)) \int \frac{1}{\left (c+\frac{d}{x^2}\right )^{3/2}} \, dx}{15 c^2}\\ &=\frac{4 d (5 b c-6 a d) x}{15 c^3 \sqrt{c+\frac{d}{x^2}}}+\frac{(5 b c-6 a d) x^3}{15 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}}-\frac{(8 d (5 b c-6 a d)) \int \frac{1}{\sqrt{c+\frac{d}{x^2}}} \, dx}{15 c^3}\\ &=\frac{4 d (5 b c-6 a d) x}{15 c^3 \sqrt{c+\frac{d}{x^2}}}-\frac{8 d (5 b c-6 a d) \sqrt{c+\frac{d}{x^2}} x}{15 c^4}+\frac{(5 b c-6 a d) x^3}{15 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^5}{5 c \sqrt{c+\frac{d}{x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0522132, size = 80, normalized size = 0.72 \[ \frac{3 a \left (-2 c^2 d x^4+c^3 x^6+8 c d^2 x^2+16 d^3\right )+5 b c \left (c^2 x^4-4 c d x^2-8 d^2\right )}{15 c^4 x \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*c*(-8*d^2 - 4*c*d*x^2 + c^2*x^4) + 3*a*(16*d^3 + 8*c*d^2*x^2 - 2*c^2*d*x^4 + c^3*x^6))/(15*c^4*Sqrt[c + d
/x^2]*x)

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 91, normalized size = 0.8 \begin{align*}{\frac{ \left ( 3\,a{x}^{6}{c}^{3}-6\,a{c}^{2}d{x}^{4}+5\,b{c}^{3}{x}^{4}+24\,ac{d}^{2}{x}^{2}-20\,b{c}^{2}d{x}^{2}+48\,a{d}^{3}-40\,bc{d}^{2} \right ) \left ( c{x}^{2}+d \right ) }{15\,{x}^{3}{c}^{4}} \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^4/(c+d/x^2)^(3/2),x)

[Out]

1/15*(3*a*c^3*x^6-6*a*c^2*d*x^4+5*b*c^3*x^4+24*a*c*d^2*x^2-20*b*c^2*d*x^2+48*a*d^3-40*b*c*d^2)*(c*x^2+d)/((c*x
^2+d)/x^2)^(3/2)/x^3/c^4

________________________________________________________________________________________

Maxima [A]  time = 0.946507, size = 173, normalized size = 1.56 \begin{align*} \frac{1}{3} \, b{\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 )} + \frac{1}{5} \, a{\left (\frac{5 \, d^{3}}{\sqrt{c + \frac{d}{x^{2}}} c^{4} x} + \frac{{\left (c + \frac{d}{x^{2}}\right )}^{\frac{5}{2}} x^{5} - 5 \,{\left (c + \frac{d}{x^{2}}\right )}^{\frac{3}{2}} d x^{3} + 15 \, \sqrt{c + \frac{d}{x^{2}}} d^{2} x}{c^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*(((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)) + 1/5*a*(5*d^3/(sq
rt(c + d/x^2)*c^4*x) + ((c + d/x^2)^(5/2)*x^5 - 5*(c + d/x^2)^(3/2)*d*x^3 + 15*sqrt(c + d/x^2)*d^2*x)/c^4)

________________________________________________________________________________________

Fricas [A]  time = 1.32175, size = 200, normalized size = 1.8 \begin{align*} \frac{{\left (3 \, a c^{3} x^{7} +{\left (5 \, b c^{3} - 6 \, a c^{2} d\right )} x^{5} - 4 \,{\left (5 \, b c^{2} d - 6 \, a c d^{2}\right )} x^{3} - 8 \,{\left (5 \, b c d^{2} - 6 \, a d^{3}\right )} x\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{15 \,{\left (c^{5} x^{2} + c^{4} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a*c^3*x^7 + (5*b*c^3 - 6*a*c^2*d)*x^5 - 4*(5*b*c^2*d - 6*a*c*d^2)*x^3 - 8*(5*b*c*d^2 - 6*a*d^3)*x)*sqr
t((c*x^2 + d)/x^2)/(c^5*x^2 + c^4*d)

________________________________________________________________________________________

Sympy [B]  time = 11.3805, size = 561, normalized size = 5.05 \begin{align*} a \left (\frac{c^{5} d^{\frac{19}{2}} x^{10} \sqrt{\frac{c x^{2}}{d} + 1}}{5 c^{7} d^{9} x^{6} + 15 c^{6} d^{10} x^{4} + 15 c^{5} d^{11} x^{2} + 5 c^{4} d^{12}} + \frac{5 c^{3} d^{\frac{23}{2}} x^{6} \sqrt{\frac{c x^{2}}{d} + 1}}{5 c^{7} d^{9} x^{6} + 15 c^{6} d^{10} x^{4} + 15 c^{5} d^{11} x^{2} + 5 c^{4} d^{12}} + \frac{30 c^{2} d^{\frac{25}{2}} x^{4} \sqrt{\frac{c x^{2}}{d} + 1}}{5 c^{7} d^{9} x^{6} + 15 c^{6} d^{10} x^{4} + 15 c^{5} d^{11} x^{2} + 5 c^{4} d^{12}} + \frac{40 c d^{\frac{27}{2}} x^{2} \sqrt{\frac{c x^{2}}{d} + 1}}{5 c^{7} d^{9} x^{6} + 15 c^{6} d^{10} x^{4} + 15 c^{5} d^{11} x^{2} + 5 c^{4} d^{12}} + \frac{16 d^{\frac{29}{2}} \sqrt{\frac{c x^{2}}{d} + 1}}{5 c^{7} d^{9} x^{6} + 15 c^{6} d^{10} x^{4} + 15 c^{5} d^{11} x^{2} + 5 c^{4} d^{12}}\right ) + b \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(c**5*d**(19/2)*x**10*sqrt(c*x**2/d + 1)/(5*c**7*d**9*x**6 + 15*c**6*d**10*x**4 + 15*c**5*d**11*x**2 + 5*c**
4*d**12) + 5*c**3*d**(23/2)*x**6*sqrt(c*x**2/d + 1)/(5*c**7*d**9*x**6 + 15*c**6*d**10*x**4 + 15*c**5*d**11*x**
2 + 5*c**4*d**12) + 30*c**2*d**(25/2)*x**4*sqrt(c*x**2/d + 1)/(5*c**7*d**9*x**6 + 15*c**6*d**10*x**4 + 15*c**5
*d**11*x**2 + 5*c**4*d**12) + 40*c*d**(27/2)*x**2*sqrt(c*x**2/d + 1)/(5*c**7*d**9*x**6 + 15*c**6*d**10*x**4 +
15*c**5*d**11*x**2 + 5*c**4*d**12) + 16*d**(29/2)*sqrt(c*x**2/d + 1)/(5*c**7*d**9*x**6 + 15*c**6*d**10*x**4 +
15*c**5*d**11*x**2 + 5*c**4*d**12)) + b*(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))

________________________________________________________________________________________

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

[Out]

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