3.670 \(\int \frac{(a+b x^2)^2}{x^6 (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac{a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac{8 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^5 \sqrt{c+d x^2}}-\frac{4 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^4 \left (c+d x^2\right )^{3/2}}-\frac{5 b^2 c^2-4 a d (5 b c-4 a d)}{5 c^3 x \left (c+d x^2\right )^{3/2}}-\frac{2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.165987, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {462, 453, 271, 192, 191} \[ -\frac{a^2}{5 c x^5 \left (c+d x^2\right )^{3/2}}-\frac{8 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^5 \sqrt{c+d x^2}}-\frac{4 d x \left (5 b^2 c^2-4 a d (5 b c-4 a d)\right )}{15 c^4 \left (c+d x^2\right )^{3/2}}-\frac{5 b^2-\frac{4 a d (5 b c-4 a d)}{c^2}}{5 c x \left (c+d x^2\right )^{3/2}}-\frac{2 a (5 b c-4 a d)}{15 c^2 x^3 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^6*(c + d*x^2)^(5/2)),x]

[Out]

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

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

Mathematica [A]  time = 0.0990494, size = 142, normalized size = 0.78 \[ \frac{-a^2 \left (48 c^2 d^2 x^4-8 c^3 d x^2+3 c^4+192 c d^3 x^6+128 d^4 x^8\right )+10 a b c x^2 \left (6 c^2 d x^2-c^3+24 c d^2 x^4+16 d^3 x^6\right )-5 b^2 c^2 x^4 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )}{15 c^5 x^5 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^6*(c + d*x^2)^(5/2)),x]

[Out]

(-5*b^2*c^2*x^4*(3*c^2 + 12*c*d*x^2 + 8*d^2*x^4) + 10*a*b*c*x^2*(-c^3 + 6*c^2*d*x^2 + 24*c*d^2*x^4 + 16*d^3*x^
6) - a^2*(3*c^4 - 8*c^3*d*x^2 + 48*c^2*d^2*x^4 + 192*c*d^3*x^6 + 128*d^4*x^8))/(15*c^5*x^5*(c + d*x^2)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 158, normalized size = 0.9 \begin{align*} -{\frac{128\,{a}^{2}{d}^{4}{x}^{8}-160\,abc{d}^{3}{x}^{8}+40\,{b}^{2}{c}^{2}{d}^{2}{x}^{8}+192\,{a}^{2}c{d}^{3}{x}^{6}-240\,ab{c}^{2}{d}^{2}{x}^{6}+60\,{b}^{2}{c}^{3}d{x}^{6}+48\,{a}^{2}{c}^{2}{d}^{2}{x}^{4}-60\,ab{c}^{3}d{x}^{4}+15\,{b}^{2}{c}^{4}{x}^{4}-8\,{a}^{2}{c}^{3}d{x}^{2}+10\,ab{c}^{4}{x}^{2}+3\,{a}^{2}{c}^{4}}{15\,{x}^{5}{c}^{5}} \left ( d{x}^{2}+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x)

[Out]

-1/15*(128*a^2*d^4*x^8-160*a*b*c*d^3*x^8+40*b^2*c^2*d^2*x^8+192*a^2*c*d^3*x^6-240*a*b*c^2*d^2*x^6+60*b^2*c^3*d
*x^6+48*a^2*c^2*d^2*x^4-60*a*b*c^3*d*x^4+15*b^2*c^4*x^4-8*a^2*c^3*d*x^2+10*a*b*c^4*x^2+3*a^2*c^4)/x^5/(d*x^2+c
)^(3/2)/c^5

________________________________________________________________________________________

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((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.7727, size = 359, normalized size = 1.96 \begin{align*} -\frac{{\left (8 \,{\left (5 \, b^{2} c^{2} d^{2} - 20 \, a b c d^{3} + 16 \, a^{2} d^{4}\right )} x^{8} + 12 \,{\left (5 \, b^{2} c^{3} d - 20 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x^{6} + 3 \, a^{2} c^{4} + 3 \,{\left (5 \, b^{2} c^{4} - 20 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} x^{4} + 2 \,{\left (5 \, a b c^{4} - 4 \, a^{2} c^{3} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \,{\left (c^{5} d^{2} x^{9} + 2 \, c^{6} d x^{7} + c^{7} x^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(8*(5*b^2*c^2*d^2 - 20*a*b*c*d^3 + 16*a^2*d^4)*x^8 + 12*(5*b^2*c^3*d - 20*a*b*c^2*d^2 + 16*a^2*c*d^3)*x^
6 + 3*a^2*c^4 + 3*(5*b^2*c^4 - 20*a*b*c^3*d + 16*a^2*c^2*d^2)*x^4 + 2*(5*a*b*c^4 - 4*a^2*c^3*d)*x^2)*sqrt(d*x^
2 + c)/(c^5*d^2*x^9 + 2*c^6*d*x^7 + c^7*x^5)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{x^{6} \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2/x**6/(d*x**2+c)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(x**6*(c + d*x**2)**(5/2)), x)

________________________________________________________________________________________

Giac [B]  time = 1.19317, size = 687, normalized size = 3.75 \begin{align*} -\frac{x{\left (\frac{{\left (5 \, b^{2} c^{6} d^{3} - 16 \, a b c^{5} d^{4} + 11 \, a^{2} c^{4} d^{5}\right )} x^{2}}{c^{9} d} + \frac{6 \,{\left (b^{2} c^{7} d^{2} - 3 \, a b c^{6} d^{3} + 2 \, a^{2} c^{5} d^{4}\right )}}{c^{9} d}\right )}}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{2 \,{\left (15 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt{d} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a b c d^{\frac{3}{2}} + 45 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a^{2} d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b^{2} c^{3} \sqrt{d} + 300 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac{3}{2}} - 240 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a^{2} c d^{\frac{5}{2}} + 90 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{2} c^{4} \sqrt{d} - 500 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac{3}{2}} + 490 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac{5}{2}} - 60 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c^{5} \sqrt{d} + 340 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac{3}{2}} - 320 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac{5}{2}} + 15 \, b^{2} c^{6} \sqrt{d} - 80 \, a b c^{5} d^{\frac{3}{2}} + 73 \, a^{2} c^{4} d^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{5} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/x^6/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

-1/3*x*((5*b^2*c^6*d^3 - 16*a*b*c^5*d^4 + 11*a^2*c^4*d^5)*x^2/(c^9*d) + 6*(b^2*c^7*d^2 - 3*a*b*c^6*d^3 + 2*a^2
*c^5*d^4)/(c^9*d))/(d*x^2 + c)^(3/2) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^2*sqrt(d) - 60*(sqrt(d)*
x - sqrt(d*x^2 + c))^8*a*b*c*d^(3/2) + 45*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(d
*x^2 + c))^6*b^2*c^3*sqrt(d) + 300*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(3/2) - 240*(sqrt(d)*x - sqrt(d*x
^2 + c))^6*a^2*c*d^(5/2) + 90*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c^4*sqrt(d) - 500*(sqrt(d)*x - sqrt(d*x^2 +
c))^4*a*b*c^3*d^(3/2) + 490*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^2*d^(5/2) - 60*(sqrt(d)*x - sqrt(d*x^2 + c))
^2*b^2*c^5*sqrt(d) + 340*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^(3/2) - 320*(sqrt(d)*x - sqrt(d*x^2 + c))^2
*a^2*c^3*d^(5/2) + 15*b^2*c^6*sqrt(d) - 80*a*b*c^5*d^(3/2) + 73*a^2*c^4*d^(5/2))/(((sqrt(d)*x - sqrt(d*x^2 + c
))^2 - c)^5*c^4)