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

Optimal. Leaf size=97 \[ -\frac{a^2}{3 c x^3 \sqrt{c+d x^2}}+\frac{x \left (3 b^2 c^2-4 a d (3 b c-2 a d)\right )}{3 c^3 \sqrt{c+d x^2}}-\frac{2 a (3 b c-2 a d)}{3 c^2 x \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0720084, antiderivative size = 98, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {462, 453, 191} \[ \frac{x \left (8 a^2 d^2-12 a b c d+3 b^2 c^2\right )}{3 c^3 \sqrt{c+d x^2}}-\frac{a^2}{3 c x^3 \sqrt{c+d x^2}}-\frac{2 a (3 b c-2 a d)}{3 c^2 x \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a^2/(3*c*x^3*Sqrt[c + d*x^2]) - (2*a*(3*b*c - 2*a*d))/(3*c^2*x*Sqrt[c + d*x^2]) + ((3*b^2*c^2 - 12*a*b*c*d +
8*a^2*d^2)*x)/(3*c^3*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 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^4 \left (c+d x^2\right )^{3/2}} \, dx &=-\frac{a^2}{3 c x^3 \sqrt{c+d x^2}}+\frac{\int \frac{2 a (3 b c-2 a d)+3 b^2 c x^2}{x^2 \left (c+d x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac{a^2}{3 c x^3 \sqrt{c+d x^2}}-\frac{2 a (3 b c-2 a d)}{3 c^2 x \sqrt{c+d x^2}}-\frac{1}{3} \left (-3 b^2+\frac{4 a d (3 b c-2 a d)}{c^2}\right ) \int \frac{1}{\left (c+d x^2\right )^{3/2}} \, dx\\ &=-\frac{a^2}{3 c x^3 \sqrt{c+d x^2}}-\frac{2 a (3 b c-2 a d)}{3 c^2 x \sqrt{c+d x^2}}+\frac{\left (3 b^2-\frac{4 a d (3 b c-2 a d)}{c^2}\right ) x}{3 c \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0236606, size = 74, normalized size = 0.76 \[ \frac{a^2 \left (-c^2+4 c d x^2+8 d^2 x^4\right )-6 a b c x^2 \left (c+2 d x^2\right )+3 b^2 c^2 x^4}{3 c^3 x^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 77, normalized size = 0.8 \begin{align*} -{\frac{-8\,{a}^{2}{d}^{2}{x}^{4}+12\,abcd{x}^{4}-3\,{b}^{2}{c}^{2}{x}^{4}-4\,{a}^{2}cd{x}^{2}+6\,a{c}^{2}b{x}^{2}+{a}^{2}{c}^{2}}{3\,{x}^{3}{c}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-8*a^2*d^2*x^4+12*a*b*c*d*x^4-3*b^2*c^2*x^4-4*a^2*c*d*x^2+6*a*b*c^2*x^2+a^2*c^2)/x^3/(d*x^2+c)^(1/2)/c^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36997, size = 173, normalized size = 1.78 \begin{align*} \frac{{\left ({\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} - a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{3 \,{\left (c^{3} d x^{5} + c^{4} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*x^4 - a^2*c^2 - 2*(3*a*b*c^2 - 2*a^2*c*d)*x^2)*sqrt(d*x^2 + c)/(c^3*
d*x^5 + c^4*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13776, size = 269, normalized size = 2.77 \begin{align*} \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{\sqrt{d x^{2} + c} c^{3}} + \frac{2 \,{\left (6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c \sqrt{d} - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a^{2} d^{\frac{3}{2}} - 12 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{2} \sqrt{d} + 12 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c d^{\frac{3}{2}} + 6 \, a b c^{3} \sqrt{d} - 5 \, a^{2} c^{2} d^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x/(sqrt(d*x^2 + c)*c^3) + 2/3*(6*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c*sqrt(d)
- 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a^2*d^(3/2) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^2*sqrt(d) + 12*(sq
rt(d)*x - sqrt(d*x^2 + c))^2*a^2*c*d^(3/2) + 6*a*b*c^3*sqrt(d) - 5*a^2*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2
+ c))^2 - c)^3*c^2)