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

Optimal. Leaf size=90 \[ -\frac{a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac{x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac{4 a x (b c-2 a d)}{3 c^3 \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0457317, antiderivative size = 90, normalized size of antiderivative = 1., 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, 378, 191} \[ -\frac{a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac{x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac{4 a x (b c-2 a d)}{3 c^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2/(c*x*(c + d*x^2)^(3/2))) + (x*(2*a*(b*c - 2*a*d) + b^2*c*x^2))/(3*c^2*(c + d*x^2)^(3/2)) + (4*a*(b*c - 2
*a*d)*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 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 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^2 \left (c+d x^2\right )^{5/2}} \, dx &=-\frac{a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac{\int \frac{2 a (b c-2 a d)+b^2 c x^2}{\left (c+d x^2\right )^{5/2}} \, dx}{c}\\ &=-\frac{a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac{x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac{(4 a (b c-2 a d)) \int \frac{1}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=-\frac{a^2}{c x \left (c+d x^2\right )^{3/2}}+\frac{x \left (2 a (b c-2 a d)+b^2 c x^2\right )}{3 c^2 \left (c+d x^2\right )^{3/2}}+\frac{4 a (b c-2 a d) x}{3 c^3 \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0250743, size = 76, normalized size = 0.84 \[ \frac{-a^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )+2 a b c x^2 \left (3 c+2 d x^2\right )+b^2 c^2 x^4}{3 c^3 x \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 78, normalized size = 0.9 \begin{align*} -{\frac{8\,{a}^{2}{d}^{2}{x}^{4}-4\,abcd{x}^{4}-{b}^{2}{c}^{2}{x}^{4}+12\,{a}^{2}cd{x}^{2}-6\,a{c}^{2}b{x}^{2}+3\,{a}^{2}{c}^{2}}{3\,x{c}^{3}} \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^2/(d*x^2+c)^(5/2),x)

[Out]

-1/3*(8*a^2*d^2*x^4-4*a*b*c*d*x^4-b^2*c^2*x^4+12*a^2*c*d*x^2-6*a*b*c^2*x^2+3*a^2*c^2)/x/(d*x^2+c)^(3/2)/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^2/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.30322, size = 188, normalized size = 2.09 \begin{align*} \frac{{\left ({\left (b^{2} c^{2} + 4 \, a b c d - 8 \, a^{2} d^{2}\right )} x^{4} - 3 \, a^{2} c^{2} + 6 \,{\left (a b c^{2} - 2 \, a^{2} c d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{3 \,{\left (c^{3} d^{2} x^{5} + 2 \, c^{4} d x^{3} + c^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*x^4 - 3*a^2*c^2 + 6*(a*b*c^2 - 2*a^2*c*d)*x^2)*sqrt(d*x^2 + c)/(c^3*d^2
*x^5 + 2*c^4*d*x^3 + c^5*x)

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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^{2} \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**2/(d*x**2+c)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.17338, size = 158, normalized size = 1.76 \begin{align*} \frac{x{\left (\frac{{\left (b^{2} c^{4} d + 4 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3}\right )} x^{2}}{c^{5} d} + \frac{6 \,{\left (a b c^{4} d - a^{2} c^{3} d^{2}\right )}}{c^{5} d}\right )}}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{2 \, a^{2} \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*((b^2*c^4*d + 4*a*b*c^3*d^2 - 5*a^2*c^2*d^3)*x^2/(c^5*d) + 6*(a*b*c^4*d - a^2*c^3*d^2)/(c^5*d))/(d*x^2 +
c)^(3/2) + 2*a^2*sqrt(d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)*c^2)