3.668 \(\int \frac{(a+b x^2)^2}{x^4 (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}+\frac{2 x \left (b^2 c^2-8 a d (b c-a d)\right )}{3 c^4 \sqrt{c+d x^2}}+\frac{x \left (b^2 c^2-8 a d (b c-a d)\right )}{3 c^3 \left (c+d x^2\right )^{3/2}}-\frac{2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.124814, antiderivative size = 130, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {462, 453, 192, 191} \[ -\frac{a^2}{3 c x^3 \left (c+d x^2\right )^{3/2}}+\frac{2 x \left (b^2 c^2-8 a d (b c-a d)\right )}{3 c^4 \sqrt{c+d x^2}}+\frac{x \left (b^2-\frac{8 a d (b c-a d)}{c^2}\right )}{3 c \left (c+d x^2\right )^{3/2}}-\frac{2 a (b c-a d)}{c^2 x \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

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

Mathematica [A]  time = 0.0715314, size = 107, normalized size = 0.82 \[ \frac{a^2 \left (6 c^2 d x^2-c^3+24 c d^2 x^4+16 d^3 x^6\right )-2 a b c x^2 \left (3 c^2+12 c d x^2+8 d^2 x^4\right )+b^2 c^2 x^4 \left (3 c+2 d x^2\right )}{3 c^4 x^3 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^4*(c + d*x^2)^(5/2)),x]

[Out]

(b^2*c^2*x^4*(3*c + 2*d*x^2) - 2*a*b*c*x^2*(3*c^2 + 12*c*d*x^2 + 8*d^2*x^4) + a^2*(-c^3 + 6*c^2*d*x^2 + 24*c*d
^2*x^4 + 16*d^3*x^6))/(3*c^4*x^3*(c + d*x^2)^(3/2))

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Maple [A]  time = 0.005, size = 116, normalized size = 0.9 \begin{align*} -{\frac{-16\,{x}^{6}{a}^{2}{d}^{3}+16\,{x}^{6}abc{d}^{2}-2\,{x}^{6}{b}^{2}{c}^{2}d-24\,{x}^{4}{a}^{2}c{d}^{2}+24\,{x}^{4}ab{c}^{2}d-3\,{x}^{4}{b}^{2}{c}^{3}-6\,{x}^{2}{a}^{2}{c}^{2}d+6\,{x}^{2}ab{c}^{3}+{a}^{2}{c}^{3}}{3\,{x}^{3}{c}^{4}} \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^4/(d*x^2+c)^(5/2),x)

[Out]

-1/3*(-16*a^2*d^3*x^6+16*a*b*c*d^2*x^6-2*b^2*c^2*d*x^6-24*a^2*c*d^2*x^4+24*a*b*c^2*d*x^4-3*b^2*c^3*x^4-6*a^2*c
^2*d*x^2+6*a*b*c^3*x^2+a^2*c^3)/x^3/(d*x^2+c)^(3/2)/c^4

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55648, size = 258, normalized size = 1.97 \begin{align*} \frac{{\left (2 \,{\left (b^{2} c^{2} d - 8 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{6} - a^{2} c^{3} + 3 \,{\left (b^{2} c^{3} - 8 \, a b c^{2} d + 8 \, a^{2} c d^{2}\right )} x^{4} - 6 \,{\left (a b c^{3} - a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{3 \,{\left (c^{4} d^{2} x^{7} + 2 \, c^{5} d x^{5} + c^{6} 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)^(5/2),x, algorithm="fricas")

[Out]

1/3*(2*(b^2*c^2*d - 8*a*b*c*d^2 + 8*a^2*d^3)*x^6 - a^2*c^3 + 3*(b^2*c^3 - 8*a*b*c^2*d + 8*a^2*c*d^2)*x^4 - 6*(
a*b*c^3 - a^2*c^2*d)*x^2)*sqrt(d*x^2 + c)/(c^4*d^2*x^7 + 2*c^5*d*x^5 + c^6*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{5}{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)**(5/2),x)

[Out]

Integral((a + b*x**2)**2/(x**4*(c + d*x**2)**(5/2)), x)

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Giac [B]  time = 1.15766, size = 348, normalized size = 2.66 \begin{align*} \frac{x{\left (\frac{2 \,{\left (b^{2} c^{5} d^{2} - 5 \, a b c^{4} d^{3} + 4 \, a^{2} c^{3} d^{4}\right )} x^{2}}{c^{7} d} + \frac{3 \,{\left (b^{2} c^{6} d - 4 \, a b c^{5} d^{2} + 3 \, a^{2} c^{4} d^{3}\right )}}{c^{7} d}\right )}}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{4 \,{\left (3 \,{\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}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{2} \sqrt{d} + 9 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} c d^{\frac{3}{2}} + 3 \, a b c^{3} \sqrt{d} - 4 \, 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^{3}} \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)^(5/2),x, algorithm="giac")

[Out]

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