3.658 \(\int \frac{(a+b x^2)^2}{x^6 (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=141 \[ -\frac{a^2}{5 c x^5 \sqrt{c+d x^2}}-\frac{2 d x \left (15 b^2 c^2-8 a d (5 b c-3 a d)\right )}{15 c^4 \sqrt{c+d x^2}}-\frac{15 b^2 c^2-8 a d (5 b c-3 a d)}{15 c^3 x \sqrt{c+d x^2}}-\frac{2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.113882, antiderivative size = 141, normalized size of antiderivative = 1., 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, 271, 191} \[ -\frac{a^2}{5 c x^5 \sqrt{c+d x^2}}-\frac{2 d x \left (15 b^2 c^2-8 a d (5 b c-3 a d)\right )}{15 c^4 \sqrt{c+d x^2}}-\frac{15 b^2-\frac{8 a d (5 b c-3 a d)}{c^2}}{15 c x \sqrt{c+d x^2}}-\frac{2 a (5 b c-3 a d)}{15 c^2 x^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/(x^6*(c + d*x^2)^(3/2)),x]

[Out]

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

Mathematica [A]  time = 0.0806853, size = 105, normalized size = 0.74 \[ \sqrt{c+d x^2} \left (\frac{-33 a^2 d^2+50 a b c d-15 b^2 c^2}{15 c^4 x}-\frac{a^2}{5 c^2 x^5}+\frac{a (9 a d-10 b c)}{15 c^3 x^3}-\frac{d x (b c-a d)^2}{c^4 \left (c+d x^2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/(x^6*(c + d*x^2)^(3/2)),x]

[Out]

Sqrt[c + d*x^2]*(-a^2/(5*c^2*x^5) + (a*(-10*b*c + 9*a*d))/(15*c^3*x^3) + (-15*b^2*c^2 + 50*a*b*c*d - 33*a^2*d^
2)/(15*c^4*x) - (d*(b*c - a*d)^2*x)/(c^4*(c + d*x^2)))

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Maple [A]  time = 0.007, size = 117, normalized size = 0.8 \begin{align*} -{\frac{48\,{x}^{6}{a}^{2}{d}^{3}-80\,{x}^{6}abc{d}^{2}+30\,{x}^{6}{b}^{2}{c}^{2}d+24\,{x}^{4}{a}^{2}c{d}^{2}-40\,{x}^{4}ab{c}^{2}d+15\,{x}^{4}{b}^{2}{c}^{3}-6\,{x}^{2}{a}^{2}{c}^{2}d+10\,{x}^{2}ab{c}^{3}+3\,{a}^{2}{c}^{3}}{15\,{x}^{5}{c}^{4}}{\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^6/(d*x^2+c)^(3/2),x)

[Out]

-1/15*(48*a^2*d^3*x^6-80*a*b*c*d^2*x^6+30*b^2*c^2*d*x^6+24*a^2*c*d^2*x^4-40*a*b*c^2*d*x^4+15*b^2*c^3*x^4-6*a^2
*c^2*d*x^2+10*a*b*c^3*x^2+3*a^2*c^3)/x^5/(d*x^2+c)^(1/2)/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^6/(d*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50501, size = 258, normalized size = 1.83 \begin{align*} -\frac{{\left (2 \,{\left (15 \, b^{2} c^{2} d - 40 \, a b c d^{2} + 24 \, a^{2} d^{3}\right )} x^{6} + 3 \, a^{2} c^{3} +{\left (15 \, b^{2} c^{3} - 40 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} x^{4} + 2 \,{\left (5 \, a b c^{3} - 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \,{\left (c^{4} d x^{7} + c^{5} 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)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(2*(15*b^2*c^2*d - 40*a*b*c*d^2 + 24*a^2*d^3)*x^6 + 3*a^2*c^3 + (15*b^2*c^3 - 40*a*b*c^2*d + 24*a^2*c*d^
2)*x^4 + 2*(5*a*b*c^3 - 3*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c)/(c^4*d*x^7 + c^5*x^5)

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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^{6} \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**6/(d*x**2+c)**(3/2),x)

[Out]

Integral((a + b*x**2)**2/(x**6*(c + d*x**2)**(3/2)), x)

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Giac [B]  time = 1.17203, size = 610, normalized size = 4.33 \begin{align*} -\frac{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x}{\sqrt{d x^{2} + c} c^{4}} + \frac{2 \,{\left (15 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} b^{2} c^{2} \sqrt{d} - 30 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{8} a b c d^{\frac{3}{2}} + 15 \,{\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} + 180 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac{3}{2}} - 90 \,{\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} - 320 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac{3}{2}} + 240 \,{\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} + 220 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac{3}{2}} - 150 \,{\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} - 50 \, a b c^{5} d^{\frac{3}{2}} + 33 \, 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^{3}} \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)^(3/2),x, algorithm="giac")

[Out]

-(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x/(sqrt(d*x^2 + c)*c^4) + 2/15*(15*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^
2*sqrt(d) - 30*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c*d^(3/2) + 15*(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) + 180*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(3/2) - 9
0*(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) - 320*(sq
rt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^3*d^(3/2) + 240*(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) + 220*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^(3/2) - 150*(sqrt(d)
*x - sqrt(d*x^2 + c))^2*a^2*c^3*d^(5/2) + 15*b^2*c^6*sqrt(d) - 50*a*b*c^5*d^(3/2) + 33*a^2*c^4*d^(5/2))/(((sqr
t(d)*x - sqrt(d*x^2 + c))^2 - c)^5*c^3)