3.728 \(\int \frac{1}{x^2 (a+b x^2) (c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=178 \[ -\frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}-\frac{\sqrt{c+d x^2} (b c-4 a d) (3 b c-2 a d)}{3 a c^3 x (b c-a d)^2}-\frac{d (7 b c-4 a d)}{3 c^2 x \sqrt{c+d x^2} (b c-a d)^2}-\frac{d}{3 c x \left (c+d x^2\right )^{3/2} (b c-a d)} \]

[Out]

-d/(3*c*(b*c - a*d)*x*(c + d*x^2)^(3/2)) - (d*(7*b*c - 4*a*d))/(3*c^2*(b*c - a*d)^2*x*Sqrt[c + d*x^2]) - ((b*c
- 4*a*d)*(3*b*c - 2*a*d)*Sqrt[c + d*x^2])/(3*a*c^3*(b*c - a*d)^2*x) - (b^3*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a
]*Sqrt[c + d*x^2])])/(a^(3/2)*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.236014, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {472, 579, 583, 12, 377, 205} \[ -\frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}-\frac{\sqrt{c+d x^2} (b c-4 a d) (3 b c-2 a d)}{3 a c^3 x (b c-a d)^2}-\frac{d (7 b c-4 a d)}{3 c^2 x \sqrt{c+d x^2} (b c-a d)^2}-\frac{d}{3 c x \left (c+d x^2\right )^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

-d/(3*c*(b*c - a*d)*x*(c + d*x^2)^(3/2)) - (d*(7*b*c - 4*a*d))/(3*c^2*(b*c - a*d)^2*x*Sqrt[c + d*x^2]) - ((b*c
- 4*a*d)*(3*b*c - 2*a*d)*Sqrt[c + d*x^2])/(3*a*c^3*(b*c - a*d)^2*x) - (b^3*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a
]*Sqrt[c + d*x^2])])/(a^(3/2)*(b*c - a*d)^(5/2))

Rule 472

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

Rule 579

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

Rule 583

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}+\frac{\int \frac{3 b c-4 a d-4 b d x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{3 c (b c-a d)}\\ &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}-\frac{d (7 b c-4 a d)}{3 c^2 (b c-a d)^2 x \sqrt{c+d x^2}}+\frac{\int \frac{(b c-4 a d) (3 b c-2 a d)-2 b d (7 b c-4 a d) x^2}{x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 c^2 (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}-\frac{d (7 b c-4 a d)}{3 c^2 (b c-a d)^2 x \sqrt{c+d x^2}}-\frac{(b c-4 a d) (3 b c-2 a d) \sqrt{c+d x^2}}{3 a c^3 (b c-a d)^2 x}-\frac{\int \frac{3 b^3 c^3}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{3 a c^3 (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}-\frac{d (7 b c-4 a d)}{3 c^2 (b c-a d)^2 x \sqrt{c+d x^2}}-\frac{(b c-4 a d) (3 b c-2 a d) \sqrt{c+d x^2}}{3 a c^3 (b c-a d)^2 x}-\frac{b^3 \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{a (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}-\frac{d (7 b c-4 a d)}{3 c^2 (b c-a d)^2 x \sqrt{c+d x^2}}-\frac{(b c-4 a d) (3 b c-2 a d) \sqrt{c+d x^2}}{3 a c^3 (b c-a d)^2 x}-\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{a (b c-a d)^2}\\ &=-\frac{d}{3 c (b c-a d) x \left (c+d x^2\right )^{3/2}}-\frac{d (7 b c-4 a d)}{3 c^2 (b c-a d)^2 x \sqrt{c+d x^2}}-\frac{(b c-4 a d) (3 b c-2 a d) \sqrt{c+d x^2}}{3 a c^3 (b c-a d)^2 x}-\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 5.25171, size = 143, normalized size = 0.8 \[ \frac{\sqrt{c+d x^2} \left (\frac{d^2 x^2 (8 b c-5 a d)}{\left (c+d x^2\right ) (b c-a d)^2}+\frac{c d^2 x^2}{\left (c+d x^2\right )^2 (b c-a d)}-\frac{3}{a}\right )}{3 c^3 x}-\frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)*(c + d*x^2)^(5/2)),x]

[Out]

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

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Maple [B]  time = 0.014, size = 1192, normalized size = 6.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}{\left (d x^{2} + c\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(5/2)*x^2), x)

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Fricas [B]  time = 5.05336, size = 1868, normalized size = 10.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3*c^5*x)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a
^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 + 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqr
t(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(3*a*b^3*c^5 - 9*a^2*b^2*c^4*d + 9*a^3*b*c^3*d^2 - 3*a^4*c^2*d^
3 + (3*a*b^3*c^3*d^2 - 17*a^2*b^2*c^2*d^3 + 22*a^3*b*c*d^4 - 8*a^4*d^5)*x^4 + 3*(2*a*b^3*c^4*d - 9*a^2*b^2*c^3
*d^2 + 11*a^3*b*c^2*d^3 - 4*a^4*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^2*b^3*c^6*d^2 - 3*a^3*b^2*c^5*d^3 + 3*a^4*b*c
^4*d^4 - a^5*c^3*d^5)*x^5 + 2*(a^2*b^3*c^7*d - 3*a^3*b^2*c^6*d^2 + 3*a^4*b*c^5*d^3 - a^5*c^4*d^4)*x^3 + (a^2*b
^3*c^8 - 3*a^3*b^2*c^7*d + 3*a^4*b*c^6*d^2 - a^5*c^5*d^3)*x), -1/6*(3*(b^3*c^3*d^2*x^5 + 2*b^3*c^4*d*x^3 + b^3
*c^5*x)*sqrt(a*b*c - a^2*d)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d
- a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + 2*(3*a*b^3*c^5 - 9*a^2*b^2*c^4*d + 9*a^3*b*c^3*d^2 - 3*a^4*c^2*d^3
+ (3*a*b^3*c^3*d^2 - 17*a^2*b^2*c^2*d^3 + 22*a^3*b*c*d^4 - 8*a^4*d^5)*x^4 + 3*(2*a*b^3*c^4*d - 9*a^2*b^2*c^3*
d^2 + 11*a^3*b*c^2*d^3 - 4*a^4*c*d^4)*x^2)*sqrt(d*x^2 + c))/((a^2*b^3*c^6*d^2 - 3*a^3*b^2*c^5*d^3 + 3*a^4*b*c^
4*d^4 - a^5*c^3*d^5)*x^5 + 2*(a^2*b^3*c^7*d - 3*a^3*b^2*c^6*d^2 + 3*a^4*b*c^5*d^3 - a^5*c^4*d^4)*x^3 + (a^2*b^
3*c^8 - 3*a^3*b^2*c^7*d + 3*a^4*b*c^6*d^2 - a^5*c^5*d^3)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b*x**2+a)/(d*x**2+c)**(5/2),x)

[Out]

Integral(1/(x**2*(a + b*x**2)*(c + d*x**2)**(5/2)), x)

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Giac [B]  time = 3.22824, size = 494, normalized size = 2.78 \begin{align*} \frac{b^{3} \sqrt{d} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt{a b c d - a^{2} d^{2}}} + \frac{{\left (\frac{{\left (8 \, b^{3} c^{5} d^{4} - 21 \, a b^{2} c^{4} d^{5} + 18 \, a^{2} b c^{3} d^{6} - 5 \, a^{3} c^{2} d^{7}\right )} x^{2}}{b^{4} c^{9} d - 4 \, a b^{3} c^{8} d^{2} + 6 \, a^{2} b^{2} c^{7} d^{3} - 4 \, a^{3} b c^{6} d^{4} + a^{4} c^{5} d^{5}} + \frac{3 \,{\left (3 \, b^{3} c^{6} d^{3} - 8 \, a b^{2} c^{5} d^{4} + 7 \, a^{2} b c^{4} d^{5} - 2 \, a^{3} c^{3} d^{6}\right )}}{b^{4} c^{9} d - 4 \, a b^{3} c^{8} d^{2} + 6 \, a^{2} b^{2} c^{7} d^{3} - 4 \, a^{3} b c^{6} d^{4} + a^{4} c^{5} d^{5}}\right )} x}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{2 \, \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )} a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

b^3*sqrt(d)*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/((a*b^2*c^2
- 2*a^2*b*c*d + a^3*d^2)*sqrt(a*b*c*d - a^2*d^2)) + 1/3*((8*b^3*c^5*d^4 - 21*a*b^2*c^4*d^5 + 18*a^2*b*c^3*d^6
- 5*a^3*c^2*d^7)*x^2/(b^4*c^9*d - 4*a*b^3*c^8*d^2 + 6*a^2*b^2*c^7*d^3 - 4*a^3*b*c^6*d^4 + a^4*c^5*d^5) + 3*(3*
b^3*c^6*d^3 - 8*a*b^2*c^5*d^4 + 7*a^2*b*c^4*d^5 - 2*a^3*c^3*d^6)/(b^4*c^9*d - 4*a*b^3*c^8*d^2 + 6*a^2*b^2*c^7*
d^3 - 4*a^3*b*c^6*d^4 + a^4*c^5*d^5))*x/(d*x^2 + c)^(3/2) + 2*sqrt(d)/(((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)*a
*c^2)