3.764 \(\int \frac{1}{x^2 (a+b x^2)^2 \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{\sqrt{c+d x^2} (3 b c-2 a d)}{2 a^2 c x (b c-a d)}-\frac{b (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} (b c-a d)^{3/2}}+\frac{b \sqrt{c+d x^2}}{2 a x \left (a+b x^2\right ) (b c-a d)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.135748, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {472, 583, 12, 377, 205} \[ -\frac{\sqrt{c+d x^2} (3 b c-2 a d)}{2 a^2 c x (b c-a d)}-\frac{b (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} (b c-a d)^{3/2}}+\frac{b \sqrt{c+d x^2}}{2 a x \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 5.15409, size = 116, normalized size = 0.79 \[ \frac{\sqrt{c+d x^2} \left (\frac{b^2 x^2}{\left (a+b x^2\right ) (a d-b c)}-\frac{2}{c}\right )}{2 a^2 x}-\frac{b (3 b c-4 a d) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 a^{5/2} (b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.013, size = 841, normalized size = 5.7 \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)^2/(d*x^2+c)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 3.36674, size = 1238, normalized size = 8.42 \begin{align*} \left [-\frac{{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{3} +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt{-a b c + a^{2} d} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} + 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt{-a b c + a^{2} d} \sqrt{d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \,{\left (2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2} +{\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{8 \,{\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{3} +{\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x\right )}}, -\frac{{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{3} +{\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{\sqrt{a b c - a^{2} d}{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} +{\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + 2 \,{\left (2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2} +{\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{4 \,{\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{3} +{\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [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(1/x**2/(b*x**2+a)**2/(d*x**2+c)**(1/2),x)

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Giac [B]  time = 3.66341, size = 535, normalized size = 3.64 \begin{align*} \frac{1}{2} \, d^{\frac{5}{2}}{\left (\frac{{\left (3 \, b^{2} c - 4 \, a b d\right )} \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^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{a b c d - a^{2} d^{2}}} + \frac{2 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b^{2} c - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a b d - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b^{2} c^{2} + 14 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c d - 8 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} d^{2} + 3 \, b^{2} c^{3} - 2 \, a b c^{2} d\right )}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{6} b - 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a d + 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{2} - 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c d - b c^{3}\right )}{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d^(5/2)*((3*b^2*c - 4*a*b*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^2*b*c*d^2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) + 2*(3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2*c - 4*(s
qrt(d)*x - sqrt(d*x^2 + c))^4*a*b*d - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^2 + 14*(sqrt(d)*x - sqrt(d*x^2 +
c))^2*a*b*c*d - 8*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^2 + 3*b^2*c^3 - 2*a*b*c^2*d)/(((sqrt(d)*x - sqrt(d*x^
2 + c))^6*b - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d + 3*(sqrt(d)*x - s
qrt(d*x^2 + c))^2*b*c^2 - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d - b*c^3)*(a^2*b*c*d^2 - a^3*d^3)))