3.719 \(\int \frac{1}{x^2 (a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=124 \[ -\frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{3/2}}-\frac{\sqrt{c+d x^2} (b c-2 a d)}{a c^2 x (b c-a d)}-\frac{d}{c x \sqrt{c+d x^2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.116737, antiderivative size = 124, 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} \[ -\frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{3/2}}-\frac{\sqrt{c+d x^2} (b c-2 a d)}{a c^2 x (b c-a d)}-\frac{d}{c x \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

Mathematica [A]  time = 5.18765, size = 102, normalized size = 0.82 \[ \frac{\frac{d^2 x^2}{b c-a d}-\frac{c+d x^2}{a}}{c^2 x \sqrt{c+d x^2}}-\frac{b^2 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} (b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.012, size = 695, normalized size = 5.6 \begin{align*} -{\frac{{b}^{2}}{2\,a \left ( ad-bc \right ) }{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{bdx}{2\,a \left ( ad-bc \right ) c}{\frac{1}{\sqrt{ \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}+{\frac{{b}^{2}}{2\,a \left ( ad-bc \right ) }\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d-2\,{\frac{d\sqrt{-ab}}{b} \left ( x+{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}+{\frac{{b}^{2}}{2\,a \left ( ad-bc \right ) }{\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{bdx}{2\,a \left ( ad-bc \right ) c}{\frac{1}{\sqrt{ \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}}}}}-{\frac{{b}^{2}}{2\,a \left ( ad-bc \right ) }\ln \left ({ \left ( -2\,{\frac{ad-bc}{b}}+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }+2\,\sqrt{-{\frac{ad-bc}{b}}}\sqrt{ \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) ^{2}d+2\,{\frac{d\sqrt{-ab}}{b} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }-{\frac{ad-bc}{b}}} \right ) \left ( x-{\frac{1}{b}\sqrt{-ab}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-ab}}}{\frac{1}{\sqrt{-{\frac{ad-bc}{b}}}}}}-{\frac{1}{acx}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-2\,{\frac{dx}{a{c}^{2}\sqrt{d{x}^{2}+c}}} \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)^(3/2),x)

[Out]

-1/2*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)^(1/2)-1/2*b/a/(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)^(1
/2)*x*d+1/2*b^2/a/(-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)))+1/2*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)^(1/2)-1/2*b/a/(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)^(1/2)*x*d-1/2*b^2/a/(-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/a/c/x/(d*x^2+c)^(1/2)-2/a*d/c^2*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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)*(d*x^2 + c)^(3/2)*x^2), x)

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Fricas [B]  time = 2.61655, size = 1139, normalized size = 9.19 \begin{align*} \left [\frac{{\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} 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 (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{4 \,{\left ({\left (a^{2} b^{2} c^{4} d - 2 \, a^{3} b c^{3} d^{2} + a^{4} c^{2} d^{3}\right )} x^{3} +{\left (a^{2} b^{2} c^{5} - 2 \, a^{3} b c^{4} d + a^{4} c^{3} d^{2}\right )} x\right )}}, -\frac{{\left (b^{2} c^{2} d x^{3} + b^{2} c^{3} 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 (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} +{\left (a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{2 \,{\left ({\left (a^{2} b^{2} c^{4} d - 2 \, a^{3} b c^{3} d^{2} + a^{4} c^{2} d^{3}\right )} x^{3} +{\left (a^{2} b^{2} c^{5} - 2 \, a^{3} b c^{4} d + a^{4} c^{3} 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)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Integral(1/(x**2*(a + b*x**2)*(c + d*x**2)**(3/2)), x)

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Giac [A]  time = 2.46304, size = 205, normalized size = 1.65 \begin{align*} -\frac{b^{2} \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 )}{\sqrt{a b c d - a^{2} d^{2}}{\left (a b c - a^{2} d\right )}} + \frac{d^{2} x}{{\left (b c^{3} - a c^{2} d\right )} \sqrt{d x^{2} + c}} + \frac{2 \, \sqrt{d}}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )} a c} \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)^(3/2),x, algorithm="giac")

[Out]

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