3.721 \(\int \frac{1}{x^4 (a+b x^2) (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=176 \[ \frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac{\sqrt{c+d x^2} (3 b c-4 a d) (2 a d+b c)}{3 a^2 c^3 x (b c-a d)}-\frac{\sqrt{c+d x^2} (b c-4 a d)}{3 a c^2 x^3 (b c-a d)}-\frac{d}{c x^3 \sqrt{c+d x^2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.218351, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac{\sqrt{c+d x^2} (3 b c-4 a d) (2 a d+b c)}{3 a^2 c^3 x (b c-a d)}-\frac{\sqrt{c+d x^2} (b c-4 a d)}{3 a c^2 x^3 (b c-a d)}-\frac{d}{c x^3 \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

Mathematica [A]  time = 5.20593, size = 124, normalized size = 0.7 \[ \frac{b^3 \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{5/2} (b c-a d)^{3/2}}+\frac{\sqrt{c+d x^2} \left (\frac{x^2 (5 a d+3 b c)}{a^2}+\frac{3 d^3 x^4}{\left (c+d x^2\right ) (a d-b c)}-\frac{c}{a}\right )}{3 c^3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.014, size = 762, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a/c/x^3/(d*x^2+c)^(1/2)+4/3/a*d/c^2/x/(d*x^2+c)^(1/2)+8/3/a*d^2/c^3*x/(d*x^2+c)^(1/2)+1/2*b^3/a^2/(-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^2/a
^2/(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^
3/a^2/(-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^3/a^2/(-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^2/a^2/(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^3/a^2/(-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)))+b/a^2/c/x/(d*x^2+c)^(1/2)+2*b/a^2*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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4), x)

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Fricas [B]  time = 3.30226, size = 1412, normalized size = 8.02 \begin{align*} \left [\frac{3 \,{\left (b^{3} c^{3} d x^{5} + b^{3} c^{4} x^{3}\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^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} -{\left (3 \, a b^{3} c^{3} d - a^{2} b^{2} c^{2} d^{2} - 10 \, a^{3} b c d^{3} + 8 \, a^{4} d^{4}\right )} x^{4} -{\left (3 \, a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d - 5 \, a^{3} b c^{2} d^{2} + 4 \, a^{4} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{12 \,{\left ({\left (a^{3} b^{2} c^{5} d - 2 \, a^{4} b c^{4} d^{2} + a^{5} c^{3} d^{3}\right )} x^{5} +{\left (a^{3} b^{2} c^{6} - 2 \, a^{4} b c^{5} d + a^{5} c^{4} d^{2}\right )} x^{3}\right )}}, \frac{3 \,{\left (b^{3} c^{3} d x^{5} + b^{3} c^{4} x^{3}\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^{2} b^{2} c^{4} - 2 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2} -{\left (3 \, a b^{3} c^{3} d - a^{2} b^{2} c^{2} d^{2} - 10 \, a^{3} b c d^{3} + 8 \, a^{4} d^{4}\right )} x^{4} -{\left (3 \, a b^{3} c^{4} - 2 \, a^{2} b^{2} c^{3} d - 5 \, a^{3} b c^{2} d^{2} + 4 \, a^{4} c d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{6 \,{\left ({\left (a^{3} b^{2} c^{5} d - 2 \, a^{4} b c^{4} d^{2} + a^{5} c^{3} d^{3}\right )} x^{5} +{\left (a^{3} b^{2} c^{6} - 2 \, a^{4} b c^{5} d + a^{5} c^{4} d^{2}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*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^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 - (3*a*b^3*c^3*d - a^2*b^2*c^2*d^2 - 10
*a^3*b*c*d^3 + 8*a^4*d^4)*x^4 - (3*a*b^3*c^4 - 2*a^2*b^2*c^3*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2)*sqrt(d*x^
2 + c))/((a^3*b^2*c^5*d - 2*a^4*b*c^4*d^2 + a^5*c^3*d^3)*x^5 + (a^3*b^2*c^6 - 2*a^4*b*c^5*d + a^5*c^4*d^2)*x^3
), 1/6*(3*(b^3*c^3*d*x^5 + b^3*c^4*x^3)*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^2*b^2*c^4 - 2*a^3*b*c^3*d + a
^4*c^2*d^2 - (3*a*b^3*c^3*d - a^2*b^2*c^2*d^2 - 10*a^3*b*c*d^3 + 8*a^4*d^4)*x^4 - (3*a*b^3*c^4 - 2*a^2*b^2*c^3
*d - 5*a^3*b*c^2*d^2 + 4*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^2*c^5*d - 2*a^4*b*c^4*d^2 + a^5*c^3*d^3)*x^5
+ (a^3*b^2*c^6 - 2*a^4*b*c^5*d + a^5*c^4*d^2)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \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**4/(b*x**2+a)/(d*x**2+c)**(3/2),x)

[Out]

Integral(1/(x**4*(a + b*x**2)*(c + d*x**2)**(3/2)), x)

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Giac [A]  time = 3.69682, size = 371, normalized size = 2.11 \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^{2} b c - a^{3} d\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{d^{3} x}{{\left (b c^{4} - a c^{3} d\right )} \sqrt{d x^{2} + c}} - \frac{2 \,{\left (3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b c \sqrt{d} + 3 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} a d^{\frac{3}{2}} - 6 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c^{2} \sqrt{d} - 12 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a c d^{\frac{3}{2}} + 3 \, b c^{3} \sqrt{d} + 5 \, a c^{2} d^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{3} a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c)^(3/2),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^2*b*c -
a^3*d)*sqrt(a*b*c*d - a^2*d^2)) - d^3*x/((b*c^4 - a*c^3*d)*sqrt(d*x^2 + c)) - 2/3*(3*(sqrt(d)*x - sqrt(d*x^2
+ c))^4*b*c*sqrt(d) + 3*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*d^(3/2) - 6*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c^2*sq
rt(d) - 12*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*c*d^(3/2) + 3*b*c^3*sqrt(d) + 5*a*c^2*d^(3/2))/(((sqrt(d)*x - sqr
t(d*x^2 + c))^2 - c)^3*a^2*c^2)