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

Optimal. Leaf size=170 \[ \frac{b^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^2 (b c-a d)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a^2 c^{3/2}}+\frac{b}{2 a \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}+\frac{d (2 a d+b c)}{2 a c \sqrt{c+d x^2} (b c-a d)^2} \]

[Out]

(d*(b*c + 2*a*d))/(2*a*c*(b*c - a*d)^2*Sqrt[c + d*x^2]) + b/(2*a*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - Ar
cTanh[Sqrt[c + d*x^2]/Sqrt[c]]/(a^2*c^(3/2)) + (b^(3/2)*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt
[b*c - a*d]])/(2*a^2*(b*c - a*d)^(5/2))

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Rubi [A]  time = 0.244308, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 103, 152, 156, 63, 208} \[ \frac{b^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^2 (b c-a d)^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a^2 c^{3/2}}+\frac{b}{2 a \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}+\frac{d (2 a d+b c)}{2 a c \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c + 2*a*d))/(2*a*c*(b*c - a*d)^2*Sqrt[c + d*x^2]) + b/(2*a*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - Ar
cTanh[Sqrt[c + d*x^2]/Sqrt[c]]/(a^2*c^(3/2)) + (b^(3/2)*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt
[b*c - a*d]])/(2*a^2*(b*c - a*d)^(5/2))

Rule 446

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.110675, size = 123, normalized size = 0.72 \[ \frac{\frac{b (2 b c-5 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )}{a (a d-b c)}+\left (\frac{2 b}{a}-\frac{2 d}{c}\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )+\frac{b}{a+b x^2}}{2 a \sqrt{c+d x^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b/(a + b*x^2) + (b*(2*b*c - 5*a*d)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^2))/(b*c - a*d)])/(a*(-(b*c) +
a*d)) + ((2*b)/a - (2*d)/c)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^2)/c])/(2*a*(b*c - a*d)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.013, size = 1672, normalized size = 9.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 21.1527, size = 4030, normalized size = 23.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2
*b*c^2*d^2)*x^2)*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d - 3*a*b
*d^2)*x^2 - 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*c - a*d)))
/(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 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*
d^3)*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c
) + 2*c)/x^2) - 4*(a*b^2*c^3 + 2*a^3*c*d^2 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^2)*sqrt(d*x^2 + c))/(a^3*b^2*c^5
- 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^4 + (a^2*b^3*c^5 - a^3*b
^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*x^2), 1/8*(8*(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a
*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sq
rt(d*x^2 + c)) - (2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 + (2*b^3*c^4 - 3*a*b^2*c^3
*d - 5*a^2*b*c^2*d^2)*x^2)*sqrt(b/(b*c - a*d))*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c
*d - 3*a*b*d^2)*x^2 - 4*(2*b^2*c^2 - 3*a*b*c*d + a^2*d^2 + (b^2*c*d - a*b*d^2)*x^2)*sqrt(d*x^2 + c)*sqrt(b/(b*
c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*(a*b^2*c^3 + 2*a^3*c*d^2 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^2)*sqrt
(d*x^2 + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*
x^4 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*x^2), -1/4*((2*a*b^2*c^4 - 5*a^2*b*c^3*d + (
2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2)*x^2)*sqrt(-b/(b*c - a*d))*a
rctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^2 + b*c)) - 2*(a*b^2*c^3 - 2*a^2
*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^
3*d^3)*x^2)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(a*b^2*c^3 + 2*a^3*c*d^2 + (a*b^2*
c^2*d + 2*a^2*b*c*d^2)*x^2)*sqrt(d*x^2 + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b^3*c^4*d - 2*a
^3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^4 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3)*x^2), -1/4*(
(2*a*b^2*c^4 - 5*a^2*b*c^3*d + (2*b^3*c^3*d - 5*a*b^2*c^2*d^2)*x^4 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*
d^2)*x^2)*sqrt(-b/(b*c - a*d))*arctan(1/2*(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-b/(b*c - a*d))/(b*d*x^
2 + b*c)) - 4*(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3
- a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - 2*(a*b^2*c^3 + 2*a^3*c
*d^2 + (a*b^2*c^2*d + 2*a^2*b*c*d^2)*x^2)*sqrt(d*x^2 + c))/(a^3*b^2*c^5 - 2*a^4*b*c^4*d + a^5*c^3*d^2 + (a^2*b
^3*c^4*d - 2*a^3*b^2*c^3*d^2 + a^4*b*c^2*d^3)*x^4 + (a^2*b^3*c^5 - a^3*b^2*c^4*d - a^4*b*c^3*d^2 + a^5*c^2*d^3
)*x^2)]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.13165, size = 319, normalized size = 1.88 \begin{align*} -\frac{1}{2} \, d^{2}{\left (\frac{{\left (2 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{2} b^{2} c^{2} d^{2} - 2 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt{-b^{2} c + a b d}} - \frac{{\left (d x^{2} + c\right )} b^{2} c + 2 \,{\left (d x^{2} + c\right )} a b d - 2 \, a b c d + 2 \, a^{2} d^{2}}{{\left (a b^{2} c^{3} d - 2 \, a^{2} b c^{2} d^{2} + a^{3} c d^{3}\right )}{\left ({\left (d x^{2} + c\right )}^{\frac{3}{2}} b - \sqrt{d x^{2} + c} b c + \sqrt{d x^{2} + c} a d\right )}} - \frac{2 \, \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d^2*((2*b^3*c - 5*a*b^2*d)*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b^2*c^2*d^2 - 2*a^3*b*c*d
^3 + a^4*d^4)*sqrt(-b^2*c + a*b*d)) - ((d*x^2 + c)*b^2*c + 2*(d*x^2 + c)*a*b*d - 2*a*b*c*d + 2*a^2*d^2)/((a*b^
2*c^3*d - 2*a^2*b*c^2*d^2 + a^3*c*d^3)*((d*x^2 + c)^(3/2)*b - sqrt(d*x^2 + c)*b*c + sqrt(d*x^2 + c)*a*d)) - 2*
arctan(sqrt(d*x^2 + c)/sqrt(-c))/(a^2*sqrt(-c)*c*d^2))