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

Optimal. Leaf size=241 \[ -\frac{d \left (3 a^2 d^2-2 a b c d+2 b^2 c^2\right )}{2 a^2 c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{b^{5/2} (4 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^3 (b c-a d)^{5/2}}+\frac{(3 a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^3 c^{5/2}}-\frac{b (2 b c-a d)}{2 a^2 c \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}-\frac{1}{2 a c x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \]

[Out]

-(d*(2*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2))/(2*a^2*c^2*(b*c - a*d)^2*Sqrt[c + d*x^2]) - (b*(2*b*c - a*d))/(2*a^2*
c*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - 1/(2*a*c*x^2*(a + b*x^2)*Sqrt[c + d*x^2]) + ((4*b*c + 3*a*d)*ArcT
anh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*a^3*c^(5/2)) - (b^(5/2)*(4*b*c - 7*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqr
t[b*c - a*d]])/(2*a^3*(b*c - a*d)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.336123, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {446, 103, 151, 152, 156, 63, 208} \[ -\frac{d \left (3 a^2 d^2-2 a b c d+2 b^2 c^2\right )}{2 a^2 c^2 \sqrt{c+d x^2} (b c-a d)^2}-\frac{b^{5/2} (4 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^2}}{\sqrt{b c-a d}}\right )}{2 a^3 (b c-a d)^{5/2}}+\frac{(3 a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 a^3 c^{5/2}}-\frac{b (2 b c-a d)}{2 a^2 c \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)}-\frac{1}{2 a c x^2 \left (a+b x^2\right ) \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(2*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2))/(2*a^2*c^2*(b*c - a*d)^2*Sqrt[c + d*x^2]) - (b*(2*b*c - a*d))/(2*a^2*
c*(b*c - a*d)*(a + b*x^2)*Sqrt[c + d*x^2]) - 1/(2*a*c*x^2*(a + b*x^2)*Sqrt[c + d*x^2]) + ((4*b*c + 3*a*d)*ArcT
anh[Sqrt[c + d*x^2]/Sqrt[c]])/(2*a^3*c^(5/2)) - (b^(5/2)*(4*b*c - 7*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqr
t[b*c - a*d]])/(2*a^3*(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 151

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] && IntegerQ[m]

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

Mathematica [C]  time = 0.115818, size = 189, normalized size = 0.78 \[ \frac{b^2 c^2 x^2 \left (a+b x^2\right ) (4 b c-7 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^2+c\right )}{b c-a d}\right )-(a d-b c) \left (x^2 \left (a+b x^2\right ) \left (3 a^2 d^2+a b c d-4 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^2}{c}+1\right )+a c \left (a^2 d+a b \left (d x^2-c\right )-2 b^2 c x^2\right )\right )}{2 a^3 c^2 x^2 \left (a+b x^2\right ) \sqrt{c+d x^2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2*(4*b*c - 7*a*d)*x^2*(a + b*x^2)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^2))/(b*c - a*d)] - (-(b*c
) + a*d)*(a*c*(a^2*d - 2*b^2*c*x^2 + a*b*(-c + d*x^2)) + (-4*b^2*c^2 + a*b*c*d + 3*a^2*d^2)*x^2*(a + b*x^2)*Hy
pergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^2)/c]))/(2*a^3*c^2*(b*c - a*d)^2*x^2*(a + b*x^2)*Sqrt[c + d*x^2])

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Maple [B]  time = 0.016, size = 1778, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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^3), x)

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Fricas [B]  time = 41.5104, size = 5106, normalized size = 21.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x^6 + (4*b^4*c^5 - 3*a*b^3*c^4*d - 7*a^2*b^2*c^3*d^2)*x^4 + (4*a*b^3*c
^5 - 7*a^2*b^2*c^4*d)*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)) - 2*((4*b^4*c^3*d - 5*a*b^3*c^2*d^2 - 2*a^2*b^2*c*d^3 + 3*a^3*b*d^4)*x
^6 + (4*b^4*c^4 - a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + a^3*b*c*d^3 + 3*a^4*d^4)*x^4 + (4*a*b^3*c^4 - 5*a^2*b^2*c^
3*d - 2*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 4*(a^2
*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (2*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x^4 + (2*a*b^3*c^
4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^3*c^5*d - 2*a^4*b^2*c^4*d^2 + a
^5*b*c^3*d^3)*x^6 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x^4 + (a^4*b^2*c^6 - 2*a^5*b*c
^5*d + a^6*c^4*d^2)*x^2), -1/8*(4*((4*b^4*c^3*d - 5*a*b^3*c^2*d^2 - 2*a^2*b^2*c*d^3 + 3*a^3*b*d^4)*x^6 + (4*b^
4*c^4 - a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + a^3*b*c*d^3 + 3*a^4*d^4)*x^4 + (4*a*b^3*c^4 - 5*a^2*b^2*c^3*d - 2*a^
3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + ((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x
^6 + (4*b^4*c^5 - 3*a*b^3*c^4*d - 7*a^2*b^2*c^3*d^2)*x^4 + (4*a*b^3*c^5 - 7*a^2*b^2*c^4*d)*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^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (2*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x^4 + (2*a
*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^3*c^5*d - 2*a^4*b^2*c^4*
d^2 + a^5*b*c^3*d^3)*x^6 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x^4 + (a^4*b^2*c^6 - 2*
a^5*b*c^5*d + a^6*c^4*d^2)*x^2), 1/4*(((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x^6 + (4*b^4*c^5 - 3*a*b^3*c^4*d - 7*a^
2*b^2*c^3*d^2)*x^4 + (4*a*b^3*c^5 - 7*a^2*b^2*c^4*d)*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*b^4*c^3*d - 5*a*b^3*c^2*d^2 - 2*a^2*b^2*c*d^3
+ 3*a^3*b*d^4)*x^6 + (4*b^4*c^4 - a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + a^3*b*c*d^3 + 3*a^4*d^4)*x^4 + (4*a*b^3*c^
4 - 5*a^2*b^2*c^3*d - 2*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*
c)/x^2) - 2*(a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (2*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x
^4 + (2*a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^3*c^5*d - 2*a^4
*b^2*c^4*d^2 + a^5*b*c^3*d^3)*x^6 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x^4 + (a^4*b^2
*c^6 - 2*a^5*b*c^5*d + a^6*c^4*d^2)*x^2), 1/4*(((4*b^4*c^4*d - 7*a*b^3*c^3*d^2)*x^6 + (4*b^4*c^5 - 3*a*b^3*c^4
*d - 7*a^2*b^2*c^3*d^2)*x^4 + (4*a*b^3*c^5 - 7*a^2*b^2*c^4*d)*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)) - 2*((4*b^4*c^3*d - 5*a*b^3*c^2*d^2 - 2*a^2
*b^2*c*d^3 + 3*a^3*b*d^4)*x^6 + (4*b^4*c^4 - a*b^3*c^3*d - 7*a^2*b^2*c^2*d^2 + a^3*b*c*d^3 + 3*a^4*d^4)*x^4 +
(4*a*b^3*c^4 - 5*a^2*b^2*c^3*d - 2*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c))
- 2*(a^2*b^2*c^4 - 2*a^3*b*c^3*d + a^4*c^2*d^2 + (2*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 + 3*a^3*b*c*d^3)*x^4 + (2
*a*b^3*c^4 - a^2*b^2*c^3*d - a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^2)*sqrt(d*x^2 + c))/((a^3*b^3*c^5*d - 2*a^4*b^2*c^
4*d^2 + a^5*b*c^3*d^3)*x^6 + (a^3*b^3*c^6 - a^4*b^2*c^5*d - a^5*b*c^4*d^2 + a^6*c^3*d^3)*x^4 + (a^4*b^2*c^6 -
2*a^5*b*c^5*d + a^6*c^4*d^2)*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**3/(b*x**2+a)**2/(d*x**2+c)**(3/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.18386, size = 510, normalized size = 2.12 \begin{align*} \frac{1}{2} \, d^{3}{\left (\frac{{\left (4 \, b^{4} c - 7 \, a b^{3} d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{3} b^{2} c^{2} d^{3} - 2 \, a^{4} b c d^{4} + a^{5} d^{5}\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \,{\left (d x^{2} + c\right )}^{2} b^{3} c^{2} - 2 \,{\left (d x^{2} + c\right )} b^{3} c^{3} - 2 \,{\left (d x^{2} + c\right )}^{2} a b^{2} c d + 3 \,{\left (d x^{2} + c\right )} a b^{2} c^{2} d + 3 \,{\left (d x^{2} + c\right )}^{2} a^{2} b d^{2} - 7 \,{\left (d x^{2} + c\right )} a^{2} b c d^{2} + 2 \, a^{2} b c^{2} d^{2} + 3 \,{\left (d x^{2} + c\right )} a^{3} d^{3} - 2 \, a^{3} c d^{3}}{{\left (a^{2} b^{2} c^{4} d^{2} - 2 \, a^{3} b c^{3} d^{3} + a^{4} c^{2} d^{4}\right )}{\left ({\left (d x^{2} + c\right )}^{\frac{5}{2}} b - 2 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b c + \sqrt{d x^{2} + c} b c^{2} +{\left (d x^{2} + c\right )}^{\frac{3}{2}} a d - \sqrt{d x^{2} + c} a c d\right )}} - \frac{{\left (4 \, b c + 3 \, a d\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c} c^{2} d^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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