3.493 \(\int \frac{1}{x (a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=172 \[ \frac{b^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{3/2}}+\frac{b}{3 a \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)}+\frac{d (2 a d+b c)}{3 a c \sqrt{c+d x^3} (b c-a d)^2} \]
[Out]
(d*(b*c + 2*a*d))/(3*a*c*(b*c - a*d)^2*Sqrt[c + d*x^3]) + b/(3*a*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) - (2
*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(3/2)) + (b^(3/2)*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3]
)/Sqrt[b*c - a*d]])/(3*a^2*(b*c - a*d)^(5/2))
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Rubi [A] time = 0.222598, antiderivative size = 172, 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} \[ \frac{b^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{5/2}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 a^2 c^{3/2}}+\frac{b}{3 a \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)}+\frac{d (2 a d+b c)}{3 a c \sqrt{c+d x^3} (b c-a d)^2} \]
Antiderivative was successfully verified.
[In]
Int[1/(x*(a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(d*(b*c + 2*a*d))/(3*a*c*(b*c - a*d)^2*Sqrt[c + d*x^3]) + b/(3*a*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3]) - (2
*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(3/2)) + (b^(3/2)*(2*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3]
)/Sqrt[b*c - a*d]])/(3*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^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\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^3\right )}{3 a (b c-a d)}\\ &=\frac{d (b c+2 a d)}{3 a c (b c-a d)^2 \sqrt{c+d x^3}}+\frac{b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{2 \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^3\right )}{3 a c (b c-a d)^2}\\ &=\frac{d (b c+2 a d)}{3 a c (b c-a d)^2 \sqrt{c+d x^3}}+\frac{b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{3 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^3\right )}{6 a^2 (b c-a d)^2}\\ &=\frac{d (b c+2 a d)}{3 a c (b c-a d)^2 \sqrt{c+d x^3}}+\frac{b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{3 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^3}\right )}{3 a^2 d (b c-a d)^2}\\ &=\frac{d (b c+2 a d)}{3 a c (b c-a d)^2 \sqrt{c+d x^3}}+\frac{b}{3 a (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{3 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^3}}{\sqrt{b c-a d}}\right )}{3 a^2 (b c-a d)^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.104378, size = 123, normalized size = 0.72 \[ \frac{\frac{b (2 b c-5 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \left (d x^3+c\right )}{b c-a d}\right )}{a (a d-b c)}+\left (\frac{2 b}{a}-\frac{2 d}{c}\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3}{c}+1\right )+\frac{b}{a+b x^3}}{3 a \sqrt{c+d x^3} (b c-a d)} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x*(a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(b/(a + b*x^3) + (b*(2*b*c - 5*a*d)*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3))/(b*c - a*d)])/(a*(-(b*c) +
a*d)) + ((2*b)/a - (2*d)/c)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^3)/c])/(3*a*(b*c - a*d)*Sqrt[c + d*x^3])
________________________________________________________________________________________
Maple [C] time = 0.01, size = 1002, normalized size = 5.8 \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^3+a)^2/(d*x^3+c)^(3/2),x)
[Out]
-1/a^2*b*(-2/3/(a*d-b*c)/((x^3+1/d*c)*d)^(1/2)-1/3*I/d^2*b*2^(1/2)*sum(1/(-a*d+b*c)/(a*d-b*c)*(-d^2*c)^(1/3)*(
1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(
-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)
))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^
2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/
d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2
/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*
(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))-b/a*(-2/3*d/(a*d-b*c)^2/((x^3
+1/d*c)*d)^(1/2)-1/3*b/(a*d-b*c)^2*(d*x^3+c)^(1/2)/(b*x^3+a)+1/2*I/d*b*2^(1/2)*sum(1/(a*d-b*c)^3*(-d^2*c)^(1/3
)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3)
)/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1
/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alp
ha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/
2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)
^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2
/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a^2*(2/3/c/((x^3+1/d*c)*
d)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{2}{\left (d x^{3} + 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^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)*x), x)
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Fricas [B] time = 3.26953, size = 3657, normalized size = 21.26 \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^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
[-1/6*((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^6 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2
*b*c^2*d^2)*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(b/(b*c -
a*d)))/(b*x^3 + a)) - 2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 +
(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^
3) - 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^3)*sqrt(d*x^3 + 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^6 + (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^3), 1/3*((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^6 + (
2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2)*x^3)*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqr
t(-b/(b*c - a*d))/(b*d*x^3 + b*c)) + ((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2*c^3 - 2*a^2*b*c^2*d
+ a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt
(c) + 2*c)/x^3) + (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^3)*sqrt(d*x^3 + 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^6 + (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^3), 1/6*(4*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^6 + a*b^2*c^3
- 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3
+ c)*sqrt(-c)/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^6 + (2*b^3*c^4 - 3*a*b^2*c
^3*d - 5*a^2*b*c^2*d^2)*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*(b*c - a*d)*sq
rt(b/(b*c - a*d)))/(b*x^3 + a)) + 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^3)*sqrt(d*x^3 +
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^6 + (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^3), 1/3*((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^6 + (2*b^3*c^4 - 3*a*b^2*c^3*d - 5*a^2*b*c^2*d^2)*x^3)*sqrt(-b/(b*c - a*d))*arctan(-sq
rt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x^3 + b*c)) + 2*((b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x
^6 + a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^3)*sqrt(-c)*arc
tan(sqrt(d*x^3 + c)*sqrt(-c)/c) + (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^3)*sqrt(d*x^3 + 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^6 + (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^3)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)
[Out]
Timed out
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Giac [A] time = 1.12252, size = 319, normalized size = 1.85 \begin{align*} -\frac{1}{3} \, d^{2}{\left (\frac{{\left (2 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{3} + 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^{3} + c\right )} b^{2} c + 2 \,{\left (d x^{3} + 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^{3} + c\right )}^{\frac{3}{2}} b - \sqrt{d x^{3} + c} b c + \sqrt{d x^{3} + c} a d\right )}} - \frac{2 \, \arctan \left (\frac{\sqrt{d x^{3} + 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^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
-1/3*d^2*((2*b^3*c - 5*a*b^2*d)*arctan(sqrt(d*x^3 + 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^3 + c)*b^2*c + 2*(d*x^3 + 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^3 + c)^(3/2)*b - sqrt(d*x^3 + c)*b*c + sqrt(d*x^3 + c)*a*d)) - 2*
arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^2*sqrt(-c)*c*d^2))