Integrand size = 29, antiderivative size = 88 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a} \left (a c^2-d^2\right )}+\frac {c \log (x)}{a c^2-d^2}-\frac {c \log \left (d+c \sqrt {a+b x^2}\right )}{a c^2-d^2} \]
c*ln(x)/(a*c^2-d^2)-c*ln(d+c*(b*x^2+a)^(1/2))/(a*c^2-d^2)+d*arctanh((b*x^2 +a)^(1/2)/a^(1/2))/(a*c^2-d^2)/a^(1/2)
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {\frac {2 d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+c \log \left (b x^2\right )-2 c \log \left (d+c \sqrt {a+b x^2}\right )}{2 a c^2-2 d^2} \]
((2*d*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a] + c*Log[b*x^2] - 2*c*Log[d + c*Sqrt[a + b*x^2]])/(2*a*c^2 - 2*d^2)
Time = 0.43 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2586, 7267, 25, 479, 452, 219, 240}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x \left (d \sqrt {a+b x^2}+a c+b c x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2586 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (b c x^2+a c+d \sqrt {b x^2+a}\right )}dx^2\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle \int -\frac {1}{\left (a-x^4\right ) \left (c \sqrt {a+b x^2}+d\right )}d\sqrt {a+b x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\left (a-x^4\right ) \left (\sqrt {b x^2+a} c+d\right )}d\sqrt {b x^2+a}\) |
| \(\Big \downarrow \) 479 |
| \(\displaystyle \frac {\int \frac {d-c \sqrt {b x^2+a}}{a-x^4}d\sqrt {b x^2+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^2}+d\right )}{a c^2-d^2}\) |
| \(\Big \downarrow \) 452 |
| \(\displaystyle \frac {d \int \frac {1}{a-x^4}d\sqrt {b x^2+a}-c \int \frac {\sqrt {b x^2+a}}{a-x^4}d\sqrt {b x^2+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^2}+d\right )}{a c^2-d^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-c \int \frac {\sqrt {b x^2+a}}{a-x^4}d\sqrt {b x^2+a}}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^2}+d\right )}{a c^2-d^2}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {\frac {d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {1}{2} c \log \left (a-x^4\right )}{a c^2-d^2}-\frac {c \log \left (c \sqrt {a+b x^2}+d\right )}{a c^2-d^2}\) |
((d*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a] + (c*Log[a - x^4])/2)/(a*c^2 - d^2) - (c*Log[d + c*Sqrt[a + b*x^2]])/(a*c^2 - d^2)
3.6.47.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log [RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)] ), x_Symbol] :> Simp[1/n Subst[Int[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c - a* d, 0] && IntegerQ[(m + 1)/n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Leaf count of result is larger than twice the leaf count of optimal. \(1696\) vs. \(2(80)=160\).
Time = 0.08 (sec) , antiderivative size = 1697, normalized size of antiderivative = 19.28
-1/2*a*c^3/(a*c^2-d^2)/d^2*ln(b*c^2*x^2+a*c^2-d^2)+c*ln(x)/(a*c^2-d^2)+1/2 *c/d^2*ln(b*c^2*x^2+a*c^2-d^2)-d*(1/a/(a*c^2-d^2)*((b*x^2+a)^(1/2)-a^(1/2) *ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))-1/2*b*c^2/a/((-a*b)^(1/2)*c^2+(-(a *c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*((b* (x-1/b*(-a*b)^(1/2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2)+(-a*b)^( 1/2)*ln(((x-1/b*(-a*b)^(1/2))*b+(-a*b)^(1/2))/b^(1/2)+(b*(x-1/b*(-a*b)^(1/ 2))^2+2*(-a*b)^(1/2)*(x-1/b*(-a*b)^(1/2)))^(1/2))/b^(1/2))-1/2*b*c^2/a/((- a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2) *b*c^2)^(1/2))*((b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/ 2)))^(1/2)-(-a*b)^(1/2)*ln(((x+1/b*(-a*b)^(1/2))*b-(-a*b)^(1/2))/b^(1/2)+( b*(x+1/b*(-a*b)^(1/2))^2-2*(-a*b)^(1/2)*(x+1/b*(-a*b)^(1/2)))^(1/2))/b^(1/ 2))+1/2*b*c^4/(a*c^2-d^2)/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/(( -a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*((b*(x-(-(a*c^2-d^2)*b*c^2)^(1 /2)/b/c^2)^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c^2-d^2)*b*c^2)^(1/2 )/b/c^2)+d^2/c^2)^(1/2)+(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*ln(((-(a*c^2-d^2)*b *c^2)^(1/2)/c^2+(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/b^(1/2)+(b*(x-(-(a *c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c ^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2))/b^(1/2)-d^2/c^2/(d^2/c^2)^(1/2 )*ln((2*d^2/c^2+2*(-(a*c^2-d^2)*b*c^2)^(1/2)/c^2*(x-(-(a*c^2-d^2)*b*c^2)^( 1/2)/b/c^2)+2*(d^2/c^2)^(1/2)*(b*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2...
Time = 0.36 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.59 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [-\frac {2 \, a c \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a c \log \left (x\right ) + a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + 2 \, \sqrt {a} d \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right )}{4 \, {\left (a^{2} c^{2} - a d^{2}\right )}}, -\frac {2 \, a c \log \left (b c^{2} x^{2} + a c^{2} - d^{2}\right ) - 4 \, a c \log \left (x\right ) + a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} + 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) - a c \log \left (-\frac {b c^{2} x^{2} + a c^{2} - 2 \, \sqrt {b x^{2} + a} c d + d^{2}}{x^{2}}\right ) + 4 \, \sqrt {-a} d \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right )}{4 \, {\left (a^{2} c^{2} - a d^{2}\right )}}\right ] \]
[-1/4*(2*a*c*log(b*c^2*x^2 + a*c^2 - d^2) - 4*a*c*log(x) + a*c*log(-(b*c^2 *x^2 + a*c^2 + 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) - a*c*log(-(b*c^2*x^2 + a *c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) + 2*sqrt(a)*d*log(-(b*x^2 - 2*sqr t(b*x^2 + a)*sqrt(a) + 2*a)/x^2))/(a^2*c^2 - a*d^2), -1/4*(2*a*c*log(b*c^2 *x^2 + a*c^2 - d^2) - 4*a*c*log(x) + a*c*log(-(b*c^2*x^2 + a*c^2 + 2*sqrt( b*x^2 + a)*c*d + d^2)/x^2) - a*c*log(-(b*c^2*x^2 + a*c^2 - 2*sqrt(b*x^2 + a)*c*d + d^2)/x^2) + 4*sqrt(-a)*d*arctan(sqrt(-a)/sqrt(b*x^2 + a)))/(a^2*c ^2 - a*d^2)]
Time = 2.93 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {b c^{2} \left (\begin {cases} \frac {\sqrt {a + b x^{2}}}{d} & \text {for}\: c = 0 \\\frac {\log {\left (c \sqrt {a + b x^{2}} + d \right )}}{c} & \text {otherwise} \end {cases}\right )}{2 \left (a c^{2} - d^{2}\right )} - \frac {b \left (- \frac {c \log {\left (- b x^{2} \right )}}{2} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}}\right )}{2 \left (a c^{2} - d^{2}\right )}\right )}{b} & \text {for}\: b \neq 0 \\\begin {cases} \frac {x^{2} \log {\left (x^{2} \right )}}{2 \sqrt {a} d x^{2} + 2 a c x^{2}} & \text {for}\: 2 \sqrt {a} d + 2 a c \neq 0 \\\tilde {\infty } x^{2} & \text {otherwise} \end {cases} & \text {otherwise} \end {cases} \]
Piecewise((2*(-b*c**2*Piecewise((sqrt(a + b*x**2)/d, Eq(c, 0)), (log(c*sqr t(a + b*x**2) + d)/c, True))/(2*(a*c**2 - d**2)) - b*(-c*log(-b*x**2)/2 + d*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a))/(2*(a*c**2 - d**2)))/b, Ne(b, 0)), (Piecewise((x**2*log(x**2)/(2*sqrt(a)*d*x**2 + 2*a*c*x**2), Ne(2*sqrt (a)*d + 2*a*c, 0)), (zoo*x**2, True)), True))
\[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\int { \frac {1}{{\left (b c x^{2} + a c + \sqrt {b x^{2} + a} d\right )} x} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {c^{2} \log \left ({\left | \sqrt {b x^{2} + a} c + d \right |}\right )}{a c^{3} - c d^{2}} + \frac {c \log \left (b x^{2}\right )}{2 \, {\left (a c^{2} - d^{2}\right )}} - \frac {d \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{{\left (a c^{2} - d^{2}\right )} \sqrt {-a}} \]
-c^2*log(abs(sqrt(b*x^2 + a)*c + d))/(a*c^3 - c*d^2) + 1/2*c*log(b*x^2)/(a *c^2 - d^2) - d*arctan(sqrt(b*x^2 + a)/sqrt(-a))/((a*c^2 - d^2)*sqrt(-a))
Time = 18.19 (sec) , antiderivative size = 1270, normalized size of antiderivative = 14.43 \[ \int \frac {1}{x \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {c\,\ln \left (x\right )}{a\,c^2-d^2}-\frac {c\,\mathrm {atan}\left (\frac {\frac {c\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}+\frac {c\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d-\frac {c\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}\right )\,1{}\mathrm {i}}{2\,\left (a\,c^2-d^2\right )}+\frac {c\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}-\frac {c\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d+\frac {c\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}\right )\,1{}\mathrm {i}}{2\,\left (a\,c^2-d^2\right )}}{\frac {c\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}+\frac {c\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d-\frac {c\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}-\frac {c\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}-\frac {c\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d+\frac {c\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}\right )}{2\,\left (a\,c^2-d^2\right )}}\right )\,1{}\mathrm {i}}{a\,c^2-d^2}-\frac {c\,\ln \left (b\,c^2\,x^2+a\,c^2-d^2\right )}{2\,a\,c^2-2\,d^2}-\frac {d\,\mathrm {atan}\left (\frac {\frac {d\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}+\frac {d\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d-\frac {d\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )\,1{}\mathrm {i}}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}+\frac {d\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}-\frac {d\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d+\frac {d\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )\,1{}\mathrm {i}}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}}{\frac {d\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}+\frac {d\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d-\frac {d\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}-\frac {d\,\left (4\,c^6\,d^2\,\sqrt {b\,x^2+a}-\frac {d\,\left (4\,c^4\,d^5-8\,a\,c^6\,d^3+4\,a^2\,c^8\,d+\frac {d\,\sqrt {b\,x^2+a}\,\left (8\,a^3\,c^{10}-8\,a^2\,c^8\,d^2-8\,a\,c^6\,d^4+8\,c^4\,d^6\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}\right )}{\sqrt {a}\,\left (2\,a\,c^2-2\,d^2\right )}}\right )\,1{}\mathrm {i}}{\sqrt {a}\,\left (a\,c^2-d^2\right )} \]
(c*log(x))/(a*c^2 - d^2) - (c*atan(((c*(4*c^6*d^2*(a + b*x^2)^(1/2) + (c*( 4*c^4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d - (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2)))*1i)/(2*(a*c^2 - d^2)) + (c*(4*c^6*d^2*(a + b*x^2)^(1/2) - (c*(4*c^ 4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d + (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c ^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2 )))*1i)/(2*(a*c^2 - d^2)))/((c*(4*c^6*d^2*(a + b*x^2)^(1/2) + (c*(4*c^4*d^ 5 - 8*a*c^6*d^3 + 4*a^2*c^8*d - (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d ^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2)))) /(2*(a*c^2 - d^2)) - (c*(4*c^6*d^2*(a + b*x^2)^(1/2) - (c*(4*c^4*d^5 - 8*a *c^6*d^3 + 4*a^2*c^8*d + (c*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d^6 - 8* a*c^6*d^4 - 8*a^2*c^8*d^2))/(2*(a*c^2 - d^2))))/(2*(a*c^2 - d^2))))/(2*(a* c^2 - d^2))))*1i)/(a*c^2 - d^2) - (c*log(a*c^2 - d^2 + b*c^2*x^2))/(2*a*c^ 2 - 2*d^2) - (d*atan(((d*(4*c^6*d^2*(a + b*x^2)^(1/2) + (d*(4*c^4*d^5 - 8* a*c^6*d^3 + 4*a^2*c^8*d - (d*(a + b*x^2)^(1/2)*(8*a^3*c^10 + 8*c^4*d^6 - 8 *a*c^6*d^4 - 8*a^2*c^8*d^2))/(a^(1/2)*(2*a*c^2 - 2*d^2))))/(a^(1/2)*(2*a*c ^2 - 2*d^2)))*1i)/(a^(1/2)*(2*a*c^2 - 2*d^2)) + (d*(4*c^6*d^2*(a + b*x^2)^ (1/2) - (d*(4*c^4*d^5 - 8*a*c^6*d^3 + 4*a^2*c^8*d + (d*(a + b*x^2)^(1/2)*( 8*a^3*c^10 + 8*c^4*d^6 - 8*a*c^6*d^4 - 8*a^2*c^8*d^2))/(a^(1/2)*(2*a*c^2 - 2*d^2))))/(a^(1/2)*(2*a*c^2 - 2*d^2)))*1i)/(a^(1/2)*(2*a*c^2 - 2*d^2))...