Integrand size = 24, antiderivative size = 211 \[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {d (3 b c-5 a d)}{6 a c^2 (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {1}{2 a c x^2 \left (c+d x^2\right )^{3/2}}-\frac {d \left (b^2 c^2-8 a b c d+5 a^2 d^2\right )}{2 a c^3 (b c-a d)^2 \sqrt {c+d x^2}}+\frac {(2 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^2 c^{7/2}}-\frac {b^{7/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a^2 (b c-a d)^{5/2}} \]
-1/6*d*(-5*a*d+3*b*c)/a/c^2/(-a*d+b*c)/(d*x^2+c)^(3/2)-1/2/a/c/x^2/(d*x^2+ c)^(3/2)+1/2*(5*a*d+2*b*c)*arctanh((d*x^2+c)^(1/2)/c^(1/2))/a^2/c^(7/2)-b^ (7/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*c)^(1/2))/a^2/(-a*d+b*c)^(5/ 2)-1/2*d*(5*a^2*d^2-8*a*b*c*d+b^2*c^2)/a/c^3/(-a*d+b*c)^2/(d*x^2+c)^(1/2)
Time = 0.67 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {-\frac {a \left (3 b^2 c^2 \left (c+d x^2\right )^2-2 a b c d \left (3 c^2+16 c d x^2+12 d^2 x^4\right )+a^2 d^2 \left (3 c^2+20 c d x^2+15 d^2 x^4\right )\right )}{c^3 (b c-a d)^2 x^2 \left (c+d x^2\right )^{3/2}}+\frac {6 b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{5/2}}+\frac {3 (2 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{7/2}}}{6 a^2} \]
(-((a*(3*b^2*c^2*(c + d*x^2)^2 - 2*a*b*c*d*(3*c^2 + 16*c*d*x^2 + 12*d^2*x^ 4) + a^2*d^2*(3*c^2 + 20*c*d*x^2 + 15*d^2*x^4)))/(c^3*(b*c - a*d)^2*x^2*(c + d*x^2)^(3/2))) + (6*b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[-(b*c ) + a*d]])/(-(b*c) + a*d)^(5/2) + (3*(2*b*c + 5*a*d)*ArcTanh[Sqrt[c + d*x^ 2]/Sqrt[c]])/c^(7/2))/(6*a^2)
Time = 0.45 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.22, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {354, 114, 27, 169, 27, 169, 27, 174, 73, 221}
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.
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {5 b d x^2+2 b c+5 a d}{2 x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2}{a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {5 b d x^2+2 b c+5 a d}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{5/2}}dx^2}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {2 \int -\frac {3 \left (b d (3 b c-5 a d) x^2+(b c-a d) (2 b c+5 a d)\right )}{2 x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx^2}{3 c (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\int \frac {b d (3 b c-5 a d) x^2+(b c-a d) (2 b c+5 a d)}{x^2 \left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx^2}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {2 d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}-\frac {2 \int -\frac {(2 b c+5 a d) (b c-a d)^2+b d \left (b^2 c^2-8 a b d c+5 a^2 d^2\right ) x^2}{2 x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{c (b c-a d)}}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {\int \frac {(2 b c+5 a d) (b c-a d)^2+b d \left (b^2 c^2-8 a b d c+5 a^2 d^2\right ) x^2}{x^2 \left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{c (b c-a d)}+\frac {2 d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {\frac {(b c-a d)^2 (5 a d+2 b c) \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{a}-\frac {2 b^4 c^3 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx^2}{a}}{c (b c-a d)}+\frac {2 d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {\frac {2 (b c-a d)^2 (5 a d+2 b c) \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{a d}-\frac {4 b^4 c^3 \int \frac {1}{\frac {b x^4}{d}+a-\frac {b c}{d}}d\sqrt {d x^2+c}}{a d}}{c (b c-a d)}+\frac {2 d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\frac {\frac {2 d \left (5 a^2 d^2-8 a b c d+b^2 c^2\right )}{c \sqrt {c+d x^2} (b c-a d)}+\frac {\frac {4 b^{7/2} c^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 (b c-a d)^2 (5 a d+2 b c) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{a \sqrt {c}}}{c (b c-a d)}}{c (b c-a d)}+\frac {2 d (3 b c-5 a d)}{3 c \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 a c}-\frac {1}{a c x^2 \left (c+d x^2\right )^{3/2}}\right )\) |
(-(1/(a*c*x^2*(c + d*x^2)^(3/2))) - ((2*d*(3*b*c - 5*a*d))/(3*c*(b*c - a*d )*(c + d*x^2)^(3/2)) + ((2*d*(b^2*c^2 - 8*a*b*c*d + 5*a^2*d^2))/(c*(b*c - a*d)*Sqrt[c + d*x^2]) + ((-2*(b*c - a*d)^2*(2*b*c + 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(a*Sqrt[c]) + (4*b^(7/2)*c^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(c*(b*c - a*d)))/(c*(b*c - a*d)))/(2*a*c))/2
3.8.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S imp[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]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 3.18 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(d^{2} \left (\frac {b^{4} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} a^{2} d^{2} \sqrt {\left (a d -b c \right ) b}}-\frac {-5 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) a d \,x^{2}-2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right ) b c \,x^{2}+\sqrt {d \,x^{2}+c}\, a \sqrt {c}}{2 x^{2} c^{\frac {7}{2}} a^{2} d^{2}}-\frac {1}{3 \left (a d -b c \right ) c^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}-\frac {2 a d -3 b c}{\left (a d -b c \right )^{2} c^{3} \sqrt {d \,x^{2}+c}}\right )\) | \(190\) |
| risch | \(\text {Expression too large to display}\) | \(1338\) |
| default | \(\text {Expression too large to display}\) | \(1548\) |
d^2*(1/(a*d-b*c)^2*b^4/a^2/d^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x^2+c)^(1/2 )/((a*d-b*c)*b)^(1/2))-1/2*(-5*arctanh((d*x^2+c)^(1/2)/c^(1/2))*a*d*x^2-2* arctanh((d*x^2+c)^(1/2)/c^(1/2))*b*c*x^2+(d*x^2+c)^(1/2)*a*c^(1/2))/x^2/c^ (7/2)/a^2/d^2-1/3/(a*d-b*c)/c^2/(d*x^2+c)^(3/2)-(2*a*d-3*b*c)/(a*d-b*c)^2/ c^3/(d*x^2+c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (181) = 362\).
Time = 1.45 (sec) , antiderivative size = 2219, normalized size of antiderivative = 10.52 \[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/12*(3*(b^3*c^4*d^2*x^6 + 2*b^3*c^5*d*x^4 + b^3*c^6*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)) + 3* ((2*b^3*c^3*d^2 + a*b^2*c^2*d^3 - 8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^ 3*c^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + (2*b^3*c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x^2)*sqrt(c)*log(-(d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + 3*(a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^4 + 2 *(3*a*b^2*c^4*d - 16*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x^2)*sqrt(d*x^2 + c)) /((a^2*b^2*c^6*d^2 - 2*a^3*b*c^5*d^3 + a^4*c^4*d^4)*x^6 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^4 + (a^2*b^2*c^8 - 2*a^3*b*c^7*d + a^4 *c^6*d^2)*x^2), -1/12*(6*((2*b^3*c^3*d^2 + a*b^2*c^2*d^3 - 8*a^2*b*c*d^4 + 5*a^3*d^5)*x^6 + 2*(2*b^3*c^4*d + a*b^2*c^3*d^2 - 8*a^2*b*c^2*d^3 + 5*a^3 *c*d^4)*x^4 + (2*b^3*c^5 + a*b^2*c^4*d - 8*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)* x^2)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^2 + c)) - 3*(b^3*c^4*d^2*x^6 + 2*b^ 3*c^5*d*x^4 + b^3*c^6*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*(3*a*b^2*c^5 - 6*a^2*b*c^4*d ...
\[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{3}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {b^{4} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {9 \, {\left (d x^{2} + c\right )} b c d^{2} + b c^{2} d^{2} - 6 \, {\left (d x^{2} + c\right )} a d^{3} - a c d^{3}}{3 \, {\left (b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}\right )} {\left (d x^{2} + c\right )}^{\frac {3}{2}}} - \frac {{\left (2 \, b c + 5 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{2} \sqrt {-c} c^{3}} - \frac {\sqrt {d x^{2} + c}}{2 \, a c^{3} x^{2}} \]
b^4*arctan(sqrt(d*x^2 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b^2*c^2 - 2*a^3*b *c*d + a^4*d^2)*sqrt(-b^2*c + a*b*d)) + 1/3*(9*(d*x^2 + c)*b*c*d^2 + b*c^2 *d^2 - 6*(d*x^2 + c)*a*d^3 - a*c*d^3)/((b^2*c^5 - 2*a*b*c^4*d + a^2*c^3*d^ 2)*(d*x^2 + c)^(3/2)) - 1/2*(2*b*c + 5*a*d)*arctan(sqrt(d*x^2 + c)/sqrt(-c ))/(a^2*sqrt(-c)*c^3) - 1/2*sqrt(d*x^2 + c)/(a*c^3*x^2)
Time = 8.14 (sec) , antiderivative size = 5409, normalized size of antiderivative = 25.64 \[ \int \frac {1}{x^3 \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
(atan(((((c + d*x^2)^(1/2)*(128*a^3*b^15*c^21*d^2 - 704*a^4*b^14*c^20*d^3 + 1040*a^5*b^13*c^19*d^4 + 1440*a^6*b^12*c^18*d^5 - 6000*a^7*b^11*c^17*d^6 + 2688*a^8*b^10*c^16*d^7 + 16864*a^9*b^9*c^15*d^8 - 41280*a^10*b^8*c^14*d ^9 + 48480*a^11*b^7*c^13*d^10 - 34240*a^12*b^6*c^12*d^11 + 14864*a^13*b^5* c^11*d^12 - 3680*a^14*b^4*c^10*d^13 + 400*a^15*b^3*c^9*d^14) + ((5*a*d + 2 *b*c)*(64*a^6*b^13*c^23*d^3 + 64*a^7*b^12*c^22*d^4 - 3648*a^8*b^11*c^21*d^ 5 + 19520*a^9*b^10*c^20*d^6 - 53632*a^10*b^9*c^19*d^7 + 92288*a^11*b^8*c^1 8*d^8 - 106624*a^12*b^7*c^17*d^9 + 84608*a^13*b^6*c^16*d^10 - 45760*a^14*b ^5*c^15*d^11 + 16192*a^15*b^4*c^14*d^12 - 3392*a^16*b^3*c^13*d^13 + 320*a^ 17*b^2*c^12*d^14 - ((c + d*x^2)^(1/2)*(5*a*d + 2*b*c)*(512*a^7*b^13*c^26*d ^2 - 5376*a^8*b^12*c^25*d^3 + 25600*a^9*b^11*c^24*d^4 - 72960*a^10*b^10*c^ 23*d^5 + 138240*a^11*b^9*c^22*d^6 - 182784*a^12*b^8*c^21*d^7 + 172032*a^13 *b^7*c^20*d^8 - 115200*a^14*b^6*c^19*d^9 + 53760*a^15*b^5*c^18*d^10 - 1664 0*a^16*b^4*c^17*d^11 + 3072*a^17*b^3*c^16*d^12 - 256*a^18*b^2*c^15*d^13))/ (4*a^2*(c^7)^(1/2))))/(4*a^2*(c^7)^(1/2)))*(5*a*d + 2*b*c)*1i)/(4*a^2*(c^7 )^(1/2)) + (((c + d*x^2)^(1/2)*(128*a^3*b^15*c^21*d^2 - 704*a^4*b^14*c^20* d^3 + 1040*a^5*b^13*c^19*d^4 + 1440*a^6*b^12*c^18*d^5 - 6000*a^7*b^11*c^17 *d^6 + 2688*a^8*b^10*c^16*d^7 + 16864*a^9*b^9*c^15*d^8 - 41280*a^10*b^8*c^ 14*d^9 + 48480*a^11*b^7*c^13*d^10 - 34240*a^12*b^6*c^12*d^11 + 14864*a^13* b^5*c^11*d^12 - 3680*a^14*b^4*c^10*d^13 + 400*a^15*b^3*c^9*d^14) - ((5*...