Integrand size = 33, antiderivative size = 198 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}-\frac {2 a^2 (A b-a B)}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]
(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(5/2)/d-(I*A -B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(5/2)/d+2*a*(2*A *b^3-a*(a^2+3*b^2)*B)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(1/2)-2/3*a^2*(A* b-B*a)/b^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.16 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 (a-i b) (a+i b) (A b+2 a B)+b (A b-a B) \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )+6 (a-i b) (a+i b) b B \tan (c+d x)+3 b B \left (i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )-(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))}{3 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}} \]
-1/3*(2*(a - I*b)*(a + I*b)*(A*b + 2*a*B) + b*(A*b - a*B)*(I*(a + I*b)*Hyp ergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)* Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I*b)]) + 6*(a - I*b)*(a + I*b)*b*B*Tan[c + d*x] + 3*b*B*(I*(a + I*b)*Hypergeometric2F1[-1 /2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] - (I*a + b)*Hypergeometric2F1[ -1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x]))/(b^2* (a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))
Time = 1.08 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4087, 25, 3042, 4111, 3042, 4022, 3042, 4020, 25, 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.
\(\displaystyle \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2 (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4087 |
\(\displaystyle \frac {\int -\frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{(a+b \tan (c+d x))^{3/2}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{(a+b \tan (c+d x))^{3/2}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan (c+d x)^2\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{(a+b \tan (c+d x))^{3/2}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4111 |
\(\displaystyle -\frac {\frac {\int \frac {b \left (A a^2+2 b B a-A b^2\right )-b \left (-B a^2+2 A b a+b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {b \left (A a^2+2 b B a-A b^2\right )-b \left (-B a^2+2 A b a+b^2 B\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} b (a-i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} b (a+i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} b (a-i b)^2 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} b (a+i b)^2 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i b (a+i b)^2 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i b (a-i b)^2 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i b (a-i b)^2 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i b (a+i b)^2 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b)^2 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {(a+i b)^2 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{3 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}-\frac {-\frac {2 a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {b (a-i b)^2 (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {b (a+i b)^2 (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}}{a^2+b^2}}{b \left (a^2+b^2\right )}\) |
(-2*a^2*(A*b - a*B))/(3*b^2*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) - (( ((a + I*b)^2*b*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b ]*d) + ((a - I*b)^2*b*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[ a + I*b]*d))/(a^2 + b^2) - (2*a*(2*A*b^3 - a*(a^2 + 3*b^2)*B))/(b*(a^2 + b ^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(b*(a^2 + b^2))
3.4.59.3.1 Defintions of rubi rules used
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_)^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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (-(B*c - A*d))*(b*c - a*d)^2*((c + d*Tan[e + f*x])^(n + 1)/(f*d^2*(n + 1)*( c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1 )*Simp[B*(b*c - a*d)^2 + A*d*(a^2*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2* c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^2 + d^2 )*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b *c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x ] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Leaf count of result is larger than twice the leaf count of optimal. \(4491\) vs. \(2(174)=348\).
Time = 0.12 (sec) , antiderivative size = 4492, normalized size of antiderivative = 22.69
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4492\) |
derivativedivides | \(\text {Expression too large to display}\) | \(12849\) |
default | \(\text {Expression too large to display}\) | \(12849\) |
A*(3/4/d*b^3/(a^2+b^2)^(7/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+ 2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a -1/d/b/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^ (1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))* a^4+1/d/b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^ 2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2 ))*a^6+4/d*b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2 +b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^( 1/2))*a^4+1/d*b^3/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2 *(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2 *a)^(1/2))*a^2+1/4/d/b/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^ (1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2* a)^(1/2)*a^5-1/2/d*b/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1 /2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a) ^(1/2)*a^3-3/4/d*b^3/(a^2+b^2)^(7/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1 /2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a) ^(1/2)*a+1/d/b/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+ b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a) ^(1/2))*a^4-1/d/b/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2* (a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2...
Leaf count of result is larger than twice the leaf count of optimal. 7218 vs. \(2 (168) = 336\).
Time = 2.23 (sec) , antiderivative size = 7218, normalized size of antiderivative = 36.45 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 27.68 (sec) , antiderivative size = 9468, normalized size of antiderivative = 47.82 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
(log((((a + b*tan(c + d*x))^(1/2)*(320*A^2*a^4*b^14*d^3 - 16*A^2*b^18*d^3 + 1024*A^2*a^6*b^12*d^3 + 1440*A^2*a^8*b^10*d^3 + 1024*A^2*a^10*b^8*d^3 + 320*A^2*a^12*b^6*d^3 - 16*A^2*a^16*b^2*d^3) + ((((320*A^4*a^2*b^8*d^4 - 16 *A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8* b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a ^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a ^8*b^2*d^4))^(1/2)*(896*A*a^6*b^15*d^4 - 32*A*b^21*d^4 - 160*A*a^2*b^19*d^ 4 - 128*A*a^4*b^17*d^4 - ((((320*A^4*a^2*b^8*d^4 - 16*A^4*b^10*d^4 - 1760* A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d^4)^(1/2) - 4*A^ 2*a^5*d^2 + 40*A^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18 *d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 1344 0*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 + 3136*A*a^8*b^13*d^4 + 4928*A*a^10*b^11*d^4 + 4480 *A*a^12*b^9*d^4 + 2432*A*a^14*b^7*d^4 + 736*A*a^16*b^5*d^4 + 96*A*a^18*b^3 *d^4))/4)*(((320*A^4*a^2*b^8*d^4 - 16*A^4*b^10*d^4 - 1760*A^4*a^4*b^6*d^4 + 1600*A^4*a^6*b^4*d^4 - 400*A^4*a^8*b^2*d^4)^(1/2) - 4*A^2*a^5*d^2 + 40*A ^2*a^3*b^2*d^2 - 20*A^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2))/4 - 96*A^3*a^3...