Integrand size = 33, antiderivative size = 167 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}-\frac {(i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 a^2 (A b-a B)}{b^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{b^2 d} \]
(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d-(I*A -B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*a^2*(A *b-B*a)/b^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)+2*B*(a+b*tan(d*x+c))^(1/2)/ b^2/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.41 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i B \left (\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}\right )+\frac {-2 A b+4 a B}{b \sqrt {a+b \tan (c+d x)}}+\frac {(A b-a B) \left ((-i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(i a+b) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )\right )}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 B \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}}{b d} \]
(I*B*(ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/Sqrt[a - I*b] - ArcT anh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]]/Sqrt[a + I*b]) + (-2*A*b + 4*a *B)/(b*Sqrt[a + b*Tan[c + d*x]]) + ((A*b - a*B)*(((-I)*a + b)*Hypergeometr ic2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a - I*b)] + (I*a + b)*Hypergeome tric2F1[-1/2, 1, 1/2, (a + b*Tan[c + d*x])/(a + I*b)]))/((a^2 + b^2)*Sqrt[ a + b*Tan[c + d*x]]) + (2*B*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]])/(b*d)
Time = 0.94 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.11, 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, 4113, 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.
\(\displaystyle \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2 (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4087 |
\(\displaystyle \frac {\int -\frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan ^2(c+d x)\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {-\left (\left (a^2+b^2\right ) B \tan (c+d x)^2\right )-b (A b-a B) \tan (c+d x)+a (A b-a B)}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle -\frac {\int \frac {b (a A+b B)-b (A b-a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {b (a A+b B)-b (A b-a B) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {1}{2} b (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} b (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {1}{2} b (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} b (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {i b (a+i b) (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i b (a-i b) (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {-\frac {i b (a+i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i b (a-i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {\frac {(a-i b) (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {(a+i b) (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b \left (a^2+b^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a^2 (A b-a B)}{b^2 d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {-\frac {2 B \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}+\frac {b (a+i b) (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {b (a-i b) (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{b \left (a^2+b^2\right )}\) |
(-2*a^2*(A*b - a*B))/(b^2*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) - (((a + I*b)*b*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + ((a - I*b)*b*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]* d) - (2*(a^2 + b^2)*B*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(b*(a^2 + b^2))
3.4.51.3.1 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[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3712\) vs. \(2(145)=290\).
Time = 0.18 (sec) , antiderivative size = 3713, normalized size of antiderivative = 22.23
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3713\) |
derivativedivides | \(\text {Expression too large to display}\) | \(7982\) |
default | \(\text {Expression too large to display}\) | \(7982\) |
A*(-1/4/d/b/(a^2+b^2)^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^3-1/ 4/d*b/(a^2+b^2)^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+1/4/d/b/(a ^2+b^2)^(5/2)*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^4-1/4/d*b^3/(a ^2+b^2)^(5/2)*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)+1/d/b/(a^2+b^2)^ (3/2)/(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^3+1/d*b/(a^2+b^ 2)^(3/2)/(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+1/d*b/(a^2+b ^2)^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^2-1/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^5+1/d*b^3/(a ^2+b^2)^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))-3/d*b^3/(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/d*b/(...
Leaf count of result is larger than twice the leaf count of optimal. 4379 vs. \(2 (139) = 278\).
Time = 0.70 (sec) , antiderivative size = 4379, normalized size of antiderivative = 26.22 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
-1/2*(((a^2*b^3 + b^5)*d*tan(d*x + c) + (a^3*b^2 + a*b^4)*d)*sqrt(-(6*A*B* a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b ^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6* (A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2* B^2 + B^4)*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*lo g(-(2*(A^3*B + A*B^3)*a^3 - 3*(A^4 - B^4)*a^2*b - 6*(A^3*B + A*B^3)*a*b^2 + (A^4 - B^4)*b^3)*sqrt(b*tan(d*x + c) + a) + ((A*a^8 + 2*B*a^7*b + 2*A*a^ 6*b^2 + 6*B*a^5*b^3 + 6*B*a^3*b^5 - 2*A*a^2*b^6 + 2*B*a*b^7 - A*b^8)*d^3*s qrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3 *B^4)*a^4*b^2 + 40*(A^3*B - A*B^3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2 *b^4 - 12*(A^3*B - A*B^3)*a*b^5 + (A^4 - 2*A^2*B^2 + B^4)*b^6)/((a^12 + 6* a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + (2*A*B^2*a^5 - (7*A^2*B - 3*B^3)*a^4*b + 2*(3*A^3 - 7*A*B^2)*a^3*b^2 + 4*(4*A^2*B - B^3)*a^2*b^3 - 2*(A^3 - 4*A*B^2)*a*b^4 - (A^2*B - B^3)*b^5)*d )*sqrt(-(6*A*B*a^2*b - 2*A*B*b^3 + (A^2 - B^2)*a^3 - 3*(A^2 - B^2)*a*b^2 + (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(4*A^2*B^2*a^6 - 12*(A^3*B - A*B^3)*a^5*b + 3*(3*A^4 - 14*A^2*B^2 + 3*B^4)*a^4*b^2 + 40*(A^3*B - A*B^ 3)*a^3*b^3 - 6*(A^4 - 8*A^2*B^2 + B^4)*a^2*b^4 - 12*(A^3*B - A*B^3)*a*b...
\[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 16.12 (sec) , antiderivative size = 5768, normalized size of antiderivative = 34.54 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
(log((((a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2*a^2*b^8*d^3 - 32*A^2*a^6*b^4*d^3 - 16*A^2*a^8*b^2*d^3) + ((((96*A^4*a^2*b^4*d^4 - 16*A^4 *b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/ (a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(64*A*a*b^11*d^ 4 - ((((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3* a^4*b^2*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^ 10*d^5 + 640*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2 *d^5))/4 + 256*A*a^3*b^9*d^4 + 384*A*a^5*b^7*d^4 + 256*A*a^7*b^5*d^4 + 64* A*a^9*b^3*d^4))/4)*(((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^ 2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^ 2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 - 8*A^3*b^9*d^2 - 24*A^3*a^2*b^7*d^2 - 24*A^3*a^4*b^5*d^2 - 8*A^3*a^6*b^3*d^2)*(((96*A^4*a^2*b^4*d^4 - 16*A^4*b ^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) - 4*A^2*a^3*d^2 + 12*A^2*a*b^2*d^2)/(a ^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 + (log((((a + b*tan(c + d*x))^(1/2)*(16*A^2*b^10*d^3 + 32*A^2*a^2*b^8*d^3 - 32*A^2*a^6*b ^4*d^3 - 16*A^2*a^8*b^2*d^3) + ((-((96*A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*a^3*d^2 - 12*A^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2)*(64*A*a*b^11*d^4 - ((-((96 *A^4*a^2*b^4*d^4 - 16*A^4*b^6*d^4 - 144*A^4*a^4*b^2*d^4)^(1/2) + 4*A^2*...