Integrand size = 36, antiderivative size = 156 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {B \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
-1/2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*B*arctan(1+2^(1/2 )*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*B*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ c))/d*2^(1/2)-1/4*B*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-2/ 3*B/d/tan(d*x+c)^(3/2)
Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.53 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {B \left (-2+3 \arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \left (-\tan ^2(c+d x)\right )^{3/4}+3 \text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \left (-\tan ^2(c+d x)\right )^{3/4}\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
(B*(-2 + 3*ArcTan[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x]^2)^(3/4) + 3*Arc Tanh[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x]^2)^(3/4)))/(3*d*Tan[c + d*x]^ (3/2))
Time = 0.41 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2011, 3042, 3955, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle B \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \int \frac {1}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 3955 |
| \(\displaystyle B \left (-\int \frac {1}{\sqrt {\tan (c+d x)}}dx-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \left (-\int \frac {1}{\sqrt {\tan (c+d x)}}dx-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle B \left (-\frac {\int \frac {1}{\sqrt {\tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle B \left (-\frac {2 \int \frac {1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle B \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}-\frac {2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
B*((-2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sq rt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]))/2))/d - 2/(3*d*Tan[c + d*x]^(3/2)))
3.5.21.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] )^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2 Int[(b*Tan[c + d*x])^(n + 2), x] , x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Time = 0.03 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {B \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(102\) |
| default | \(\frac {B \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(102\) |
1/d*B*(-2/3/tan(d*x+c)^(3/2)-1/4*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+t an(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan (d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.33 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {3 \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (B \sqrt {\tan \left (d x + c\right )} + d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}\right ) \tan \left (d x + c\right )^{2} + 3 i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (B \sqrt {\tan \left (d x + c\right )} + i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}\right ) \tan \left (d x + c\right )^{2} - 3 i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (B \sqrt {\tan \left (d x + c\right )} - i \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}\right ) \tan \left (d x + c\right )^{2} - 3 \, d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (B \sqrt {\tan \left (d x + c\right )} - d \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}}\right ) \tan \left (d x + c\right )^{2} + 4 \, B \sqrt {\tan \left (d x + c\right )}}{6 \, d \tan \left (d x + c\right )^{2}} \]
-1/6*(3*d*(-B^4/d^4)^(1/4)*log(B*sqrt(tan(d*x + c)) + d*(-B^4/d^4)^(1/4))* tan(d*x + c)^2 + 3*I*d*(-B^4/d^4)^(1/4)*log(B*sqrt(tan(d*x + c)) + I*d*(-B ^4/d^4)^(1/4))*tan(d*x + c)^2 - 3*I*d*(-B^4/d^4)^(1/4)*log(B*sqrt(tan(d*x + c)) - I*d*(-B^4/d^4)^(1/4))*tan(d*x + c)^2 - 3*d*(-B^4/d^4)^(1/4)*log(B* sqrt(tan(d*x + c)) - d*(-B^4/d^4)^(1/4))*tan(d*x + c)^2 + 4*B*sqrt(tan(d*x + c)))/(d*tan(d*x + c)^2)
\[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=B \int \frac {1}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Time = 0.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {6 \, \sqrt {2} B \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} B \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} B \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} B \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, B}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]
-1/12*(6*sqrt(2)*B*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 6*sqrt(2)*B*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt (2)*B*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*B*log (-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*B/tan(d*x + c)^(3/2)) /d
Timed out. \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Timed out} \]
Time = 13.57 (sec) , antiderivative size = 16545, normalized size of antiderivative = 106.06 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]
atan(((tan(c + d*x)^(1/2)*(64*B^4*a^9*b^9*d^5 + 32*B^4*a^13*b^5*d^5) - (-( (64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^ (1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*( (tan(c + d*x)^(1/2)*(512*B^2*a^8*b^10*d^7 + 448*B^2*a^12*b^6*d^7 - 128*B^2 *a^14*b^4*d^7 - 64*B^2*a^16*b^2*d^7) - (-((64*B^4*a^6*b^2*d^4 - B^4*a^4*(1 6*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^ 4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(tan(c + d*x)^(1/2)*(-((64*B^4*a^ 6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8* B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(512*a^9*b^ 9*d^9 + 512*a^11*b^7*d^9 - 512*a^13*b^5*d^9 - 512*a^15*b^3*d^9) - 512*B*a^ 8*b^10*d^8 - 512*B*a^10*b^8*d^8 + 384*B*a^12*b^6*d^8 + 256*B*a^14*b^4*d^8 - 128*B*a^16*b^2*d^8))*(-((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b ^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) - 384*B^3*a^9*b^9*d^6 + 32*B^3*a^13*b^5*d^6 + 32* B^3*a^15*b^3*d^6))*(-((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d ^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2* a^2*b^2*d^4)))^(1/2)*1i + (tan(c + d*x)^(1/2)*(64*B^4*a^9*b^9*d^5 + 32*B^4 *a^13*b^5*d^5) - (-((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^ 2*b^2*d^4)))^(1/2)*((tan(c + d*x)^(1/2)*(512*B^2*a^8*b^10*d^7 + 448*B^2...