3.5.19 \(\int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx\) [419]

Optimal. Leaf size=138 \[ -\frac {B \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d} \]

[Out]

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*l
n(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)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {21, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {B \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {B \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {B \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*B + b*B*Tan[c + d*x])/(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])),x]

[Out]

-((B*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d)) + (B*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2
]*d) - (B*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (B*Log[1 + Sqrt[2]*Sqrt[Tan[c +
d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(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])

Rule 210

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])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps

\begin {align*} \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx &=B \int \frac {1}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {(2 B) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {B \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {B \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {B \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {B \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {B \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {B \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\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 {B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {B \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=-\frac {B \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \tan ^{-1}\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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 110, normalized size = 0.80 \begin {gather*} \frac {B \left (-2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{2 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*B + b*B*Tan[c + d*x])/(Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])),x]

[Out]

(B*(-2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] - Log[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/(2*Sqrt[2]*d)

________________________________________________________________________________________

Maple [A]
time = 0.07, size = 90, normalized size = 0.65

method	result	size
derivativedivides	\(\frac {B \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\)	\(90\)
default	\(\frac {B \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/4/d*B*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(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)))

________________________________________________________________________________________

Maxima [A]
time = 0.53, size = 112, normalized size = 0.81 \begin {gather*} \frac {2 \, \sqrt {2} B \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} B \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} B \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} B \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*B*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*B*arctan(-1/2*sqrt(2)*(sqrt(
2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*B*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*B*log(-sq
rt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/d

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (110) = 220\).
time = 2.07, size = 501, normalized size = 3.63 \begin {gather*} -\sqrt {2} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} B d^{3} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} - \sqrt {2} d^{3} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}} \sqrt {\frac {\sqrt {2} B d \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + d^{2} \sqrt {\frac {B^{4}}{d^{4}}} \cos \left (d x + c\right ) + B^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} + B^{4}}{B^{4}}\right ) - \sqrt {2} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} B d^{3} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} - \sqrt {2} d^{3} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}} \sqrt {-\frac {\sqrt {2} B d \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - d^{2} \sqrt {\frac {B^{4}}{d^{4}}} \cos \left (d x + c\right ) - B^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} - B^{4}}{B^{4}}\right ) + \frac {1}{4} \, \sqrt {2} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {2} B d \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + d^{2} \sqrt {\frac {B^{4}}{d^{4}}} \cos \left (d x + c\right ) + B^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) - \frac {1}{4} \, \sqrt {2} \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} B d \left (\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - d^{2} \sqrt {\frac {B^{4}}{d^{4}}} \cos \left (d x + c\right ) - B^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-sqrt(2)*(B^4/d^4)^(1/4)*arctan(-(sqrt(2)*B*d^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c)) - sqrt(2)*d^3*
(B^4/d^4)^(3/4)*sqrt((sqrt(2)*B*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) + d^2*sqrt(B^4/
d^4)*cos(d*x + c) + B^2*sin(d*x + c))/cos(d*x + c)) + B^4)/B^4) - sqrt(2)*(B^4/d^4)^(1/4)*arctan(-(sqrt(2)*B*d
^3*(B^4/d^4)^(3/4)*sqrt(sin(d*x + c)/cos(d*x + c)) - sqrt(2)*d^3*(B^4/d^4)^(3/4)*sqrt(-(sqrt(2)*B*d*(B^4/d^4)^
(1/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - d^2*sqrt(B^4/d^4)*cos(d*x + c) - B^2*sin(d*x + c))/cos(d*
x + c)) - B^4)/B^4) + 1/4*sqrt(2)*(B^4/d^4)^(1/4)*log((sqrt(2)*B*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x +
c))*cos(d*x + c) + d^2*sqrt(B^4/d^4)*cos(d*x + c) + B^2*sin(d*x + c))/cos(d*x + c)) - 1/4*sqrt(2)*(B^4/d^4)^(
1/4)*log(-(sqrt(2)*B*d*(B^4/d^4)^(1/4)*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c) - d^2*sqrt(B^4/d^4)*cos(d*
x + c) - B^2*sin(d*x + c))/cos(d*x + c))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} B \int \frac {1}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*tan(d*x+c))/tan(d*x+c)**(1/2)/(a+b*tan(d*x+c)),x)

[Out]

B*Integral(1/sqrt(tan(c + d*x)), x)

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*B+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 11.12, size = 2500, normalized size = 18.12 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*a + B*b*tan(c + d*x))/(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))),x)

[Out]

atan(((((32*(13*B^3*a^2*b^7*d^2 + B^3*a^4*b^5*d^2))/d^5 + (((32*(12*B*a*b^8*d^4 + 24*B*a^3*b^6*d^4 + 12*B*a^5*
b^4*d^4))/d^5 - (32*tan(c + d*x)^(1/2)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d
^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16*b^9*d^4 + 16*a^2*b^7*d^4 - 1
6*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^
4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(20*B^2*
a^3*b^6*d^2 + 2*B^2*a^5*b^4*d^2 - 14*B^2*a*b^8*d^2))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4
*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2))*(((64*B^4*a^
2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^
4 + 2*a^2*b^2*d^4)))^(1/2) - (32*tan(c + d*x)^(1/2)*(B^4*b^9 - 2*B^4*a^2*b^7))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^
4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*
d^4)))^(1/2)*1i - (((32*(13*B^3*a^2*b^7*d^2 + B^3*a^4*b^5*d^2))/d^5 + (((32*(12*B*a*b^8*d^4 + 24*B*a^3*b^6*d^4
+ 12*B*a^5*b^4*d^4))/d^5 + (32*tan(c + d*x)^(1/2)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 +
32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16*b^9*d^4 + 16*a^2
*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 3
2*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) - (32*tan(c + d*x)^(1
/2)*(20*B^2*a^3*b^6*d^2 + 2*B^2*a^5*b^4*d^2 - 14*B^2*a*b^8*d^2))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*
d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2))*
(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*
d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(B^4*b^9 - 2*B^4*a^2*b^7))/d^4)*(((64*B^4*a^2*
b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4
+ 2*a^2*b^2*d^4)))^(1/2)*1i)/((((32*(13*B^3*a^2*b^7*d^2 + B^3*a^4*b^5*d^2))/d^5 + (((32*(12*B*a*b^8*d^4 + 24*B
*a^3*b^6*d^4 + 12*B*a^5*b^4*d^4))/d^5 - (32*tan(c + d*x)^(1/2)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 1
6*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16*b^9*
d^4 + 16*a^2*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16
*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan
(c + d*x)^(1/2)*(20*B^2*a^3*b^6*d^2 + 2*B^2*a^5*b^4*d^2 - 14*B^2*a*b^8*d^2))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*
b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^
4)))^(1/2))*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^
2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) - (32*tan(c + d*x)^(1/2)*(B^4*b^9 - 2*B^4*a^2*b^7))/d^4)*((
(64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^
4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (((32*(13*B^3*a^2*b^7*d^2 + B^3*a^4*b^5*d^2))/d^5 + (((32*(12*B*a*b^8*d
^4 + 24*B*a^3*b^6*d^4 + 12*B*a^5*b^4*d^4))/d^5 + (32*tan(c + d*x)^(1/2)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^
4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)
*(16*b^9*d^4 + 16*a^2*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4
*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)
- (32*tan(c + d*x)^(1/2)*(20*B^2*a^3*b^6*d^2 + 2*B^2*a^5*b^4*d^2 - 14*B^2*a*b^8*d^2))/d^4)*(((64*B^4*a^2*b^6*d
^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a
^2*b^2*d^4)))^(1/2))*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2
*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(B^4*b^9 - 2*B^4*a^2*b^7)
)/d^4)*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(1
6*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (64*B^5*a*b^8)/d^5))*(((64*B^4*a^2*b^6*d^4 - B^4*b^4*(16*a^4*d
^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b^3*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*2i
- atan(((((32*(5*B^3*a^4*b^5 + B^3*a^6*b^3))/d^3 - (((32*(16*B*a*b^8*d^2 + 28*B*a^3*b^6*d^2 + 8*B*a^5*b^4*d^2
- 4*B*a^7*b^2*d^2))/d^3 - (32*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 + 3
2*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)*(16*b^9*d^4 + 16*a^2*
b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4...

________________________________________________________________________________________