3.5.21 \(\int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [421]

Optimal. Leaf size=156 \[ \frac {B \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[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*
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)

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Rubi [A]
time = 0.06, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {21, 3555, 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 {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\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])/(Tan[c + d*x]^(5/2)*(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) - (2*B)/(3*d*Tan[c + d*x]^(3/2))

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 3555

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

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)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx &=B \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}-B \int \frac {1}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\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}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {(2 B) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\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 {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\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 {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\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}-\frac {2 B}{3 d \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 36, normalized size = 0.23 \begin {gather*} -\frac {2 B \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*B + b*B*Tan[c + d*x])/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]

[Out]

(-2*B*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2])/(3*d*Tan[c + d*x]^(3/2))

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Maple [A]
time = 0.05, size = 102, normalized size = 0.65

method	result	size
derivativedivides	\(\frac {B \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}\right )}{d}\)	\(102\)
default	\(\frac {B \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}\right )}{d}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*B+b*B*tan(d*x+c))/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*B*(-1/4*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*arc
tan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))-2/3/tan(d*x+c)^(3/2))

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Maxima [A]
time = 0.50, size = 124, normalized size = 0.79 \begin {gather*} -\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} \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)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-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)*(sqr
t(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*l
og(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*B/tan(d*x + c)^(3/2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (124) = 248\).
time = 1.32, size = 615, normalized size = 3.94 \begin {gather*} \frac {8 \, B \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} + 12 \, {\left (\sqrt {2} d \cos \left (d x + c\right )^{2} - \sqrt {2} d\right )} \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 ) + 12 \, {\left (\sqrt {2} d \cos \left (d x + c\right )^{2} - \sqrt {2} d\right )} \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 ) - 3 \, {\left (\sqrt {2} d \cos \left (d x + c\right )^{2} - \sqrt {2} d\right )} \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 ) + 3 \, {\left (\sqrt {2} d \cos \left (d x + c\right )^{2} - \sqrt {2} d\right )} \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 )}{12 \, {\left (d \cos \left (d x + c\right )^{2} - d\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)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(8*B*sqrt(sin(d*x + c)/cos(d*x + c))*cos(d*x + c)^2 + 12*(sqrt(2)*d*cos(d*x + c)^2 - sqrt(2)*d)*(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)*sq
rt((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) + 12*(sqrt(2)*d*cos(d*x + c)^2 - sqrt(2)*d)*(B^4/d^4)^(1/4)*arct
an(-(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) - 3*(sqrt(2)*d*cos(d*x + c)^2 - sqrt(2)*d)*(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)) + 3*(sqrt(2)*d*cos(d*x + c)^2 - sqrt(2)*d)*(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)))/(d*cos(d*x + c)^2 - d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} B \int \frac {1}{\tan ^{\frac {5}{2}}{\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)**(5/2)/(a+b*tan(d*x+c)),x)

[Out]

B*Integral(tan(c + d*x)**(-5/2), x)

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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)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 11.59, size = 2500, normalized size = 16.03 \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)^(5/2)*(a + b*tan(c + d*x))),x)

[Out]

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*(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)*(-((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) - 3
84*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*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*(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)*(-((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 - 38
4*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*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*(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)*(-((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) - (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*(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)*(-((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^1
1*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) + 64*B^5*a^10*b^8*d^4))*(-
((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)*2i + atan...

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