3.2.90 \(\int \frac {a+b \sec (e+f x)}{(c+d \sec (e+f x))^2} \, dx\) [190]

Optimal. Leaf size=123 \[ \frac {a x}{c^2}+\frac {2 \left (b c^3-2 a c^2 d+a d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} f}-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))} \]

[Out]

a*x/c^2+2*(-2*a*c^2*d+a*d^3+b*c^3)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/c^2/(c-d)^(3/2)/(c+d)^(
3/2)/f-d*(-a*d+b*c)*tan(f*x+e)/c/(c^2-d^2)/f/(c+d*sec(f*x+e))

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Rubi [A]
time = 0.18, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4008, 4004, 3916, 2738, 214} \begin {gather*} -\frac {d (b c-a d) \tan (e+f x)}{c f \left (c^2-d^2\right ) (c+d \sec (e+f x))}+\frac {2 \left (-2 a c^2 d+a d^3+b c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 f (c-d)^{3/2} (c+d)^{3/2}}+\frac {a x}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x])/(c + d*Sec[e + f*x])^2,x]

[Out]

(a*x)/c^2 + (2*(b*c^3 - 2*a*c^2*d + a*d^3)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^2*(c - d)^(
3/2)*(c + d)^(3/2)*f) - (d*(b*c - a*d)*Tan[e + f*x])/(c*(c^2 - d^2)*f*(c + d*Sec[e + f*x]))

Rule 214

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]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4008

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[b*(b
*c - a*d)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {a+b \sec (e+f x)}{(c+d \sec (e+f x))^2} \, dx &=-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}-\frac {\int \frac {-a \left (c^2-d^2\right )-c (b c-a d) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c \left (c^2-d^2\right )}\\ &=\frac {a x}{c^2}-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {\left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right ) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{c^2 \left (c^2-d^2\right )}\\ &=\frac {a x}{c^2}-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {\left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right ) \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{c^2 d \left (c^2-d^2\right )}\\ &=\frac {a x}{c^2}-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}+\frac {\left (2 \left (c^2 (b c-a d)-a d \left (c^2-d^2\right )\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {c}{d}+\left (1-\frac {c}{d}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{c^2 d \left (c^2-d^2\right ) f}\\ &=\frac {a x}{c^2}+\frac {2 \left (b c^3-2 a c^2 d+a d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} f}-\frac {d (b c-a d) \tan (e+f x)}{c \left (c^2-d^2\right ) f (c+d \sec (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 155, normalized size = 1.26 \begin {gather*} \frac {-\frac {2 \left (b c^3+a d \left (-2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+\frac {a d \left (c^2-d^2\right ) (e+f x)+a c \left (c^2-d^2\right ) (e+f x) \cos (e+f x)-c d (b c-a d) \sin (e+f x)}{d+c \cos (e+f x)}}{c^2 (c-d) (c+d) f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[e + f*x])/(c + d*Sec[e + f*x])^2,x]

[Out]

((-2*(b*c^3 + a*d*(-2*c^2 + d^2))*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] + (a*d
*(c^2 - d^2)*(e + f*x) + a*c*(c^2 - d^2)*(e + f*x)*Cos[e + f*x] - c*d*(b*c - a*d)*Sin[e + f*x])/(d + c*Cos[e +
f*x]))/(c^2*(c - d)*(c + d)*f)

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Maple [A]
time = 0.22, size = 168, normalized size = 1.37

method	result	size
derivativedivides	\(\frac {\frac {2 a \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{2}}+\frac {-\frac {2 d \left (a d -b c \right ) c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) \left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-d \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c -d \right )}-\frac {2 \left (2 a \,c^{2} d -a \,d^{3}-b \,c^{3}\right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c +d \right ) \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{c^{2}}}{f}\)	\(168\)
default	\(\frac {\frac {2 a \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{2}}+\frac {-\frac {2 d \left (a d -b c \right ) c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) \left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-d \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c -d \right )}-\frac {2 \left (2 a \,c^{2} d -a \,d^{3}-b \,c^{3}\right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c +d \right ) \left (c -d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{c^{2}}}{f}\)	\(168\)
risch	\(\frac {a x}{c^{2}}-\frac {2 i d \left (-a d +b c \right ) \left (d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}{c^{2} \left (c^{2}-d^{2}\right ) f \left ({\mathrm e}^{2 i \left (f x +e \right )} c +2 d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a d}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a \,d^{3}}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f \,c^{2}}-\frac {c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i c^{2}-i d^{2}-\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a d}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) a \,d^{3}}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f \,c^{2}}+\frac {c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right ) b}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}\)	\(583\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e))/(c+d*sec(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(2*a/c^2*arctan(tan(1/2*f*x+1/2*e))+2/c^2*(-d*(a*d-b*c)*c/(c^2-d^2)*tan(1/2*f*x+1/2*e)/(c*tan(1/2*f*x+1/2*
e)^2-d*tan(1/2*f*x+1/2*e)^2-c-d)-(2*a*c^2*d-a*d^3-b*c^3)/(c+d)/(c-d)/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tan(1/2
*f*x+1/2*e)/((c+d)*(c-d))^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c^2-4*d^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (117) = 234\).
time = 4.42, size = 577, normalized size = 4.69 \begin {gather*} \left [\frac {2 \, {\left (a c^{5} - 2 \, a c^{3} d^{2} + a c d^{4}\right )} f x \cos \left (f x + e\right ) + 2 \, {\left (a c^{4} d - 2 \, a c^{2} d^{3} + a d^{5}\right )} f x - {\left (b c^{3} d - 2 \, a c^{2} d^{2} + a d^{4} + {\left (b c^{4} - 2 \, a c^{3} d + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) - 2 \, {\left (b c^{4} d - a c^{3} d^{2} - b c^{2} d^{3} + a c d^{4}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{7} - 2 \, c^{5} d^{2} + c^{3} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} d - 2 \, c^{4} d^{3} + c^{2} d^{5}\right )} f\right )}}, \frac {{\left (a c^{5} - 2 \, a c^{3} d^{2} + a c d^{4}\right )} f x \cos \left (f x + e\right ) + {\left (a c^{4} d - 2 \, a c^{2} d^{3} + a d^{5}\right )} f x + {\left (b c^{3} d - 2 \, a c^{2} d^{2} + a d^{4} + {\left (b c^{4} - 2 \, a c^{3} d + a c d^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) - {\left (b c^{4} d - a c^{3} d^{2} - b c^{2} d^{3} + a c d^{4}\right )} \sin \left (f x + e\right )}{{\left (c^{7} - 2 \, c^{5} d^{2} + c^{3} d^{4}\right )} f \cos \left (f x + e\right ) + {\left (c^{6} d - 2 \, c^{4} d^{3} + c^{2} d^{5}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/2*(2*(a*c^5 - 2*a*c^3*d^2 + a*c*d^4)*f*x*cos(f*x + e) + 2*(a*c^4*d - 2*a*c^2*d^3 + a*d^5)*f*x - (b*c^3*d -
2*a*c^2*d^2 + a*d^4 + (b*c^4 - 2*a*c^3*d + a*c*d^3)*cos(f*x + e))*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c
^2 - 2*d^2)*cos(f*x + e)^2 - 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x +
e)^2 + 2*c*d*cos(f*x + e) + d^2)) - 2*(b*c^4*d - a*c^3*d^2 - b*c^2*d^3 + a*c*d^4)*sin(f*x + e))/((c^7 - 2*c^5
*d^2 + c^3*d^4)*f*cos(f*x + e) + (c^6*d - 2*c^4*d^3 + c^2*d^5)*f), ((a*c^5 - 2*a*c^3*d^2 + a*c*d^4)*f*x*cos(f*
x + e) + (a*c^4*d - 2*a*c^2*d^3 + a*d^5)*f*x + (b*c^3*d - 2*a*c^2*d^2 + a*d^4 + (b*c^4 - 2*a*c^3*d + a*c*d^3)*
cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) - (b*
c^4*d - a*c^3*d^2 - b*c^2*d^3 + a*c*d^4)*sin(f*x + e))/((c^7 - 2*c^5*d^2 + c^3*d^4)*f*cos(f*x + e) + (c^6*d -
2*c^4*d^3 + c^2*d^5)*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \sec {\left (e + f x \right )}}{\left (c + d \sec {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e))/(c+d*sec(f*x+e))**2,x)

[Out]

Integral((a + b*sec(e + f*x))/(c + d*sec(e + f*x))**2, x)

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Giac [A]
time = 0.51, size = 201, normalized size = 1.63 \begin {gather*} \frac {\frac {2 \, {\left (b c^{3} - 2 \, a c^{2} d + a d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, c + 2 \, d\right ) + \arctan \left (-\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )}}{{\left (c^{4} - c^{2} d^{2}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {{\left (f x + e\right )} a}{c^{2}} + \frac {2 \, {\left (b c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c^{3} - c d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e))/(c+d*sec(f*x+e))^2,x, algorithm="giac")

[Out]

(2*(b*c^3 - 2*a*c^2*d + a*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(-2*c + 2*d) + arctan(-(c*tan(1/2*f*x + 1/
2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/((c^4 - c^2*d^2)*sqrt(-c^2 + d^2)) + (f*x + e)*a/c^2 + 2*(b*
c*d*tan(1/2*f*x + 1/2*e) - a*d^2*tan(1/2*f*x + 1/2*e))/((c^3 - c*d^2)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*
x + 1/2*e)^2 - c - d)))/f

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Mupad [B]
time = 9.33, size = 2500, normalized size = 20.33 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x))/(c + d/cos(e + f*x))^2,x)

[Out]

(2*a*atan(((a*((a*((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*
d + b*c^8*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) - (a*tan(e/2 + (f*x)/2)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c
^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2)*32i)/(c^2*(c^4*d + c^5 - c^2*d^3 - c^3*d^2)))*1i)/c^2 + (32*tan(e/2 + (f*x)/2)
*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 -
4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2)))/c^2 - (a*((a*((32*(a*c^4*d^5 - b*c^9 - a*c^9
- 3*a*c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) +
(a*tan(e/2 + (f*x)/2)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2)*32i)/(c^2*(c^4*d
+ c^5 - c^2*d^3 - c^3*d^2)))*1i)/c^2 - (32*tan(e/2 + (f*x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2
*a^2*c^5*d - 5*a^2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*
d^3 - c^3*d^2)))/c^2)/((64*(a^3*d^5 + a*b^2*c^5 - a^2*b*c^5 - a^3*c*d^4 + 2*a^3*c^4*d - 3*a^3*c^2*d^3 + 2*a^3*
c^3*d^2 + a^2*b*c^2*d^3 + a^2*b*c^3*d^2 - 3*a^2*b*c^4*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) + (a*((a*((32*(a*c
^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^5*d + c^6
- c^3*d^3 - c^4*d^2) - (a*tan(e/2 + (f*x)/2)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c^6*d^4 - 4*c^7*d^3 - 2*c^8*
d^2)*32i)/(c^2*(c^4*d + c^5 - c^2*d^3 - c^3*d^2)))*1i)/c^2 + (32*tan(e/2 + (f*x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2
*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3
))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2))*1i)/c^2 + (a*((a*((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6*d^3 + a*c^7*d
^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) + (a*tan(e/2 + (f*x)/2)*(
2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2)*32i)/(c^2*(c^4*d + c^5 - c^2*d^3 - c^3*d^
2)))*1i)/c^2 - (32*tan(e/2 + (f*x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^2*c^2*d
^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2))*1i)/c^2)
))/(c^2*f) + (atan(((((32*tan(e/2 + (f*x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^
2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2) +
(((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^
5*d + c^6 - c^3*d^3 - c^4*d^2) - (32*tan(e/2 + (f*x)/2)*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d
)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2))/((c^4*d + c^5 - c^2*d^3 - c^3*d^2)*(c
^8 - c^2*d^6 + 3*c^4*d^4 - 3*c^6*d^2)))*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d))/(c^8 - c^2*d^
6 + 3*c^4*d^4 - 3*c^6*d^2))*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d)*1i)/(c^8 - c^2*d^6 + 3*c^4
*d^4 - 3*c^6*d^2) + (((32*tan(e/2 + (f*x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^
2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2) -
(((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^
5*d + c^6 - c^3*d^3 - c^4*d^2) + (32*tan(e/2 + (f*x)/2)*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d
)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*c^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2))/((c^4*d + c^5 - c^2*d^3 - c^3*d^2)*(c
^8 - c^2*d^6 + 3*c^4*d^4 - 3*c^6*d^2)))*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d))/(c^8 - c^2*d^
6 + 3*c^4*d^4 - 3*c^6*d^2))*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d)*1i)/(c^8 - c^2*d^6 + 3*c^4
*d^4 - 3*c^6*d^2))/((64*(a^3*d^5 + a*b^2*c^5 - a^2*b*c^5 - a^3*c*d^4 + 2*a^3*c^4*d - 3*a^3*c^2*d^3 + 2*a^3*c^3
*d^2 + a^2*b*c^2*d^3 + a^2*b*c^3*d^2 - 3*a^2*b*c^4*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) + (((32*tan(e/2 + (f*
x)/2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d
^2 - 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2) + (((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*
c^6*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*c^8*d + b*c^8*d))/(c^5*d + c^6 - c^3*d^3 - c^4*d^2) - (32*ta
n(e/2 + (f*x)/2)*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d)*(2*c^9*d - 2*c^4*d^6 + 2*c^5*d^5 + 4*
c^6*d^4 - 4*c^7*d^3 - 2*c^8*d^2))/((c^4*d + c^5 - c^2*d^3 - c^3*d^2)*(c^8 - c^2*d^6 + 3*c^4*d^4 - 3*c^6*d^2)))
*((c + d)^3*(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d))/(c^8 - c^2*d^6 + 3*c^4*d^4 - 3*c^6*d^2))*((c + d)^3*
(c - d)^3)^(1/2)*(a*d^3 + b*c^3 - 2*a*c^2*d))/(c^8 - c^2*d^6 + 3*c^4*d^4 - 3*c^6*d^2) - (((32*tan(e/2 + (f*x)/
2)*(a^2*c^6 + 2*a^2*d^6 + b^2*c^6 - 2*a^2*c*d^5 - 2*a^2*c^5*d - 5*a^2*c^2*d^4 + 4*a^2*c^3*d^3 + 3*a^2*c^4*d^2
- 4*a*b*c^5*d + 2*a*b*c^3*d^3))/(c^4*d + c^5 - c^2*d^3 - c^3*d^2) - (((32*(a*c^4*d^5 - b*c^9 - a*c^9 - 3*a*c^6
*d^3 + a*c^7*d^2 - b*c^6*d^3 + b*c^7*d^2 + 2*a*...

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