3.1.11 \(\int \frac {(c+d \sec (e+f x))^2}{a+b \cos (e+f x)} \, dx\) [11]

Optimal. Leaf size=103 \[ \frac {2 (a c-b d)^2 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} f}+\frac {d (2 a c-b d) \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {d^2 \tan (e+f x)}{a f} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2907, 3031, 2738, 211, 3855, 3852, 8} \begin {gather*} \frac {2 (a c-b d)^2 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^2 f \sqrt {a-b} \sqrt {a+b}}+\frac {d (2 a c-b d) \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac {d^2 \tan (e+f x)}{a f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[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 2907

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3031

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.92, size = 135, normalized size = 1.31 \begin {gather*} \frac {-\frac {2 (a c-b d)^2 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+d \left (-\left ((2 a c-b d) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )+a d \tan (e+f x)\right )}{a^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.35, size = 165, normalized size = 1.60

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a^{2} c^{2}-2 c d a b +b^{2} d^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{2}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d \left (2 a c -b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2}}-\frac {d^{2}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d \left (2 a c -b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2}}}{f}\)	\(165\)
default	\(\frac {\frac {2 \left (a^{2} c^{2}-2 c d a b +b^{2} d^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{2}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {d \left (2 a c -b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2}}-\frac {d^{2}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d \left (2 a c -b d \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2}}}{f}\)	\(165\)
risch	\(\frac {2 i d^{2}}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c}{a f}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b}{a^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c^{2}}{\sqrt {-a^{2}+b^{2}}\, f}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c d b}{\sqrt {-a^{2}+b^{2}}\, f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2} d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c^{2}}{\sqrt {-a^{2}+b^{2}}\, f}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c d b}{\sqrt {-a^{2}+b^{2}}\, f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2} d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{2}}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c}{a f}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b}{a^{2} f}\)	\(570\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((c+d*sec(f*x+e))^2/(a+b*cos(f*x+e)),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*b^2-4*a^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. 227 vs. \(2 (97) = 194\).
time = 3.77, size = 527, normalized size = 5.12 \begin {gather*} \left [\frac {2 \, {\left (a^{3} - a b^{2}\right )} d^{2} \sin \left (f x + e\right ) - {\left (a^{2} c^{2} - 2 \, a b c d + b^{2} d^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (f x + e\right ) \log \left (\frac {2 \, a b \cos \left (f x + e\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (f x + e\right ) + b\right )} \sin \left (f x + e\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + a^{2}}\right ) + {\left (2 \, {\left (a^{3} - a b^{2}\right )} c d - {\left (a^{2} b - b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (a^{3} - a b^{2}\right )} c d - {\left (a^{2} b - b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} f \cos \left (f x + e\right )}, \frac {2 \, {\left (a^{3} - a b^{2}\right )} d^{2} \sin \left (f x + e\right ) + 2 \, {\left (a^{2} c^{2} - 2 \, a b c d + b^{2} d^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + {\left (2 \, {\left (a^{3} - a b^{2}\right )} c d - {\left (a^{2} b - b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (a^{3} - a b^{2}\right )} c d - {\left (a^{2} b - b^{3}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} f \cos \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (97) = 194\).
time = 0.49, size = 203, normalized size = 1.97 \begin {gather*} -\frac {\frac {2 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a} - \frac {{\left (2 \, a c d - b d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac {{\left (2 \, a c d - b d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (a^{2} c^{2} - 2 \, a b c d + b^{2} d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 7.47, size = 2500, normalized size = 24.27 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((-(a + b)*(a - b))^(1/2)*((32*tan(e/2 + (f*x)/2)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 - 3*a^
2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d^2 - 16*a^2*b^3*c*d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b*c^2
*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^2*c^2*d^2 + 8*a*b^4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 + ((-(a +
b)*(a - b))^(1/2)*((32*(a^7*c^2 - 2*a^6*b*c^2 - a^6*b*d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2*a^7
*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 + (32*tan(e/2 + (f*x)/2)*(-(a + b)*(a - b))^(1/2)*(a*c - b*d)^2*(2*a^
6*b + 2*a^4*b^3 - 4*a^5*b^2))/(a^2*(a^4 - a^2*b^2)))*(a*c - b*d)^2)/(a^4 - a^2*b^2))*(a*c - b*d)^2*1i)/(a^4 -
a^2*b^2) + ((-(a + b)*(a - b))^(1/2)*((32*tan(e/2 + (f*x)/2)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 -
3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d^2 - 16*a^2*b^3*c*d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b
*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^2*c^2*d^2 + 8*a*b^4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 - ((-
(a + b)*(a - b))^(1/2)*((32*(a^7*c^2 - 2*a^6*b*c^2 - a^6*b*d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2
*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 - (32*tan(e/2 + (f*x)/2)*(-(a + b)*(a - b))^(1/2)*(a*c - b*d)^2*(
2*a^6*b + 2*a^4*b^3 - 4*a^5*b^2))/(a^2*(a^4 - a^2*b^2)))*(a*c - b*d)^2)/(a^4 - a^2*b^2))*(a*c - b*d)^2*1i)/(a^
4 - a^2*b^2))/((64*(a*b^4*d^6 - b^5*d^6 - 2*a^5*c^5*d + 4*a^5*c^4*d^2 - a*b^4*c^2*d^4 - 6*a^2*b^3*c*d^5 - 12*a
^4*b*c^3*d^3 + a^4*b*c^4*d^2 - 12*a^2*b^3*c^2*d^4 + 4*a^2*b^3*c^3*d^3 + 13*a^3*b^2*c^2*d^4 + 8*a^3*b^2*c^3*d^3
- 5*a^3*b^2*c^4*d^2 + 6*a*b^4*c*d^5 + 2*a^4*b*c^5*d))/a^3 - ((-(a + b)*(a - b))^(1/2)*((32*tan(e/2 + (f*x)/2)
*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 - 3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d^2 - 16*a^2*b^3*c*d
^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^2*c^2*d^2 + 8*a*b^4
*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 + ((-(a + b)*(a - b))^(1/2)*((32*(a^7*c^2 - 2*a^6*b*c^2 - a^6*b*d
^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 + (32*tan(e/2 +
(f*x)/2)*(-(a + b)*(a - b))^(1/2)*(a*c - b*d)^2*(2*a^6*b + 2*a^4*b^3 - 4*a^5*b^2))/(a^2*(a^4 - a^2*b^2)))*(a*
c - b*d)^2)/(a^4 - a^2*b^2))*(a*c - b*d)^2)/(a^4 - a^2*b^2) + ((-(a + b)*(a - b))^(1/2)*((32*tan(e/2 + (f*x)/2
)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 - 3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d^2 - 16*a^2*b^3*c*
d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^2*c^2*d^2 + 8*a*b^
4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 - ((-(a + b)*(a - b))^(1/2)*((32*(a^7*c^2 - 2*a^6*b*c^2 - a^6*b*
d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 - (32*tan(e/2
+ (f*x)/2)*(-(a + b)*(a - b))^(1/2)*(a*c - b*d)^2*(2*a^6*b + 2*a^4*b^3 - 4*a^5*b^2))/(a^2*(a^4 - a^2*b^2)))*(a
*c - b*d)^2)/(a^4 - a^2*b^2))*(a*c - b*d)^2)/(a^4 - a^2*b^2)))*(-(a + b)*(a - b))^(1/2)*(a*c - b*d)^2*2i)/(f*(
a^4 - a^2*b^2)) - (2*d^2*tan(e/2 + (f*x)/2))/(a*f*(tan(e/2 + (f*x)/2)^2 - 1)) + (d*atan(((d*(2*a*c - b*d)*((32
*tan(e/2 + (f*x)/2)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 - 3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d
^2 - 16*a^2*b^3*c*d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^
2*c^2*d^2 + 8*a*b^4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 + (d*(2*a*c - b*d)*((32*(a^7*c^2 - 2*a^6*b*c^2
- a^6*b*d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 + (32
*d*tan(e/2 + (f*x)/2)*(2*a*c - b*d)*(2*a^6*b + 2*a^4*b^3 - 4*a^5*b^2))/a^4))/a^2)*1i)/a^2 + (d*(2*a*c - b*d)*(
(32*tan(e/2 + (f*x)/2)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a*b^4*d^4 - 3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^
2*d^2 - 16*a^2*b^3*c*d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d - 12*a^4*b*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3
*b^2*c^2*d^2 + 8*a*b^4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d))/a^2 - (d*(2*a*c - b*d)*((32*(a^7*c^2 - 2*a^6*b*
c^2 - a^6*b*d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 + 2*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 -
(32*d*tan(e/2 + (f*x)/2)*(2*a*c - b*d)*(2*a^6*b + 2*a^4*b^3 - 4*a^5*b^2))/a^4))/a^2)*1i)/a^2)/((64*(a*b^4*d^6
- b^5*d^6 - 2*a^5*c^5*d + 4*a^5*c^4*d^2 - a*b^4*c^2*d^4 - 6*a^2*b^3*c*d^5 - 12*a^4*b*c^3*d^3 + a^4*b*c^4*d^2 -
12*a^2*b^3*c^2*d^4 + 4*a^2*b^3*c^3*d^3 + 13*a^3*b^2*c^2*d^4 + 8*a^3*b^2*c^3*d^3 - 5*a^3*b^2*c^4*d^2 + 6*a*b^4
*c*d^5 + 2*a^4*b*c^5*d))/a^3 - (d*(2*a*c - b*d)*((32*tan(e/2 + (f*x)/2)*(a^5*c^4 - 2*b^5*d^4 - a^4*b*c^4 + 4*a
*b^4*d^4 - 3*a^2*b^3*d^4 + a^3*b^2*d^4 + 4*a^5*c^2*d^2 - 16*a^2*b^3*c*d^3 + 12*a^3*b^2*c*d^3 + 4*a^3*b^2*c^3*d
- 12*a^4*b*c^2*d^2 - 10*a^2*b^3*c^2*d^2 + 18*a^3*b^2*c^2*d^2 + 8*a*b^4*c*d^3 - 4*a^4*b*c*d^3 - 4*a^4*b*c^3*d)
)/a^2 + (d*(2*a*c - b*d)*((32*(a^7*c^2 - 2*a^6*b*c^2 - a^6*b*d^2 + a^5*b^2*c^2 - a^4*b^3*d^2 + 2*a^5*b^2*d^2 +
2*a^7*c*d - 4*a^6*b*c*d + 2*a^5*b^2*c*d))/a^3 + (32*d*tan(e/2 + (f*x)/2)*(2*a*c - b*d)*(2*a^6*b + 2*a^4*b^3 -
4*a^5*b^2))/a^4))/a^2))/a^2 + (d*(2*a*c - b*d)...

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