3.3.54 \(\int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{a+b \sec (e+f x)} \, dx\) [254]

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

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4073, 3031, 2738, 214, 3855, 3852, 8} \begin {gather*} \frac {d (2 b c-a d) \tanh ^{-1}(\sin (e+f x))}{b^2 f}+\frac {2 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{b^2 f \sqrt {a-b} \sqrt {a+b}}+\frac {d^2 \tan (e+f x)}{b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 4073

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[1/g^(m + n), Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d
+ c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && Inte
gerQ[n]

Rubi steps

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

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Mathematica [A]
time = 0.83, size = 135, normalized size = 1.31 \begin {gather*} \frac {-\frac {2 (b c-a 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 b c-a 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 )+b d \tan (e+f x)\right )}{b^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.37, size = 165, normalized size = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {d^{2}}{b \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d \left (a d -2 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b^{2}}-\frac {d^{2}}{b \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d \left (a d -2 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b^{2}}-\frac {2 \left (-a^{2} d^{2}+2 a b d c -b^{2} c^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{f}\)	\(165\)
default	\(\frac {-\frac {d^{2}}{b \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {d \left (a d -2 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{b^{2}}-\frac {d^{2}}{b \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {d \left (a d -2 b c \right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{b^{2}}-\frac {2 \left (-a^{2} d^{2}+2 a b d c -b^{2} c^{2}\right ) \arctanh \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{f}\)	\(165\)
risch	\(\frac {2 i d^{2}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a}{b^{2} f}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c}{b f}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) a}{b^{2} f}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c}{b f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2} d^{2}}{\sqrt {a^{2}-b^{2}}\, f \,b^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a d c}{\sqrt {a^{2}-b^{2}}\, f b}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) c^{2}}{\sqrt {a^{2}-b^{2}}\, f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a^{2} d^{2}}{\sqrt {a^{2}-b^{2}}\, f \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) a d c}{\sqrt {a^{2}-b^{2}}\, f b}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right ) c^{2}}{\sqrt {a^{2}-b^{2}}\, f}\)	\(564\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-d^2/b/(tan(1/2*f*x+1/2*e)-1)+d*(a*d-2*b*c)/b^2*ln(tan(1/2*f*x+1/2*e)-1)-d^2/b/(tan(1/2*f*x+1/2*e)+1)-d*(
a*d-2*b*c)/b^2*ln(tan(1/2*f*x+1/2*e)+1)-2/b^2*(-a^2*d^2+2*a*b*c*d-b^2*c^2)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*t
an(1/2*f*x+1/2*e)/((a+b)*(a-b))^(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(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(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*a^2-4*b^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. 240 vs. \(2 (97) = 194\).
time = 8.62, size = 540, normalized size = 5.24 \begin {gather*} \left [\frac {2 \, {\left (a^{2} b - b^{3}\right )} d^{2} \sin \left (f x + e\right ) + {\left (b^{2} c^{2} - 2 \, a b c d + a^{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 (a^{2} - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + b^{2}}\right ) + {\left (2 \, {\left (a^{2} b - b^{3}\right )} c d - {\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (a^{2} b - b^{3}\right )} c d - {\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (f x + e\right )}, \frac {2 \, {\left (a^{2} b - b^{3}\right )} d^{2} \sin \left (f x + e\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (f x + e\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + {\left (2 \, {\left (a^{2} b - b^{3}\right )} c d - {\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, {\left (a^{2} b - b^{3}\right )} c d - {\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a^2*b - b^3)*d^2*sin(f*x + e) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a^2 - b^2)*cos(f*x + e)*log((2*a
*b*cos(f*x + e) - (a^2 - 2*b^2)*cos(f*x + e)^2 + 2*sqrt(a^2 - b^2)*(b*cos(f*x + e) + a)*sin(f*x + e) + 2*a^2 -
b^2)/(a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + b^2)) + (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e
)*log(sin(f*x + e) + 1) - (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e)*log(-sin(f*x + e) + 1))/((a^2
*b^2 - b^4)*f*cos(f*x + e)), 1/2*(2*(a^2*b - b^3)*d^2*sin(f*x + e) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-a
^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2)*sin(f*x + e)))*cos(f*x + e) + (2*(a^2*b -
b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e)*log(sin(f*x + e) + 1) - (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*
cos(f*x + e)*log(-sin(f*x + e) + 1))/((a^2*b^2 - b^4)*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} \sec {\left (e + f x \right )}}{a + b \sec {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (94) = 188\).
time = 0.51, size = 195, normalized size = 1.89 \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 )} b} - \frac {{\left (2 \, b c d - a d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{b^{2}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{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}} b^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(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)*b) - (2*b*c*d - a*d^2)*log(abs(tan(1/2*f*x + 1/2*e)
+ 1))/b^2 + (2*b*c*d - a*d^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^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)*b^2))/f

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Mupad [B]
time = 7.32, 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/(cos(e + f*x)*(a + b/cos(e + f*x))),x)

[Out]

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

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