3.263 \(\int \frac{\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=231 \[ -\frac{a^4 \sin (c+d x)}{b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 a^3 \left (a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 a^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

[Out]

ArcTanh[Sin[c + d*x]]/(b^2*d) + (2*a^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a
+ b)^(5/2)*d) - (2*a^3*(a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^2*(
a + b)^(5/2)*d) - Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) - Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d
*x])) - (a^4*Sin[c + d*x])/(b*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

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Rubi [A]  time = 0.43695, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4397, 2897, 2648, 2664, 12, 2659, 208, 3770} \[ -\frac{a^4 \sin (c+d x)}{b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 a^3 \left (a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^2 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 a^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

ArcTanh[Sin[c + d*x]]/(b^2*d) + (2*a^3*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a
+ b)^(5/2)*d) - (2*a^3*(a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^2*(
a + b)^(5/2)*d) - Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) - Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d
*x])) - (a^4*Sin[c + d*x])/(b*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2659

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 208

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

Rule 3770

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{\sec ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx &=\int \frac{\csc ^2(c+d x) \sec (c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=-\int \left (-\frac{1}{2 (a-b)^2 (-1-\cos (c+d x))}-\frac{1}{2 (a+b)^2 (1-\cos (c+d x))}+\frac{a^3}{b \left (a^2-b^2\right ) (-b-a \cos (c+d x))^2}+\frac{-a^5+3 a^3 b^2}{b^2 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))}-\frac{\sec (c+d x)}{b^2}\right ) \, dx\\ &=\frac{\int \frac{1}{-1-\cos (c+d x)} \, dx}{2 (a-b)^2}+\frac{\int \sec (c+d x) \, dx}{b^2}+\frac{\int \frac{1}{1-\cos (c+d x)} \, dx}{2 (a+b)^2}+\frac{\left (a^3 \left (a^2-3 b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac{a^3 \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac{a^4 \sin (c+d x)}{b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{a^3 \int \frac{b}{-b-a \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )^2}+\frac{\left (2 a^3 \left (a^2-3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac{2 a^3 \left (a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^2 (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac{a^4 \sin (c+d x)}{b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{a^3 \int \frac{1}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d}-\frac{2 a^3 \left (a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^2 (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac{a^4 \sin (c+d x)}{b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a-b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^2 d}+\frac{2 a^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac{2 a^3 \left (a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^2 (a+b)^{5/2} d}-\frac{\sin (c+d x)}{2 (a+b)^2 d (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 (a-b)^2 d (1+\cos (c+d x))}-\frac{a^4 \sin (c+d x)}{b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.00197, size = 196, normalized size = 0.85 \[ -\frac{-\frac{4 \left (a^5-4 a^3 b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2}}+\frac{2 a^4 \sin (c+d x)}{b (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}+\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}+\frac{2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{b^2}-\frac{2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{b^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

-((-4*(a^5 - 4*a^3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)) + Cot[(c
+ d*x)/2]/(a + b)^2 + (2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^2 - (2*Log[Cos[(c + d*x)/2] + Sin[(c + d
*x)/2]])/b^2 + (2*a^4*Sin[c + d*x])/((a - b)^2*b*(a + b)^2*(b + a*Cos[c + d*x])) + Tan[(c + d*x)/2]/(a - b)^2)
/(2*d)

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Maple [A]  time = 0.186, size = 276, normalized size = 1.2 \begin{align*} -{\frac{1}{2\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{{a}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}b \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-2\,{\frac{{a}^{5}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+8\,{\frac{{a}^{3}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{1}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{d{b}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

[Out]

-1/2/d/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)+2/d*a^4/(a-b)^2/(a+b)^2/b*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a
-tan(1/2*d*x+1/2*c)^2*b-a-b)-2/d*a^5/(a-b)^2/(a+b)^2/b^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/
((a+b)*(a-b))^(1/2))+8/d*a^3/(a-b)^2/(a+b)^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b)
)^(1/2))+1/d/b^2*ln(tan(1/2*d*x+1/2*c)+1)-1/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/(a+b)^2/tan(1/2*d*x+1/2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54428, size = 1906, normalized size = 8.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*a^6*b - 2*b^7 + (a^5*b - 4*a^3*b^3 + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d
*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a
^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2))*sin(d*x + c) - 2*(a^6*b + a^4*b^3 - 2*a^2*b^5)*cos(d*x + c)^2 -
(a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(sin(d*x + c) +
1)*sin(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*
log(-sin(d*x + c) + 1)*sin(d*x + c) + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^7*b^2 - 3*a^5*b^4 + 3
*a^3*b^6 - a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d)*sin(d*x + c)), -1/2*(2*a^6*b - 2
*b^7 + 2*(a^5*b - 4*a^3*b^3 + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos
(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*sin(d*x + c) - 2*(a^6*b + a^4*b^3 - 2*a^2*b^5)*cos(d*x + c)^2 - (a^
6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1)*
sin(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*cos(d*x + c))*log(
-sin(d*x + c) + 1)*sin(d*x + c) + 2*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*cos(d*x + c))/(((a^7*b^2 - 3*a^5*b^4 + 3*a^3
*b^6 - a*b^8)*d*cos(d*x + c) + (a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d)*sin(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a \sin{\left (c + d x \right )} + b \tan{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a*sin(d*x+c)+b*tan(d*x+c))**2,x)

[Out]

Integral(sec(c + d*x)**3/(a*sin(c + d*x) + b*tan(c + d*x))**2, x)

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Giac [A]  time = 1.29365, size = 478, normalized size = 2.07 \begin{align*} \frac{\frac{4 \,{\left (a^{5} - 4 \, a^{3} b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{3} b - a^{2} b^{2} - a b^{3} + b^{4}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}} + \frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac{2 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(4*(a^5 - 4*a^3*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b
*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*b^2 - 2*a^2*b^4 + b^6)*sqrt(-a^2 + b^2)) - tan(1/2*d*x + 1/2*c
)/(a^2 - 2*a*b + b^2) + (4*a^4*tan(1/2*d*x + 1/2*c)^2 - a^3*b*tan(1/2*d*x + 1/2*c)^2 + 3*a^2*b^2*tan(1/2*d*x +
1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c)^2 + b^4*tan(1/2*d*x + 1/2*c)^2 + a^3*b - a^2*b^2 - a*b^3 + b^4)/((a^4
*b - 2*a^2*b^3 + b^5)*(a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) - b*tan(1/
2*d*x + 1/2*c))) + 2*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 - 2*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2)/d