3.308 \(\int \frac{\tan ^4(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=150 \[ \frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (a \cos (c+d x)+b)}+\frac{2 \sqrt{a-b} \sqrt{a+b} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 b^3 d}+\frac{x}{a^2}-\frac{2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{\tan (c+d x)}{b d (a \cos (c+d x)+b)} \]

[Out]

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

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Rubi [A]  time = 0.325731, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3898, 2890, 3057, 2659, 208, 3770} \[ \frac{\left (2 a^2-b^2\right ) \sin (c+d x)}{a b^2 d (a \cos (c+d x)+b)}+\frac{2 \sqrt{a-b} \sqrt{a+b} \left (2 a^2+b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 b^3 d}+\frac{x}{a^2}-\frac{2 a \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{\tan (c+d x)}{b d (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

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

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rule 2890

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b*d*(n + 1)*(m + 1)), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1)*Simp[a^2*(n + 1)
*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m
+ n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[((a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n
+ 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*b*d^2*f*(n + 1)*(m + 1)), x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

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

Mathematica [B]  time = 1.57522, size = 327, normalized size = 2.18 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b) \left (\frac{\left (a^2-b^2\right ) \sin (c+d x)}{a b^2}+\frac{2 \left (a^2 b^2-2 a^4+b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 b^3 \sqrt{a^2-b^2}}+\frac{(c+d x) (a \cos (c+d x)+b)}{a^2}+\frac{\sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{b^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{b^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{2 a (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{b^3}-\frac{2 a (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{b^3}\right )}{d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sec[c + d*x])^2,x]

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]^2*(((c + d*x)*(b + a*Cos[c + d*x]))/a^2 + (2*(-2*a^4 + a^2*b^2 + b^4)*ArcTa
nh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2*b^3*Sqrt[a^2 - b^2]) + (2*a*(b + a*
Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^3 - (2*a*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]])/b^3 + ((b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(b^2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))
+ ((b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(b^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + ((a^2 - b^2)*Sin[c + d
*x])/(a*b^2)))/(d*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.075, size = 353, normalized size = 2.4 \begin{align*} 2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{d{b}^{2} \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{\tan \left ( 1/2\,dx+c/2 \right ) }{ad \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 ) }}+4\,{\frac{{a}^{2}}{d{b}^{3}\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 ) }-2\,{\frac{1}{db\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 ) }-2\,{\frac{b}{d{a}^{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}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{a\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{b}^{3}}}-{\frac{1}{d{b}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{a\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x)

[Out]

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

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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(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.18755, size = 1374, normalized size = 9.16 \begin{align*} \left [\frac{2 \, a b^{3} d x \cos \left (d x + c\right )^{2} + 2 \, b^{4} d x \cos \left (d x + c\right ) +{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \,{\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (a^{2} b^{2} +{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{3} b^{3} d \cos \left (d x + c\right )^{2} + a^{2} b^{4} d \cos \left (d x + c\right )\right )}}, \frac{a b^{3} d x \cos \left (d x + c\right )^{2} + b^{4} d x \cos \left (d x + c\right ) +{\left ({\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) -{\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} \cos \left (d x + c\right )^{2} + a^{3} b \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} b^{2} +{\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{a^{3} b^{3} d \cos \left (d x + c\right )^{2} + a^{2} b^{4} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(2*a*b^3*d*x*cos(d*x + c)^2 + 2*b^4*d*x*cos(d*x + c) + ((2*a^3 + a*b^2)*cos(d*x + c)^2 + (2*a^2*b + b^3)*
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*co
s(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)) - 2*(a^4*cos(d*x
+ c)^2 + a^3*b*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*log(-sin(d*x
+ c) + 1) + 2*(a^2*b^2 + (2*a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 + a^2*b^4*d*c
os(d*x + c)), (a*b^3*d*x*cos(d*x + c)^2 + b^4*d*x*cos(d*x + c) + ((2*a^3 + a*b^2)*cos(d*x + c)^2 + (2*a^2*b +
b^3)*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)))
- (a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^4*cos(d*x + c)^2 + a^3*b*cos(d*x + c))*
log(-sin(d*x + c) + 1) + (a^2*b^2 + (2*a^3*b - a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 +
a^2*b^4*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sec(c + d*x))**2, x)

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Giac [B]  time = 2.60112, size = 397, normalized size = 2.65 \begin{align*} \frac{\frac{d x + c}{a^{2}} - \frac{2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac{2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac{2 \,{\left (2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )} a b^{2}} + \frac{2 \,{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\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 )}}{\sqrt{-a^{2} + b^{2}} a^{2} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

((d*x + c)/a^2 - 2*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + 2*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - 2*(
2*a^2*tan(1/2*d*x + 1/2*c)^3 - a*b*tan(1/2*d*x + 1/2*c)^3 - b^2*tan(1/2*d*x + 1/2*c)^3 - 2*a^2*tan(1/2*d*x + 1
/2*c) - a*b*tan(1/2*d*x + 1/2*c) + b^2*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*
c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b)*a*b^2) + 2*(2*a^4 - a^2*b^2 - b^4)*(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)))/(sqrt(-a^2 +
b^2)*a^2*b^3))/d