3.507 \(\int \frac{\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=188 \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

[Out]

ArcTanh[Sin[c + d*x]]/(b^3*d) - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a +
b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^2) - (a^2*(2*a^2 - 5*b^2)*Tan[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

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Rubi [A]  time = 0.409982, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3845, 4080, 3998, 3770, 3831, 2659, 208} \[ -\frac{a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{a^2 \tan (c+d x) \sec (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac{\tanh ^{-1}(\sin (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^3,x]

[Out]

ArcTanh[Sin[c + d*x]]/(b^3*d) - (a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a +
b]])/((a - b)^(5/2)*b^3*(a + b)^(5/2)*d) - (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[c +
d*x])^2) - (a^2*(2*a^2 - 5*b^2)*Tan[c + d*x])/(2*b^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4080

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f
*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e +
f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

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

Rule 3831

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

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+b \sec (c+d x))^3} \, dx &=-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec (c+d x) \left (a^2-2 a b \sec (c+d x)-2 \left (a^2-b^2\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (a b \left (a^2-4 b^2\right )+2 \left (a^2-b^2\right )^2 \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \sec (c+d x) \, dx}{b^3}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (a \left (2 a^4-5 a^2 b^2+6 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 \left (a^2-b^2\right )^2 d}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{a \left (2 a^4-5 a^2 b^2+6 b^4\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^3 (a+b)^{5/2} d}-\frac{a^2 \sec (c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a^2 \left (2 a^2-5 b^2\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.26872, size = 194, normalized size = 1.03 \[ \frac{-\frac{a^2 b \sin (c+d x) \left (a \left (2 a^2-5 b^2\right ) \cos (c+d x)+3 b \left (a^2-2 b^2\right )\right )}{(a-b)^2 (a+b)^2 (a \cos (c+d x)+b)^2}+\frac{2 a \left (-5 a^2 b^2+2 a^4+6 b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^3,x]

[Out]

((2*a*(2*a^4 - 5*a^2*b^2 + 6*b^4)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - 2*
Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (a^2*b*(3*b*(a^2 - 2*b
^2) + a*(2*a^2 - 5*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])^2))/(2*b^3*d)

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Maple [B]  time = 0.066, size = 685, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+b*sec(d*x+c))^3,x)

[Out]

2/d*a^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1
/d*a^3/b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-6/d*
a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-2/d*a^4/b
^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-1/d*a^3/b/(t
an(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)+6/d*a^2/(tan(1/2*
d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-2/d*a^5/b^3/(a^4-2*a^2*b
^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+5/d*a^3/b/(a^4-2*a^2*b^2+b^4
)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-6/d*a*b/(a^4-2*a^2*b^2+b^4)/((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^3*ln(tan(1/2*d*x+1/2*c)+1)-1/d/b^3*l
n(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(sec(d*x+c)^4/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.56752, size = 2466, normalized size = 13.12 \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)^4/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*((2*a^5*b^2 - 5*a^3*b^4 + 6*a*b^6 + (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4)*cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b
^3 + 6*a^2*b^5)*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)) +
2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b
- 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^
8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x
+ c))*log(-sin(d*x + c) + 1) - 2*(3*a^6*b^2 - 9*a^4*b^4 + 6*a^2*b^6 + (2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(d
*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6
+ 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d), -1/2*((2*a^5*b^2 - 5*a^3*b
^4 + 6*a*b^6 + (2*a^7 - 5*a^5*b^2 + 6*a^3*b^4)*cos(d*x + c)^2 + 2*(2*a^6*b - 5*a^4*b^3 + 6*a^2*b^5)*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^6*b^2 - 3*
a^4*b^4 + 3*a^2*b^6 - b^8 + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*
a^3*b^5 - a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8 + (a^8 - 3*a^6*b
^2 + 3*a^4*b^4 - a^2*b^6)*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*cos(d*x + c))*log(-sin(d*
x + c) + 1) + (3*a^6*b^2 - 9*a^4*b^4 + 6*a^2*b^6 + (2*a^7*b - 7*a^5*b^3 + 5*a^3*b^5)*cos(d*x + c))*sin(d*x + c
))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10
)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+b*sec(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*sec(c + d*x))**3, x)

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Giac [A]  time = 1.4412, size = 468, normalized size = 2.49 \begin{align*} \frac{\frac{{\left (2 \, a^{5} - 5 \, a^{3} b^{2} + 6 \, a 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 )}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

((2*a^5 - 5*a^3*b^2 + 6*a*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)))/((a^4*b^3 - 2*a^2*b^5 + b^7)*sqrt(-a^2 + b^2)) + (2*a^5*tan(1/
2*d*x + 1/2*c)^3 - 3*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*b^3*tan(1/2*d*x +
1/2*c)^3 - 2*a^5*tan(1/2*d*x + 1/2*c) - 3*a^4*b*tan(1/2*d*x + 1/2*c) + 5*a^3*b^2*tan(1/2*d*x + 1/2*c) + 6*a^2
*b^3*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 -
a - b)^2) + log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 - log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3)/d