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

Optimal. Leaf size=193 \[ -\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (4 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}-\frac{b \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a d} \]

[Out]

((3*a^4 + 4*a^2*b^2 + 8*b^4)*x)/(8*a^5) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqr
t[a - b]*Sqrt[a + b]*d) - (b*(2*a^2 + 3*b^2)*Sin[c + d*x])/(3*a^4*d) + ((3*a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d
*x])/(8*a^3*d) - (b*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.686522, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3853, 4104, 3919, 3831, 2659, 208} \[ -\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 a^3 d}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (4 a^2 b^2+3 a^4+8 b^4\right )}{8 a^5}-\frac{b \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d}+\frac{\sin (c+d x) \cos ^3(c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^4 + 4*a^2*b^2 + 8*b^4)*x)/(8*a^5) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*Sqr
t[a - b]*Sqrt[a + b]*d) - (b*(2*a^2 + 3*b^2)*Sin[c + d*x])/(3*a^4*d) + ((3*a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d
*x])/(8*a^3*d) - (b*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d) + (Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d)

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 4104

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

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{\cos ^4(c+d x)}{a+b \sec (c+d x)} \, dx &=\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac{\int \frac{\cos ^3(c+d x) \left (-4 b+3 a \sec (c+d x)+3 b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{4 a}\\ &=-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\int \frac{\cos ^2(c+d x) \left (-3 \left (3 a^2+4 b^2\right )-a b \sec (c+d x)+8 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{12 a^2}\\ &=\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac{\int \frac{\cos (c+d x) \left (-8 b \left (2 a^2+3 b^2\right )+a \left (9 a^2-4 b^2\right ) \sec (c+d x)+3 b \left (3 a^2+4 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{24 a^3}\\ &=-\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\int \frac{-3 \left (3 a^4+4 a^2 b^2+8 b^4\right )-3 a b \left (3 a^2+4 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{24 a^4}\\ &=\frac{\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{b^5 \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^5}\\ &=\frac{\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{b^4 \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^5}\\ &=\frac{\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\left (2 b^4\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 )}{a^5 d}\\ &=\frac{\left (3 a^4+4 a^2 b^2+8 b^4\right ) x}{8 a^5}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 \sqrt{a-b} \sqrt{a+b} d}-\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x)}{3 a^4 d}+\frac{\left (3 a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 a^3 d}-\frac{b \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d}+\frac{\cos ^3(c+d x) \sin (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.563935, size = 153, normalized size = 0.79 \[ \frac{12 \left (4 a^2 b^2+3 a^4+8 b^4\right ) (c+d x)-24 a b \left (3 a^2+4 b^2\right ) \sin (c+d x)+24 a^2 \left (a^2+b^2\right ) \sin (2 (c+d x))+\frac{192 b^5 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-8 a^3 b \sin (3 (c+d x))+3 a^4 \sin (4 (c+d x))}{96 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^4 + 4*a^2*b^2 + 8*b^4)*(c + d*x) + (192*b^5*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqr
t[a^2 - b^2] - 24*a*b*(3*a^2 + 4*b^2)*Sin[c + d*x] + 24*a^2*(a^2 + b^2)*Sin[2*(c + d*x)] - 8*a^3*b*Sin[3*(c +
d*x)] + 3*a^4*Sin[4*(c + d*x)])/(96*a^5*d)

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Maple [B]  time = 0.083, size = 672, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)
^7*b-1/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*b^2-2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*
x+1/2*c)^7*b^3+3/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-10/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*t
an(1/2*d*x+1/2*c)^5*b-6/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*b^3-1/d/a^3/(1+tan(1/2*d*x+1/2*c
)^2)^4*tan(1/2*d*x+1/2*c)^5*b^2-3/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3+1/d/a^3/(1+tan(1/2*d*x
+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^2-10/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b-6/d/a^4/(1+
tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*b^3+5/4/d/a/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+1/d/a^3
/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^2-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b-2/d
/a^4/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)*b^3+3/4/d/a*arctan(tan(1/2*d*x+1/2*c))+1/d/a^3*arctan(tan(1
/2*d*x+1/2*c))*b^2+2/d/a^5*arctan(tan(1/2*d*x+1/2*c))*b^4-2/d*b^5/a^5/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/
2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98027, size = 1057, normalized size = 5.48 \begin{align*} \left [\frac{12 \, \sqrt{a^{2} - b^{2}} b^{5} \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 ) + 3 \,{\left (3 \, a^{6} + a^{4} b^{2} + 4 \, a^{2} b^{4} - 8 \, b^{6}\right )} d x -{\left (16 \, a^{5} b + 8 \, a^{3} b^{3} - 24 \, a b^{5} - 6 \,{\left (a^{6} - a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (a^{5} b - a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, a^{6} + a^{4} b^{2} - 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{7} - a^{5} b^{2}\right )} d}, -\frac{24 \, \sqrt{-a^{2} + b^{2}} b^{5} \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 ) - 3 \,{\left (3 \, a^{6} + a^{4} b^{2} + 4 \, a^{2} b^{4} - 8 \, b^{6}\right )} d x +{\left (16 \, a^{5} b + 8 \, a^{3} b^{3} - 24 \, a b^{5} - 6 \,{\left (a^{6} - a^{4} b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left (a^{5} b - a^{3} b^{3}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (3 \, a^{6} + a^{4} b^{2} - 4 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \,{\left (a^{7} - a^{5} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(12*sqrt(a^2 - b^2)*b^5*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)) + 3*(3*a^6 + a^4*
b^2 + 4*a^2*b^4 - 8*b^6)*d*x - (16*a^5*b + 8*a^3*b^3 - 24*a*b^5 - 6*(a^6 - a^4*b^2)*cos(d*x + c)^3 + 8*(a^5*b
- a^3*b^3)*cos(d*x + c)^2 - 3*(3*a^6 + a^4*b^2 - 4*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5*b^2)*d), -
1/24*(24*sqrt(-a^2 + b^2)*b^5*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - 3*(3
*a^6 + a^4*b^2 + 4*a^2*b^4 - 8*b^6)*d*x + (16*a^5*b + 8*a^3*b^3 - 24*a*b^5 - 6*(a^6 - a^4*b^2)*cos(d*x + c)^3
+ 8*(a^5*b - a^3*b^3)*cos(d*x + c)^2 - 3*(3*a^6 + a^4*b^2 - 4*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^7 - a^5
*b^2)*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.28458, size = 531, normalized size = 2.75 \begin{align*} -\frac{\frac{48 \,{\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 )} b^{5}}{\sqrt{-a^{2} + b^{2}} a^{5}} - \frac{3 \,{\left (3 \, a^{4} + 4 \, a^{2} b^{2} + 8 \, b^{4}\right )}{\left (d x + c\right )}}{a^{5}} + \frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{4} a^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(48*(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)))*b^5/(sqrt(-a^2 + b^2)*a^5) - 3*(3*a^4 + 4*a^2*b^2 + 8*b^4)*(d*x + c)/a^5 + 2*(15*
a^3*tan(1/2*d*x + 1/2*c)^7 + 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*b^3*tan(1/
2*d*x + 1/2*c)^7 - 9*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*a*b^2*tan(1/2*d*x + 1/2
*c)^5 + 72*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*
b^2*tan(1/2*d*x + 1/2*c)^3 + 72*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*a^3*tan(1/2*d*x + 1/2*c) + 24*a^2*b*tan(1/2*d*
x + 1/2*c) - 12*a*b^2*tan(1/2*d*x + 1/2*c) + 24*b^3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^4)
)/d