3.476 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=244 \[ -\frac{a \left (-121 a^2 b^2+4 a^4-128 b^4\right ) \tan (c+d x)}{60 b d}+\frac{\left (36 a^2 b^2+8 a^4+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (4 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}-\frac{a \left (4 a^2-53 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac{\left (-178 a^2 b^2+8 a^4-75 b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac{a \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

[Out]

((8*a^4 + 36*a^2*b^2 + 5*b^4)*ArcTanh[Sin[c + d*x]])/(16*d) - (a*(4*a^4 - 121*a^2*b^2 - 128*b^4)*Tan[c + d*x])
/(60*b*d) - ((8*a^4 - 178*a^2*b^2 - 75*b^4)*Sec[c + d*x]*Tan[c + d*x])/(240*d) - (a*(4*a^2 - 53*b^2)*(a + b*Se
c[c + d*x])^2*Tan[c + d*x])/(120*b*d) - ((4*a^2 - 25*b^2)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b*d) - (a*
(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + ((a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*d)

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Rubi [A]  time = 0.450268, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3840, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac{a \left (-121 a^2 b^2+4 a^4-128 b^4\right ) \tan (c+d x)}{60 b d}+\frac{\left (36 a^2 b^2+8 a^4+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (4 a^2-25 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^3}{120 b d}-\frac{a \left (4 a^2-53 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{120 b d}-\frac{\left (-178 a^2 b^2+8 a^4-75 b^4\right ) \tan (c+d x) \sec (c+d x)}{240 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^5}{6 b d}-\frac{a \tan (c+d x) (a+b \sec (c+d x))^4}{30 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4 + 36*a^2*b^2 + 5*b^4)*ArcTanh[Sin[c + d*x]])/(16*d) - (a*(4*a^4 - 121*a^2*b^2 - 128*b^4)*Tan[c + d*x])
/(60*b*d) - ((8*a^4 - 178*a^2*b^2 - 75*b^4)*Sec[c + d*x]*Tan[c + d*x])/(240*d) - (a*(4*a^2 - 53*b^2)*(a + b*Se
c[c + d*x])^2*Tan[c + d*x])/(120*b*d) - ((4*a^2 - 25*b^2)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(120*b*d) - (a*
(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(30*b*d) + ((a + b*Sec[c + d*x])^5*Tan[c + d*x])/(6*b*d)

Rule 3840

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4002

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

Rule 3997

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

Rule 3787

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

Rule 3770

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

Rule 3767

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 8

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

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (5 b-a \sec (c+d x)) (a+b \sec (c+d x))^4 \, dx}{6 b}\\ &=-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^3 \left (21 a b-\left (4 a^2-25 b^2\right ) \sec (c+d x)\right ) \, dx}{30 b}\\ &=-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (3 b \left (24 a^2+25 b^2\right )-3 a \left (4 a^2-53 b^2\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=-\frac{a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (3 a b \left (64 a^2+181 b^2\right )-3 \left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x)\right ) \, dx}{360 b}\\ &=-\frac{\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac{a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\int \sec (c+d x) \left (45 b \left (8 a^4+36 a^2 b^2+5 b^4\right )-12 a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \sec (c+d x)\right ) \, dx}{720 b}\\ &=-\frac{\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac{a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}-\frac{\left (a \left (4 a^4-121 a^2 b^2-128 b^4\right )\right ) \int \sec ^2(c+d x) \, dx}{60 b}+\frac{1}{16} \left (8 a^4+36 a^2 b^2+5 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac{a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}+\frac{\left (a \left (4 a^4-121 a^2 b^2-128 b^4\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{60 b d}\\ &=\frac{\left (8 a^4+36 a^2 b^2+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \tan (c+d x)}{60 b d}-\frac{\left (8 a^4-178 a^2 b^2-75 b^4\right ) \sec (c+d x) \tan (c+d x)}{240 d}-\frac{a \left (4 a^2-53 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \sec (c+d x))^3 \tan (c+d x)}{120 b d}-\frac{a (a+b \sec (c+d x))^4 \tan (c+d x)}{30 b d}+\frac{(a+b \sec (c+d x))^5 \tan (c+d x)}{6 b d}\\ \end{align*}

Mathematica [A]  time = 0.901929, size = 154, normalized size = 0.63 \[ \frac{15 \left (36 a^2 b^2+8 a^4+5 b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (64 a b \left (5 \left (a^2+2 b^2\right ) \tan ^2(c+d x)+15 \left (a^2+b^2\right )+3 b^2 \tan ^4(c+d x)\right )+10 b^2 \left (36 a^2+5 b^2\right ) \sec ^3(c+d x)+15 \left (36 a^2 b^2+8 a^4+5 b^4\right ) \sec (c+d x)+40 b^4 \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*Sec[c +
d*x] + 10*b^2*(36*a^2 + 5*b^2)*Sec[c + d*x]^3 + 40*b^4*Sec[c + d*x]^5 + 64*a*b*(15*(a^2 + b^2) + 5*(a^2 + 2*b
^2)*Tan[c + d*x]^2 + 3*b^2*Tan[c + d*x]^4)))/(240*d)

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Maple [A]  time = 0.036, size = 302, normalized size = 1.2 \begin{align*}{\frac{{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{8\,{a}^{3}b\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,{a}^{3}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{3\,{a}^{2}{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{9\,{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{9\,{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{32\,a{b}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{4\,a{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{16\,a{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^4*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+8/3/d*a^3*b*tan(d*x+c)+4/3/d*a^3*b*tan(d*x
+c)*sec(d*x+c)^2+3/2/d*a^2*b^2*tan(d*x+c)*sec(d*x+c)^3+9/4/d*a^2*b^2*sec(d*x+c)*tan(d*x+c)+9/4/d*a^2*b^2*ln(se
c(d*x+c)+tan(d*x+c))+32/15*a*b^3*tan(d*x+c)/d+4/5/d*a*b^3*tan(d*x+c)*sec(d*x+c)^4+16/15/d*a*b^3*tan(d*x+c)*sec
(d*x+c)^2+1/6/d*b^4*tan(d*x+c)*sec(d*x+c)^5+5/24/d*b^4*tan(d*x+c)*sec(d*x+c)^3+5/16/d*b^4*sec(d*x+c)*tan(d*x+c
)+5/16/d*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.1976, size = 371, normalized size = 1.52 \begin{align*} \frac{640 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} b + 128 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a b^{3} - 5 \, b^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, a^{2} b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(640*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3*b + 128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x +
c))*a*b^3 - 5*b^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c
)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*a^2*b^2*(2*(3*sin(d*x
+ c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 120*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 1.83403, size = 528, normalized size = 2.16 \begin{align*} \frac{15 \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (128 \,{\left (5 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 192 \, a b^{3} \cos \left (d x + c\right ) + 15 \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 40 \, b^{4} + 64 \,{\left (5 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \,{\left (36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*
cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(128*(5*a^3*b + 4*a*b^3)*cos(d*x + c)^5 + 192*a*b^3*cos(d*x + c) + 1
5*(8*a^4 + 36*a^2*b^2 + 5*b^4)*cos(d*x + c)^4 + 40*b^4 + 64*(5*a^3*b + 4*a*b^3)*cos(d*x + c)^3 + 10*(36*a^2*b^
2 + 5*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.3573, size = 799, normalized size = 3.27 \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+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*lo
g(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(120*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 90
0*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 165*b^4*tan(1/2*d*x + 1/2*c)^11 - 360*
a^4*tan(1/2*d*x + 1/2*c)^9 + 3520*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 2240*a*
b^3*tan(1/2*d*x + 1/2*c)^9 + 25*b^4*tan(1/2*d*x + 1/2*c)^9 + 240*a^4*tan(1/2*d*x + 1/2*c)^7 - 5760*a^3*b*tan(1
/2*d*x + 1/2*c)^7 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 4992*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 450*b^4*tan(1/2*d
*x + 1/2*c)^7 + 240*a^4*tan(1/2*d*x + 1/2*c)^5 + 5760*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 360*a^2*b^2*tan(1/2*d*x +
1/2*c)^5 + 4992*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 450*b^4*tan(1/2*d*x + 1/2*c)^5 - 360*a^4*tan(1/2*d*x + 1/2*c)^
3 - 3520*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 2240*a*b^3*tan(1/2*d*x + 1/2*c)^
3 + 25*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*a^4*tan(1/2*d*x + 1/2*c) + 960*a^3*b*tan(1/2*d*x + 1/2*c) + 900*a^2*b^
2*tan(1/2*d*x + 1/2*c) + 960*a*b^3*tan(1/2*d*x + 1/2*c) + 165*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
2 - 1)^6)/d