3.54 \(\int \csc ^6(c+d x) (a+a \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=165 \[ \frac{152 a^3 \tan (c+d x)}{15 d}+\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{76 a^6 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{a^6 \tan (c+d x) \sec (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \tan (c+d x) \sec (c+d x)}{15 d (a-a \cos (c+d x))^2} \]

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (152*a^3*Tan[c + d*x])/(15*d) + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d
) - (a^6*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a - a*Cos[c + d*x])^3) - (11*a^5*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a
- a*Cos[c + d*x])^2) - (76*a^6*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 - a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.435617, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3872, 2869, 2766, 2978, 2748, 3768, 3770, 3767, 8} \[ \frac{152 a^3 \tan (c+d x)}{15 d}+\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{2 d}-\frac{76 a^6 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{a^6 \tan (c+d x) \sec (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \tan (c+d x) \sec (c+d x)}{15 d (a-a \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (152*a^3*Tan[c + d*x])/(15*d) + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(2*d
) - (a^6*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a - a*Cos[c + d*x])^3) - (11*a^5*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a
- a*Cos[c + d*x])^2) - (76*a^6*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 - a^3*Cos[c + d*x]))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2869

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

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 \csc ^6(c+d x) (a+a \sec (c+d x))^3 \, dx &=-\int (-a-a \cos (c+d x))^3 \csc ^6(c+d x) \sec ^3(c+d x) \, dx\\ &=-\left (a^6 \int \frac{\sec ^3(c+d x)}{(-a+a \cos (c+d x))^3} \, dx\right )\\ &=-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{1}{5} a^4 \int \frac{(-7 a-4 a \cos (c+d x)) \sec ^3(c+d x)}{(-a+a \cos (c+d x))^2} \, dx\\ &=-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac{1}{15} a^2 \int \frac{\left (43 a^2+33 a^2 \cos (c+d x)\right ) \sec ^3(c+d x)}{-a+a \cos (c+d x)} \, dx\\ &=-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac{76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}-\frac{1}{15} \int \left (-195 a^3-152 a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac{76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}+\frac{1}{15} \left (152 a^3\right ) \int \sec ^2(c+d x) \, dx+\left (13 a^3\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac{76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}+\frac{1}{2} \left (13 a^3\right ) \int \sec (c+d x) \, dx-\frac{\left (152 a^3\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{152 a^3 \tan (c+d x)}{15 d}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{2 d}-\frac{a^6 \sec (c+d x) \tan (c+d x)}{5 d (a-a \cos (c+d x))^3}-\frac{11 a^5 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))^2}-\frac{76 a^4 \sec (c+d x) \tan (c+d x)}{15 d (a-a \cos (c+d x))}\\ \end{align*}

Mathematica [B]  time = 1.15463, size = 353, normalized size = 2.14 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (24960 \cos ^2(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\csc \left (\frac{c}{2}\right ) \sec (c) \left (4329 \sin \left (c-\frac{d x}{2}\right )-1989 \sin \left (c+\frac{d x}{2}\right )-3575 \sin \left (2 c+\frac{d x}{2}\right )+475 \sin \left (c+\frac{3 d x}{2}\right )+2005 \sin \left (2 c+\frac{3 d x}{2}\right )+2275 \sin \left (3 c+\frac{3 d x}{2}\right )-2673 \sin \left (c+\frac{5 d x}{2}\right )+105 \sin \left (2 c+\frac{5 d x}{2}\right )-1593 \sin \left (3 c+\frac{5 d x}{2}\right )-975 \sin \left (4 c+\frac{5 d x}{2}\right )+1325 \sin \left (2 c+\frac{7 d x}{2}\right )-255 \sin \left (3 c+\frac{7 d x}{2}\right )+875 \sin \left (4 c+\frac{7 d x}{2}\right )+195 \sin \left (5 c+\frac{7 d x}{2}\right )-304 \sin \left (3 c+\frac{9 d x}{2}\right )+90 \sin \left (4 c+\frac{9 d x}{2}\right )-214 \sin \left (5 c+\frac{9 d x}{2}\right )-1235 \sin \left (\frac{d x}{2}\right )+3805 \sin \left (\frac{3 d x}{2}\right )\right ) \csc ^5\left (\frac{1}{2} (c+d x)\right )\right )}{30720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + a*Sec[c + d*x])^3,x]

[Out]

-(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*Sec[c + d*x]^2*(24960*Cos[c + d*x]^2*(Log[Cos[(c + d*x)/2] - Sin
[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Csc[c/2]*Csc[(c + d*x)/2]^5*Sec[c]*(-1235*Sin[(d*
x)/2] + 3805*Sin[(3*d*x)/2] + 4329*Sin[c - (d*x)/2] - 1989*Sin[c + (d*x)/2] - 3575*Sin[2*c + (d*x)/2] + 475*Si
n[c + (3*d*x)/2] + 2005*Sin[2*c + (3*d*x)/2] + 2275*Sin[3*c + (3*d*x)/2] - 2673*Sin[c + (5*d*x)/2] + 105*Sin[2
*c + (5*d*x)/2] - 1593*Sin[3*c + (5*d*x)/2] - 975*Sin[4*c + (5*d*x)/2] + 1325*Sin[2*c + (7*d*x)/2] - 255*Sin[3
*c + (7*d*x)/2] + 875*Sin[4*c + (7*d*x)/2] + 195*Sin[5*c + (7*d*x)/2] - 304*Sin[3*c + (9*d*x)/2] + 90*Sin[4*c
+ (9*d*x)/2] - 214*Sin[5*c + (9*d*x)/2])))/(30720*d)

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Maple [A]  time = 0.074, size = 274, normalized size = 1.7 \begin{align*} -{\frac{152\,{a}^{3}\cot \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{3}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{3\,{a}^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}}}-{\frac{{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{13\,{a}^{3}}{2\,d\sin \left ( dx+c \right ) }}+{\frac{13\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{3\,{a}^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) }}-{\frac{6\,{a}^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{24\,{a}^{3}}{5\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{{a}^{3}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,{a}^{3}}{15\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{7\,{a}^{3}}{6\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x)

[Out]

-152/15*a^3*cot(d*x+c)/d-1/5/d*a^3*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^3*cot(d*x+c)*csc(d*x+c)^2-3/5/d*a^3/sin(d*
x+c)^5-1/d*a^3/sin(d*x+c)^3-13/2/d*a^3/sin(d*x+c)+13/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))-3/5/d*a^3/sin(d*x+c)^5/
cos(d*x+c)-6/5/d*a^3/sin(d*x+c)^3/cos(d*x+c)+24/5/d*a^3/sin(d*x+c)/cos(d*x+c)-1/5/d*a^3/sin(d*x+c)^5/cos(d*x+c
)^2-7/15/d*a^3/sin(d*x+c)^3/cos(d*x+c)^2+7/6/d*a^3/sin(d*x+c)/cos(d*x+c)^2

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Maxima [A]  time = 1.0469, size = 308, normalized size = 1.87 \begin{align*} -\frac{a^{3}{\left (\frac{2 \,{\left (105 \, \sin \left (d x + c\right )^{6} - 70 \, \sin \left (d x + c\right )^{4} - 14 \, \sin \left (d x + c\right )^{2} - 6\right )}}{\sin \left (d x + c\right )^{7} - \sin \left (d x + c\right )^{5}} - 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{4} + 5 \, \sin \left (d x + c\right )^{2} + 3\right )}}{\sin \left (d x + c\right )^{5}} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a^{3}{\left (\frac{15 \, \tan \left (d x + c\right )^{4} + 5 \, \tan \left (d x + c\right )^{2} + 1}{\tan \left (d x + c\right )^{5}} - 5 \, \tan \left (d x + c\right )\right )} + \frac{4 \,{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(a^3*(2*(105*sin(d*x + c)^6 - 70*sin(d*x + c)^4 - 14*sin(d*x + c)^2 - 6)/(sin(d*x + c)^7 - sin(d*x + c)^
5) - 105*log(sin(d*x + c) + 1) + 105*log(sin(d*x + c) - 1)) + 6*a^3*(2*(15*sin(d*x + c)^4 + 5*sin(d*x + c)^2 +
3)/sin(d*x + c)^5 - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) + 36*a^3*((15*tan(d*x + c)^4 + 5*tan
(d*x + c)^2 + 1)/tan(d*x + c)^5 - 5*tan(d*x + c)) + 4*(15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 + 3)*a^3/tan(d*x
+ c)^5)/d

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Fricas [A]  time = 1.76661, size = 575, normalized size = 3.48 \begin{align*} -\frac{608 \, a^{3} \cos \left (d x + c\right )^{5} - 826 \, a^{3} \cos \left (d x + c\right )^{4} - 476 \, a^{3} \cos \left (d x + c\right )^{3} + 868 \, a^{3} \cos \left (d x + c\right )^{2} - 120 \, a^{3} \cos \left (d x + c\right ) - 30 \, a^{3} - 195 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 195 \,{\left (a^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{3} \cos \left (d x + c\right )^{3} + a^{3} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{60 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/60*(608*a^3*cos(d*x + c)^5 - 826*a^3*cos(d*x + c)^4 - 476*a^3*cos(d*x + c)^3 + 868*a^3*cos(d*x + c)^2 - 120
*a^3*cos(d*x + c) - 30*a^3 - 195*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2)*log(sin(d*x
+ c) + 1)*sin(d*x + c) + 195*(a^3*cos(d*x + c)^4 - 2*a^3*cos(d*x + c)^3 + a^3*cos(d*x + c)^2)*log(-sin(d*x + c
) + 1)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3 + d*cos(d*x + c)^2)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**6*(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.38286, size = 190, normalized size = 1.15 \begin{align*} \frac{390 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 390 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{60 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, a^{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 )}^{2}} - \frac{465 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 40 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 390*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 60*(5*a^3*tan(
1/2*d*x + 1/2*c)^3 - 7*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 - (465*a^3*tan(1/2*d*x + 1/2*c
)^4 + 40*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3)/tan(1/2*d*x + 1/2*c)^5)/d