3.293 \(\int \frac{\cot ^5(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=234 \[ -\frac{b^6 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^3}+\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 d (a+b)^3}+\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sec (c+d x)+1)}{16 d (a-b)^3}-\frac{5 a+7 b}{16 d (a+b)^2 (1-\sec (c+d x))}-\frac{5 a-7 b}{16 d (a-b)^2 (\sec (c+d x)+1)}-\frac{1}{16 d (a+b) (1-\sec (c+d x))^2}-\frac{1}{16 d (a-b) (\sec (c+d x)+1)^2}+\frac{\log (\cos (c+d x))}{a d} \]

[Out]

Log[Cos[c + d*x]]/(a*d) + ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sec[c + d*x]])/(16*(a + b)^3*d) + ((8*a^2 - 21*a*
b + 15*b^2)*Log[1 + Sec[c + d*x]])/(16*(a - b)^3*d) - (b^6*Log[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)^3*d) - 1/(1
6*(a + b)*d*(1 - Sec[c + d*x])^2) - (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sec[c + d*x])) - 1/(16*(a - b)*d*(1 + Sec
[c + d*x])^2) - (5*a - 7*b)/(16*(a - b)^2*d*(1 + Sec[c + d*x]))

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Rubi [A]  time = 0.294619, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3885, 894} \[ -\frac{b^6 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^3}+\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 d (a+b)^3}+\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (\sec (c+d x)+1)}{16 d (a-b)^3}-\frac{5 a+7 b}{16 d (a+b)^2 (1-\sec (c+d x))}-\frac{5 a-7 b}{16 d (a-b)^2 (\sec (c+d x)+1)}-\frac{1}{16 d (a+b) (1-\sec (c+d x))^2}-\frac{1}{16 d (a-b) (\sec (c+d x)+1)^2}+\frac{\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

Log[Cos[c + d*x]]/(a*d) + ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sec[c + d*x]])/(16*(a + b)^3*d) + ((8*a^2 - 21*a*
b + 15*b^2)*Log[1 + Sec[c + d*x]])/(16*(a - b)^3*d) - (b^6*Log[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)^3*d) - 1/(1
6*(a + b)*d*(1 - Sec[c + d*x])^2) - (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sec[c + d*x])) - 1/(16*(a - b)*d*(1 + Sec
[c + d*x])^2) - (5*a - 7*b)/(16*(a - b)^2*d*(1 + Sec[c + d*x]))

Rule 3885

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \frac{\cot ^5(c+d x)}{a+b \sec (c+d x)} \, dx &=-\frac{b^6 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac{b^6 \operatorname{Subst}\left (\int \left (\frac{1}{8 b^4 (a+b) (b-x)^3}+\frac{5 a+7 b}{16 b^5 (a+b)^2 (b-x)^2}+\frac{8 a^2+21 a b+15 b^2}{16 b^6 (a+b)^3 (b-x)}+\frac{1}{a b^6 x}+\frac{1}{a (a-b)^3 (a+b)^3 (a+x)}+\frac{1}{8 b^4 (-a+b) (b+x)^3}+\frac{-5 a+7 b}{16 (a-b)^2 b^5 (b+x)^2}+\frac{8 a^2-21 a b+15 b^2}{16 b^6 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac{\log (\cos (c+d x))}{a d}+\frac{\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 (a+b)^3 d}+\frac{\left (8 a^2-21 a b+15 b^2\right ) \log (1+\sec (c+d x))}{16 (a-b)^3 d}-\frac{b^6 \log (a+b \sec (c+d x))}{a \left (a^2-b^2\right )^3 d}-\frac{1}{16 (a+b) d (1-\sec (c+d x))^2}-\frac{5 a+7 b}{16 (a+b)^2 d (1-\sec (c+d x))}-\frac{1}{16 (a-b) d (1+\sec (c+d x))^2}-\frac{5 a-7 b}{16 (a-b)^2 d (1+\sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 6.2418, size = 625, normalized size = 2.67 \[ \frac{2 i \left (-3 a^3 b^2+a^5+3 a b^4\right ) (c+d x) \sec (c+d x) (a \cos (c+d x)+b)}{d (a-b)^3 (a+b)^3 (a+b \sec (c+d x))}+\frac{\left (-8 a^2+21 a b-15 b^2\right ) \sec (c+d x) \log \left (\cos ^2\left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{16 d (b-a)^3 (a+b \sec (c+d x))}+\frac{b^6 \sec (c+d x) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{a d \left (b^2-a^2\right )^3 (a+b \sec (c+d x))}-\frac{i \left (-8 a^2+21 a b-15 b^2\right ) \tan ^{-1}(\tan (c+d x)) \sec (c+d x) (a \cos (c+d x)+b)}{8 d (b-a)^3 (a+b \sec (c+d x))}-\frac{i \left (8 a^2+21 a b+15 b^2\right ) \tan ^{-1}(\tan (c+d x)) \sec (c+d x) (a \cos (c+d x)+b)}{8 d (a+b)^3 (a+b \sec (c+d x))}+\frac{\left (8 a^2+21 a b+15 b^2\right ) \sec (c+d x) \log \left (\sin ^2\left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{16 d (a+b)^3 (a+b \sec (c+d x))}+\frac{\sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{64 d (b-a) (a+b \sec (c+d x))}+\frac{(7 a-9 b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{32 d (b-a)^2 (a+b \sec (c+d x))}-\frac{\csc ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{64 d (a+b) (a+b \sec (c+d x))}+\frac{(7 a+9 b) \csc ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{32 d (a+b)^2 (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5/(a + b*Sec[c + d*x]),x]

[Out]

((2*I)*(a^5 - 3*a^3*b^2 + 3*a*b^4)*(c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x])/((a - b)^3*(a + b)^3*d*(a + b*
Sec[c + d*x])) - ((I/8)*(-8*a^2 + 21*a*b - 15*b^2)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x])*Sec[c + d*x])/((-
a + b)^3*d*(a + b*Sec[c + d*x])) - ((I/8)*(8*a^2 + 21*a*b + 15*b^2)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x])*
Sec[c + d*x])/((a + b)^3*d*(a + b*Sec[c + d*x])) + ((7*a + 9*b)*(b + a*Cos[c + d*x])*Csc[(c + d*x)/2]^2*Sec[c
+ d*x])/(32*(a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Csc[(c + d*x)/2]^4*Sec[c + d*x])/(64*(a
+ b)*d*(a + b*Sec[c + d*x])) + ((-8*a^2 + 21*a*b - 15*b^2)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2]^2]*Sec[c
+ d*x])/(16*(-a + b)^3*d*(a + b*Sec[c + d*x])) + (b^6*(b + a*Cos[c + d*x])*Log[b + a*Cos[c + d*x]]*Sec[c + d*x
])/(a*(-a^2 + b^2)^3*d*(a + b*Sec[c + d*x])) + ((8*a^2 + 21*a*b + 15*b^2)*(b + a*Cos[c + d*x])*Log[Sin[(c + d*
x)/2]^2]*Sec[c + d*x])/(16*(a + b)^3*d*(a + b*Sec[c + d*x])) + ((7*a - 9*b)*(b + a*Cos[c + d*x])*Sec[(c + d*x)
/2]^2*Sec[c + d*x])/(32*(-a + b)^2*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sec[c +
d*x])/(64*(-a + b)*d*(a + b*Sec[c + d*x]))

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Maple [A]  time = 0.076, size = 308, normalized size = 1.3 \begin{align*} -{\frac{{b}^{6}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}a}}-{\frac{1}{2\,d \left ( 8\,a-8\,b \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}+{\frac{7\,a}{16\,d \left ( a-b \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{9\,b}{16\,d \left ( a-b \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) }}+{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ){a}^{2}}{2\,d \left ( a-b \right ) ^{3}}}-{\frac{21\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ) ab}{16\,d \left ( a-b \right ) ^{3}}}+{\frac{15\,\ln \left ( \cos \left ( dx+c \right ) +1 \right ){b}^{2}}{16\,d \left ( a-b \right ) ^{3}}}-{\frac{1}{2\,d \left ( 8\,a+8\,b \right ) \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{7\,a}{16\,d \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{9\,b}{16\,d \left ( a+b \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ){a}^{2}}{2\,d \left ( a+b \right ) ^{3}}}+{\frac{21\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ) ab}{16\,d \left ( a+b \right ) ^{3}}}+{\frac{15\,\ln \left ( -1+\cos \left ( dx+c \right ) \right ){b}^{2}}{16\,d \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5/(a+b*sec(d*x+c)),x)

[Out]

-1/d*b^6/(a+b)^3/(a-b)^3/a*ln(b+a*cos(d*x+c))-1/2/d/(8*a-8*b)/(cos(d*x+c)+1)^2+7/16/d/(a-b)^2/(cos(d*x+c)+1)*a
-9/16/d/(a-b)^2/(cos(d*x+c)+1)*b+1/2/d/(a-b)^3*ln(cos(d*x+c)+1)*a^2-21/16/d/(a-b)^3*ln(cos(d*x+c)+1)*a*b+15/16
/d/(a-b)^3*ln(cos(d*x+c)+1)*b^2-1/2/d/(8*a+8*b)/(-1+cos(d*x+c))^2-7/16/d/(a+b)^2/(-1+cos(d*x+c))*a-9/16/d/(a+b
)^2/(-1+cos(d*x+c))*b+1/2/d/(a+b)^3*ln(-1+cos(d*x+c))*a^2+21/16/d/(a+b)^3*ln(-1+cos(d*x+c))*a*b+15/16/d/(a+b)^
3*ln(-1+cos(d*x+c))*b^2

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Maxima [A]  time = 1.01366, size = 390, normalized size = 1.67 \begin{align*} -\frac{\frac{16 \, b^{6} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac{{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{2 \,{\left ({\left (5 \, a^{2} b - 9 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} -{\left (3 \, a^{2} b - 7 \, b^{3}\right )} \cos \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(16*b^6*log(a*cos(d*x + c) + b)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - (8*a^2 - 21*a*b + 15*b^2)*log(co
s(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2
*b + 3*a*b^2 + b^3) - 2*((5*a^2*b - 9*b^3)*cos(d*x + c)^3 + 6*a^3 - 10*a*b^2 - 4*(2*a^3 - 3*a*b^2)*cos(d*x + c
)^2 - (3*a^2*b - 7*b^3)*cos(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4
- 2*a^2*b^2 + b^4)*cos(d*x + c)^2))/d

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Fricas [B]  time = 1.88919, size = 1312, normalized size = 5.61 \begin{align*} \frac{12 \, a^{6} - 32 \, a^{4} b^{2} + 20 \, a^{2} b^{4} + 2 \,{\left (5 \, a^{5} b - 14 \, a^{3} b^{3} + 9 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 8 \,{\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right ) - 16 \,{\left (b^{6} \cos \left (d x + c\right )^{4} - 2 \, b^{6} \cos \left (d x + c\right )^{2} + b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) +{\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5} +{\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5} +{\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{16 \,{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(12*a^6 - 32*a^4*b^2 + 20*a^2*b^4 + 2*(5*a^5*b - 14*a^3*b^3 + 9*a*b^5)*cos(d*x + c)^3 - 8*(2*a^6 - 5*a^4*
b^2 + 3*a^2*b^4)*cos(d*x + c)^2 - 2*(3*a^5*b - 10*a^3*b^3 + 7*a*b^5)*cos(d*x + c) - 16*(b^6*cos(d*x + c)^4 - 2
*b^6*cos(d*x + c)^2 + b^6)*log(a*cos(d*x + c) + b) + (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 +
15*a*b^5 + (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^4 - 2*(8*a^6 + 3*
a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (8*a^6
- 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 - 15*a*b^5 + (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*
a^2*b^4 - 15*a*b^5)*cos(d*x + c)^4 - 2*(8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 - 15*a*b^5)*cos
(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^4 - 2*(a^7 -
3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\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(cot(d*x+c)**5/(a+b*sec(d*x+c)),x)

[Out]

Integral(cot(c + d*x)**5/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 1.49062, size = 639, normalized size = 2.73 \begin{align*} -\frac{\frac{64 \, b^{6} \log \left ({\left | a + b + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac{4 \,{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac{\frac{12 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{16 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{a^{2} - 2 \, a b + b^{2}} + \frac{{\left (a^{2} + 2 \, a b + b^{2} + \frac{12 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{28 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{16 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{48 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{126 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{90 \, b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} + \frac{64 \, \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/64*(64*b^6*log(abs(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
)))/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - 4*(8*a^2 + 21*a*b + 15*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x
+ c) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + (12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 16*b*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - b*(cos(d*x + c) - 1)^2/(cos(d*x + c)
+ 1)^2)/(a^2 - 2*a*b + b^2) + (a^2 + 2*a*b + b^2 + 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 28*a*b*(cos(
d*x + c) - 1)/(cos(d*x + c) + 1) + 16*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 48*a^2*(cos(d*x + c) - 1)^2/
(cos(d*x + c) + 1)^2 + 126*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 90*b^2*(cos(d*x + c) - 1)^2/(cos(d*
x + c) + 1)^2)*(cos(d*x + c) + 1)^2/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(d*x + c) - 1)^2) + 64*log(abs(-(cos(
d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a)/d