\(\int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [306]

Optimal result

Integrand size = 21, antiderivative size = 278 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (4 a^2+13 a b+12 b^2\right ) \log (1-\sec (c+d x))}{8 (a+b)^4 d}+\frac {\left (4 a^2-13 a b+12 b^2\right ) \log (1+\sec (c+d x))}{8 (a-b)^4 d}-\frac {b^6 \left (7 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 \left (a^2-b^2\right )^4 d}-\frac {1}{16 (a+b)^2 d (1-\sec (c+d x))^2}-\frac {5 a+9 b}{16 (a+b)^3 d (1-\sec (c+d x))}-\frac {1}{16 (a-b)^2 d (1+\sec (c+d x))^2}-\frac {5 a-9 b}{16 (a-b)^3 d (1+\sec (c+d x))}+\frac {b^6}{a \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \]

[Out]

ln(cos(d*x+c))/a^2/d+1/8*(4*a^2+13*a*b+12*b^2)*ln(1-sec(d*x+c))/(a+b)^4/d+1/8*(4*a^2-13*a*b+12*b^2)*ln(1+sec(d
*x+c))/(a-b)^4/d-b^6*(7*a^2-b^2)*ln(a+b*sec(d*x+c))/a^2/(a^2-b^2)^4/d-1/16/(a+b)^2/d/(1-sec(d*x+c))^2+1/16*(-5
*a-9*b)/(a+b)^3/d/(1-sec(d*x+c))-1/16/(a-b)^2/d/(1+sec(d*x+c))^2+1/16*(-5*a+9*b)/(a-b)^3/d/(1+sec(d*x+c))+b^6/
a/(a^2-b^2)^3/d/(a+b*sec(d*x+c))

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3970, 908} \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (4 a^2+13 a b+12 b^2\right ) \log (1-\sec (c+d x))}{8 d (a+b)^4}+\frac {\left (4 a^2-13 a b+12 b^2\right ) \log (\sec (c+d x)+1)}{8 d (a-b)^4}+\frac {b^6}{a d \left (a^2-b^2\right )^3 (a+b \sec (c+d x))}-\frac {b^6 \left (7 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 d \left (a^2-b^2\right )^4}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {5 a+9 b}{16 d (a+b)^3 (1-\sec (c+d x))}-\frac {5 a-9 b}{16 d (a-b)^3 (\sec (c+d x)+1)}-\frac {1}{16 d (a+b)^2 (1-\sec (c+d x))^2}-\frac {1}{16 d (a-b)^2 (\sec (c+d x)+1)^2} \]

[In]

Int[Cot[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]

[Out]

Log[Cos[c + d*x]]/(a^2*d) + ((4*a^2 + 13*a*b + 12*b^2)*Log[1 - Sec[c + d*x]])/(8*(a + b)^4*d) + ((4*a^2 - 13*a
*b + 12*b^2)*Log[1 + Sec[c + d*x]])/(8*(a - b)^4*d) - (b^6*(7*a^2 - b^2)*Log[a + b*Sec[c + d*x]])/(a^2*(a^2 -
b^2)^4*d) - 1/(16*(a + b)^2*d*(1 - Sec[c + d*x])^2) - (5*a + 9*b)/(16*(a + b)^3*d*(1 - Sec[c + d*x])) - 1/(16*
(a - b)^2*d*(1 + Sec[c + d*x])^2) - (5*a - 9*b)/(16*(a - b)^3*d*(1 + Sec[c + d*x])) + b^6/(a*(a^2 - b^2)^3*d*(
a + b*Sec[c + d*x]))

Rule 908

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]))

Rule 3970

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {b^6 \text {Subst}\left (\int \frac {1}{x (a+x)^2 \left (b^2-x^2\right )^3} \, dx,x,b \sec (c+d x)\right )}{d} \\ & = -\frac {b^6 \text {Subst}\left (\int \left (\frac {1}{8 b^4 (a+b)^2 (b-x)^3}+\frac {5 a+9 b}{16 b^5 (a+b)^3 (b-x)^2}+\frac {4 a^2+13 a b+12 b^2}{8 b^6 (a+b)^4 (b-x)}+\frac {1}{a^2 b^6 x}+\frac {1}{a (a-b)^3 (a+b)^3 (a+x)^2}+\frac {7 a^2-b^2}{a^2 (a-b)^4 (a+b)^4 (a+x)}-\frac {1}{8 (a-b)^2 b^4 (b+x)^3}+\frac {-5 a+9 b}{16 (a-b)^3 b^5 (b+x)^2}+\frac {-4 a^2+13 a b-12 b^2}{8 (a-b)^4 b^6 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d} \\ & = \frac {\log (\cos (c+d x))}{a^2 d}+\frac {\left (4 a^2+13 a b+12 b^2\right ) \log (1-\sec (c+d x))}{8 (a+b)^4 d}+\frac {\left (4 a^2-13 a b+12 b^2\right ) \log (1+\sec (c+d x))}{8 (a-b)^4 d}-\frac {b^6 \left (7 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 \left (a^2-b^2\right )^4 d}-\frac {1}{16 (a+b)^2 d (1-\sec (c+d x))^2}-\frac {5 a+9 b}{16 (a+b)^3 d (1-\sec (c+d x))}-\frac {1}{16 (a-b)^2 d (1+\sec (c+d x))^2}-\frac {5 a-9 b}{16 (a-b)^3 d (1+\sec (c+d x))}+\frac {b^6}{a \left (a^2-b^2\right )^3 d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.45 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {b^6 \left (-\frac {\log (\cos (c+d x))}{a^2 b^6}-\frac {\left (4 a^2+13 a b+12 b^2\right ) \log (1-\sec (c+d x))}{8 b^6 (a+b)^4}-\frac {\left (4 a^2-13 a b+12 b^2\right ) \log (1+\sec (c+d x))}{8 (a-b)^4 b^6}+\frac {\left (7 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 (a-b)^4 (a+b)^4}+\frac {1}{16 b^4 (a+b)^2 (b-b \sec (c+d x))^2}+\frac {5 a+9 b}{16 b^5 (a+b)^3 (b-b \sec (c+d x))}-\frac {1}{a (a-b)^3 (a+b)^3 (a+b \sec (c+d x))}+\frac {1}{16 (a-b)^2 b^4 (b+b \sec (c+d x))^2}+\frac {5 a-9 b}{16 (a-b)^3 b^5 (b+b \sec (c+d x))}\right )}{d} \]

[In]

Integrate[Cot[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]

[Out]

-((b^6*(-(Log[Cos[c + d*x]]/(a^2*b^6)) - ((4*a^2 + 13*a*b + 12*b^2)*Log[1 - Sec[c + d*x]])/(8*b^6*(a + b)^4) -
((4*a^2 - 13*a*b + 12*b^2)*Log[1 + Sec[c + d*x]])/(8*(a - b)^4*b^6) + ((7*a^2 - b^2)*Log[a + b*Sec[c + d*x]])
/(a^2*(a - b)^4*(a + b)^4) + 1/(16*b^4*(a + b)^2*(b - b*Sec[c + d*x])^2) + (5*a + 9*b)/(16*b^5*(a + b)^3*(b -
b*Sec[c + d*x])) - 1/(a*(a - b)^3*(a + b)^3*(a + b*Sec[c + d*x])) + 1/(16*(a - b)^2*b^4*(b + b*Sec[c + d*x])^2
) + (5*a - 9*b)/(16*(a - b)^3*b^5*(b + b*Sec[c + d*x]))))/d)

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {-\frac {1}{16 \left (a -b \right )^{2} \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {-7 a +11 b}{16 \left (a -b \right )^{3} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (4 a^{2}-13 a b +12 b^{2}\right ) \ln \left (\cos \left (d x +c \right )+1\right )}{8 \left (a -b \right )^{4}}-\frac {b^{7}}{a^{2} \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}-\frac {b^{6} \left (7 a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4} a^{2}}-\frac {1}{16 \left (a +b \right )^{2} \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {7 a +11 b}{16 \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{8 \left (a +b \right )^{4}}}{d}\)	\(230\)
default	\(\frac {-\frac {1}{16 \left (a -b \right )^{2} \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {-7 a +11 b}{16 \left (a -b \right )^{3} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (4 a^{2}-13 a b +12 b^{2}\right ) \ln \left (\cos \left (d x +c \right )+1\right )}{8 \left (a -b \right )^{4}}-\frac {b^{7}}{a^{2} \left (a +b \right )^{3} \left (a -b \right )^{3} \left (b +a \cos \left (d x +c \right )\right )}-\frac {b^{6} \left (7 a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{4} \left (a -b \right )^{4} a^{2}}-\frac {1}{16 \left (a +b \right )^{2} \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {7 a +11 b}{16 \left (a +b \right )^{3} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (4 a^{2}+13 a b +12 b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{8 \left (a +b \right )^{4}}}{d}\)	\(230\)
risch	\(\text {Expression too large to display}\)	\(1577\)

[In]

int(cot(d*x+c)^5/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/16/(a-b)^2/(cos(d*x+c)+1)^2-1/16*(-7*a+11*b)/(a-b)^3/(cos(d*x+c)+1)+1/8*(4*a^2-13*a*b+12*b^2)/(a-b)^4*
ln(cos(d*x+c)+1)-b^7/a^2/(a+b)^3/(a-b)^3/(b+a*cos(d*x+c))-b^6*(7*a^2-b^2)/(a+b)^4/(a-b)^4/a^2*ln(b+a*cos(d*x+c
))-1/16/(a+b)^2/(cos(d*x+c)-1)^2-1/16*(7*a+11*b)/(a+b)^3/(cos(d*x+c)-1)+1/8*(4*a^2+13*a*b+12*b^2)/(a+b)^4*ln(c
os(d*x+c)-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1378 vs. \(2 (262) = 524\).

Time = 0.62 (sec) , antiderivative size = 1378, normalized size of antiderivative = 4.96 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/8*(6*a^8*b - 18*a^6*b^3 + 2*a^4*b^5 + 2*a^2*b^7 + 8*b^9 + 2*(5*a^8*b - 18*a^6*b^3 + 13*a^4*b^5 - 4*a^2*b^7 +
4*b^9)*cos(d*x + c)^4 - 2*(4*a^9 - 15*a^7*b^2 + 18*a^5*b^4 - 7*a^3*b^6)*cos(d*x + c)^3 - 2*(7*a^8*b - 24*a^6*
b^3 + 11*a^4*b^5 - 2*a^2*b^7 + 8*b^9)*cos(d*x + c)^2 + 6*(a^9 - 4*a^7*b^2 + 5*a^5*b^4 - 2*a^3*b^6)*cos(d*x + c
) - 8*(7*a^2*b^7 - b^9 + (7*a^3*b^6 - a*b^8)*cos(d*x + c)^5 + (7*a^2*b^7 - b^9)*cos(d*x + c)^4 - 2*(7*a^3*b^6
- a*b^8)*cos(d*x + c)^3 - 2*(7*a^2*b^7 - b^9)*cos(d*x + c)^2 + (7*a^3*b^6 - a*b^8)*cos(d*x + c))*log(a*cos(d*x
+ c) + b) + (4*a^8*b + 3*a^7*b^2 - 16*a^6*b^3 - 14*a^5*b^4 + 24*a^4*b^5 + 35*a^3*b^6 + 12*a^2*b^7 + (4*a^9 +
3*a^8*b - 16*a^7*b^2 - 14*a^6*b^3 + 24*a^5*b^4 + 35*a^4*b^5 + 12*a^3*b^6)*cos(d*x + c)^5 + (4*a^8*b + 3*a^7*b^
2 - 16*a^6*b^3 - 14*a^5*b^4 + 24*a^4*b^5 + 35*a^3*b^6 + 12*a^2*b^7)*cos(d*x + c)^4 - 2*(4*a^9 + 3*a^8*b - 16*a
^7*b^2 - 14*a^6*b^3 + 24*a^5*b^4 + 35*a^4*b^5 + 12*a^3*b^6)*cos(d*x + c)^3 - 2*(4*a^8*b + 3*a^7*b^2 - 16*a^6*b
^3 - 14*a^5*b^4 + 24*a^4*b^5 + 35*a^3*b^6 + 12*a^2*b^7)*cos(d*x + c)^2 + (4*a^9 + 3*a^8*b - 16*a^7*b^2 - 14*a^
6*b^3 + 24*a^5*b^4 + 35*a^4*b^5 + 12*a^3*b^6)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (4*a^8*b - 3*a^7*b^2
- 16*a^6*b^3 + 14*a^5*b^4 + 24*a^4*b^5 - 35*a^3*b^6 + 12*a^2*b^7 + (4*a^9 - 3*a^8*b - 16*a^7*b^2 + 14*a^6*b^3
+ 24*a^5*b^4 - 35*a^4*b^5 + 12*a^3*b^6)*cos(d*x + c)^5 + (4*a^8*b - 3*a^7*b^2 - 16*a^6*b^3 + 14*a^5*b^4 + 24*
a^4*b^5 - 35*a^3*b^6 + 12*a^2*b^7)*cos(d*x + c)^4 - 2*(4*a^9 - 3*a^8*b - 16*a^7*b^2 + 14*a^6*b^3 + 24*a^5*b^4
- 35*a^4*b^5 + 12*a^3*b^6)*cos(d*x + c)^3 - 2*(4*a^8*b - 3*a^7*b^2 - 16*a^6*b^3 + 14*a^5*b^4 + 24*a^4*b^5 - 35
*a^3*b^6 + 12*a^2*b^7)*cos(d*x + c)^2 + (4*a^9 - 3*a^8*b - 16*a^7*b^2 + 14*a^6*b^3 + 24*a^5*b^4 - 35*a^4*b^5 +
12*a^3*b^6)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)
*d*cos(d*x + c)^5 + (a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d*cos(d*x + c)^4 - 2*(a^11 - 4*a^9*
b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c)^3 - 2*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*
b^9)*d*cos(d*x + c)^2 + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8)*d*cos(d*x + c) + (a^10*b - 4*a^8*
b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2*b^9)*d)

Sympy [F]

\[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(cot(d*x+c)**5/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**5/(a + b*sec(c + d*x))**2, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (262) = 524\).

Time = 0.21 (sec) , antiderivative size = 558, normalized size of antiderivative = 2.01 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {8 \, {\left (7 \, a^{2} b^{6} - b^{8}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{10} - 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} - 4 \, a^{4} b^{6} + a^{2} b^{8}} - \frac {{\left (4 \, a^{2} - 13 \, a b + 12 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {{\left (4 \, a^{2} + 13 \, a b + 12 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}} - \frac {2 \, {\left (3 \, a^{6} b - 6 \, a^{4} b^{3} - 5 \, a^{2} b^{5} - 4 \, b^{7} + {\left (5 \, a^{6} b - 13 \, a^{4} b^{3} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{4} - {\left (4 \, a^{7} - 11 \, a^{5} b^{2} + 7 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (7 \, a^{6} b - 17 \, a^{4} b^{3} - 6 \, a^{2} b^{5} - 8 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )}}{a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7} + {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} \cos \left (d x + c\right )^{5} + {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} \cos \left (d x + c\right )}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/8*(8*(7*a^2*b^6 - b^8)*log(a*cos(d*x + c) + b)/(a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8) - (4*a^
2 - 13*a*b + 12*b^2)*log(cos(d*x + c) + 1)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (4*a^2 + 13*a*b + 12*
b^2)*log(cos(d*x + c) - 1)/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4) - 2*(3*a^6*b - 6*a^4*b^3 - 5*a^2*b^5 -
4*b^7 + (5*a^6*b - 13*a^4*b^3 - 4*b^7)*cos(d*x + c)^4 - (4*a^7 - 11*a^5*b^2 + 7*a^3*b^4)*cos(d*x + c)^3 - (7*a
^6*b - 17*a^4*b^3 - 6*a^2*b^5 - 8*b^7)*cos(d*x + c)^2 + 3*(a^7 - 3*a^5*b^2 + 2*a^3*b^4)*cos(d*x + c))/(a^8*b -
3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7 + (a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^5 + (a^8*b - 3*a^6*b^3
+ 3*a^4*b^5 - a^2*b^7)*cos(d*x + c)^4 - 2*(a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)^3 - 2*(a^8*b -
3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*cos(d*x + c)^2 + (a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*cos(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (262) = 524\).

Time = 0.43 (sec) , antiderivative size = 795, normalized size of antiderivative = 2.86 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/64*(8*(4*a^2 + 13*a*b + 12*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^4 + 4*a^3*b + 6*a^2*b^2
+ 4*a*b^3 + b^4) - 64*(7*a^2*b^6 - b^8)*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^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^6 + a^2*b^8) - (12*a^2*(cos(d*x + c) -
1)/(cos(d*x + c) + 1) - 32*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 20*b^2*(cos(d*x + c) - 1)/(cos(d*x + c
) + 1) + a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 2*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b^2
*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - (a^2 + 2*a*b + b^2 +
12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 32*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 20*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 + 156*a*b*(cos(d*x + c) - 1)^
2/(cos(d*x + c) + 1)^2 + 144*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/((a^4 + 4*a^3
*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(cos(d*x + c) - 1)^2) + 64*(7*a^3*b^6 + 5*a^2*b^7 - 3*a*b^8 - b^9 + 7*a^3*b^6*
(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 7*a^2*b^7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - a*b^8*(cos(d*x + c)
- 1)/(cos(d*x + c) + 1) + b^9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((a^10 - 4*a^8*b^2 + 6*a^6*b^4 - 4*a^4*b^
6 + a^2*b^8)*(a + b + a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))) - 64
*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1))/a^2)/d

Mupad [B] (verification not implemented)

Time = 15.76 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.69 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {a^3-3\,a^2\,b+3\,a\,b^2-b^3}{4\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-13\,a^4+20\,a^3\,b+18\,a^2\,b^2-44\,a\,b^3+19\,b^4\right )}{4\,{\left (a+b\right )}^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^7-10\,a^6\,b+5\,a^5\,b^2+20\,a^4\,b^3-35\,a^3\,b^4+22\,a^2\,b^5-5\,a\,b^6+32\,b^7\right )}{a\,{\left (a+b\right )}^3\,\left (a-b\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (16\,a^4-64\,a^3\,b+96\,a^2\,b^2-64\,a\,b^3+16\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a^4-32\,a^3\,b+32\,a\,b^3-16\,b^4\right )\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,d\,{\left (a-b\right )}^2}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {16\,a^2+32\,a\,b-48\,b^2}{512\,{\left (a-b\right )}^4}-\frac {7}{32\,{\left (a-b\right )}^2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,a^2+13\,a\,b+12\,b^2\right )}{d\,\left (4\,a^4+16\,a^3\,b+24\,a^2\,b^2+16\,a\,b^3+4\,b^4\right )}-\frac {b^6\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (7\,a^2-b^2\right )}{a^2\,d\,{\left (a^2-b^2\right )}^4} \]

[In]

int(cot(c + d*x)^5/(a + b/cos(c + d*x))^2,x)

[Out]

((3*a*b^2 - 3*a^2*b + a^3 - b^3)/(4*(a + b)) + (tan(c/2 + (d*x)/2)^2*(20*a^3*b - 44*a*b^3 - 13*a^4 + 19*b^4 +
18*a^2*b^2))/(4*(a + b)^2) + (tan(c/2 + (d*x)/2)^4*(3*a^7 - 10*a^6*b - 5*a*b^6 + 32*b^7 + 22*a^2*b^5 - 35*a^3*
b^4 + 20*a^4*b^3 + 5*a^5*b^2))/(a*(a + b)^3*(a - b)))/(d*(tan(c/2 + (d*x)/2)^6*(16*a^4 - 64*a^3*b - 64*a*b^3 +
16*b^4 + 96*a^2*b^2) - tan(c/2 + (d*x)/2)^4*(32*a*b^3 - 32*a^3*b + 16*a^4 - 16*b^4))) - tan(c/2 + (d*x)/2)^4/
(64*d*(a - b)^2) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a^2*d) - (tan(c/2 + (d*x)/2)^2*((32*a*b + 16*a^2 - 48*b^2)/(
512*(a - b)^4) - 7/(32*(a - b)^2)))/d + (log(tan(c/2 + (d*x)/2))*(13*a*b + 4*a^2 + 12*b^2))/(d*(16*a*b^3 + 16*
a^3*b + 4*a^4 + 4*b^4 + 24*a^2*b^2)) - (b^6*log(a + b - a*tan(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(7*a^
2 - b^2))/(a^2*d*(a^2 - b^2)^4)