\(\int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [198]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {7}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d-1/5/d/(a+a*sec(d*x+c))^(5/2)-1/2/a/d/(a+a*sec(d*x+c))^(3/2
)-1/8*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)-7/4/a^2/d/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3965, 87, 157, 162, 65, 213} \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {7}{4 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {1}{2 a d (a \sec (c+d x)+a)^{3/2}}-\frac {1}{5 d (a \sec (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])
]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + a*Sec[c + d*x])^(5/2)) - 1/(2*a*d*(a + a*Sec[c + d*x])^(3/2)) - 7/(4*a^2
*d*Sqrt[a + a*Sec[c + d*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 87

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[f*((e + f*x)^(p +
1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[(b*d*e - b*c*f - a*d*f - b*
d*f*x)*((e + f*x)^(p + 1)/((a + b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3965

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

Rubi steps \begin{align*} \text {integral}& = \frac {a^2 \text {Subst}\left (\int \frac {1}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {\text {Subst}\left (\int \frac {2 a^2-a^2 x}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{2 a d} \\ & = -\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {\text {Subst}\left (\int \frac {-6 a^4+\frac {9 a^4 x}{2}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{6 a^4 d} \\ & = -\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {7}{4 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\text {Subst}\left (\int \frac {6 a^6-\frac {21 a^6 x}{4}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{6 a^7 d} \\ & = -\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {7}{4 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac {\text {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{8 a d} \\ & = -\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {7}{4 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{a^3 d}+\frac {\text {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{4 a^2 d} \\ & = \frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac {1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac {7}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.42 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1+\sec (c+d x))\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},1+\sec (c+d x)\right )}{5 d (a (1+\sec (c+d x)))^{5/2}} \]

[In]

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

[Out]

(Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sec[c + d*x])/2] - 2*Hypergeometric2F1[-5/2, 1, -3/2, 1 + Sec[c + d*x]]
)/(5*d*(a*(1 + Sec[c + d*x]))^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(377\) vs. \(2(115)=230\).

Time = 1.73 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.62

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-15 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}+15 \left (1-\cos \left (d x +c \right )\right )^{6} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{6}+21 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-87 \left (1-\cos \left (d x +c \right )\right )^{4} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{4}-35 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}+269 \left (1-\cos \left (d x +c \right )\right )^{2} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{2}+840 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}{2}\right )+105 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )-932 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )}{840 d \,a^{3}}\)	\(378\)

[In]

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

[Out]

-1/840/d/a^3*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-15*((1-c
os(d*x+c))^2*csc(d*x+c)^2-1)^(7/2)+15*(1-cos(d*x+c))^6*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*csc(d*x+c)^6+21
*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)-87*(1-cos(d*x+c))^4*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*csc(d*x+c
)^4-35*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)+269*(1-cos(d*x+c))^2*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*cs
c(d*x+c)^2+840*2^(1/2)*arctan(1/2*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+105*arctan(1/((1-cos(d*x+c)
)^2*csc(d*x+c)^2-1)^(1/2))-932*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (115) = 230\).

Time = 0.36 (sec) , antiderivative size = 573, normalized size of antiderivative = 3.98 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) + 40 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, {\left (49 \, \cos \left (d x + c\right )^{3} + 80 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{80 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {5 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 40 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) - 2 \, {\left (49 \, \cos \left (d x + c\right )^{3} + 80 \, \cos \left (d x + c\right )^{2} + 35 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{40 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

[In]

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

[Out]

[1/80*(5*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-(2*sqrt(2)*sqrt(a)*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - 3*a*cos(d*x + c) - a)/(cos(d*x + c) - 1)) + 40*(cos(d*x + c
)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-8*a*cos(d*x + c)^2 - 4*(2*cos(d*x + c)^2 + cos(d*x +
c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) - 4*(49*cos(d*x + c)^3 + 80*cos(d
*x + c)^2 + 35*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x
+ c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/40*(5*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) +
1)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a))
- 40*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a)) - 2*(49*cos(d*x + c)^3 + 80*cos(d*x + c)^2 + 35*cos(d
*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos
(d*x + c) + a^3*d)]

Sympy [F]

\[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\cot {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(cot(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.93 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.43 \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {5 \, \sqrt {2} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {80 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left ({\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{20} + 5 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{21} + 35 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{22}\right )}}{a^{25} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{40 \, d} \]

[In]

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

[Out]

1/40*(5*sqrt(2)*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(cos(d*x + c))) - 80*arc
tan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(cos(d*x + c))) - sqrt(2)*((a*t
an(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^20 + 5*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2
)*a^21 + 35*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^22)/(a^25*sgn(cos(d*x + c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\mathrm {cot}\left (c+d\,x\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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