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

Optimal. Leaf size=262 \[ \frac{199 a^2}{288 d (a \sec (c+d x)+a)^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{9/2}}-\frac{761}{512 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{263 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{512 \sqrt{2} a^{5/2} d}+\frac{135 a}{448 d (a \sec (c+d x)+a)^{7/2}}+\frac{7}{640 d (a \sec (c+d x)+a)^{5/2}}-\frac{83}{256 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - (263*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(512*Sqrt[2]*a^(5/2)*d) + (199*a^2)/(288*d*(a + a*Sec[c + d*x])^(9/2)) - a^2/(4*d*(1 - Sec[c + d*x])^2
*(a + a*Sec[c + d*x])^(9/2)) - (21*a^2)/(16*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(9/2)) + (135*a)/(448*d*
(a + a*Sec[c + d*x])^(7/2)) + 7/(640*d*(a + a*Sec[c + d*x])^(5/2)) - 83/(256*a*d*(a + a*Sec[c + d*x])^(3/2)) -
761/(512*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.226566, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3880, 103, 151, 152, 156, 63, 207} \[ \frac{199 a^2}{288 d (a \sec (c+d x)+a)^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{9/2}}-\frac{761}{512 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{263 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{512 \sqrt{2} a^{5/2} d}+\frac{135 a}{448 d (a \sec (c+d x)+a)^{7/2}}+\frac{7}{640 d (a \sec (c+d x)+a)^{5/2}}-\frac{83}{256 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5/(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) - (263*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(512*Sqrt[2]*a^(5/2)*d) + (199*a^2)/(288*d*(a + a*Sec[c + d*x])^(9/2)) - a^2/(4*d*(1 - Sec[c + d*x])^2
*(a + a*Sec[c + d*x])^(9/2)) - (21*a^2)/(16*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(9/2)) + (135*a)/(448*d*
(a + a*Sec[c + d*x])^(7/2)) + 7/(640*d*(a + a*Sec[c + d*x])^(5/2)) - 83/(256*a*d*(a + a*Sec[c + d*x])^(3/2)) -
761/(512*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 3880

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]

Rule 103

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

Rule 151

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] && IntegerQ[m]

Rule 152

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 156

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 63

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 207

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

Rubi steps

\begin{align*} \int \frac{\cot ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{a^6 \operatorname{Subst}\left (\int \frac{1}{x (-a+a x)^3 (a+a x)^{11/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{a^3 \operatorname{Subst}\left (\int \frac{4 a^2+\frac{13 a^2 x}{2}}{x (-a+a x)^2 (a+a x)^{11/2}} \, dx,x,\sec (c+d x)\right )}{4 d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{\operatorname{Subst}\left (\int \frac{8 a^4+\frac{231 a^4 x}{4}}{x (-a+a x) (a+a x)^{11/2}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}-\frac{\operatorname{Subst}\left (\int \frac{-72 a^6-\frac{1791 a^6 x}{8}}{x (-a+a x) (a+a x)^{9/2}} \, dx,x,\sec (c+d x)\right )}{72 a^3 d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{\operatorname{Subst}\left (\int \frac{504 a^8+\frac{8505 a^8 x}{16}}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{504 a^6 d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{-2520 a^{10}-\frac{2205 a^{10} x}{32}}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{2520 a^9 d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{83}{256 a d (a+a \sec (c+d x))^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{7560 a^{12}-\frac{235305 a^{12} x}{64}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{7560 a^{12} d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{83}{256 a d (a+a \sec (c+d x))^{3/2}}-\frac{761}{512 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{-7560 a^{14}+\frac{719145 a^{14} x}{128}}{x (-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{7560 a^{15} d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{83}{256 a d (a+a \sec (c+d x))^{3/2}}-\frac{761}{512 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac{263 \operatorname{Subst}\left (\int \frac{1}{(-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{1024 a d}\\ &=\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{83}{256 a d (a+a \sec (c+d x))^{3/2}}-\frac{761}{512 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{a^3 d}+\frac{263 \operatorname{Subst}\left (\int \frac{1}{-2 a+x^2} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{512 a^2 d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{263 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{512 \sqrt{2} a^{5/2} d}+\frac{199 a^2}{288 d (a+a \sec (c+d x))^{9/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{9/2}}-\frac{21 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{9/2}}+\frac{135 a}{448 d (a+a \sec (c+d x))^{7/2}}+\frac{7}{640 d (a+a \sec (c+d x))^{5/2}}-\frac{83}{256 a d (a+a \sec (c+d x))^{3/2}}-\frac{761}{512 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.290965, size = 99, normalized size = 0.38 \[ \frac{\cot ^4(c+d x) \left (263 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{9}{2},1,-\frac{7}{2},\frac{1}{2} (\sec (c+d x)+1)\right )-64 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{9}{2},1,-\frac{7}{2},\sec (c+d x)+1\right )+378 \sec (c+d x)-450\right )}{288 d (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^4*(-450 + 263*Hypergeometric2F1[-9/2, 1, -7/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x])^2 - 64*
Hypergeometric2F1[-9/2, 1, -7/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])^2 + 378*Sec[c + d*x]))/(288*d*(a*(1 + S
ec[c + d*x]))^(5/2))

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Maple [B]  time = 0.342, size = 986, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/322560/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^5*(322560*cos(d*x+c)^7*2^(
1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+82845*cos(d
*x+c)^7*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+967680*2^(1/2)*cos
(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+248535
*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+322560*cos(d
*x+c)^5*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+
764402*cos(d*x+c)^7+82845*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2))-1612800*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2))+1183040*cos(d*x+c)^6-414225*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan
(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-1612800*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arc
tan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-807214*cos(d*x+c)^5-414225*cos(d*x+c)^3*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+322560*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2224080*cos(d*x+c)^4+82845*cos
(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+967680*2^(1/2)*c
os(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-378378
*cos(d*x+c)^3+248535*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2))+322560*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))+1063440*cos(d*x+c)^2+82845*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2))+479430*cos(d*x+c))/sin(d*x+c)^14

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

[Out]

Timed out

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

[Out]

Timed out

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Giac [A]  time = 9.52446, size = 541, normalized size = 2.06 \begin{align*} -\frac{\frac{82845 \, \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 (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{645120 \, \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 (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{315 \, \sqrt{2}{\left (33 \,{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} - 31 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a\right )}}{a^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}} - \frac{8 \,{\left (35 \, \sqrt{2}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{56} - 225 \, \sqrt{2}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{57} + 1008 \, \sqrt{2}{\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^{58} + 4410 \, \sqrt{2}{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a^{59} + 31185 \, \sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{60}\right )}}{a^{63} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{322560 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/322560*(82845*sqrt(2)*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x +
1/2*c)^2 - 1)) - 645120*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan
(1/2*d*x + 1/2*c)^2 - 1)) - 315*sqrt(2)*(33*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 31*sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a)*a)/(a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*tan(1/2*d*x + 1/2*c)^4) - 8*(35*sqrt(2)*(a*tan(1/2*d*x +
1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^56 - 225*sqrt(2)*(a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-
a*tan(1/2*d*x + 1/2*c)^2 + a)*a^57 + 1008*sqrt(2)*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c
)^2 + a)*a^58 + 4410*sqrt(2)*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^59 + 31185*sqrt(2)*sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a)*a^60)/(a^63*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d