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

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

[Out]

2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+199/288*a^2/d/(a+a*sec(d*x+c))^(9/2)-1/4*a^2/d/(1-sec(d*x+
c))^2/(a+a*sec(d*x+c))^(9/2)-21/16*a^2/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(9/2)+135/448*a/d/(a+a*sec(d*x+c))^(7
/2)+7/640/d/(a+a*sec(d*x+c))^(5/2)-83/256/a/d/(a+a*sec(d*x+c))^(3/2)-263/1024*arctanh(1/2*(a+a*sec(d*x+c))^(1/
2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)-761/512/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 262, normalized size of antiderivative = 1.00, 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 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 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 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])

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]

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*}

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Mathematica [C]  time = 0.32, size = 99, normalized size = 0.38 \[ \frac {\cot ^4(c+d x) \left (263 (\sec (c+d x)-1)^2 \, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};\frac {1}{2} (\sec (c+d x)+1)\right )-64 (\sec (c+d x)-1)^2 \, _2F_1\left (-\frac {9}{2},1;-\frac {7}{2};\sec (c+d x)+1\right )+378 \sec (c+d x)-450\right )}{288 d (a (\sec (c+d x)+1))^{5/2}} \]

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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fricas [B]  time = 0.69, size = 905, normalized size = 3.45 \[ \text {result too large to display} \]

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]

[1/645120*(82845*sqrt(2)*(cos(d*x + c)^7 + 3*cos(d*x + c)^6 + cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 5*cos(d*x +
c)^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)) + 322560*(cos(d*x + c)^7 + 3*cos(d*x + c)^6 + c
os(d*x + c)^5 - 5*cos(d*x + c)^4 - 5*cos(d*x + c)^3 + cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-8*a*co
s(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*(382201*cos(d*x + c)^7 + 591520*cos(d*x + c)^6 - 403607*cos(d*x + c)^5 - 1112040*cos(d*x + c)
^4 - 189189*cos(d*x + c)^3 + 531720*cos(d*x + c)^2 + 239715*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c)))/(a^3*d*cos(d*x + c)^7 + 3*a^3*d*cos(d*x + c)^6 + a^3*d*cos(d*x + c)^5 - 5*a^3*d*cos(d*x + c)^4 - 5*a^3*d*
cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/322560*(82845*sqrt(2)*(cos(d*x + c)^7
+ 3*cos(d*x + c)^6 + cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 5*cos(d*x + c)^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)
) - 322560*(cos(d*x + c)^7 + 3*cos(d*x + c)^6 + cos(d*x + c)^5 - 5*cos(d*x + c)^4 - 5*cos(d*x + c)^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*(382201*cos(d*x + c)^7 + 591520*cos(d*x + c)^6 - 403607*cos(d*x + c)^5 - 1112040*c
os(d*x + c)^4 - 189189*cos(d*x + c)^3 + 531720*cos(d*x + c)^2 + 239715*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c)))/(a^3*d*cos(d*x + c)^7 + 3*a^3*d*cos(d*x + c)^6 + a^3*d*cos(d*x + c)^5 - 5*a^3*d*cos(d*x + c)^4
- 5*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

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giac [A]  time = 2.67, size = 392, normalized size = 1.50 \[ -\frac {\frac {82845 \, \sqrt {2} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {645120 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {315 \, {\left (33 \, \sqrt {2} {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} - 31 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a\right )}}{a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}} - \frac {8 \, \sqrt {2} {\left (35 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{56} - 225 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{57} + 1008 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{58} + 4410 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{59} + 31185 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{60}\right )}}{a^{63} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{322560 \, d} \]

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*(33*sqrt(2)*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 31*sqrt(2)*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*sqrt(2)*(35*(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*(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*(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*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^59 + 31185*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^60)/(a^6
3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.49, size = 986, normalized size = 3.76 \[ \text {result too large to display} \]

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*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^5*(322560*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^7*2^(1/2)+82845*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^7+967680*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^6*2^(1/2)+248535*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^6+322560*cos(d*x+c
)^5*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+8284
5*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-1612800*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)+764402*cos(d*x+c)^7-414225*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*arctan(1/(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2))-1612800*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*2^(1/2))+1183040*cos(d*x+c)^6-414225*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arc
tan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+322560*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^2-807214*cos(d*x+c)^5+82845*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+967680*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-2224080*cos(d*x+c)^4+248535*cos
(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+322560*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-378378*cos(d*x+c)^3
+82845*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+1063440*cos(d*x+c)^
2+479430*cos(d*x+c))/sin(d*x+c)^14/a^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^5}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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]

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

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