3.152 \(\int \cot ^5(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=171 \[ \frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {71 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}+\frac {7 a^2}{32 d \sqrt {a \sec (c+d x)+a}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a \sec (c+d x)+a}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a \sec (c+d x)+a}} \]

[Out]

2*a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-71/64*a^(3/2)*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a
^(1/2))*2^(1/2)/d+7/32*a^2/d/(a+a*sec(d*x+c))^(1/2)-1/4*a^2/d/(1-sec(d*x+c))^2/(a+a*sec(d*x+c))^(1/2)-13/16*a^
2/d/(1-sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {7 a^2}{32 d \sqrt {a \sec (c+d x)+a}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a \sec (c+d x)+a}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a \sec (c+d x)+a}}+\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {71 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(2*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (71*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(32*Sqrt[2]*d) + (7*a^2)/(32*d*Sqrt[a + a*Sec[c + d*x]]) - a^2/(4*d*(1 - Sec[c + d*x])^2*Sqrt[a +
a*Sec[c + d*x]]) - (13*a^2)/(16*d*(1 - Sec[c + d*x])*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 \cot ^5(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac {a^6 \operatorname {Subst}\left (\int \frac {1}{x (-a+a x)^3 (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {a^3 \operatorname {Subst}\left (\int \frac {4 a^2+\frac {5 a^2 x}{2}}{x (-a+a x)^2 (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{4 d}\\ &=-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}+\frac {\operatorname {Subst}\left (\int \frac {8 a^4+\frac {39 a^4 x}{4}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=\frac {7 a^2}{32 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {-8 a^6-\frac {7 a^6 x}{8}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{8 a^3 d}\\ &=\frac {7 a^2}{32 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (71 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{64 d}\\ &=\frac {7 a^2}{32 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}+\frac {\left (71 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{32 d}\\ &=\frac {2 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {71 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} d}+\frac {7 a^2}{32 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2}{4 d (1-\sec (c+d x))^2 \sqrt {a+a \sec (c+d x)}}-\frac {13 a^2}{16 d (1-\sec (c+d x)) \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 104, normalized size = 0.61 \[ \frac {a^2 \left (71 (\sec (c+d x)-1)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {1}{2} (\sec (c+d x)+1)\right )-64 (\sec (c+d x)-1)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sec (c+d x)+1\right )+26 \sec (c+d x)-34\right )}{32 d (\sec (c+d x)-1)^2 \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(a^2*(-34 + 71*Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x])^2 - 64*Hypergeometric
2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])^2 + 26*Sec[c + d*x]))/(32*d*(-1 + Sec[c + d*x])^2*Sqrt
[a*(1 + Sec[c + d*x])])

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fricas [B]  time = 0.85, size = 589, normalized size = 3.44 \[ \left [\frac {64 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\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 ) + 71 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{3} - \sqrt {2} a \cos \left (d x + c\right )^{2} - \sqrt {2} a \cos \left (d x + c\right ) + \sqrt {2} a\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 ) - 4 \, {\left (27 \, a \cos \left (d x + c\right )^{3} - 12 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{128 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )}}, \frac {71 \, {\left (\sqrt {2} a \cos \left (d x + c\right )^{3} - \sqrt {2} a \cos \left (d x + c\right )^{2} - \sqrt {2} a \cos \left (d x + c\right ) + \sqrt {2} a\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 ) - 64 \, {\left (a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) + a\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 (27 \, a \cos \left (d x + c\right )^{3} - 12 \, a \cos \left (d x + c\right )^{2} - 7 \, a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{64 \, {\left (d \cos \left (d x + c\right )^{3} - d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/128*(64*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*sqrt(a)*log(-8*a*cos(d*x + c)^2 - 4*(2*c
os(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) + 71*(sq
rt(2)*a*cos(d*x + c)^3 - sqrt(2)*a*cos(d*x + c)^2 - sqrt(2)*a*cos(d*x + c) + sqrt(2)*a)*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)) - 4
*(27*a*cos(d*x + c)^3 - 12*a*cos(d*x + c)^2 - 7*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*co
s(d*x + c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d), 1/64*(71*(sqrt(2)*a*cos(d*x + c)^3 - sqrt(2)*a*cos(d*x
+ c)^2 - sqrt(2)*a*cos(d*x + c) + sqrt(2)*a)*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)) - 64*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*sqr
t(-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*(27*a
*cos(d*x + c)^3 - 12*a*cos(d*x + c)^2 - 7*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x
+ c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d)]

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giac [A]  time = 1.16, size = 211, normalized size = 1.23 \[ \frac {\frac {71 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - \frac {128 \, a^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - 8 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \frac {17 \, \sqrt {2} {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 15 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/64*(71*sqrt(2)*a^2*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))*sgn(cos(d*x + c))/sqrt(-a) - 128*a^2
*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))*sgn(cos(d*x + c))/sqrt(-a) - 8*sqrt(2)*sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a)*a*sgn(cos(d*x + c)) - (17*sqrt(2)*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^2*sgn
(cos(d*x + c)) - 15*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^3*sgn(cos(d*x + c)))/(a^2*tan(1/2*d*x + 1/2*
c)^4))/d

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maple [B]  time = 1.27, size = 502, normalized size = 2.94 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (64 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+71 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-64 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-71 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-64 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+54 \left (\cos ^{3}\left (d x +c \right )\right )-71 \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+64 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}-24 \left (\cos ^{2}\left (d x +c \right )\right )+71 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-14 \cos \left (d x +c \right )\right ) a}{64 d \sin \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/64/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(1+cos(d*x+c))^2*(64*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))+71*cos(d*x+c)^3*(-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))-64*(-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)^2-71*(-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)^2-64*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))+54*cos(d*x+c)^3-71*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))+64*(-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)-24*cos(d*x+c)^2+71*(-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))-14*cos(d*x+c))/sin(d*x+c)^6*a

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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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(a + a/cos(c + d*x))^(3/2),x)

[Out]

int(cot(c + d*x)^5*(a + a/cos(c + d*x))^(3/2), x)

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sympy [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))**(3/2),x)

[Out]

Timed out

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