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

Optimal. Leaf size=200 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}+\frac {51}{32 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {5 a}{28 d (a \sec (c+d x)+a)^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{7/2}}+\frac {3}{40 d (a \sec (c+d x)+a)^{5/2}}+\frac {19}{48 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d-5/28*a/d/(a+a*sec(d*x+c))^(7/2)+1/2*a/d/(1-sec(d*x+c))/(a
+a*sec(d*x+c))^(7/2)+3/40/d/(a+a*sec(d*x+c))^(5/2)+19/48/a/d/(a+a*sec(d*x+c))^(3/2)+13/64*arctanh(1/2*(a+a*sec
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+51/32/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3880, 103, 152, 156, 63, 207} \[ \frac {51}{32 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}-\frac {5 a}{28 d (a \sec (c+d x)+a)^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{7/2}}+\frac {3}{40 d (a \sec (c+d x)+a)^{5/2}}+\frac {19}{48 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(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) + (13*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(32*Sqrt[2]*a^(5/2)*d) - (5*a)/(28*d*(a + a*Sec[c + d*x])^(7/2)) + a/(2*d*(1 - Sec[c + d*x])*(a + a*Se
c[c + d*x])^(7/2)) + 3/(40*d*(a + a*Sec[c + d*x])^(5/2)) + 19/(48*a*d*(a + a*Sec[c + d*x])^(3/2)) + 51/(32*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 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 ^3(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac {a^4 \operatorname {Subst}\left (\int \frac {1}{x (-a+a x)^2 (a+a x)^{9/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}-\frac {a \operatorname {Subst}\left (\int \frac {2 a^2+\frac {9 a^2 x}{2}}{x (-a+a x) (a+a x)^{9/2}} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {\operatorname {Subst}\left (\int \frac {-14 a^4-\frac {35 a^4 x}{4}}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{14 a^2 d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}-\frac {\operatorname {Subst}\left (\int \frac {70 a^6-\frac {105 a^6 x}{8}}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{70 a^5 d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}+\frac {19}{48 a d (a+a \sec (c+d x))^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {-210 a^8+\frac {1995 a^8 x}{16}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{210 a^8 d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}+\frac {19}{48 a d (a+a \sec (c+d x))^{3/2}}+\frac {51}{32 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {210 a^{10}-\frac {5355 a^{10} x}{32}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{210 a^{11} d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}+\frac {19}{48 a d (a+a \sec (c+d x))^{3/2}}+\frac {51}{32 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 {13 \operatorname {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{64 a d}\\ &=-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}+\frac {19}{48 a d (a+a \sec (c+d x))^{3/2}}+\frac {51}{32 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 {13 \operatorname {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{32 a^2 d}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {13 \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}-\frac {5 a}{28 d (a+a \sec (c+d x))^{7/2}}+\frac {a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{7/2}}+\frac {3}{40 d (a+a \sec (c+d x))^{5/2}}+\frac {19}{48 a d (a+a \sec (c+d x))^{3/2}}+\frac {51}{32 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 90, normalized size = 0.45 \[ \frac {a \left (-13 (\sec (c+d x)-1) \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};\frac {1}{2} (\sec (c+d x)+1)\right )+8 (\sec (c+d x)-1) \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};\sec (c+d x)+1\right )-14\right )}{28 d (\sec (c+d x)-1) (a (\sec (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-14 - 13*Hypergeometric2F1[-7/2, 1, -5/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 8*Hypergeometric2F1[
-7/2, 1, -5/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])))/(28*d*(-1 + Sec[c + d*x])*(a*(1 + Sec[c + d*x]))^(7/2))

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fricas [B]  time = 0.61, size = 748, normalized size = 3.74 \[ \left [\frac {1365 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 ) + 6720 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 (8017 \, \cos \left (d x + c\right )^{5} + 12640 \, \cos \left (d x + c\right )^{4} - 1582 \, \cos \left (d x + c\right )^{3} - 12040 \, \cos \left (d x + c\right )^{2} - 5355 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{13440 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, 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 {1365 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 ) - 6720 \, {\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 (8017 \, \cos \left (d x + c\right )^{5} + 12640 \, \cos \left (d x + c\right )^{4} - 1582 \, \cos \left (d x + c\right )^{3} - 12040 \, \cos \left (d x + c\right )^{2} - 5355 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{6720 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, 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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/13440*(1365*sqrt(2)*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*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)) + 6720*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*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*(8017*cos(d*x + c)^5 + 12640*cos(d*x + c)^4 - 1582*cos(d
*x + c)^3 - 12040*cos(d*x + c)^2 - 5355*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^3*d*cos(d*x
+ c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*a^3*d*cos(d*x + c)^2 - 3*a^3*d*cos(d*x + c) - a^3
*d), -1/6720*(1365*sqrt(2)*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*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)) - 6720*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*cos(d*x + c)^3 - 2*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*
(8017*cos(d*x + c)^5 + 12640*cos(d*x + c)^4 - 1582*cos(d*x + c)^3 - 12040*cos(d*x + c)^2 - 5355*cos(d*x + c))*
sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)
^3 - 2*a^3*d*cos(d*x + c)^2 - 3*a^3*d*cos(d*x + c) - a^3*d)]

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giac [B]  time = 2.02, size = 323, normalized size = 1.62 \[ \frac {\frac {1365 \, \sqrt {2} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {13440 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {105 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {2 \, \sqrt {2} {\left (15 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{36} - 84 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{37} - 385 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{38} - 2730 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{39}\right )}}{a^{42} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6720*(1365*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)) - 13440*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)) + 105*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*ta
n(1/2*d*x + 1/2*c)^2) + 2*sqrt(2)*(15*(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^3
6 - 84*(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^37 - 385*(-a*tan(1/2*d*x + 1/2*c
)^2 + a)^(3/2)*a^38 - 2730*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^39)/(a^42*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.57, size = 744, normalized size = 3.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6720/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))*(-1+cos(d*x+c))^4*(6720*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))+1365*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))+20160*(-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))+4095*(-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))+13440*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))+16034*cos(d*x+
c)^5+2730*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))-134
40*(-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+25280*cos(d*x+c)^4-2730*(-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)^2-20160*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))-3164*cos(d*x+c)^3-4095*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))-6720*(-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)-24080*cos(d*x+c)^2-1365*(-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))-10710*cos(d*x+c))/sin(d*x+c)^10/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)^3/(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 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^3}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2), x)

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