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

Optimal. Leaf size=215 \[ -\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac {71 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {7 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{32 a^2 d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{16 a^2 d}-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{32 a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3887, 472, 579, 583, 522, 203} \[ \frac {7 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{32 a^2 d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}+\frac {71 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{32 \sqrt {2} a^{3/2} d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{16 a^2 d}-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a \sec (c+d x)+a}}{32 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + (71*ArcTan[(Sqrt[a]*Tan[c + d*x])/(
Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(32*Sqrt[2]*a^(3/2)*d) + (7*Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/(32*a^2
*d) - (13*Cos[c + d*x]*Cot[c + d*x]*Sec[(c + d*x)/2]^2*Sqrt[a + a*Sec[c + d*x]])/(32*a^2*d) - (Cos[c + d*x]^2*
Cot[c + d*x]*Sec[(c + d*x)/2]^4*Sqrt[a + a*Sec[c + d*x]])/(16*a^2*d)

Rule 203

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

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

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

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 3887

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

Rubi steps

\begin {align*} \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}\\ &=-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{16 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {3 a-5 a^2 x^2}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^3 d}\\ &=-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{16 a^2 d}-\frac {\operatorname {Subst}\left (\int \frac {-7 a^2-39 a^3 x^2}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^4 d}\\ &=\frac {7 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{16 a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {57 a^3-7 a^4 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{32 a^4 d}\\ &=\frac {7 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{16 a^2 d}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a d}-\frac {71 \operatorname {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{32 a d}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {71 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {7 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {13 \cos (c+d x) \cot (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {\cos ^2(c+d x) \cot (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a+a \sec (c+d x)}}{16 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 24.65, size = 5578, normalized size = 25.94 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.66, size = 603, normalized size = 2.80 \[ \left [-\frac {71 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \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 ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 64 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (d x + c\right )^{3} - 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 )}} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right ) + 1}\right ) \sin \left (d x + c\right ) + 4 \, {\left (27 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} - 7 \, \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 (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )}, -\frac {71 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 64 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \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 ) \sin \left (d x + c\right )}{2 \, a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right ) - a}\right ) \sin \left (d x + c\right ) + 2 \, {\left (27 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} - 7 \, \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 (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/128*(71*sqrt(2)*(cos(d*x + c)^2 + 2*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)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 +
2*cos(d*x + c) + 1))*sin(d*x + c) + 64*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log(-(8*a*cos(d*x + c)^3
- 4*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos
(d*x + c) + a)/(cos(d*x + c) + 1))*sin(d*x + c) + 4*(27*cos(d*x + c)^3 + 12*cos(d*x + c)^2 - 7*cos(d*x + c))*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c)))/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c)),
-1/64*(71*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) + 64*(cos(d*x + c)^2 + 2*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)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos
(d*x + c) - a))*sin(d*x + c) + 2*(27*cos(d*x + c)^3 + 12*cos(d*x + c)^2 - 7*cos(d*x + c))*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c)))/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c))]

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giac [A]  time = 1.49, size = 164, normalized size = 0.76 \[ \frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {17 \, \sqrt {2}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {16 \, \sqrt {2}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} \sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 -
1)) - 17*sqrt(2)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c) + 16*sqrt(2)/(((sqrt(-a)*tan(1/2*
d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.33, size = 542, normalized size = 2.52 \[ \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 )^{2} \left (64 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+128 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+71 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+64 \sqrt {2}\, \sin \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+142 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )+71 \sin \left (d x +c \right ) \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-54 \left (\cos ^{3}\left (d x +c \right )\right )-24 \left (\cos ^{2}\left (d x +c \right )\right )+14 \cos \left (d x +c \right )\right )}{64 d \sin \left (d x +c \right )^{5} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(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))^2*(64*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)+128*co
s(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+71*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+64*2^(1/2)*sin(d*x+c)*arctanh(1/2*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+142*cos(d*x+
c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d
*x+c)-1)/sin(d*x+c))+71*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x
+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-54*cos(d*x+c)^3-24*cos(d*x+c)^2+14*cos(d*x+c))/sin(d*x+c)^5/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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