3.205 \(\int \frac {\cot ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=355 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {9683 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {5587 \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{6144 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{128 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{256 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{1024 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{2048 a^4 d}-\frac {1491 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4096 a^3 d} \]

[Out]

2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+5587/6144*cot(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)/a^
4/d-1527/2048*cos(d*x+c)*cot(d*x+c)^3*sec(1/2*d*x+1/2*c)^2*(a+a*sec(d*x+c))^(3/2)/a^4/d-145/1024*cos(d*x+c)^2*
cot(d*x+c)^3*sec(1/2*d*x+1/2*c)^4*(a+a*sec(d*x+c))^(3/2)/a^4/d-9/256*cos(d*x+c)^3*cot(d*x+c)^3*sec(1/2*d*x+1/2
*c)^6*(a+a*sec(d*x+c))^(3/2)/a^4/d-1/128*cos(d*x+c)^4*cot(d*x+c)^3*sec(1/2*d*x+1/2*c)^8*(a+a*sec(d*x+c))^(3/2)
/a^4/d-9683/8192*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^(5/2)/d-1491/4096*cot
(d*x+c)*(a+a*sec(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 0.33, antiderivative size = 355, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {5587 \cot ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{6144 a^4 d}-\frac {1491 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4096 a^3 d}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {9683 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{128 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{256 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{1024 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \sec (c+d x)+a)^{3/2}}{2048 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - (9683*ArcTan[(Sqrt[a]*Tan[c + d*x])/
(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(4096*Sqrt[2]*a^(5/2)*d) - (1491*Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/(
4096*a^3*d) + (5587*Cot[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2))/(6144*a^4*d) - (1527*Cos[c + d*x]*Cot[c + d*x]^
3*Sec[(c + d*x)/2]^2*(a + a*Sec[c + d*x])^(3/2))/(2048*a^4*d) - (145*Cos[c + d*x]^2*Cot[c + d*x]^3*Sec[(c + d*
x)/2]^4*(a + a*Sec[c + d*x])^(3/2))/(1024*a^4*d) - (9*Cos[c + d*x]^3*Cot[c + d*x]^3*Sec[(c + d*x)/2]^6*(a + a*
Sec[c + d*x])^(3/2))/(256*a^4*d) - (Cos[c + d*x]^4*Cot[c + d*x]^3*Sec[(c + d*x)/2]^8*(a + a*Sec[c + d*x])^(3/2
))/(128*a^4*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 ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )^5} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^4 d}\\ &=-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}-\frac {\operatorname {Subst}\left (\int \frac {5 a-11 a^2 x^2}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )^4} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 a^5 d}\\ &=-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}-\frac {\operatorname {Subst}\left (\int \frac {-51 a^2-243 a^3 x^2}{x^4 \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 )}{96 a^6 d}\\ &=-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}-\frac {\operatorname {Subst}\left (\int \frac {-1509 a^3-3045 a^4 x^2}{x^4 \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 )}{768 a^7 d}\\ &=-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{2048 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}-\frac {\operatorname {Subst}\left (\int \frac {-16761 a^4-22905 a^5 x^2}{x^4 \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 )}{3072 a^8 d}\\ &=\frac {5587 \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{6144 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{2048 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}+\frac {\operatorname {Subst}\left (\int \frac {-13419 a^5-50283 a^6 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 )}{18432 a^8 d}\\ &=-\frac {1491 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4096 a^3 d}+\frac {5587 \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{6144 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{2048 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}-\frac {\operatorname {Subst}\left (\int \frac {60309 a^6-13419 a^7 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 )}{36864 a^8 d}\\ &=-\frac {1491 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4096 a^3 d}+\frac {5587 \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{6144 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{2048 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 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^2 d}+\frac {9683 \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 )}{4096 a^2 d}\\ &=\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {9683 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {1491 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4096 a^3 d}+\frac {5587 \cot ^3(c+d x) (a+a \sec (c+d x))^{3/2}}{6144 a^4 d}-\frac {1527 \cos (c+d x) \cot ^3(c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{2048 a^4 d}-\frac {145 \cos ^2(c+d x) \cot ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{1024 a^4 d}-\frac {9 \cos ^3(c+d x) \cot ^3(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{256 a^4 d}-\frac {\cos ^4(c+d x) \cot ^3(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+a \sec (c+d x))^{3/2}}{128 a^4 d}\\ \end {align*}

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Mathematica [C]  time = 24.16, size = 5646, normalized size = 15.90 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

Result too large to show

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fricas [A]  time = 0.71, size = 868, normalized size = 2.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/49152*(29049*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)*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) + 24576*(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*c
os(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*(14629*cos(d*x + c)^6 + 14233*cos(d*x + c)^5
- 14058*cos(d*x + c)^4 - 17426*cos(d*x + c)^3 + 2245*cos(d*x + c)^2 + 4473*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)*sin(d*x + c)), 1/24576*(29049*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*cos(d*x + c) + a
)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) + 24576*(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)*sin(d*x + c)/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*sin(d*x + c) + 2*(14629*cos
(d*x + c)^6 + 14233*cos(d*x + c)^5 - 14058*cos(d*x + c)^4 - 17426*cos(d*x + c)^3 + 2245*cos(d*x + c)^2 + 4473*
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)*sin(d*x + c))]

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giac [A]  time = 2.69, size = 331, normalized size = 0.93 \[ -\frac {3 \, {\left (2 \, {\left (4 \, {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {19 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {369 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {2989 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {512 \, \sqrt {2} {\left (12 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 21 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + 11 \, a^{2}\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}^{3} \sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{24576 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/24576*(3*(2*(4*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 19*sqrt(2)/(a^3*sg
n(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 + 369*sqrt(2)/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*ta
n(1/2*d*x + 1/2*c)^2 - 2989*sqrt(2)/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)
*tan(1/2*d*x + 1/2*c) + 512*sqrt(2)*(12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^
4 - 21*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + 11*a^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)^3*sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.56, size = 1066, normalized size = 3.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/24576/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))*(-1+cos(d*x+c))^4*(24576*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*cos(d*x+c)^5*sin(d*x+c)*2^(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))+73728*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)*2^(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))+29049*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x
+c)^5*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))+49152*cos(d*x
+c)^3*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))+87147*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*sin(d*x+c)*ln(-(-(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-49152*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)+58098*cos(d*x+c)^3*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))-29258*cos(d*x+c)^6-73728*cos(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))-58098*(
-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)-28466*cos(d*x+c)^5-24576*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)-87147*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))+28116*cos(d*x+c)^4-29049*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)+34852*cos(d*x+c)^3-4490*cos(d*x+c)^2-8946*cos(d*x+c))/sin(d*x
+c)^11/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)^4/(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 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^4}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cot(c + d*x)^4/(a + a/cos(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(cot(c + d*x)**4/(a*(sec(c + d*x) + 1))**(5/2), x)

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