3.781 \(\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ -\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(-1/8+1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2
)/a^(5/2)/d+89/20*cot(d*x+c)^(3/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-361/60*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(
1/2)/a^3/d+707/60*I*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/5*cot(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c))^
(5/2)+7/10*cot(d*x+c)^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.81, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4241, 3559, 3596, 3598, 12, 3544, 205} \[ -\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/(a^(5/2)*d) + Cot[c + d*x]^(3/2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (7*Cot[c + d*x]^(3/2))/(
10*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (89*Cot[c + d*x]^(3/2))/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) + (((707*
I)/60)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d) - (361*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d
*x]])/(60*a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {13 a}{2}-4 i a \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {141 a^2}{4}-\frac {63}{2} i a^2 \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1083 a^3}{8}-\frac {267}{2} i a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {2121 i a^4}{16}-\frac {1083}{8} a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{45 a^7}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {45 a^5 \sqrt {a+i a \tan (c+d x)}}{32 \sqrt {\tan (c+d x)}} \, dx}{45 a^8}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}+\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {7 \cot ^{\frac {3}{2}}(c+d x)}{10 a d (a+i a \tan (c+d x))^{3/2}}+\frac {89 \cot ^{\frac {3}{2}}(c+d x)}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {707 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}-\frac {361 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 2.06, size = 199, normalized size = 0.77 \[ \frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\cot (c+d x)} \left (33 e^{2 i (c+d x)}+348 e^{4 i (c+d x)}-1527 e^{6 i (c+d x)}+983 e^{8 i (c+d x)}+15 e^{5 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+3\right )}{60 \sqrt {2} a^3 d \left (-1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/60)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(3 + 33*E^((2*I)*(c + d*x)) + 348*E^((4*I)*(c
+ d*x)) - 1527*E^((6*I)*(c + d*x)) + 983*E^((8*I)*(c + d*x)) + 15*E^((5*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)
))^(3/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*a^3*d*E^((6*I)*
(c + d*x))*(-1 + E^((2*I)*(c + d*x))))

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fricas [B]  time = 0.94, size = 427, normalized size = 1.66 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (983 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 1527 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 348 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 33 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} - 30 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} \log \left ({\left (\sqrt {2} {\left (8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 30 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} \log \left ({\left (\sqrt {2} {\left (-8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(
983*I*e^(8*I*d*x + 8*I*c) - 1527*I*e^(6*I*d*x + 6*I*c) + 348*I*e^(4*I*d*x + 4*I*c) + 33*I*e^(2*I*d*x + 2*I*c)
+ 3*I) - 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-1/8*I/(a^5*d^2))*log((sqrt(2)*(8*I*a
^3*d*e^(2*I*d*x + 2*I*c) - 8*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2
*I*d*x + 2*I*c) - 1))*sqrt(-1/8*I/(a^5*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 30*(a^3*d*e^(7*I*d*x
+ 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-1/8*I/(a^5*d^2))*log((sqrt(2)*(-8*I*a^3*d*e^(2*I*d*x + 2*I*c) + 8
*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-
1/8*I/(a^5*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5
*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(d*x + c)^(5/2)/(I*a*tan(d*x + c) + a)^(5/2), x)

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maple [B]  time = 2.64, size = 499, normalized size = 1.93 \[ \frac {\left (-\frac {1}{120}-\frac {i}{120}\right ) \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \left (48 i \left (\cos ^{7}\left (d x +c \right )\right )+267 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-48 \left (\cos ^{7}\left (d x +c \right )\right )+48 \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )-707 i \sin \left (d x +c \right )+48 i \left (\cos ^{6}\left (d x +c \right )\right ) \sin \left (d x +c \right )+48 i \left (\cos ^{5}\left (d x +c \right )\right )+15 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}-48 \left (\cos ^{5}\left (d x +c \right )\right )+72 \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-15 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+72 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-361 i \cos \left (d x +c \right )-225 \left (\cos ^{3}\left (d x +c \right )\right )+267 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+15 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+225 i \left (\cos ^{3}\left (d x +c \right )\right )+15 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}+361 \cos \left (d x +c \right )-707 \sin \left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/120-1/120*I)/d*(cos(d*x+c)/sin(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)*(48
*I*cos(d*x+c)^7+267*I*cos(d*x+c)^2*sin(d*x+c)-48*cos(d*x+c)^7+48*cos(d*x+c)^6*sin(d*x+c)-707*I*sin(d*x+c)+48*I
*cos(d*x+c)^6*sin(d*x+c)+48*I*cos(d*x+c)^5+15*I*cos(d*x+c)*sin(d*x+c)*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-48*cos(d*x+c)^5+72*cos(d*x+c)^4*sin(d*x+c)-1
5*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)
^2*2^(1/2)+72*I*cos(d*x+c)^4*sin(d*x+c)-361*I*cos(d*x+c)-225*cos(d*x+c)^3+267*cos(d*x+c)^2*sin(d*x+c)+15*arcta
n((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+225*I*cos
(d*x+c)^3+15*I*sin(d*x+c)*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*2^(1/2))+361*cos(d*x+c)-707*sin(d*x+c))/cos(d*x+c)^2/a^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(5/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cot(c + d*x)^(5/2)/(a + a*tan(c + d*x)*1i)^(5/2), 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)**(5/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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