3.763 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=176 \[ -\frac {(4+4 i) a^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

(-4-4*I)*a^(5/2)*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)/d+4*a^2*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*I*a*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d-
2/5*cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(5/2)/d

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Rubi [A]  time = 0.35, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4241, 3548, 3545, 3544, 205} \[ \frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {(4+4 i) a^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-4 - 4*I)*a^(5/2)*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]])/d + (4*a^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (((2*I)/3)*a*Cot[c + d*x]^
(3/2)*(a + I*a*Tan[c + d*x])^(3/2))/d - (2*Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2))/(5*d)

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 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), 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] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3548

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*m*(c^2 + d^2)), x] + Dist[a/(a*c - b*d), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*
d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -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 \cot ^{\frac {7}{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 {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\left (2 a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\left (4 i a^2 \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\\ &=\frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {\left (8 a^4 \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 )}{d}\\ &=-\frac {(4+4 i) a^{5/2} \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)}}{d}+\frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.94, size = 161, normalized size = 0.91 \[ \frac {16 a^2 e^{i (c+d x)} \cos ^2(c+d x) \sqrt {\cot (c+d x)} \left (e^{i (c+d x)} \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{15 d \left (-1+e^{4 i (c+d x)}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(16*a^2*E^(I*(c + d*x))*(E^(I*(c + d*x))*(15 - 35*E^((2*I)*(c + d*x)) + 26*E^((4*I)*(c + d*x))) - 15*(-1 + E^(
(2*I)*(c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Cos[c + d*x]^2*Sqrt[Cot[c + d
*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*(-1 + E^((4*I)*(c + d*x)))^2)

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fricas [B]  time = 1.12, size = 409, normalized size = 2.32 \[ \frac {16 \, \sqrt {2} {\left (26 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 15 \, \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 15 \, \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/60*(16*sqrt(2)*(26*a^2*e^(5*I*d*x + 5*I*c) - 35*a^2*e^(3*I*d*x + 3*I*c) + 15*a^2*e^(I*d*x + I*c))*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)) - 15*sqrt(128*I*a^5/d^2)*(
d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(128*I*a^
5/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/a^2) + 15*sqrt(128*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2
*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(128*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/a^2))
/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(7/2), x)

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maple [B]  time = 1.87, size = 1149, normalized size = 6.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/15/d*(60*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+
c)^2*sin(d*x+c)+30*I*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1
/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*cos(d*x+c)^2*sin(d*x+c)+60*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-60*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)+1)*sin(d*x+c)-60*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)+1)*cos(d*x+c)^2*sin(d*x+c)-60*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)-30*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x
+c)-sin(d*x+c)+1))*cos(d*x+c)^2*sin(d*x+c)-38*I*2^(1/2)*sin(d*x+c)-60*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arc
tan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*sin(d*x+c)-11*I*cos(d*x+c)*sin(d*x+c)*2^(1/2)+52*I*cos(d*x+c
)^2*sin(d*x+c)*2^(1/2)+52*2^(1/2)*cos(d*x+c)^3-30*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+60*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)+1)*sin(d*x+c)+60*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)-1)*sin(d*x+c)+30*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d
*x+c)+1))*sin(d*x+c)-41*cos(d*x+c)^2*2^(1/2)-49*cos(d*x+c)*2^(1/2)+38*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(7/2)*(
a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3*2^(1/2)*a^2

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maxima [B]  time = 0.81, size = 1261, normalized size = 7.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((900*I + 900)*a^2*cos(3*d*x +
3*c) - (930*I + 930)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(3*d*x + 3*c) - (930*I - 930)*a^2*sin(d*x + c))*c
os(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*a^2*cos(3*d*x + 3*c) + (930*I - 930)
*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(3*d*x + 3*c) - (930*I + 930)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((-(900*I - 900)*a^2*cos(2*d*x + 2*c)^2 - (900*I - 900)*a^2*sin
(2*d*x + 2*c)^2 + (1800*I - 1800)*a^2*cos(2*d*x + 2*c) - (900*I - 900)*a^2)*arctan2(2*(cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*
sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + (-(450*I + 450)*a^2*cos(2*d*x + 2*c)^2 - (450*I + 450)
*a^2*sin(2*d*x + 2*c)^2 + (900*I + 900)*a^2*cos(2*d*x + 2*c) - (450*I + 450)*a^2)*log(4*cos(d*x + c)^2 + 4*sin
(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))))*(cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (((900*I + 900)*a^2*cos(5*d*x + 5*c
) - (750*I + 750)*a^2*cos(3*d*x + 3*c) + (210*I + 210)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(5*d*x + 5*c) -
(750*I - 750)*a^2*sin(3*d*x + 3*c) + (210*I - 210)*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) - 1)) + ((-(240*I + 240)*a^2*cos(d*x + c) - (240*I - 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (2
40*I + 240)*a^2*cos(d*x + c) + (-(240*I + 240)*a^2*cos(d*x + c) - (240*I - 240)*a^2*sin(d*x + c))*sin(2*d*x +
2*c)^2 - (240*I - 240)*a^2*sin(d*x + c) + ((480*I + 480)*a^2*cos(d*x + c) + (480*I - 480)*a^2*sin(d*x + c))*co
s(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*a^2*cos(5*d*x + 5*c
) + (750*I - 750)*a^2*cos(3*d*x + 3*c) - (210*I - 210)*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(5*d*x + 5*c) -
(750*I + 750)*a^2*sin(3*d*x + 3*c) + (210*I + 210)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) - 1)) + (((240*I - 240)*a^2*cos(d*x + c) - (240*I + 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (24
0*I - 240)*a^2*cos(d*x + c) + ((240*I - 240)*a^2*cos(d*x + c) - (240*I + 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*
c)^2 - (240*I + 240)*a^2*sin(d*x + c) + (-(480*I - 480)*a^2*cos(d*x + c) + (480*I + 480)*a^2*sin(d*x + c))*cos
(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cot(c + d*x)^(7/2)*(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)**(7/2)*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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