3.757 \(\int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=218 \[ -\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{(2+2 i) a^{3/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d} \]

[Out]

((-2 - 2*I)*a^(3/2)*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 - (((2*I)/5)*a^2*Cot[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a^2*Cot[c + d
*x]^(5/2))/(5*d*Sqrt[a + I*a*Tan[c + d*x]]) + (12*a*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (((
4*I)/5)*a*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.652138, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4241, 3553, 3596, 3598, 12, 3544, 205} \[ -\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{(2+2 i) a^{3/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-2 - 2*I)*a^(3/2)*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 - (((2*I)/5)*a^2*Cot[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a^2*Cot[c + d
*x]^(5/2))/(5*d*Sqrt[a + I*a*Tan[c + d*x]]) + (12*a*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (((
4*I)/5)*a*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

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]

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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] && GtQ[m, 1] && LtQ[
n, -1] && (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 12

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

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 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]

Rubi steps

\begin{align*} \int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{1}{5} \left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{-\frac{9 i a^2}{2}+\frac{11}{2} a^2 \tan (c+d x)}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\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 (-3 i a^3+2 a^3 \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a^2}\\ &=-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{9 a^4}{2}+3 i a^4 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^3}\\ &=-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{\left (8 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{15 i a^5 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\left (2 i a \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{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{\left (4 a^3 \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{(2+2 i) a^{3/2} \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{2 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{a+i a \tan (c+d x)}}+\frac{12 a \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{4 i a \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.64099, size = 161, normalized size = 0.74 \[ \frac{4 a \cos (c+d x) \sqrt{\cot (c+d x)} \left (e^{i (c+d x)} \left (-10 e^{2 i (c+d x)}+9 e^{4 i (c+d x)}+5\right )-5 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{5 d \left (-1+e^{2 i (c+d x)}\right )^2 \left (1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(4*a*(E^(I*(c + d*x))*(5 - 10*E^((2*I)*(c + d*x)) + 9*E^((4*I)*(c + d*x))) - 5*(-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]*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c
+ d*x]])/(5*d*(-1 + E^((2*I)*(c + d*x)))^2*(1 + E^((2*I)*(c + d*x))))

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Maple [B]  time = 0.419, size = 1147, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

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))^(3/2),x)

[Out]

-1/5/d*2^(1/2)*a*(5*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x
+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*c
os(d*x+c)^2*sin(d*x+c)-5*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*si
n(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-
1))*sin(d*x+c)-2*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-10*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)
/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)-5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(-(((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2
^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-10*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arcta
n(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-10*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ar
ctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-10*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1
)/sin(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)+10*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)+9*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+10*I*((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)+9*2^(1/2)*cos(d*x+c)^3
-6*I*2^(1/2)*sin(d*x+c)+5*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*
2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-s
in(d*x+c)+1))+10*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)
+1)+10*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-7*2^(1
/2)*cos(d*x+c)^2-8*2^(1/2)*cos(d*x+c)+6*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(7/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/co
s(d*x+c))^(1/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3

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Maxima [B]  time = 2.53381, size = 1584, normalized size = 7.27 \begin{align*} \text{result too large to display} \end{align*}

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))^(3/2),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)*(((450*I + 450)*a*cos(3*d*x + 3*
c) - (480*I + 480)*a*cos(d*x + c) + (450*I - 450)*a*sin(3*d*x + 3*c) - (480*I - 480)*a*sin(d*x + c))*cos(3/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(450*I - 450)*a*cos(3*d*x + 3*c) + (480*I - 480)*a*cos(d*x
+ c) + (450*I + 450)*a*sin(3*d*x + 3*c) - (480*I + 480)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) - 1)))*sqrt(a) + ((-(450*I - 450)*a*cos(2*d*x + 2*c)^2 - (450*I - 450)*a*sin(2*d*x + 2*c)^2 + (9
00*I - 900)*a*cos(2*d*x + 2*c) - (450*I - 450)*a)*arctan2(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)*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)) + (-(225*I + 225)*a*cos(2*d*x + 2*c)^2 - (225*I + 225)*a*sin(2*d*x + 2*c)^2 + (450
*I + 450)*a*cos(2*d*x + 2*c) - (225*I + 225)*a)*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) + (((450*I + 450)*a*cos(5*d*x + 5*c) - (150*I + 150)*a*cos(3*d*x + 3*c)
+ (60*I + 60)*a*cos(d*x + c) + (450*I - 450)*a*sin(5*d*x + 5*c) - (150*I - 150)*a*sin(3*d*x + 3*c) + (60*I -
60)*a*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((-(90*I + 90)*a*cos(d*x + c) -
(90*I - 90)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(90*I + 90)*a*cos(d*x + c) - (90*I - 90)*a*sin(d*x + c))*s
in(2*d*x + 2*c)^2 + ((180*I + 180)*a*cos(d*x + c) + (180*I - 180)*a*sin(d*x + c))*cos(2*d*x + 2*c) - (90*I + 9
0)*a*cos(d*x + c) - (90*I - 90)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(
450*I - 450)*a*cos(5*d*x + 5*c) + (150*I - 150)*a*cos(3*d*x + 3*c) - (60*I - 60)*a*cos(d*x + c) + (450*I + 450
)*a*sin(5*d*x + 5*c) - (150*I + 150)*a*sin(3*d*x + 3*c) + (60*I + 60)*a*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) - 1)) + (((90*I - 90)*a*cos(d*x + c) - (90*I + 90)*a*sin(d*x + c))*cos(2*d*x + 2*c)
^2 + ((90*I - 90)*a*cos(d*x + c) - (90*I + 90)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(180*I - 180)*a*cos(d*x
+ c) + (180*I + 180)*a*sin(d*x + c))*cos(2*d*x + 2*c) + (90*I - 90)*a*cos(d*x + c) - (90*I + 90)*a*sin(d*x + 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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Fricas [B]  time = 1.43012, size = 1135, normalized size = 5.21 \begin{align*} \frac{4 \, \sqrt{2}{\left (9 \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 10 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 5 \, a\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}} e^{\left (i \, d x + i \, c\right )} - 5 \,{\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 )} \sqrt{\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} - a\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}} e^{\left (i \, d x + i \, c\right )} + \sqrt{\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right ) + 5 \,{\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 )} \sqrt{\frac{8 i \, a^{3}}{d^{2}}} \log \left (\frac{{\left (2 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} - a\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}} e^{\left (i \, d x + i \, c\right )} - \sqrt{\frac{8 i \, a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a}\right )}{10 \,{\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 )}} \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/10*(4*sqrt(2)*(9*a*e^(4*I*d*x + 4*I*c) - 10*a*e^(2*I*d*x + 2*I*c) + 5*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((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) - 5*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^
(2*I*d*x + 2*I*c) + d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) - a)*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) + sqrt(8*I*a^3/d^2)*d*
e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) + 5*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(8*
I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) - a)*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) - sqrt(8*I*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d
*x - 2*I*c)/a))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(7/2), x)