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

Optimal. Leaf size=174 \[ -\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

((-1 - I)*Sqrt[a]*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 + (26*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (((2*I)/15)*Cot[c + d*x]^(
3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*Cot[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(5*d)

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Rubi [A]  time = 0.487873, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4241, 3561, 3598, 12, 3544, 205} \[ -\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 - I)*Sqrt[a]*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 + (26*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (((2*I)/15)*Cot[c + d*x]^(
3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*Cot[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])/(5*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 3561

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

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) \sqrt{a+i a \tan (c+d x)} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\left (\frac{i a}{2}-2 a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 \cot ^{\frac{5}{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{13 a^2}{4}-\frac{1}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=\frac{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 \cot ^{\frac{5}{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^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^3}\\ &=\frac{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\left (i \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{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\frac{\left (2 a^2 \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{(1+i) \sqrt{a} \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{26 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 i \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.19497, size = 149, normalized size = 0.86 \[ \frac{e^{-i (c+d x)} \sqrt{\cot (c+d x)} \left (30 e^{i (c+d x)}-40 e^{3 i (c+d x)}+34 e^{5 i (c+d x)}-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{15 d \left (-1+e^{2 i (c+d x)}\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((30*E^(I*(c + d*x)) - 40*E^((3*I)*(c + d*x)) + 34*E^((5*I)*(c + d*x)) - 15*(-1 + E^((2*I)*(c + d*x)))^(5/2)*A
rcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*E
^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^2)

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Maple [B]  time = 0.459, size = 1146, normalized size = 6.6 \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))^(1/2),x)

[Out]

-1/30/d*2^(1/2)*(15*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^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)+sin(d*x+c)-1))-2*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-30*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arc
tan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-30*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-30*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan((
(cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-30*((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)-15*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^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)-sin(d*x+c)+1))-15*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*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))+30*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arc
tan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+34*I*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+30*I*((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)+34*2^(1/2)*cos(d*x+
c)^3-26*I*2^(1/2)*sin(d*x+c)+30*((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)+30*((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)+15*((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)-sin(d*x+c)+1))-
32*2^(1/2)*cos(d*x+c)^2-28*2^(1/2)*cos(d*x+c)+26*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

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Maxima [B]  time = 2.41315, size = 1521, normalized size = 8.74 \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))^(1/2),x, algorithm="maxima")

[Out]

1/900*(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)*cos(3*d*x + 3*c)
- (1170*I + 1170)*cos(d*x + c) + (900*I - 900)*sin(3*d*x + 3*c) - (1170*I - 1170)*sin(d*x + c))*cos(3/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*cos(3*d*x + 3*c) + (1170*I - 1170)*cos(d*x + c)
+ (900*I + 900)*sin(3*d*x + 3*c) - (1170*I + 1170)*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)*cos(2*d*x + 2*c)^2 - (900*I - 900)*sin(2*d*x + 2*c)^2 + (1800*I - 1800
)*cos(2*d*x + 2*c) - 900*I + 900)*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*co
s(d*x + c)) + (-(450*I + 450)*cos(2*d*x + 2*c)^2 - (450*I + 450)*sin(2*d*x + 2*c)^2 + (900*I + 900)*cos(2*d*x
+ 2*c) - 450*I - 450)*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(si
n(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)*cos(5*d*x + 5*c) + (150*I + 150)*cos(3*d*x + 3*c) + (390*I + 390)*cos(d*x + c)
+ (900*I - 900)*sin(5*d*x + 5*c) + (150*I - 150)*sin(3*d*x + 3*c) + (390*I - 390)*sin(d*x + c))*cos(5/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((240*I + 240)*cos(d*x + c) + (240*I - 240)*sin(d*x + c))*cos(2*
d*x + 2*c)^2 + ((240*I + 240)*cos(d*x + c) + (240*I - 240)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (-(480*I + 480)*
cos(d*x + c) - (480*I - 480)*sin(d*x + c))*cos(2*d*x + 2*c) + (240*I + 240)*cos(d*x + c) + (240*I - 240)*sin(d
*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*cos(5*d*x + 5*c) - (150*I
- 150)*cos(3*d*x + 3*c) - (390*I - 390)*cos(d*x + c) + (900*I + 900)*sin(5*d*x + 5*c) + (150*I + 150)*sin(3*d*
x + 3*c) + (390*I + 390)*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((-(240*I -
240)*cos(d*x + c) + (240*I + 240)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(240*I - 240)*cos(d*x + c) + (240*I + 2
40)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((480*I - 480)*cos(d*x + c) - (480*I + 480)*sin(d*x + c))*cos(2*d*x + 2
*c) - (240*I - 240)*cos(d*x + c) + (240*I + 240)*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 = 1.40631, size = 1095, normalized size = 6.29 \begin{align*} \frac{4 \, \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 (17 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 20 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 15\right )} e^{\left (i \, d x + i \, c\right )} - 15 \,{\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{2 i \, a}{d^{2}}} \log \left ({\left (\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 (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )} + d \sqrt{\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 15 \,{\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{2 i \, a}{d^{2}}} \log \left ({\left (\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 (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )} - d \sqrt{\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{30 \,{\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))^(1/2),x, algorithm="fricas")

[Out]

1/30*(4*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))*
(17*e^(4*I*d*x + 4*I*c) - 20*e^(2*I*d*x + 2*I*c) + 15)*e^(I*d*x + I*c) - 15*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*
I*d*x + 2*I*c) + d)*sqrt(2*I*a/d^2)*log((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))*(e^(2*I*d*x + 2*I*c) - 1)*e^(I*d*x + I*c) + d*sqrt(2*I*a/d^2)*e^(2*I*d*x + 2*
I*c))*e^(-2*I*d*x - 2*I*c)) + 15*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(2*I*a/d^2)*log((sq
rt(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))*(e^(2*I*d*
x + 2*I*c) - 1)*e^(I*d*x + I*c) - d*sqrt(2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)))/(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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \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))^(1/2),x, algorithm="giac")

[Out]

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