3.537 \(\int \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=155 \[ -\frac {2 (3 B+i A) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {(1+i) \sqrt {a} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

(1+I)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)*cot(d*x+c)^(1/2)*tan(d*
x+c)^(1/2)/d-2/3*A*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*(I*A+3*B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c
))^(1/2)/d

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Rubi [A]  time = 0.48, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4241, 3598, 12, 3544, 205} \[ -\frac {2 (3 B+i A) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {(1+i) \sqrt {a} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + I)*Sqrt[a]*(I*A + B)*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*A + 3*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(3*d) - (2*A*Co
t[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(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 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 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \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)} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{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 {1}{2} a (i A+3 B)-a A \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a}\\ &=-\frac {2 (i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {3 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {2 (i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\left ((A-i B) \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+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (2 i a^2 (A-i B) \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} (i A+B) \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+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 2.59, size = 162, normalized size = 1.05 \[ -\frac {e^{-i (c+d x)} \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-3 i (A-i B) \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+4 i A e^{3 i (c+d x)}+6 B e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )}{3 d \left (-1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(((4*I)*A*E^((3*I)*(c + d*x)) + 6*B*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x))) - (3*I)*(A - I*B)*(-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[a +
I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x))))

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fricas [B]  time = 0.48, size = 425, normalized size = 2.74 \[ \frac {\sqrt {2} {\left ({\left (-16 i \, A - 24 \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + 24 \, B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {{\left (-8 i \, A^{2} - 16 \, A B + 8 i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {{\left (-8 i \, A^{2} - 16 \, A B + 8 i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {{\left (-8 i \, A^{2} - 16 \, A B + 8 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {{\left (-8 i \, A^{2} - 16 \, A B + 8 i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*((-16*I*A - 24*B)*e^(3*I*d*x + 3*I*c) + 24*B*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)) - 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((-8*I*A^2 - 1
6*A*B + 8*I*B^2)*a/d^2)*log(-(sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a/d^2)*sq
rt(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)) + 4*(A - I*B)*a*e^
(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) + 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*
a/d^2)*log((sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((-8*I*A^2 - 16*A*B + 8*I*B^2)*a/d^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)) - 4*(A - I*B)*a*e^(I*d*x + I*c))*e^(
-I*d*x - I*c)/(I*A + B)))/(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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

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))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 4.23, size = 2016, normalized size = 13.01 \[ \text {result too large to display} \]

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))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

-1/6/d*(4*A*2^(1/2)*cos(d*x+c)^2-2*A*2^(1/2)*cos(d*x+c)-6*B*sin(d*x+c)*2^(1/2)+3*I*A*((-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+co
s(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)^2-6*I*B*arctan(2^(1/2)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-3*I*B*((-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)^2-6*I*B*((-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+4*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)+6*I
*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+6*I*A*
((-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)^2-2*A*2^(1/
2)-6*I*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-6*I*A*((-1+co
s(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)-3*I*A*((-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))-2*I*A*sin(d*x+c)*2^(1/2)+6*I*B*arctan(2^(1/
2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+3*I*B*((-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))+6*I*B*((-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)+6*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+3*A*((-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)+c
os(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)^2+6*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)-1)*cos(d*x+c)^2+6*B*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)*cos(d*x+c)^2+6*B*((-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+3*B*((-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)^2-6*I*B*cos(d*x+c)^2*2^(1/2)+6*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-6*A*arctan(2^(1/2)*((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-3*A*((-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))-6*A*((-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)-6*B*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x
+c))^(1/2)-6*B*((-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)-3*B*((-
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))+6*I*B*2^(1/2))*(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)/(I*sin(d*x+c)+cos(d*x+c)-1)/co
s(d*x+c)^2*2^(1/2)

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maxima [B]  time = 1.84, size = 1145, normalized size = 7.39 \[ \text {result too large to display} \]

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))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/36*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((((36*I - 36)*A + (36*I + 36)*B
)*cos(3*d*x + 3*c) + ((12*I - 12)*A - (36*I + 36)*B)*cos(d*x + c) + (-(36*I + 36)*A + (36*I - 36)*B)*sin(3*d*x
+ 3*c) + (-(12*I + 12)*A - (36*I - 36)*B)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) -
1)) + (((36*I + 36)*A - (36*I - 36)*B)*cos(3*d*x + 3*c) + ((12*I + 12)*A + (36*I - 36)*B)*cos(d*x + c) + ((36*
I - 36)*A + (36*I + 36)*B)*sin(3*d*x + 3*c) + ((12*I - 12)*A - (36*I + 36)*B)*sin(d*x + c))*sin(3/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((((36*I + 36)*A - (36*I - 36)*B)*cos(2*d*x + 2*c)^2 + ((36*
I + 36)*A - (36*I - 36)*B)*sin(2*d*x + 2*c)^2 + (-(72*I + 72)*A + (72*I - 72)*B)*cos(2*d*x + 2*c) + (36*I + 36
)*A - (36*I - 36)*B)*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)) +
((-(18*I - 18)*A - (18*I + 18)*B)*cos(2*d*x + 2*c)^2 + (-(18*I - 18)*A - (18*I + 18)*B)*sin(2*d*x + 2*c)^2 +
((36*I - 36)*A + (36*I + 36)*B)*cos(2*d*x + 2*c) - (18*I - 18)*A - (18*I + 18)*B)*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) + (((((12*I - 12)*A + (36*I + 36)*B)*
cos(d*x + c) + (-(12*I + 12)*A + (36*I - 36)*B)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (((12*I - 12)*A + (36*I + 3
6)*B)*cos(d*x + c) + (-(12*I + 12)*A + (36*I - 36)*B)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((-(24*I - 24)*A - (7
2*I + 72)*B)*cos(d*x + c) + ((24*I + 24)*A - (72*I - 72)*B)*sin(d*x + c))*cos(2*d*x + 2*c) + ((12*I - 12)*A +
(36*I + 36)*B)*cos(d*x + c) + (-(12*I + 12)*A + (36*I - 36)*B)*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) - 1)) + ((((12*I + 12)*A - (36*I - 36)*B)*cos(d*x + c) + ((12*I - 12)*A + (36*I + 36)*B)*sin
(d*x + c))*cos(2*d*x + 2*c)^2 + (((12*I + 12)*A - (36*I - 36)*B)*cos(d*x + c) + ((12*I - 12)*A + (36*I + 36)*B
)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((-(24*I + 24)*A + (72*I - 72)*B)*cos(d*x + c) + (-(24*I - 24)*A - (72*I
+ 72)*B)*sin(d*x + c))*cos(2*d*x + 2*c) + ((12*I + 12)*A - (36*I - 36)*B)*cos(d*x + c) + ((12*I - 12)*A + (36*
I + 36)*B)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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