\(\int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) [512]

Optimal result

Integrand size = 36, antiderivative size = 130 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

4*(-1)^(1/4)*a^2*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2/15*a^2*(5*A-7*I*B)/d/cot(d*x+c)^(3/2)+2/5*I*
B*(I*a^2+a^2*cot(d*x+c))/d/cot(d*x+c)^(5/2)+4*a^2*(I*A+B)/d/cot(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3662, 3674, 3672, 3610, 3614, 214} \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (B+i A)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a^2*(5*A - (7*I)*B))/(15*d*Cot[c +
d*x]^(3/2)) + (4*a^2*(I*A + B))/(d*Sqrt[Cot[c + d*x]]) + (((2*I)/5)*B*(I*a^2 + a^2*Cot[c + d*x]))/(d*Cot[c + d
*x]^(5/2))

Rule 214

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

Rule 3610

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

Rule 3614

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

Rule 3662

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3672

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

Rule 3674

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((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 - 1)*(c
+ d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
- 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(i a+a \cot (c+d x))^2 (B+A \cot (c+d x))}{\cot ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(i a+a \cot (c+d x)) \left (\frac {1}{2} a (5 i A+7 B)+\frac {1}{2} a (5 A-3 i B) \cot (c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {5 a^2 (i A+B)+5 a^2 (A-i B) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {5 a^2 (A-i B)-5 a^2 (i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx \\ & = -\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {\left (20 a^4 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{-5 a^2 (A-i B)-5 a^2 (i A+B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = \frac {4 \sqrt [4]{-1} a^2 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.87 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\frac {2 a^2 \left (30 (i A+B)+\frac {30 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{\sqrt {\tan (c+d x)}}-5 (A-2 i B) \tan (c+d x)-3 B \tan ^2(c+d x)\right )}{15 d \sqrt {\cot (c+d x)}} \]

[In]

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

[Out]

(2*a^2*(30*(I*A + B) + (30*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/Sqrt[Tan[c + d*x]] - 5*
(A - (2*I)*B)*Tan[c + d*x] - 3*B*Tan[c + d*x]^2))/(15*d*Sqrt[Cot[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (109 ) = 218\).

Time = 0.38 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.87

method	result	size
derivativedivides	\(-\frac {a^{2} \left (-\frac {2 \left (2 i A +2 B \right )}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 \left (2 i B -A \right )}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B}{5 \cot \left (d x +c \right )^{\frac {5}{2}}}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(243\)
default	\(-\frac {a^{2} \left (-\frac {2 \left (2 i A +2 B \right )}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 \left (2 i B -A \right )}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B}{5 \cot \left (d x +c \right )^{\frac {5}{2}}}+\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-2 i A -2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(243\)

[In]

int((a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-a^2/d*(-2*(2*I*A+2*B)/cot(d*x+c)^(1/2)-2/3*(2*I*B-A)/cot(d*x+c)^(3/2)+2/5*B/cot(d*x+c)^(5/2)+1/4*(2*A-2*I*B)*
2^(1/2)*(ln((1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2
)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2)))+1/4*(-2*I*A-2*B)*2^(1/2)*(ln((1+cot(d*x+c)-2^(1/2)*
cot(d*x+c)^(1/2))/(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^
(1/2)*cot(d*x+c)^(1/2))))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (104) = 208\).

Time = 0.26 (sec) , antiderivative size = 505, normalized size of antiderivative = 3.88 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, {\left ({\left (35 \, A - 43 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (25 \, A - 11 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (35 \, A - 31 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (25 \, A - 23 i \, B\right )} a^{2}\right )} \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 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

-1/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) + d)*log(2*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(I*d*e^
(2*I*d*x + 2*I*c) - I*d)*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)/((-
I*A - B)*a^2)) - 15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) +
3*d*e^(2*I*d*x + 2*I*c) + d)*log(2*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^
2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*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)/((-I*A - B)*a^2)) - 2*((35*A - 43*I*B)*a^2*e^(6*I*d*x + 6*I*c) + (25*A - 11*I*B)*a^2*e^(4*I*d*x + 4*I*
c) - (35*A - 31*I*B)*a^2*e^(2*I*d*x + 2*I*c) - (25*A - 23*I*B)*a^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d
*x + 2*I*c) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=- a^{2} \left (\int \left (- \frac {A}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {A \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {B \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {B \tan ^{3}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {2 i A \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

-a**2*(Integral(-A/sqrt(cot(c + d*x)), x) + Integral(A*tan(c + d*x)**2/sqrt(cot(c + d*x)), x) + Integral(-B*ta
n(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(B*tan(c + d*x)**3/sqrt(cot(c + d*x)), x) + Integral(-2*I*A*tan(c
+ d*x)/sqrt(cot(c + d*x)), x) + Integral(-2*I*B*tan(c + d*x)**2/sqrt(cot(c + d*x)), x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.53 \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {4 \, {\left (3 \, B a^{2} + \frac {5 \, {\left (A - 2 i \, B\right )} a^{2}}{\tan \left (d x + c\right )} - \frac {30 \, {\left (i \, A + B\right )} a^{2}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} + 15 \, {\left (2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2}}{30 \, d} \]

[In]

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

[Out]

-1/30*(4*(3*B*a^2 + 5*(A - 2*I*B)*a^2/tan(d*x + c) - 30*(I*A + B)*a^2/tan(d*x + c)^2)*tan(d*x + c)^(5/2) + 15*
(2*sqrt(2)*(-(I - 1)*A - (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(-(I - 1)
*A - (I + 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(s
qrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)/sqrt(tan(d*x +
c)) + 1/tan(d*x + c) + 1))*a^2)/d

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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