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

Optimal. Leaf size=130 \[ -\frac{2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (B+i A)}{d \sqrt{\cot (c+d x)}}+\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.371507, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3581, 3593, 3591, 3529, 3533, 208} \[ -\frac{2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (B+i A)}{d \sqrt{\cot (c+d x)}}+\frac{4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[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 3581

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 3593

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] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3591

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 3529

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 3533

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 208

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

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt{\cot (c+d x)}} \, dx &=\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 ) \operatorname{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) \tanh ^{-1}\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]  time = 7.07377, size = 133, normalized size = 1.02 \[ \frac{a^2 \left (\sec ^2(c+d x) (-5 (A-2 i B) \sin (2 (c+d x))+(33 B+30 i A) \cos (2 (c+d x))+30 i A+27 B)-\frac{60 i (A-i B) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}\right )}{15 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Sec[c + d*x]^2*((30*I)*A + 27*B + ((30*I)*A + 33*B)*Cos[2*(c + d*x)] - 5*(A - (2*I)*B)*Sin[2*(c + d*x)])
- ((60*I)*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/Sqrt[I*Tan[c + d*x]]
))/(15*d*Sqrt[Cot[c + d*x]])

________________________________________________________________________________________

Maple [C]  time = 0.481, size = 971, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^2/d*2^(1/2)*(cos(d*x+c)-1)*(30*I*A*2^(1/2)*cos(d*x+c)^3-10*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)-30*I*A*2^(
1/2)*cos(d*x+c)^2+30*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/
2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+30*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^
(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)-30*B*sin(d*x+c)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x
+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-30*I*A*sin(d*x+c)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d
*x+c))^(1/2),1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(co
s(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+10*I*B*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2-30*I*B*EllipticPi
((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-5
*A*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2+30*I*A*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1
/2*2^(1/2))*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+33*B*2^(1/2)*cos(d*x+c)^3+5*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)
-33*B*cos(d*x+c)^2*2^(1/2)-3*B*cos(d*x+c)*2^(1/2)+3*B*2^(1/2))*(cos(d*x+c)+1)^2/cos(d*x+c)^2/sin(d*x+c)^4/(cos
(d*x+c)/sin(d*x+c))^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.55116, size = 270, normalized size = 2.08 \begin{align*} -\frac{4 \,{\left (3 \, B a^{2} + \frac{{\left (5 \, A - 10 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 - 10*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(
sqrt(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

________________________________________________________________________________________

Fricas [B]  time = 1.63894, size = 1361, normalized size = 10.47 \begin{align*} -\frac{15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 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{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 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{{\left (4 \,{\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (16 i \, A^{2} + 32 \, A B - 16 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 (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \,{\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}}}{60 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*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(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*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)/((2*I*A + 2*B)*a^2)) - 15*sqrt((16*I*A^2 + 32*A*B - 16*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(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 +
32*A*B - 16*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)/((2*I*A + 2*B)*a^2)) - 8*((35*A - 43*I*B)*a^2*e^(6*I*d*x + 6*I*c) + (25*A - 1
1*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{A}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int - \frac{A \tan ^{2}{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{B \tan{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int - \frac{B \tan ^{3}{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{2 i A \tan{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{2 i B \tan ^{2}{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*tan(
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))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)