\(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [517]

Optimal result

Integrand size = 36, antiderivative size = 148 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {16 a^3 (5 A-6 i B)}{15 d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

8*(-1)^(1/4)*a^3*(I*A+B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d+2/5*I*a*B*(I*a+a*cot(d*x+c))^2/d/cot(d*x+c)^(5
/2)-2/15*(5*A-9*I*B)*(I*a^3+a^3*cot(d*x+c))/d/cot(d*x+c)^(3/2)-16/15*a^3*(5*A-6*I*B)/d/cot(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3662, 3674, 3672, 3614, 214} \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (B+i A) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 (5 A-9 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {16 a^3 (5 A-6 i B)}{15 d \sqrt {\cot (c+d x)}}+\frac {2 i a B (a \cot (c+d x)+i a)^2}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[In]

Int[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B))/(15*d*Sqrt[Co
t[c + d*x]]) + (((2*I)/5)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c + d*x]^(5/2)) - (2*(5*A - (9*I)*B)*(I*a^3 + a
^3*Cot[c + d*x]))/(15*d*Cot[c + d*x]^(3/2))

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 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))^3 (B+A \cot (c+d x))}{\cot ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(i a+a \cot (c+d x))^2 \left (\frac {1}{2} a (5 i A+9 B)+\frac {1}{2} a (5 A-i B) \cot (c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {(i a+a \cot (c+d x)) \left (2 a^2 (5 i A+6 B)+a^2 (5 A-3 i B) \cot (c+d x)\right )}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \\ & = -\frac {16 a^3 (5 A-6 i B)}{15 d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4}{15} \int \frac {15 a^3 (i A+B)+15 a^3 (A-i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx \\ & = -\frac {16 a^3 (5 A-6 i B)}{15 d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (120 a^6 (i A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-15 a^3 (i A+B)+15 a^3 (A-i B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = \frac {8 \sqrt [4]{-1} a^3 (i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {16 a^3 (5 A-6 i B)}{15 d \sqrt {\cot (c+d x)}}+\frac {2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.94 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 i a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-60 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} \left (-45 i A-60 B+5 (A-3 i B) \tan (c+d x)+3 B \tan ^2(c+d x)\right )\right )}{15 d} \]

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(((-2*I)/15)*a^3*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-60*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c
+ d*x]]] + Sqrt[Tan[c + d*x]]*((-45*I)*A - 60*B + 5*(A - (3*I)*B)*Tan[c + d*x] + 3*B*Tan[c + d*x]^2)))/d

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.64

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {-\frac {2 i A}{3}-2 B}{\cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {8 i B -6 A}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{5 \cot \left (d x +c \right )^{\frac {5}{2}}}\right )}{d}\)	\(243\)
default	\(\frac {a^{3} \left (-\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}+\frac {-\frac {2 i A}{3}-2 B}{\cot \left (d x +c \right )^{\frac {3}{2}}}+\frac {8 i B -6 A}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{5 \cot \left (d x +c \right )^{\frac {5}{2}}}\right )}{d}\)	\(243\)

[In]

int(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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. 502 vs. \(2 (118) = 236\).

Time = 0.27 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.39 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (15 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{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^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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^{3}}\right ) - 15 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{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^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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^{3}}\right ) - 2 \, {\left ({\left (-25 i \, A - 39 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (-10 i \, A - 9 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (25 i \, A + 33 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (5 i \, A + 6 \, B\right )} a^{3}\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 )}}{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(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

2/15*(15*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/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^3*e^(2*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(2*I*
d*x + 2*I*c) - 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^3)) - 15*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/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^3*e^(2*I*d*x + 2*I*c) - sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(
2*I*d*x + 2*I*c) - 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^3)) - 2*((-25*I*A - 39*B)*a^3*e^(6*I*d*x + 6*I*c) + 2*(-10*I*A - 9*B)*a^3*e^(4*I*d*x + 4*I*c) + (25*I*
A + 33*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(5*I*A + 6*B)*a^3)*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]

\[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=- i a^{3} \left (\int i A \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 A \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int A \tan ^{3}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 B \tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int B \tan ^{4}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 i A \tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int i B \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 i B \tan ^{3}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx\right ) \]

[In]

integrate(cot(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

-I*a**3*(Integral(I*A*sqrt(cot(c + d*x)), x) + Integral(-3*A*tan(c + d*x)*sqrt(cot(c + d*x)), x) + Integral(A*
tan(c + d*x)**3*sqrt(cot(c + d*x)), x) + Integral(-3*B*tan(c + d*x)**2*sqrt(cot(c + d*x)), x) + Integral(B*tan
(c + d*x)**4*sqrt(cot(c + d*x)), x) + Integral(-3*I*A*tan(c + d*x)**2*sqrt(cot(c + d*x)), x) + Integral(I*B*ta
n(c + d*x)*sqrt(cot(c + d*x)), x) + Integral(-3*I*B*tan(c + d*x)**3*sqrt(cot(c + d*x)), x))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.37 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {15 \, {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} - 2 \, {\left (3 i \, B a^{3} - \frac {5 \, {\left (-i \, A - 3 \, B\right )} a^{3}}{\tan \left (d x + c\right )} + \frac {15 \, {\left (3 \, A - 4 i \, B\right )} a^{3}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{15 \, d} \]

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/15*(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(ta
n(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 - 2*(3*I*B*a^3 - 5*(-I*A - 3*B)*a^3/tan(d*x + c) + 15*(3*A - 4*I*B)*a^3
/tan(d*x + c)^2)*tan(d*x + c)^(5/2))/d

Giac [F]

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

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*sqrt(cot(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

[In]

int(cot(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(cot(c + d*x)^(1/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)