\(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [81]

Optimal result

Integrand size = 36, antiderivative size = 213 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {a^{3/2} (23 i A+22 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {2 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {a (7 i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \] Output:

1/8*a^(3/2)*(23*I*A+22*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-2*2^ 
(1/2)*a^(3/2)*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2) 
)/d+1/8*a*(9*A-10*I*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/12*a*(7*I*A 
+6*B)*cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d-1/3*a*A*cot(d*x+c)^3*(a+I*a* 
tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 3.32 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.74 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {-3 a^{3/2} (23 i A+22 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+48 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a \cot (c+d x) \left (-27 A+30 i B+2 (7 i A+6 B) \cot (c+d x)+8 A \cot ^2(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{24 d} \] Input:

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]) 
,x]
 

Output:

-1/24*(-3*a^(3/2)*((23*I)*A + 22*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqr 
t[a]] + 48*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(S 
qrt[2]*Sqrt[a])] + a*Cot[c + d*x]*(-27*A + (30*I)*B + 2*((7*I)*A + 6*B)*Co 
t[c + d*x] + 8*A*Cot[c + d*x]^2)*Sqrt[a + I*a*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.10, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4076, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^4}dx\)

\(\Big \downarrow \) 4076

\(\displaystyle \frac {1}{3} \int \frac {1}{2} \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (a (7 i A+6 B)-a (5 A-6 i B) \tan (c+d x))dx-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (a (7 i A+6 B)-a (5 A-6 i B) \tan (c+d x))dx-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \frac {\sqrt {i \tan (c+d x) a+a} (a (7 i A+6 B)-a (5 A-6 i B) \tan (c+d x))}{\tan (c+d x)^3}dx-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{6} \left (\frac {\int -\frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(7 i A+6 B) \tan (c+d x) a^2\right )dx}{2 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (-\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(7 i A+6 B) \tan (c+d x) a^2\right )dx}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (-\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(7 i A+6 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^2}dx}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (23 i A+22 B)-a^3 (9 A-10 i B) \tan (c+d x)\right )dx}{a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (23 i A+22 B)-a^3 (9 A-10 i B) \tan (c+d x)\right )dx}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (23 i A+22 B)-a^3 (9 A-10 i B) \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {a^2 (22 B+23 i A) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx-32 a^3 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {a^2 (22 B+23 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-32 a^3 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {64 i a^4 (A-i B) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+a^2 (22 B+23 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {a^2 (22 B+23 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {32 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {a^4 (22 B+23 i A) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {32 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {32 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 i a^3 (22 B+23 i A) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {32 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{7/2} (22 B+23 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (9 A-10 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+7 i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\)

Input:

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
 

Output:

-1/3*(a*A*Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (-1/2*(a*((7*I)*A 
 + 6*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (3*(((-2*a^(7/2)*(( 
23*I)*A + 22*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((32*I)*S 
qrt[2]*a^(7/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[ 
a])])/d)/(2*a) - (a^2*(9*A - (10*I)*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d 
*x]])/d))/(4*a))/6
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4076

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-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] - Simp[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 4081

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(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] - Simp[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] /; Fr 
eeQ[{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 4082

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

rule 4083

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( 
A*b + a*B)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A 
*d)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T 
an[e + f*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] && NeQ[A*b + a*B, 0]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {2 i a^{4} \left (-\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {i \left (\left (\frac {5 i B}{8}-\frac {9 A}{16}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}+\left (-i a B +\frac {5}{6} a A \right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {3}{8} i B \,a^{2}-\frac {7}{16} A \,a^{2}\right ) \sqrt {a +i a \tan \left (d x +c \right )}\right )}{a^{3} \tan \left (d x +c \right )^{3}}+\frac {\left (-22 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(179\)
default	\(\frac {2 i a^{4} \left (-\frac {\left (-2 i B +2 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {i \left (\left (\frac {5 i B}{8}-\frac {9 A}{16}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}+\left (-i a B +\frac {5}{6} a A \right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {3}{8} i B \,a^{2}-\frac {7}{16} A \,a^{2}\right ) \sqrt {a +i a \tan \left (d x +c \right )}\right )}{a^{3} \tan \left (d x +c \right )^{3}}+\frac {\left (-22 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(179\)

Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

2*I/d*a^4*(-1/2*(-2*I*B+2*A)/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c) 
)^(1/2)*2^(1/2)/a^(1/2))+1/a^2*(-I*((5/8*I*B-9/16*A)*(a+I*a*tan(d*x+c))^(5 
/2)+(-I*a*B+5/6*a*A)*(a+I*a*tan(d*x+c))^(3/2)+(3/8*I*B*a^2-7/16*A*a^2)*(a+ 
I*a*tan(d*x+c))^(1/2))/a^3/tan(d*x+c)^3+1/16*(-22*I*B+23*A)/a^(1/2)*arctan 
h((a+I*a*tan(d*x+c))^(1/2)/a^(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. 856 vs. \(2 (166) = 332\).

Time = 0.15 (sec) , antiderivative size = 856, normalized size of antiderivative = 4.02 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

1/96*(96*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/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*((-I*A - 
 B)*a^2*e^(I*d*x + I*c) + sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(2*I*d 
*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I* 
A - B)*a)) - 96*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/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*( 
(-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e 
^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c 
)/((-I*A - B)*a)) - 3*sqrt(-(529*A^2 - 1012*I*A*B - 484*B^2)*a^3/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(-16*(3*(-23*I*A - 22*B)*a^2*e^(2*I*d*x + 2*I*c) + (-23*I*A - 22*B)*a 
^2 + 2*sqrt(2)*sqrt(-(529*A^2 - 1012*I*A*B - 484*B^2)*a^3/d^2)*(d*e^(3*I*d 
*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2* 
I*d*x - 2*I*c)/(23*I*A + 22*B)) + 3*sqrt(-(529*A^2 - 1012*I*A*B - 484*B^2) 
*a^3/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(-16*(3*(-23*I*A - 22*B)*a^2*e^(2*I*d*x + 2*I*c) + (-23 
*I*A - 22*B)*a^2 - 2*sqrt(2)*sqrt(-(529*A^2 - 1012*I*A*B - 484*B^2)*a^3/d^ 
2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) 
 + 1)))*e^(-2*I*d*x - 2*I*c)/(23*I*A + 22*B)) - 4*sqrt(2)*(7*(-7*I*A - 6*B 
)*a*e^(7*I*d*x + 7*I*c) - (11*I*A - 18*B)*a*e^(5*I*d*x + 5*I*c) - (-17*...
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {i \, a^{3} {\left (\frac {48 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {3 \, {\left (23 \, A - 22 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (9 \, A - 10 i \, B\right )} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (5 \, A - 6 i \, B\right )} a + 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (7 \, A - 6 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - a^{4}}\right )}}{48 \, d} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

1/48*I*a^3*(48*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x 
+ c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(3/2) - 3*(23 
*A - 22*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x 
+ c) + a) + sqrt(a)))/a^(3/2) + 2*(3*(I*a*tan(d*x + c) + a)^(5/2)*(9*A - 1 
0*I*B) - 8*(I*a*tan(d*x + c) + a)^(3/2)*(5*A - 6*I*B)*a + 3*sqrt(I*a*tan(d 
*x + c) + a)*(7*A - 6*I*B)*a^2)/((I*a*tan(d*x + c) + a)^3*a - 3*(I*a*tan(d 
*x + c) + a)^2*a^2 + 3*(I*a*tan(d*x + c) + a)*a^3 - a^4))/d
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algori 
thm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar 
gument Ty
 

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 3084, normalized size of antiderivative = 14.48 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

int(cot(c + d*x)^4*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

2*atanh((6*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((249*B^2*a^3)/(128*d^2) - (1 
041*A^2*a^3)/(512*d^2) - ((289*A^4*a^18)/(64*d^4) + (49*B^4*a^18)/(4*d^4) 
+ (101*A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + (A^3*B*a^18*51i) 
/(8*d^4))^(1/2)/(64*a^6) + (A*B*a^3*509i)/(128*d^2))^(1/2)*((289*A^4*a^18) 
/(64*d^4) + (49*B^4*a^18)/(4*d^4) + (101*A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^ 
18*21i)/(2*d^4) + (A^3*B*a^18*51i)/(8*d^4))^(1/2))/((A^3*a^11*d*663i)/32 + 
 (133*B^3*a^11*d)/4 + (A*B^2*a^11*d*387i)/8 + (89*A^2*B*a^11*d)/16 + (A*a^ 
2*d^3*((289*A^4*a^18)/(64*d^4) + (49*B^4*a^18)/(4*d^4) + (101*A^2*B^2*a^18 
)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + (A^3*B*a^18*51i)/(8*d^4))^(1/2)*7i) 
/4 + (3*B*a^2*d^3*((289*A^4*a^18)/(64*d^4) + (49*B^4*a^18)/(4*d^4) + (101* 
A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + (A^3*B*a^18*51i)/(8*d^4 
))^(1/2))/2) + (17*A^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((249*B^2*a^3 
)/(128*d^2) - (1041*A^2*a^3)/(512*d^2) - ((289*A^4*a^18)/(64*d^4) + (49*B^ 
4*a^18)/(4*d^4) + (101*A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + 
(A^3*B*a^18*51i)/(8*d^4))^(1/2)/(64*a^6) + (A*B*a^3*509i)/(128*d^2))^(1/2) 
)/(4*((A^3*a^8*d*663i)/32 + (133*B^3*a^8*d)/4 + (A*B^2*a^8*d*387i)/8 + (89 
*A^2*B*a^8*d)/16 + (A*d^3*((289*A^4*a^18)/(64*d^4) + (49*B^4*a^18)/(4*d^4) 
 + (101*A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + (A^3*B*a^18*51i 
)/(8*d^4))^(1/2)*7i)/(4*a) + (3*B*d^3*((289*A^4*a^18)/(64*d^4) + (49*B^4*a 
^18)/(4*d^4) + (101*A^2*B^2*a^18)/(8*d^4) + (A*B^3*a^18*21i)/(2*d^4) + ...
 

Reduce [F]

\[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}d x \right ) b i +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) a i +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
 

Output:

sqrt(a)*a*(int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4*tan(c + d*x)**2,x) 
*b*i + int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4*tan(c + d*x),x)*a*i + 
int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4*tan(c + d*x),x)*b + int(sqrt( 
tan(c + d*x)*i + 1)*cot(c + d*x)**4,x)*a)