3.1.96 \(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [96]

3.1.96.1 Optimal result

Integrand size = 36, antiderivative size = 219 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(11 A+4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 \sqrt {a} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {(7 i A-8 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d}-\frac {(3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a d} \]

output

1/4*(11*A+4*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-1/2*( 
A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^( 
1/2)+(A+I*B)*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/4*(7*I*A-8*B)*cot(d 
*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a/d-1/2*(3*A+2*I*B)*cot(d*x+c)^2*(a+I*a*tan 
(d*x+c))^(1/2)/a/d
 

3.1.96.2 Mathematica [A] (verified)

Time = 4.42 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.68 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {(11 A+4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}+\frac {-7 A-8 i B+i (A+4 i B) \cot (c+d x)-2 A \cot ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}}}{4 d} \]

input

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

output

(((11*A + (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/Sqrt[a] - 
(2*Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] 
)/Sqrt[a] + (-7*A - (8*I)*B + I*(A + (4*I)*B)*Cot[c + d*x] - 2*A*Cot[c + d 
*x]^2)/Sqrt[a + I*a*Tan[c + d*x]])/(4*d)
 

3.1.96.3 Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4079, 27, 3042, 4081, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

\(\displaystyle \frac {\frac {\int -\cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((7 i A-8 B) a^2+3 (3 A+2 i B) \tan (c+d x) a^2\right )dx}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((7 i A-8 B) a^2+3 (3 A+2 i B) \tan (c+d x) a^2\right )dx}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((7 i A-8 B) a^2+3 (3 A+2 i B) \tan (c+d x) a^2\right )}{\tan (c+d x)^2}dx}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {-\frac {\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (11 A+4 i B)-a^3 (7 i A-8 B) \tan (c+d x)\right )dx}{a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (11 A+4 i B)-a^3 (7 i A-8 B) \tan (c+d x)\right )dx}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (11 A+4 i B)-a^3 (7 i A-8 B) \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {-\frac {\frac {4 a^3 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a^2 (11 A+4 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {4 a^3 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a^2 (11 A+4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {-\frac {\frac {a^2 (11 A+4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {8 i a^4 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {\frac {a^2 (11 A+4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {4 i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {-\frac {\frac {\frac {a^4 (11 A+4 i B) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {4 i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {\frac {-\frac {2 i a^3 (11 A+4 i B) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {4 i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {-\frac {2 a^{7/2} (11 A+4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {4 i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (-8 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a}-\frac {a (3 A+2 i B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}+\frac {(A+i B) \cot ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\)

input

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

output

((A + I*B)*Cot[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (-((a*(3*A + ( 
2*I)*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d) - (((-2*a^(7/2)*(11* 
A + (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d - ((4*I)*Sqrt[ 
2]*a^(7/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] 
)/d)/(2*a) - (a^2*((7*I)*A - 8*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) 
/d)/(2*a))/(2*a^2)
 

3.1.96.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4079

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

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]
 

3.1.96.4 Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {i B +A}{2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {7}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}-\frac {3 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {5}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (4 i B +11 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )}{d}\)	\(171\)
default	\(\frac {2 a^{3} \left (-\frac {i B +A}{2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {7}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}-\frac {3 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {5}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (4 i B +11 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )}{d}\)	\(171\)

input

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

output

2/d*a^3*(-1/2/a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/4*(A-I*B)/a^(7/2)*2^( 
1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/a^3*(-((-1/2* 
I*B-3/8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2*I*a*B+5/8*a*A)*(a+I*a*tan(d*x+c)) 
^(1/2))/a^2/tan(d*x+c)^2+1/8*(11*A+4*I*B)/a^(1/2)*arctanh((a+I*a*tan(d*x+c 
))^(1/2)/a^(1/2))))
 

3.1.96.5 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. 835 vs. \(2 (172) = 344\).

Time = 0.28 (sec) , antiderivative size = 835, normalized size of antiderivative = 3.81 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

1/16*(4*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d 
*e^(I*d*x + I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*((-I*A - B)*a 
*e^(I*d*x + I*c) + (I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d*x 
+ 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A 
+ B)) - 4*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a 
*d*e^(I*d*x + I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*((-I*A - B) 
*a*e^(I*d*x + I*c) + (-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2*I*d 
*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I 
*A + B)) + (a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I 
*d*x + I*c))*sqrt((121*A^2 + 88*I*A*B - 16*B^2)/(a*d^2))*log(-16*(3*(11*I* 
A - 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (11*I*A - 4*B)*a^2 + 2*sqrt(2)*(I*a^2*d 
*e^(3*I*d*x + 3*I*c) + I*a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c 
) + 1))*sqrt((121*A^2 + 88*I*A*B - 16*B^2)/(a*d^2)))*e^(-2*I*d*x - 2*I*c)/ 
(-11*I*A + 4*B)) - (a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^(3*I*d*x + 3*I*c) + 
a*d*e^(I*d*x + I*c))*sqrt((121*A^2 + 88*I*A*B - 16*B^2)/(a*d^2))*log(-16*( 
3*(11*I*A - 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (11*I*A - 4*B)*a^2 + 2*sqrt(2)* 
(-I*a^2*d*e^(3*I*d*x + 3*I*c) - I*a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d* 
x + 2*I*c) + 1))*sqrt((121*A^2 + 88*I*A*B - 16*B^2)/(a*d^2)))*e^(-2*I*d*x 
- 2*I*c)/(-11*I*A + 4*B)) - 4*sqrt(2)*(3*(A + 2*I*B)*e^(6*I*d*x + 6*I*c) - 
 2*(3*A + I*B)*e^(4*I*d*x + 4*I*c) - (7*A + 6*I*B)*e^(2*I*d*x + 2*I*c) ...
 

3.1.96.6 Sympy [F]

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

input

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

output

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

3.1.96.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.06 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {a^{2} {\left (\frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (7 \, A + 8 i \, B\right )} - {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (13 \, A + 12 i \, B\right )} a + 4 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}} - \frac {2 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {{\left (11 \, A + 4 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{8 \, d} \]

input

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

output

-1/8*a^2*(2*((I*a*tan(d*x + c) + a)^2*(7*A + 8*I*B) - (I*a*tan(d*x + c) + 
a)*(13*A + 12*I*B)*a + 4*(A + I*B)*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a^2 
- 2*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + sqrt(I*a*tan(d*x + c) + a)*a^4) - 2 
*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sq 
rt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2) + (11*A + 4*I*B)*log( 
(sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt( 
a)))/a^(5/2))/d
 

3.1.96.8 Giac [F]

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

input

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

output

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

3.1.96.9 Mupad [B] (verification not implemented)

Time = 9.58 (sec) , antiderivative size = 3037, normalized size of antiderivative = 13.87 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]

input

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

output

2*atanh((12*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((129*A^2)/(128*a*d^2) - ((1 
2769*A^4*a^10)/(4*d^4) + (16*B^4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A* 
B^3*a^10*416i)/d^4 + (A^3*B*a^10*5876i)/d^4)^(1/2)/(64*a^6) - (3*B^2)/(16* 
a*d^2) + (A*B*9i)/(16*a*d^2))^(1/2)*((12769*A^4*a^10)/(4*d^4) + (16*B^4*a^ 
10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*416i)/d^4 + (A^3*B*a^10*58 
76i)/d^4)^(1/2))/(B^3*a^5*d*8i - (1469*A^3*a^5*d)/2 + 9*A*d^3*((12769*A^4* 
a^10)/(4*d^4) + (16*B^4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A*B^3*a^10* 
416i)/d^4 + (A^3*B*a^10*5876i)/d^4)^(1/2) + B*d^3*((12769*A^4*a^10)/(4*d^4 
) + (16*B^4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*416i)/d^4 + 
(A^3*B*a^10*5876i)/d^4)^(1/2)*6i + 156*A*B^2*a^5*d - A^2*B*a^5*d*789i) - ( 
226*A^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((129*A^2)/(128*a*d^2) - ((1 
2769*A^4*a^10)/(4*d^4) + (16*B^4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A* 
B^3*a^10*416i)/d^4 + (A^3*B*a^10*5876i)/d^4)^(1/2)/(64*a^6) - (3*B^2)/(16* 
a*d^2) + (A*B*9i)/(16*a*d^2))^(1/2))/(B^3*a^2*d*8i - (1469*A^3*a^2*d)/2 + 
156*A*B^2*a^2*d - A^2*B*a^2*d*789i + (9*A*d^3*((12769*A^4*a^10)/(4*d^4) + 
(16*B^4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*416i)/d^4 + (A^3 
*B*a^10*5876i)/d^4)^(1/2))/a^3 + (B*d^3*((12769*A^4*a^10)/(4*d^4) + (16*B^ 
4*a^10)/d^4 - (3156*A^2*B^2*a^10)/d^4 - (A*B^3*a^10*416i)/d^4 + (A^3*B*a^1 
0*5876i)/d^4)^(1/2)*6i)/a^3) + (16*B^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/ 
2)*((129*A^2)/(128*a*d^2) - ((12769*A^4*a^10)/(4*d^4) + (16*B^4*a^10)/d...