\(\int \frac {\cot ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [769]

Optimal result

Integrand size = 28, antiderivative size = 140 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}+\frac {\sqrt {\cot (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d} \] Output:

(1/2+1/2*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2 
))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/a^(1/2)/d+cot(d*x+c)^(1/2)/d/(a+I*a*t 
an(d*x+c))^(1/2)-3*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {-6 i-4 \cot (c+d x)}{\sqrt {a+i a \tan (c+d x)}}}{2 d \sqrt {\cot (c+d x)}} \] Input:

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

Output:

((I*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
d*x]]])/Sqrt[I*a*Tan[c + d*x]] + (-6*I - 4*Cot[c + d*x])/Sqrt[a + I*a*Tan[ 
c + d*x]])/(2*d*Sqrt[Cot[c + d*x]])
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4729, 3042, 4042, 27, 3042, 4081, 27, 3042, 4027, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (3 a-2 i a \tan (c+d x))}{\tan (c+d x)^{3/2}}dx}{2 a^2}+\frac {1}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\right )\)

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {(1+i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {6 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{2 a^2}+\frac {1}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}\right )\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(1/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I* 
a*Tan[c + d*x]]) + (((1 + I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + 
 d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (6*a*Sqrt[a + I*a*Tan[c + d*x]])/ 
(d*Sqrt[Tan[c + d*x]]))/(2*a^2))
 

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 218

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

rule 3042

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

rule 4027

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

rule 4042

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

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 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \cot \left (d x +c \right )^{\frac {3}{2}} \left (\sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right ) \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \left (\tan \left (d x +c \right )+\tan \left (d x +c \right ) \sec \left (d x +c \right )\right )+i \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, \sqrt {-\csc \left (d x +c \right )+\cot \left (d x +c \right )}\right ) \sqrt {2}\, \tan \left (d x +c \right )^{2}+i \left (-2 \tan \left (d x +c \right )+3 \tan \left (d x +c \right )^{2}\right )+2 \tan \left (d x +c \right )+3 \tan \left (d x +c \right )^{2}\right )}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(194\)

Input:

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

Output:

(-1/2-1/2*I)/d*cot(d*x+c)^(3/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(2^(1/2)*arctan 
((1/2+1/2*I)*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2))*(-csc(d*x+c)+cot(d*x+ 
c))^(1/2)*(tan(d*x+c)+tan(d*x+c)*sec(d*x+c))+I*(-csc(d*x+c)+cot(d*x+c))^(1 
/2)*arctan((1/2+1/2*I)*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))^(1/2))*2^(1/2)*tan 
(d*x+c)^2+I*(-2*tan(d*x+c)+3*tan(d*x+c)^2)+2*tan(d*x+c)+3*tan(d*x+c)^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. 331 vs. \(2 (106) = 212\).

Time = 0.12 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.36 \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (2 \, {\left (\sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i}{a d^{2}}} + 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-2 \, {\left (\sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (5 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \] Input:

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

Output:

1/4*(a*d*sqrt(2*I/(a*d^2))*e^(I*d*x + I*c)*log(2*(sqrt(2)*(a*d*e^(2*I*d*x 
+ 2*I*c) - a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I 
*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(2*I/(a*d^2)) + 2*I*a*e^(I*d*x + I 
*c))*e^(-I*d*x - I*c)) - a*d*sqrt(2*I/(a*d^2))*e^(I*d*x + I*c)*log(-2*(sqr 
t(2)*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr 
t((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(2*I/(a*d^2)) 
 - 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 2*sqrt(2)*sqrt(a/(e^(2*I*d*x 
 + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1) 
)*(5*e^(2*I*d*x + 2*I*c) - 1))*e^(-I*d*x - I*c)/(a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(cot(d*x + c)^(3/2)/sqrt(I*a*tan(d*x + c) + a), x)
 

Giac [F(-2)]

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

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{3/2}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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