\(\int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\) [270]

Optimal result

Integrand size = 28, antiderivative size = 82 \[ \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{2},1,\frac {4}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \sqrt [3]{\tan (c+d x)}}{a d \sqrt {a+i a \tan (c+d x)}} \] Output:

3*AppellF1(1/3,5/2,1,4/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2 
)*tan(d*x+c)^(1/3)/a/d/(a+I*a*tan(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [F]

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

Integrate[1/(Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2)),x]
 

Output:

Integrate[1/(Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2)), x]
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4047, 25, 27, 148, 27, 937, 936}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4047

\(\displaystyle \frac {i a^2 \int -\frac {1}{a \tan ^{\frac {2}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i a^2 \int \frac {1}{a \tan ^{\frac {2}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {i a \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 148

\(\displaystyle \frac {3 a^2 \int \frac {1}{a \left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 a \int \frac {1}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{d}\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {3 \sqrt {a^3 \tan ^3(c+d x)+1} \int \frac {1}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (a^3 \tan ^3(c+d x)+1\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{a d \sqrt {a^4 \tan ^3(c+d x)+a}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {3 i \tan (c+d x) \sqrt {a^3 \tan ^3(c+d x)+1} \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {5}{2},\frac {4}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{d \sqrt {a^4 \tan ^3(c+d x)+a}}\)

Input:

Int[1/(Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2)),x]
 

Output:

((3*I)*AppellF1[1/3, 1, 5/2, 4/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x]^3 
)]*Tan[c + d*x]*Sqrt[1 + a^3*Tan[c + d*x]^3])/(d*Sqrt[a + a^4*Tan[c + d*x] 
^3])
 

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 148

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), 
x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1) - 1)*(c 
 + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, 
 d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
 

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) 
  Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 3042

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

rule 4047

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

Maple [F]

\[\int \frac {1}{\tan \left (d x +c \right )^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionUnable to transpose Error: Bad Argument ValueDegree mismat 
ch inside
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\int \frac {1}{\tan \left (d x +c \right )^{\frac {5}{3}} \sqrt {\tan \left (d x +c \right ) i +1}\, i +\tan \left (d x +c \right )^{\frac {2}{3}} \sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a} \] Input:

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

Output:

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