\(\int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\) [271]

Optimal result

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

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

Mathematica [F]

\[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx \] Input:

Integrate[1/(Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x])^(3/2)),x]
 

Output:

Integrate[1/(Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x])^(3/2)), x]
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07, 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, 1013, 1012}

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.

\(\displaystyle \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x)^{4/3} (a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4047

\(\displaystyle \frac {i a^2 \int -\frac {1}{a \tan ^{\frac {4}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i a^2 \int \frac {1}{a \tan ^{\frac {4}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {i a \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

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

Input:

Int[1/(Tan[c + d*x]^(4/3)*(a + I*a*Tan[c + d*x])^(3/2)),x]
 

Output:

((3*I)*AppellF1[-1/3, 1, 5/2, 2/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x]^ 
3)]*Cot[c + d*x]*Sqrt[1 + a^3*Tan[c + d*x]^3])/(a^2*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 1012

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

rule 1013

Int[((e_.)*(x_))^(m_.)*((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[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, 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]

\[\int \frac {1}{\tan \left (d x +c \right )^{\frac {4}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(1/tan(d*x+c)^(4/3)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas 
")
 

Output:

1/36*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*((-I*e^(2*I*d*x + 2*I*c) + 
 I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-421*I*e^(8*I*d*x + 8*I*c) + 228*I*e 
^(7*I*d*x + 7*I*c) - 703*I*e^(6*I*d*x + 6*I*c) + 444*I*e^(5*I*d*x + 5*I*c) 
 - 131*I*e^(4*I*d*x + 4*I*c) + 204*I*e^(3*I*d*x + 3*I*c) + 163*I*e^(2*I*d* 
x + 2*I*c) - 12*I*e^(I*d*x + I*c) + 12*I) + 36*(a^2*d*e^(7*I*d*x + 7*I*c) 
- 4*a^2*d*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^(5*I*d*x + 5*I*c) + 4*a^2*d*e^(4 
*I*d*x + 4*I*c) - 4*a^2*d*e^(3*I*d*x + 3*I*c))*integral(1/108*sqrt(2)*sqrt 
(a/(e^(2*I*d*x + 2*I*c) + 1))*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 
2*I*c) + 1))^(2/3)*(27*I*e^(5*I*d*x + 5*I*c) - 4530*I*e^(4*I*d*x + 4*I*c) 
- 286*I*e^(3*I*d*x + 3*I*c) - 2380*I*e^(2*I*d*x + 2*I*c) - 313*I*e^(I*d*x 
+ I*c) + 2150*I)/(a^2*d*e^(5*I*d*x + 5*I*c) - 6*a^2*d*e^(4*I*d*x + 4*I*c) 
+ 11*a^2*d*e^(3*I*d*x + 3*I*c) - 2*a^2*d*e^(2*I*d*x + 2*I*c) - 12*a^2*d*e^ 
(I*d*x + I*c) + 8*a^2*d), x))/(a^2*d*e^(7*I*d*x + 7*I*c) - 4*a^2*d*e^(6*I* 
d*x + 6*I*c) + 3*a^2*d*e^(5*I*d*x + 5*I*c) + 4*a^2*d*e^(4*I*d*x + 4*I*c) - 
 4*a^2*d*e^(3*I*d*x + 3*I*c))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/tan(d*x+c)**(4/3)/(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/tan(d*x+c)^(4/3)/(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. \[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/tan(d*x+c)^(4/3)/(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. \[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{4/3}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {4}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{\frac {13}{3}} i +\tan \left (d x +c \right )^{\frac {10}{3}}+\tan \left (d x +c \right )^{\frac {7}{3}} i +\tan \left (d x +c \right )^{\frac {4}{3}}}d x -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{\frac {10}{3}} i +\tan \left (d x +c \right )^{\frac {7}{3}}+\tan \left (d x +c \right )^{\frac {4}{3}} i +\tan \left (d x +c \right )^{\frac {1}{3}}}d x \right ) i \right )}{a^{2}} \] Input:

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

Output:

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