\(\int \frac {\tan ^{\frac {2}{3}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [267]

Optimal result

Integrand size = 28, antiderivative size = 84 \[ \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {5}{3},\frac {5}{2},1,\frac {8}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{\frac {5}{3}}(c+d x)}{5 a d \sqrt {a+i a \tan (c+d x)}} \] Output:

3/5*AppellF1(5/3,5/2,1,8/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1 
/2)*tan(d*x+c)^(5/3)/a/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4047

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {3 i a^4 \tan ^5(c+d x) \sqrt {a^3 \tan ^3(c+d x)+1} \operatorname {AppellF1}\left (\frac {5}{3},1,\frac {5}{2},\frac {8}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{5 d \sqrt {a^4 \tan ^3(c+d x)+a}}\)

Input:

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

Output:

(((3*I)/5)*a^4*AppellF1[5/3, 1, 5/2, 8/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c 
+ d*x]^3)]*Tan[c + d*x]^5*Sqrt[1 + a^3*Tan[c + d*x]^3])/(d*Sqrt[a + a^4*Ta 
n[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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(tan(d*x+c)^(2/3)/(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)*(7*I*e^(6*I*d*x + 6*I*c) - 12*I*e^(5* 
I*d*x + 5*I*c) + 26*I*e^(4*I*d*x + 4*I*c) - 24*I*e^(3*I*d*x + 3*I*c) + 31* 
I*e^(2*I*d*x + 2*I*c) - 12*I*e^(I*d*x + I*c) + 12*I) + 36*(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))*in 
tegral(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) + 2 
10*I*e^(4*I*d*x + 4*I*c) - 344*I*e^(3*I*d*x + 3*I*c) + 400*I*e^(2*I*d*x + 
2*I*c) - 317*I*e^(I*d*x + I*c) + 190*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^(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]

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

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

Output:

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

Maxima [F(-2)]

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

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

integrate(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 ValueDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*( - 48*tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1)*i - 939343107 
37612800000000000*int(( - tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1)*tan 
(c + d*x)**2)/(1252457476501504000000000*tan(c + d*x)**3*a**2*i + 12524574 
76501504000000000*tan(c + d*x)**2*a**2 + 1252457476501504000000000*tan(c + 
 d*x)*a**2*i + 1252457476501504000000000*a**2),x)*tan(c + d*x)**2*a**2*d - 
 93934310737612800000000000*int(( - tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)* 
i + 1)*tan(c + d*x)**2)/(1252457476501504000000000*tan(c + d*x)**3*a**2*i 
+ 1252457476501504000000000*tan(c + d*x)**2*a**2 + 12524574765015040000000 
00*tan(c + d*x)*a**2*i + 1252457476501504000000000*a**2),x)*a**2*d + 15029 
4897180180480000000000*int(( - tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1 
)*tan(c + d*x))/(1252457476501504000000000*tan(c + d*x)**3*a**2*i + 125245 
7476501504000000000*tan(c + d*x)**2*a**2 + 1252457476501504000000000*tan(c 
 + d*x)*a**2*i + 1252457476501504000000000*a**2),x)*tan(c + d*x)**2*a**2*d 
*i + 150294897180180480000000000*int(( - tan(c + d*x)**(2/3)*sqrt(tan(c + 
d*x)*i + 1)*tan(c + d*x))/(1252457476501504000000000*tan(c + d*x)**3*a**2* 
i + 1252457476501504000000000*tan(c + d*x)**2*a**2 + 125245747650150400000 
0000*tan(c + d*x)*a**2*i + 1252457476501504000000000*a**2),x)*a**2*d*i - 7 
5147448590090240000000000*int(( - tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i 
+ 1))/(1252457476501504000000000*tan(c + d*x)**4*a**2*i + 1252457476501504 
000000000*tan(c + d*x)**3*a**2 + 1252457476501504000000000*tan(c + d*x)...