\(\int \frac {\sqrt {\tan (c+d x)}}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [289]

Optimal result

Integrand size = 28, antiderivative size = 89 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3 i \operatorname {AppellF1}\left (-\frac {1}{3},-\frac {1}{2},1,\frac {2}{3},1+i \tan (c+d x),\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt {\tan (c+d x)}}{2 d \sqrt {-i \tan (c+d x)} \sqrt [3]{a+i a \tan (c+d x)}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16, 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, 395, 394}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\tan (c+d x)}}{\sqrt [3]{a+i a \tan (c+d x)}}dx\)

\(\Big \downarrow \) 4047

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

-(3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*sqrt((-I*e^(2*I*d*x + 2*I* 
c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(-I*e^(4*I*d*x + 4*I*c) - 2*I*e^(2*I*d* 
x + 2*I*c) - I)*e^(4/3*I*d*x + 4/3*I*c) - (a*d*e^(4*I*d*x + 4*I*c) - 4*a*d 
*e^(3*I*d*x + 3*I*c) + 4*a*d*e^(2*I*d*x + 2*I*c))*integral(1/2*2^(2/3)*(a/ 
(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I 
*d*x + 2*I*c) + 1))*(-I*e^(6*I*d*x + 6*I*c) + 30*I*e^(5*I*d*x + 5*I*c) - 8 
*I*e^(4*I*d*x + 4*I*c) + 24*I*e^(3*I*d*x + 3*I*c) - 11*I*e^(2*I*d*x + 2*I* 
c) - 6*I*e^(I*d*x + I*c) - 4*I)*e^(4/3*I*d*x + 4/3*I*c)/(a*d*e^(7*I*d*x + 
7*I*c) - 6*a*d*e^(6*I*d*x + 6*I*c) + 11*a*d*e^(5*I*d*x + 5*I*c) - 2*a*d*e^ 
(4*I*d*x + 4*I*c) - 12*a*d*e^(3*I*d*x + 3*I*c) + 8*a*d*e^(2*I*d*x + 2*I*c) 
), x))/(a*d*e^(4*I*d*x + 4*I*c) - 4*a*d*e^(3*I*d*x + 3*I*c) + 4*a*d*e^(2*I 
*d*x + 2*I*c))
 

Sympy [F]

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

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

Output:

Integral(sqrt(tan(c + d*x))/(I*a*(tan(c + d*x) - I))**(1/3), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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