\(\int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx\) [334]

Optimal result

Integrand size = 26, antiderivative size = 79 \[ \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},1-m,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^m}{d} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4047, 25, 27, 148, 27, 334, 333}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^m}{\sqrt {\tan (c+d x)}}dx\)

\(\Big \downarrow \) 4047

\(\displaystyle \frac {i a^2 \int -\frac {(i \tan (c+d x) a+a)^{m-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i a^2 \int \frac {(i \tan (c+d x) a+a)^{m-1}}{a \sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {i a \int \frac {(i \tan (c+d x) a+a)^{m-1}}{\sqrt {\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

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

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 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt((-I*e^ 
(2*I*d*x + 2*I*c) + I)/(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), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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