\(\int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^m \, dx\) [333]

Optimal result

Integrand size = 26, antiderivative size = 81 \[ \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^m \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {3}{2},1-m,1,\frac {5}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^m}{3 d} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.27, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4047

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 395

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

\(\Big \downarrow \) 394

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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