\(\int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\) [1186]

Optimal result

Integrand size = 30, antiderivative size = 133 \[ \int (a+i a \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\frac {2 a \operatorname {AppellF1}\left (\frac {5}{2},1-m,1,\frac {7}{2},\frac {c+d \tan (e+f x)}{c+i d},\frac {c+d \tan (e+f x)}{c-i d}\right ) \left (-\frac {d (1+i \tan (e+f x))}{i c-d}\right )^{1-m} (a+i a \tan (e+f x))^{-1+m} (c+d \tan (e+f x))^{5/2}}{5 (i c+d) f} \] Output:

2/5*a*AppellF1(5/2,1,1-m,7/2,(c+d*tan(f*x+e))/(c-I*d),(c+d*tan(f*x+e))/(c+ 
I*d))*(-d*(1+I*tan(f*x+e))/(I*c-d))^(1-m)*(a+I*a*tan(f*x+e))^(-1+m)*(c+d*t 
an(f*x+e))^(5/2)/(I*c+d)/f
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4047, 25, 27, 154, 153}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4047

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 154

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

\(\Big \downarrow \) 153

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

Input:

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

Output:

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

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 153

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp 
lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c 
 - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( 
b*c - a*d)], 0] &&  !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, 
a + b*x])
 

rule 154

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[p] &&  !G 
tQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((d*tan(f*x + e) + c)^(3/2)*(I*a*tan(f*x + e) + a)^m, x)
 

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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