\(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^m \, dx\) [327]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m}-\frac {i (a+i a \tan (c+d x))^{1+m}}{a d (1+m)} \] Output:

1/2*I*hypergeom([1, m],[1+m],1/2+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^m/d/ 
m-I*(a+I*a*tan(d*x+c))^(1+m)/a/d/(1+m)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m \left (i (1+m) \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right )+2 m (-i+\tan (c+d x))\right )}{2 d m (1+m)} \] Input:

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

Output:

((a + I*a*Tan[c + d*x])^m*(I*(1 + m)*Hypergeometric2F1[1, m, 1 + m, (1 + I 
*Tan[c + d*x])/2] + 2*m*(-I + Tan[c + d*x])))/(2*d*m*(1 + m))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4026, 25, 3042, 3962, 78}

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.

\(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^mdx\)

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {i (a+i a \tan (c+d x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {i \tan (c+d x) a+a}{2 a}\right )}{2 d m}-\frac {i (a+i a \tan (c+d x))^{m+1}}{a d (m+1)}\)

Input:

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

Output:

((I/2)*Hypergeometric2F1[1, m, 1 + m, (a + I*a*Tan[c + d*x])/(2*a)]*(a + I 
*a*Tan[c + d*x])^m)/(d*m) - (I*(a + I*a*Tan[c + d*x])^(1 + m))/(a*d*(1 + m 
))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3962

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-b/d   S 
ubst[Int[(a + x)^(n - 1)/(a - x), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b 
, c, d, n}, x] && EqQ[a^2 + b^2, 0]
 

rule 4026

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( 
m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* 
x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ 
[m, -1] &&  !(EqQ[m, 2] && EqQ[a, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(tan(d*x+c)^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*(e^(4*I*d* 
x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c) + 1)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d* 
x + 2*I*c) + 1), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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