\(\int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx\) [325]

Optimal result

Integrand size = 24, antiderivative size = 205 \[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {2 i (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}-\frac {i \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m}-\frac {i m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d \left (6+5 m+m^2\right )}+\frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (3+m)}+\frac {i \left (6+3 m+m^2\right ) (a+i a \tan (c+d x))^{1+m}}{a d (1+m) (2+m) (3+m)} \] Output:

2*I*(a+I*a*tan(d*x+c))^m/d/(m^2+5*m+6)-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*m*tan(d*x+c)^2*(a+I*a*tan(d*x+c 
))^m/d/(m^2+5*m+6)+tan(d*x+c)^3*(a+I*a*tan(d*x+c))^m/d/(3+m)+I*(m^2+3*m+6) 
*(a+I*a*tan(d*x+c))^(1+m)/a/d/(1+m)/(2+m)/(3+m)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.66 \[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m \left (-i \left (6+11 m+6 m^2+m^3\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right )-2 m \sec ^2(c+d x) \left (3+m+\left (5+4 m+m^2\right ) \cos (2 (c+d x))-i (1+m) \sin (2 (c+d x))\right ) (-i+\tan (c+d x))\right )}{2 d m (1+m) (2+m) (3+m)} \] Input:

Integrate[Tan[c + d*x]^4*(a + I*a*Tan[c + d*x])^m,x]
 

Output:

((a + I*a*Tan[c + d*x])^m*((-I)*(6 + 11*m + 6*m^2 + m^3)*Hypergeometric2F1 
[1, m, 1 + m, (1 + I*Tan[c + d*x])/2] - 2*m*Sec[c + d*x]^2*(3 + m + (5 + 4 
*m + m^2)*Cos[2*(c + d*x)] - I*(1 + m)*Sin[2*(c + d*x)])*(-I + Tan[c + d*x 
])))/(2*d*m*(1 + m)*(2 + m)*(3 + m))
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4043, 3042, 4080, 25, 3042, 4075, 3042, 4010, 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.

\(\displaystyle \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^4 (a+i a \tan (c+d x))^mdx\)

\(\Big \downarrow \) 4043

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4080

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {\int -\tan (c+d x) (i \tan (c+d x) a+a)^m \left (2 i a^2 m-a^2 \left (m^2+3 m+6\right ) \tan (c+d x)\right )dx}{a (m+2)}+\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^m \left (2 i a^2 m-a^2 \left (m^2+3 m+6\right ) \tan (c+d x)\right )dx}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {\int \tan (c+d x) (i \tan (c+d x) a+a)^m \left (2 i a^2 m-a^2 \left (m^2+3 m+6\right ) \tan (c+d x)\right )dx}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {\int (i \tan (c+d x) a+a)^m \left (\left (m^2+3 m+6\right ) a^2+2 i m \tan (c+d x) a^2\right )dx+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {\int (i \tan (c+d x) a+a)^m \left (\left (m^2+3 m+6\right ) a^2+2 i m \tan (c+d x) a^2\right )dx+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {a^2 (m+2) (m+3) \int (i \tan (c+d x) a+a)^mdx+\frac {2 i a^2 (a+i a \tan (c+d x))^m}{d}+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {a^2 (m+2) (m+3) \int (i \tan (c+d x) a+a)^mdx+\frac {2 i a^2 (a+i a \tan (c+d x))^m}{d}+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 3962

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {-\frac {i a^3 (m+2) (m+3) \int \frac {(i \tan (c+d x) a+a)^{m-1}}{a-i a \tan (c+d x)}d(i a \tan (c+d x))}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^m}{d}+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {\tan ^3(c+d x) (a+i a \tan (c+d x))^m}{d (m+3)}-\frac {\frac {i a m \tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {-\frac {i a^2 (m+2) (m+3) (a+i a \tan (c+d x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {i \tan (c+d x) a+a}{2 a}\right )}{2 d m}+\frac {2 i a^2 (a+i a \tan (c+d x))^m}{d}+\frac {i a \left (m^2+3 m+6\right ) (a+i a \tan (c+d x))^{m+1}}{d (m+1)}}{a (m+2)}}{a (m+3)}\)

Input:

Int[Tan[c + d*x]^4*(a + I*a*Tan[c + d*x])^m,x]
 

Output:

(Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^m)/(d*(3 + m)) - ((I*a*m*Tan[c + d* 
x]^2*(a + I*a*Tan[c + d*x])^m)/(d*(2 + m)) - (((2*I)*a^2*(a + I*a*Tan[c + 
d*x])^m)/d - ((I/2)*a^2*(2 + m)*(3 + m)*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*(6 + 3*m 
 + m^2)*(a + I*a*Tan[c + d*x])^(1 + m))/(d*(1 + m)))/(a*(2 + m)))/(a*(3 + 
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 4010

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

rule 4043

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

rule 4075

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

rule 4080

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 
1/(a*(m + n))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim 
p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T 
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c 
 - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
 

Maple [F]

\[\int \tan \left (d x +c \right )^{4} \left (a +i a \tan \left (d x +c \right )\right )^{m}d x\]

Input:

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^m,x)
 

Output:

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^m,x)
 

Fricas [F]

\[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate(tan(d*x+c)^4*(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^(8*I*d*x 
 + 8*I*c) - 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) - 4*e^(2*I*d*x + 
 2*I*c) + 1)/(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 
 4*I*c) + 4*e^(2*I*d*x + 2*I*c) + 1), x)
 

Sympy [F]

\[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{m} \tan ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate(tan(d*x+c)**4*(a+I*a*tan(d*x+c))**m,x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**m*tan(c + d*x)**4, x)
 

Maxima [F]

\[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate(tan(d*x+c)^4*(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)^4, x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{4} \,d x } \] Input:

integrate(tan(d*x+c)^4*(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)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^m \,d x \] Input:

int(tan(c + d*x)^4*(a + a*tan(c + d*x)*1i)^m,x)
 

Output:

int(tan(c + d*x)^4*(a + a*tan(c + d*x)*1i)^m, x)
 

Reduce [F]

\[ \int \tan ^4(c+d x) (a+i a \tan (c+d x))^m \, dx=\int \left (\tan \left (d x +c \right ) a i +a \right )^{m} \tan \left (d x +c \right )^{4}d x \] Input:

int(tan(d*x+c)^4*(a+I*a*tan(d*x+c))^m,x)
 

Output:

int((tan(c + d*x)*a*i + a)**m*tan(c + d*x)**4,x)