\(\int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^3} \, dx\) [1057]

Optimal result

Integrand size = 31, antiderivative size = 52 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^3} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (4,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{16 c^3 f m} \] Output:

-1/16*I*hypergeom([4, m],[1+m],1/2+1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/ 
c^3/f/m
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^3} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (4,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{16 c^3 f m} \] Input:

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

Output:

((-1/16*I)*Hypergeometric2F1[4, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a 
*Tan[e + f*x])^m)/(c^3*f*m)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4005, 3042, 3968, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \cos ^6(e+f x) (i \tan (e+f x) a+a)^{m+3}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(i \tan (e+f x) a+a)^{m+3}}{\sec (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i a^4 \int \frac {(i \tan (e+f x) a+a)^{m-1}}{(a-i a \tan (e+f x))^4}d(i a \tan (e+f x))}{c^3 f}\)

\(\Big \downarrow \) 78

\(\displaystyle -\frac {i (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (4,m,m+1,\frac {i \tan (e+f x) a+a}{2 a}\right )}{16 c^3 f m}\)

Input:

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

Output:

((-1/16*I)*Hypergeometric2F1[4, m, 1 + m, (a + I*a*Tan[e + f*x])/(2*a)]*(a 
 + I*a*Tan[e + f*x])^m)/(c^3*f*m)
 

Defintions of rubi rules used

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 3968

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

rule 4005

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(1/8*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*(e^(6*I 
*f*x + 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1)/c^3, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^3} \, dx=\frac {-\left (\tan \left (f x +e \right ) a i +a \right )^{m} i -\left (\int \frac {\left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) f m -4 \left (\int \frac {\left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) f i m +6 \left (\int \frac {\left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) f m +4 \left (\int \frac {\left (\tan \left (f x +e \right ) a i +a \right )^{m} \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} i -3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i +1}d x \right ) f i m}{c^{3} f m} \] Input:

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

Output:

( - (tan(e + f*x)*a*i + a)**m*i - int(((tan(e + f*x)*a*i + a)**m*tan(e + f 
*x)**4)/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)* 
f*m - 4*int(((tan(e + f*x)*a*i + a)**m*tan(e + f*x)**3)/(tan(e + f*x)**3*i 
 - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*f*i*m + 6*int(((tan(e + f* 
x)*a*i + a)**m*tan(e + f*x)**2)/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3 
*tan(e + f*x)*i + 1),x)*f*m + 4*int(((tan(e + f*x)*a*i + a)**m*tan(e + f*x 
))/(tan(e + f*x)**3*i - 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i + 1),x)*f*i*m 
)/(c**3*f*m)