\(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\) [211]

Optimal result

Integrand size = 34, antiderivative size = 386 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\left (3-4 m+m^2\right ) \left (i B \left (1-m^2\right )-A \left (1-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+m)}-\frac {\left (i B \left (1+3 m-m^2\right )-A \left (13-7 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {(2-m) \left (i B \left (2+2 m-m^2\right )-A \left (8-6 m+m^2\right )\right ) \tan ^{1+m}(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(2-m) m \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{48 a^4 d (2+m)}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(i B (1-m)+A (5-m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \] Output:

-1/48*(m^2-4*m+3)*(I*B*(-m^2+1)-A*(m^2-4*m+1))*hypergeom([1, 1/2+1/2*m],[3 
/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(1+m)/a^4/d/(1+m)-1/48*(I*B*(-m^2+3*m+ 
1)-A*(m^2-7*m+13))*tan(d*x+c)^(1+m)/a^4/d/(1+I*tan(d*x+c))^2-1/48*(2-m)*(I 
*B*(-m^2+2*m+2)-A*(m^2-6*m+8))*tan(d*x+c)^(1+m)/a^4/d/(1+I*tan(d*x+c))+1/4 
8*(2-m)*m*(B*(-m^2+2*m+2)+I*A*(m^2-6*m+8))*hypergeom([1, 1+1/2*m],[2+1/2*m 
],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/a^4/d/(2+m)+1/8*(A+I*B)*tan(d*x+c)^(1+m) 
/d/(a+I*a*tan(d*x+c))^4+1/24*(I*B*(1-m)+A*(5-m))*tan(d*x+c)^(1+m)/a/d/(a+I 
*a*tan(d*x+c))^3
 

Mathematica [A] (verified)

Time = 2.29 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.72 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\tan ^{1+m}(c+d x) \left (\frac {\left (3-4 m+m^2\right ) \left (i B \left (-1+m^2\right )+A \left (1-4 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )}{1+m}+\frac {(-2+m) m \left (-i A \left (8-6 m+m^2\right )+B \left (-2-2 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{2+m}+\frac {6 (A+i B)}{(-i+\tan (c+d x))^4}+\frac {2 (-i A (-5+m)+B (-1+m))}{(-i+\tan (c+d x))^3}-\frac {A \left (13-7 m+m^2\right )+i B \left (-1-3 m+m^2\right )}{(-i+\tan (c+d x))^2}+\frac {(-2+m) \left (B \left (2+2 m-m^2\right )+i A \left (8-6 m+m^2\right )\right )}{-i+\tan (c+d x)}\right )}{48 a^4 d} \] Input:

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

Output:

(Tan[c + d*x]^(1 + m)*(((3 - 4*m + m^2)*(I*B*(-1 + m^2) + A*(1 - 4*m + m^2 
))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2])/(1 + m) + 
((-2 + m)*m*((-I)*A*(8 - 6*m + m^2) + B*(-2 - 2*m + m^2))*Hypergeometric2F 
1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(2 + m) + (6*(A 
+ I*B))/(-I + Tan[c + d*x])^4 + (2*((-I)*A*(-5 + m) + B*(-1 + m)))/(-I + T 
an[c + d*x])^3 - (A*(13 - 7*m + m^2) + I*B*(-1 - 3*m + m^2))/(-I + Tan[c + 
 d*x])^2 + ((-2 + m)*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2)))/(-I + Tan[ 
c + d*x])))/(48*a^4*d)
 

Rubi [A] (verified)

Time = 2.03 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4079, 3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4021, 3042, 3957, 278}

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.

\(\displaystyle \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^m (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4}dx\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\int \frac {\tan ^m(c+d x) (a (A (7-m)-i B (m+1))-a (i A-B) (3-m) \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^m (a (A (7-m)-i B (m+1))-a (i A-B) (3-m) \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int -\frac {2 \tan ^m(c+d x) \left (a^2 \left (i B \left (-m^2+3 m+4\right )-A \left (m^2-7 m+16\right )\right )-a^2 (B (1-m)-i A (5-m)) (2-m) \tan (c+d x)\right )}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}+\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^m(c+d x) \left (a^2 \left (i B \left (-m^2+3 m+4\right )-A \left (m^2-7 m+16\right )\right )-a^2 (B (1-m)-i A (5-m)) (2-m) \tan (c+d x)\right )}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x)^m \left (a^2 \left (i B \left (-m^2+3 m+4\right )-A \left (m^2-7 m+16\right )\right )-a^2 (B (1-m)-i A (5-m)) (2-m) \tan (c+d x)\right )}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int -\frac {2 \tan ^m(c+d x) \left (a^3 \left (A \left (-m^3+8 m^2-20 m+19\right )-i B \left (m^3-4 m^2+2 m+7\right )\right )-a^3 (1-m) \left (B \left (-m^2+3 m+1\right )+i A \left (m^2-7 m+13\right )\right ) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {\int \frac {\tan ^m(c+d x) \left (a^3 \left (A \left (-m^3+8 m^2-20 m+19\right )-i B \left (m^3-4 m^2+2 m+7\right )\right )-a^3 (1-m) \left (B \left (-m^2+3 m+1\right )+i A \left (m^2-7 m+13\right )\right ) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x)^m \left (a^3 \left (A \left (-m^3+8 m^2-20 m+19\right )-i B \left (m^3-4 m^2+2 m+7\right )\right )-a^3 (1-m) \left (B \left (-m^2+3 m+1\right )+i A \left (m^2-7 m+13\right )\right ) \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {\frac {\int -2 \tan ^m(c+d x) \left (a^4 \left (m^2-4 m+3\right ) \left (i B \left (1-m^2\right )-A \left (m^2-4 m+1\right )\right )-a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan (c+d x)\right )dx}{2 a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {\int \tan ^m(c+d x) \left (a^4 \left (m^2-4 m+3\right ) \left (i B \left (1-m^2\right )-A \left (m^2-4 m+1\right )\right )-a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan (c+d x)\right )dx}{a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {\int \tan (c+d x)^m \left (a^4 \left (m^2-4 m+3\right ) \left (i B \left (1-m^2\right )-A \left (m^2-4 m+1\right )\right )-a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan (c+d x)\right )dx}{a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {a^4 \left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \int \tan ^m(c+d x)dx-a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \int \tan ^{m+1}(c+d x)dx}{a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {a^4 \left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \int \tan (c+d x)^mdx-a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \int \tan (c+d x)^{m+1}dx}{a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {\frac {a^4 \left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \int \frac {\tan ^m(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \int \frac {\tan ^{m+1}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}}{a^2}-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\frac {a (A (5-m)+i B (1-m)) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\frac {\left (-A \left (m^2-7 m+13\right )+i B \left (-m^2+3 m+1\right )\right ) \tan ^{m+1}(c+d x)}{2 d (1+i \tan (c+d x))^2}-\frac {-\frac {a^3 (2-m) \left (-A \left (m^2-6 m+8\right )+i B \left (-m^2+2 m+2\right )\right ) \tan ^{m+1}(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\frac {a^4 \left (m^2-4 m+3\right ) \left (-A \left (m^2-4 m+1\right )+i B \left (1-m^2\right )\right ) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {a^4 (2-m) m \left (B \left (-m^2+2 m+2\right )+i A \left (m^2-6 m+8\right )\right ) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d (m+2)}}{a^2}}{2 a^2}}{3 a^2}}{8 a^2}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

Input:

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

Output:

((A + I*B)*Tan[c + d*x]^(1 + m))/(8*d*(a + I*a*Tan[c + d*x])^4) + ((a*(I*B 
*(1 - m) + A*(5 - m))*Tan[c + d*x]^(1 + m))/(3*d*(a + I*a*Tan[c + d*x])^3) 
 - (((I*B*(1 + 3*m - m^2) - A*(13 - 7*m + m^2))*Tan[c + d*x]^(1 + m))/(2*d 
*(1 + I*Tan[c + d*x])^2) - (-((a^3*(2 - m)*(I*B*(2 + 2*m - m^2) - A*(8 - 6 
*m + m^2))*Tan[c + d*x]^(1 + m))/(d*(a + I*a*Tan[c + d*x]))) - ((a^4*(3 - 
4*m + m^2)*(I*B*(1 - m^2) - A*(1 - 4*m + m^2))*Hypergeometric2F1[1, (1 + m 
)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m))/(d*(1 + m)) - (a^4* 
(2 - m)*m*(B*(2 + 2*m - m^2) + I*A*(8 - 6*m + m^2))*Hypergeometric2F1[1, ( 
2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(d*(2 + m)))/a 
^2)/(2*a^2))/(3*a^2))/(8*a^2)
 

Defintions of rubi rules used

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 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4021

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

rule 4079

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[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm= 
"fricas")
 

Output:

integral(1/16*((A - I*B)*e^(8*I*d*x + 8*I*c) + 2*(2*A - I*B)*e^(6*I*d*x + 
6*I*c) + 6*A*e^(4*I*d*x + 4*I*c) + 2*(2*A + I*B)*e^(2*I*d*x + 2*I*c) + A + 
 I*B)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(-8*I*d 
*x - 8*I*c)/a^4, x)
 

Sympy [F]

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

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

Output:

(Integral(A*tan(c + d*x)**m/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan 
(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x) + Integral(B*tan(c + d*x)*tan(c + 
 d*x)**m/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I* 
tan(c + d*x) + 1), x))/a**4
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm= 
"maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm= 
"giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{1,[0 
,1,0]%%%} / %%%{1,[0,0,4]%%%} Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(tan(c + d*x)**m/(tan(c + d*x)**4 - 4*tan(c + d*x)**3*i - 6*tan(c + d* 
x)**2 + 4*tan(c + d*x)*i + 1),x)*a + int((tan(c + d*x)**m*tan(c + d*x))/(t 
an(c + d*x)**4 - 4*tan(c + d*x)**3*i - 6*tan(c + d*x)**2 + 4*tan(c + d*x)* 
i + 1),x)*b)/a**4