\(\int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) [479]

Optimal result

Integrand size = 31, antiderivative size = 403 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {b \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) (3+m) (4+m)}+\frac {\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac {b^2 \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan ^{2+m}(c+d x)}{d (2+m) (3+m) (4+m)}+\frac {\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{d (2+m)}+\frac {b (A b (4+m)+a B (7+m)) \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^2}{d (3+m) (4+m)}+\frac {b B \tan ^{1+m}(c+d x) (a+b \tan (c+d x))^3}{d (4+m)} \] Output:

-b*(A*b^3*(m^2+7*m+12)+4*a*b^2*B*(m^2+7*m+12)-2*a^3*B*(m^2+8*m+19)-a^2*A*b 
*(5*m^2+37*m+68))*tan(d*x+c)^(1+m)/d/(1+m)/(3+m)/(4+m)+(A*a^4-6*A*a^2*b^2+ 
A*b^4-4*B*a^3*b+4*B*a*b^3)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c 
)^2)*tan(d*x+c)^(1+m)/d/(1+m)+b^2*(2*a*A*b*(4+m)^2-b^2*B*(m^2+7*m+12)+a^2* 
B*(m^2+9*m+26))*tan(d*x+c)^(2+m)/d/(2+m)/(3+m)/(4+m)+(4*A*a^3*b-4*A*a*b^3+ 
B*a^4-6*B*a^2*b^2+B*b^4)*hypergeom([1, 1+1/2*m],[2+1/2*m],-tan(d*x+c)^2)*t 
an(d*x+c)^(2+m)/d/(2+m)+b*(A*b*(4+m)+a*B*(7+m))*tan(d*x+c)^(1+m)*(a+b*tan( 
d*x+c))^2/d/(3+m)/(4+m)+b*B*tan(d*x+c)^(1+m)*(a+b*tan(d*x+c))^3/d/(4+m)
 

Mathematica [A] (verified)

Time = 3.11 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.88 \[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {\tan ^{1+m}(c+d x) \left (-b (2+m) \left (A b^3 \left (12+7 m+m^2\right )+4 a b^2 B \left (12+7 m+m^2\right )-2 a^3 B \left (19+8 m+m^2\right )-a^2 A b \left (68+37 m+5 m^2\right )\right )+\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) (2+m) (3+m) (4+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right )+b^2 (1+m) \left (2 a A b (4+m)^2-b^2 B \left (12+7 m+m^2\right )+a^2 B \left (26+9 m+m^2\right )\right ) \tan (c+d x)+\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) (1+m) (3+m) (4+m) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)+b (1+m) (2+m) (A b (4+m)+a B (7+m)) (a+b \tan (c+d x))^2+b B (1+m) (2+m) (3+m) (a+b \tan (c+d x))^3\right )}{d (1+m) (2+m) (3+m) (4+m)} \] Input:

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

Output:

(Tan[c + d*x]^(1 + m)*(-(b*(2 + m)*(A*b^3*(12 + 7*m + m^2) + 4*a*b^2*B*(12 
 + 7*m + m^2) - 2*a^3*B*(19 + 8*m + m^2) - a^2*A*b*(68 + 37*m + 5*m^2))) + 
 (a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3*B)*(2 + m)*(3 + m)*(4 
+ m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2] + b^2*(1 
+ m)*(2*a*A*b*(4 + m)^2 - b^2*B*(12 + 7*m + m^2) + a^2*B*(26 + 9*m + m^2)) 
*Tan[c + d*x] + (4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*(1 + 
 m)*(3 + m)*(4 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d* 
x]^2]*Tan[c + d*x] + b*(1 + m)*(2 + m)*(A*b*(4 + m) + a*B*(7 + m))*(a + b* 
Tan[c + d*x])^2 + b*B*(1 + m)*(2 + m)*(3 + m)*(a + b*Tan[c + d*x])^3))/(d* 
(1 + m)*(2 + m)*(3 + m)*(4 + m))
 

Rubi [A] (verified)

Time = 2.37 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4090, 25, 3042, 4130, 3042, 4120, 25, 3042, 4113, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^m (a+b \tan (c+d x))^4 (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\int \tan ^m(c+d x) (a+b \tan (c+d x)) \left (-b \left (B \left (m^2+9 m+26\right ) a^2+2 A b (m+4)^2 a-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^2(c+d x)-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+3) (m+4) \tan (c+d x)+a (a (m+3) (b B (m+1)-a A (m+4))+b (m+1) (A b (m+4)+a B (m+7)))\right )dx}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\int \tan (c+d x)^m (a+b \tan (c+d x)) \left (-b \left (B \left (m^2+9 m+26\right ) a^2+2 A b (m+4)^2 a-b^2 B \left (m^2+7 m+12\right )\right ) \tan (c+d x)^2-\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) (m+3) (m+4) \tan (c+d x)+a (a (m+3) (b B (m+1)-a A (m+4))+b (m+1) (A b (m+4)+a B (m+7)))\right )dx}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {-\frac {\int -\tan ^m(c+d x) \left ((m+2) (a (m+3) (b B (m+1)-a A (m+4))+b (m+1) (A b (m+4)+a B (m+7))) a^2+b (m+2) \left (-2 B \left (m^2+8 m+19\right ) a^3-A b \left (5 m^2+37 m+68\right ) a^2+4 b^2 B \left (m^2+7 m+12\right ) a+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^2(c+d x)-\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) (m+2) (m+3) (m+4) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {\int \tan ^m(c+d x) \left ((m+2) (a (m+3) (b B (m+1)-a A (m+4))+b (m+1) (A b (m+4)+a B (m+7))) a^2+b (m+2) \left (-2 B \left (m^2+8 m+19\right ) a^3-A b \left (5 m^2+37 m+68\right ) a^2+4 b^2 B \left (m^2+7 m+12\right ) a+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^2(c+d x)-\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) (m+2) (m+3) (m+4) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {\int \tan (c+d x)^m \left ((m+2) (a (m+3) (b B (m+1)-a A (m+4))+b (m+1) (A b (m+4)+a B (m+7))) a^2+b (m+2) \left (-2 B \left (m^2+8 m+19\right ) a^3-A b \left (5 m^2+37 m+68\right ) a^2+4 b^2 B \left (m^2+7 m+12\right ) a+A b^3 \left (m^2+7 m+12\right )\right ) \tan (c+d x)^2-\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) (m+2) (m+3) (m+4) \tan (c+d x)\right )dx}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {\int \tan ^m(c+d x) \left (-\left (\left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) (m+2) (m+3) (m+4)\right )-\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) (m+2) (m+3) \tan (c+d x) (m+4)\right )dx+\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {\int \tan (c+d x)^m \left (-\left (\left (A a^4-4 b B a^3-6 A b^2 a^2+4 b^3 B a+A b^4\right ) (m+2) (m+3) (m+4)\right )-\left (B a^4+4 A b a^3-6 b^2 B a^2-4 A b^3 a+b^4 B\right ) (m+2) (m+3) \tan (c+d x) (m+4)\right )dx+\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 4021

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {-(m+2) (m+3) (m+4) \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int \tan ^{m+1}(c+d x)dx-(m+2) (m+3) (m+4) \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \int \tan ^m(c+d x)dx+\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {-(m+2) (m+3) (m+4) \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \int \tan (c+d x)^mdx-(m+2) (m+3) (m+4) \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int \tan (c+d x)^{m+1}dx+\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {-\frac {(m+2) (m+3) (m+4) \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \int \frac {\tan ^{m+1}(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}-\frac {(m+2) (m+3) (m+4) \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \int \frac {\tan ^m(c+d x)}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {b B \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^3}{d (m+4)}-\frac {\frac {\frac {\frac {b (m+2) \left (-2 a^3 B \left (m^2+8 m+19\right )-a^2 A b \left (5 m^2+37 m+68\right )+4 a b^2 B \left (m^2+7 m+12\right )+A b^3 \left (m^2+7 m+12\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1)}-\frac {(m+2) (m+3) (m+4) \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right ) \tan ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d (m+1)}-\frac {(m+3) (m+4) \left (a^4 B+4 a^3 A b-6 a^2 b^2 B-4 a A b^3+b^4 B\right ) \tan ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},-\tan ^2(c+d x)\right )}{d}}{m+2}-\frac {b^2 \left (a^2 B \left (m^2+9 m+26\right )+2 a A b (m+4)^2-b^2 B \left (m^2+7 m+12\right )\right ) \tan ^{m+2}(c+d x)}{d (m+2)}}{m+3}-\frac {b (a B (m+7)+A b (m+4)) \tan ^{m+1}(c+d x) (a+b \tan (c+d x))^2}{d (m+3)}}{m+4}\)

Input:

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

Output:

(b*B*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^3)/(d*(4 + m)) - (-((b*(A*b 
*(4 + m) + a*B*(7 + m))*Tan[c + d*x]^(1 + m)*(a + b*Tan[c + d*x])^2)/(d*(3 
 + m))) + (-((b^2*(2*a*A*b*(4 + m)^2 - b^2*B*(12 + 7*m + m^2) + a^2*B*(26 
+ 9*m + m^2))*Tan[c + d*x]^(2 + m))/(d*(2 + m))) + ((b*(2 + m)*(A*b^3*(12 
+ 7*m + m^2) + 4*a*b^2*B*(12 + 7*m + m^2) - 2*a^3*B*(19 + 8*m + m^2) - a^2 
*A*b*(68 + 37*m + 5*m^2))*Tan[c + d*x]^(1 + m))/(d*(1 + m)) - ((a^4*A - 6* 
a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3*B)*(2 + m)*(3 + m)*(4 + m)*Hyperge 
ometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(1 + m)) 
/(d*(1 + m)) - ((4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*(3 + 
 m)*(4 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Ta 
n[c + d*x]^(2 + m))/d)/(2 + m))/(3 + m))/(4 + m)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4090

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

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4120

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) 
*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 
 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2))   Int[(c + d*Tan[e + f*x])^n*Si 
mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* 
d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && 
  !LtQ[n, -1]
 

rule 4130

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="f 
ricas")
 

Output:

integral((B*b^4*tan(d*x + c)^5 + A*a^4 + (4*B*a*b^3 + A*b^4)*tan(d*x + c)^ 
4 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*tan(d*x + c)^3 + 2*(2*B*a^3*b + 3*A*a^2*b^ 
2)*tan(d*x + c)^2 + (B*a^4 + 4*A*a^3*b)*tan(d*x + c))*tan(d*x + c)^m, x)
 

Sympy [F]

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

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

Output:

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**4*tan(c + d*x)**m, x)
 

Maxima [F]

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

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="m 
axima")
                                                                                    
                                                                                    
 

Output:

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^4*tan(d*x + c)^m, x)
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^m*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="g 
iac")
 

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,[0,0,5]%%%} Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \tan ^m(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

(tan(c + d*x)**m*tan(c + d*x)**4*b**5*m**2 + 2*tan(c + d*x)**m*tan(c + d*x 
)**4*b**5*m + 10*tan(c + d*x)**m*tan(c + d*x)**2*a**2*b**3*m**2 + 40*tan(c 
 + d*x)**m*tan(c + d*x)**2*a**2*b**3*m - tan(c + d*x)**m*tan(c + d*x)**2*b 
**5*m**2 - 4*tan(c + d*x)**m*tan(c + d*x)**2*b**5*m + 5*tan(c + d*x)**m*a* 
*4*b*m**2 + 30*tan(c + d*x)**m*a**4*b*m + 40*tan(c + d*x)**m*a**4*b - 10*t 
an(c + d*x)**m*a**2*b**3*m**2 - 60*tan(c + d*x)**m*a**2*b**3*m - 80*tan(c 
+ d*x)**m*a**2*b**3 + tan(c + d*x)**m*b**5*m**2 + 6*tan(c + d*x)**m*b**5*m 
 + 8*tan(c + d*x)**m*b**5 + int(tan(c + d*x)**m,x)*a**5*d*m**3 + 6*int(tan 
(c + d*x)**m,x)*a**5*d*m**2 + 8*int(tan(c + d*x)**m,x)*a**5*d*m - 5*int(ta 
n(c + d*x)**m/tan(c + d*x),x)*a**4*b*d*m**3 - 30*int(tan(c + d*x)**m/tan(c 
 + d*x),x)*a**4*b*d*m**2 - 40*int(tan(c + d*x)**m/tan(c + d*x),x)*a**4*b*d 
*m + 10*int(tan(c + d*x)**m/tan(c + d*x),x)*a**2*b**3*d*m**3 + 60*int(tan( 
c + d*x)**m/tan(c + d*x),x)*a**2*b**3*d*m**2 + 80*int(tan(c + d*x)**m/tan( 
c + d*x),x)*a**2*b**3*d*m - int(tan(c + d*x)**m/tan(c + d*x),x)*b**5*d*m** 
3 - 6*int(tan(c + d*x)**m/tan(c + d*x),x)*b**5*d*m**2 - 8*int(tan(c + d*x) 
**m/tan(c + d*x),x)*b**5*d*m + 5*int(tan(c + d*x)**m*tan(c + d*x)**4,x)*a* 
b**4*d*m**3 + 30*int(tan(c + d*x)**m*tan(c + d*x)**4,x)*a*b**4*d*m**2 + 40 
*int(tan(c + d*x)**m*tan(c + d*x)**4,x)*a*b**4*d*m + 10*int(tan(c + d*x)** 
m*tan(c + d*x)**2,x)*a**3*b**2*d*m**3 + 60*int(tan(c + d*x)**m*tan(c + d*x 
)**2,x)*a**3*b**2*d*m**2 + 80*int(tan(c + d*x)**m*tan(c + d*x)**2,x)*a*...