\(\int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [633]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [F(-1)]
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 349 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{105 a d}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{105 d}-\frac {2 a (10 A b+7 a B) \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{35 d}-\frac {2 a A \cot ^{\frac {7}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{7 d} \] Output:

(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c 
))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+(I*a+b)^(5/2)*(I*A+B)*arctan 
h((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)* 
tan(d*x+c)^(1/2)/d+2/105*(245*A*a^2*b-15*A*b^3+105*B*a^3-161*B*a*b^2)*cot( 
d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/a/d+2/105*(35*A*a^2-45*A*b^2-77*B*a*b) 
*cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^(1/2)/d-2/35*a*(10*A*b+7*B*a)*cot(d*x+c 
)^(5/2)*(a+b*tan(d*x+c))^(1/2)/d-2/7*a*A*cot(d*x+c)^(7/2)*(a+b*tan(d*x+c)) 
^(3/2)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 33.05 (sec) , antiderivative size = 5276, normalized size of antiderivative = 15.12 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Result too large to show} \] Input:

Integrate[Cot[c + d*x]^(9/2)*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x 
]),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 4.27 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.06, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4729, 3042, 4088, 27, 3042, 4128, 27, 3042, 4132, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

\(\displaystyle \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot (c+d x)^{9/2} (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{9/2}}dx\)

\(\Big \downarrow \) 4088

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan ^2(c+d x)-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{2 \tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan ^2(c+d x)-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan (c+d x)^2-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{\tan (c+d x)^{7/2}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4128

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {2}{5} \int -\frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan ^2(c+d x)+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {1}{5} \int \frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan ^2(c+d x)+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {1}{5} \int \frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan (c+d x)^2+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4132

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \int -\frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}+\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan (c+d x)^2-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4132

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \int \frac {105 \left (\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {105 \int \frac {\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {105 \int \frac {\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4099

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {1}{2} a^2 (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a^2 (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {1}{2} a^2 (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a^2 (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\right )\)

\(\Big \downarrow \) 4098

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a^2 (a+i b)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}\right )\right )\right )\)

\(\Big \downarrow \) 104

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a^2 (a+i b)^3 (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}\right )\right )\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a^2 (a+i b)^3 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}\right )\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a+i b)^3 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a^2 (a-i b)^3 (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a}}{3 a}\right )\right )\right )\)

Input:

Int[Cot[c + d*x]^(9/2)*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*a*A*(a + b*Tan[c + d*x])^(3/2)) 
/(7*d*Tan[c + d*x]^(7/2)) + ((-2*a*(10*A*b + 7*a*B)*Sqrt[a + b*Tan[c + d*x 
]])/(5*d*Tan[c + d*x]^(5/2)) + ((2*(35*a^2*A - 45*A*b^2 - 77*a*b*B)*Sqrt[a 
 + b*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - ((-105*((a^2*(a + I*b)^3*(A 
 + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] 
])/(Sqrt[I*a - b]*d) + (a^2*(a - I*b)^3*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*S 
qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d)))/a - (2*( 
245*a^2*A*b - 15*A*b^3 + 105*a^3*B - 161*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]] 
)/(d*Sqrt[Tan[c + d*x]]))/(3*a))/5)/7)
 

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 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4088
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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x 
])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) 
  Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* 
(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n 
 + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ 
e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n 
 + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & 
& LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4098
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[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099
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[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4128
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[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4132
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[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4729
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 
Maple [F(-1)]

Timed out.

hanged

Input:

int(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
 

Output:

int(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18996 vs. \(2 (289) = 578\).

Time = 4.34 (sec) , antiderivative size = 18996, normalized size of antiderivative = 54.43 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(9/2)*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {9}{2}} \,d x } \] Input:

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="maxima")
 

Output:

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(9/ 
2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algo 
rithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{9/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^(9/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)
 

Output:

int(cot(c + d*x)^(9/2)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \cot ^{\frac {9}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \cot \left (d x +c \right )^{\frac {9}{2}} \left (a +\tan \left (d x +c \right ) b \right )^{\frac {5}{2}} \left (A +B \tan \left (d x +c \right )\right )d x \] Input:

int(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
 

Output:

int(cot(d*x+c)^(9/2)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)