Integrand size = 33, antiderivative size = 277 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {\left (40 a^2 A b-5 A b^3+16 a^3 B-30 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{8 \sqrt {a} d}-\frac {(a-i b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {\left (8 a^2 A-11 A b^2-18 a b B\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{8 d}-\frac {a (3 A b+2 a B) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{4 d}-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d} \]
-(a-I*b)^(5/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+ I*b)^(5/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+1/8*(40 *A*a^2*b-5*A*b^3+16*B*a^3-30*B*a*b^2)*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/ 2))/d/a^(1/2)+1/8*(8*A*a^2-11*A*b^2-18*B*a*b)*cot(d*x+c)*(a+b*tan(d*x+c))^ (1/2)/d-1/4*a*(3*A*b+2*B*a)*cot(d*x+c)^2*(a+b*tan(d*x+c))^(1/2)/d-1/3*a*A* cot(d*x+c)^3*(a+b*tan(d*x+c))^(3/2)/d
Time = 6.49 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.98 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {2 b B \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2}{3} \left (\frac {3 A b^2 \cot ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{5 d}-\frac {2}{5} \left (\frac {\left (6 A b^2-5 a (a A-2 b B)\right ) \cot ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{4 d}-\frac {\frac {15 a \left (13 a A b+6 a^2 B-8 b^2 B\right ) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{16 d}-\frac {-\frac {-\frac {45 a^{5/2} \left (40 a^2 A b-5 A b^3+16 a^3 B-30 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{16 d}+\frac {i \sqrt {a-i b} \left (\frac {45}{2} i a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {45}{2} a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(-a+i b) d}-\frac {i \sqrt {a+i b} \left (-\frac {45}{2} i a^3 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-\frac {45}{2} a^3 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(-a-i b) d}}{a}+\frac {45 a^2 \left (8 a^2 A-11 A b^2-18 a b B\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{16 d}}{2 a}}{3 a}\right )\right ) \]
(-2*b*B*Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(3/2))/(3*d) - (2*((3*A*b^2*Co t[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(5*d) - (2*(((6*A*b^2 - 5*a*(a*A - 2*b*B))*Cot[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(4*d) - ((15*a*(13*a*A*b + 6*a^2*B - 8*b^2*B)*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(16*d) - (-( ((-45*a^(5/2)*(40*a^2*A*b - 5*A*b^3 + 16*a^3*B - 30*a*b^2*B)*ArcTanh[Sqrt[ a + b*Tan[c + d*x]]/Sqrt[a]])/(16*d) + (I*Sqrt[a - I*b]*(((45*I)/2)*a^3*(3 *a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B) - (45*a^3*(a^3*A - 3*a*A*b^2 - 3*a^2 *b*B + b^3*B))/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/((-a + I*b)*d) - (I*Sqrt[a + I*b]*(((-45*I)/2)*a^3*(3*a^2*A*b - A*b^3 + a^3*B - 3 *a*b^2*B) - (45*a^3*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B))/2)*ArcTanh[Sq rt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/((-a - I*b)*d))/a) + (45*a^2*(8*a^2 *A - 11*A*b^2 - 18*a*b*B)*Cot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(16*d))/( 2*a))/(3*a)))/5))/3
Time = 2.46 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4132, 27, 3042, 4136, 27, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 221}
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.
\(\displaystyle \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^4}dx\) |
\(\Big \downarrow \) 4088 |
\(\displaystyle \frac {1}{3} \int \frac {3}{2} \cot ^3(c+d x) \sqrt {a+b \tan (c+d x)} \left (-b (a A-2 b B) \tan ^2(c+d x)-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)\right )dx-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \cot ^3(c+d x) \sqrt {a+b \tan (c+d x)} \left (-b (a A-2 b B) \tan ^2(c+d x)-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)\right )dx-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (a A-2 b B) \tan (c+d x)^2-2 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (3 A b+2 a B)\right )}{\tan (c+d x)^3}dx-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4128 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {\cot ^2(c+d x) \left (b \left (6 B a^2+13 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-18 b B a-11 A b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int \frac {\cot ^2(c+d x) \left (b \left (6 B a^2+13 A b a-8 b^2 B\right ) \tan ^2(c+d x)+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-18 b B a-11 A b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int \frac {b \left (6 B a^2+13 A b a-8 b^2 B\right ) \tan (c+d x)^2+8 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (8 A a^2-18 b B a-11 A b^2\right )}{\tan (c+d x)^2 \sqrt {a+b \tan (c+d x)}}dx-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\int -\frac {\cot (c+d x) \left (-a b \left (8 A a^2-18 b B a-11 A b^2\right ) \tan ^2(c+d x)-16 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (16 B a^3+40 A b a^2-30 b^2 B a-5 A b^3\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{a}+\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {\cot (c+d x) \left (-a b \left (8 A a^2-18 b B a-11 A b^2\right ) \tan ^2(c+d x)-16 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (16 B a^3+40 A b a^2-30 b^2 B a-5 A b^3\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {-a b \left (8 A a^2-18 b B a-11 A b^2\right ) \tan (c+d x)^2-16 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (16 B a^3+40 A b a^2-30 b^2 B a-5 A b^3\right )}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx}{2 a}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4136 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int -\frac {16 \left (a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx+a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\cot (c+d x) \left (\tan ^2(c+d x)+1\right )}{\sqrt {a+b \tan (c+d x)}}dx-16 \int \frac {a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \int \frac {a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+a \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{2 a}\right )-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\right )-\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {1}{2} a (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {1}{2} a (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {i a (a-i b)^3 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i a (a+i b)^3 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {i a (a+i b)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i a (a-i b)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {a (a-i b)^3 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {a (a+i b)^3 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-16 \left (\frac {a (a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {a (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\frac {a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}-16 \left (\frac {a (a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {a (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {\frac {2 a \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}-16 \left (\frac {a (a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {a (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a A \cot ^3(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {1}{2} \left (-\frac {a (2 a B+3 A b) \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {\left (8 a^2 A-18 a b B-11 A b^2\right ) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{d}-\frac {-\frac {2 \sqrt {a} \left (16 a^3 B+40 a^2 A b-30 a b^2 B-5 A b^3\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}-16 \left (\frac {a (a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {a (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}\right )}{2 a}\right )\right )\) |
-1/3*(a*A*Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(3/2))/d + (-1/2*(a*(3*A*b + 2*a*B)*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/d + (-1/2*(-16*((a*(a - I *b)^(5/2)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + (a*(a + I*b)^( 5/2)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d) - (2*Sqrt[a]*(40*a^2 *A*b - 5*A*b^3 + 16*a^3*B - 30*a*b^2*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/S qrt[a]])/d)/a + ((8*a^2*A - 11*A*b^2 - 18*a*b*B)*Cot[c + d*x]*Sqrt[a + b*T an[c + d*x]])/d)/4)/2
3.4.38.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
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])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
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]
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])))
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*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[n, 0] && !LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(2667\) vs. \(2(239)=478\).
Time = 0.25 (sec) , antiderivative size = 2668, normalized size of antiderivative = 9.63
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2668\) |
default | \(\text {Expression too large to display}\) | \(2668\) |
-11/8/d/tan(d*x+c)^3*(a+b*tan(d*x+c))^(5/2)*A+2/d*a^(5/2)*arctanh((a+b*tan (d*x+c))^(1/2)/a^(1/2))*B+1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b *tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^ (1/2))*B*a^3+1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1 /2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A- 1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2* (a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A-1/4/d*b^2*ln( (a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b ^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/4/d*b^2*ln(b*tan(d*x+c)+a+(a+ b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*B*(2*(a ^2+b^2)^(1/2)+2*a)^(1/2)+3/4/d*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2 )+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/ 2)*a^2-1/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^( 1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a^3-3/4/d* ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^ 2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-3/d*b/(2*(a^2+b^2)^(1/2) -2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2) )/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a^2+3/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/ 2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2 +b^2)^(1/2)-2*a)^(1/2))*B*a+1/d*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arcta...
Leaf count of result is larger than twice the leaf count of optimal. 4974 vs. \(2 (233) = 466\).
Time = 25.11 (sec) , antiderivative size = 9965, normalized size of antiderivative = 35.97 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
Timed out. \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Time = 11.68 (sec) , antiderivative size = 33949, normalized size of antiderivative = 122.56 \[ \int \cot ^4(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
(atan((((((a + b*tan(c + d*x))^(1/2)*(153*A^4*b^20 + 128*B^4*b^20 + 231*A^ 2*B^2*b^20 - 7*A^4*a^2*b^18 + 9895*A^4*a^4*b^16 - 27465*A^4*a^6*b^14 + 263 20*A^4*a^8*b^12 - 832*A^4*a^10*b^10 + 128*A^4*a^12*b^8 - 132*B^4*a^2*b^18 + 16380*B^4*a^4*b^16 - 25596*B^4*a^6*b^14 + 21060*B^4*a^8*b^12 - 4032*B^4* a^10*b^10 + 384*B^4*a^12*b^8 + 6811*A^2*B^2*a^2*b^18 - 61315*A^2*B^2*a^4*b ^16 + 184661*A^2*B^2*a^6*b^14 - 121620*A^2*B^2*a^8*b^12 + 23296*A^2*B^2*a^ 10*b^10 - 300*A*B^3*a*b^19 + 600*A^3*B*a*b^19 + 17860*A*B^3*a^3*b^17 - 917 00*A*B^3*a^5*b^15 + 110172*A*B^3*a^7*b^13 - 43520*A*B^3*a^9*b^11 + 4352*A* B^3*a^11*b^9 - 12860*A^3*B*a^3*b^17 + 79680*A^3*B*a^5*b^15 - 126700*A^3*B* a^7*b^13 + 40960*A^3*B*a^9*b^11 - 1280*A^3*B*a^11*b^9))/(64*d^4) + (((3225 *A^3*a^3*b^15*d^2 - 1088*A^3*a^5*b^13*d^2 - 3984*A^3*a^7*b^11*d^2 + 736*A^ 3*a^9*b^9*d^2 + 1854*B^3*a^2*b^16*d^2 - 3456*B^3*a^4*b^14*d^2 - 2910*B^3*a ^6*b^12*d^2 + 2304*B^3*a^8*b^10*d^2 - 96*B^3*a^10*b^8*d^2 + (295*A^2*B*b^1 8*d^2)/2 - 407*A^3*a*b^17*d^2 + 1178*A*B^2*a*b^17*d^2 - 10572*A*B^2*a^3*b^ 15*d^2 + 1930*A*B^2*a^5*b^13*d^2 + 11472*A*B^2*a^7*b^11*d^2 - 2208*A*B^2*a ^9*b^9*d^2 - 4716*A^2*B*a^2*b^16*d^2 + (22193*A^2*B*a^4*b^14*d^2)/2 + 8568 *A^2*B*a^6*b^12*d^2 - 7104*A^2*B*a^8*b^10*d^2 + 288*A^2*B*a^10*b^8*d^2)/(1 6*d^5) + ((((a + b*tan(c + d*x))^(1/2)*(320*A^2*a^3*b^12*d^2 - 19456*A^2*a ^5*b^10*d^2 + 1280*A^2*a^7*b^8*d^2 - 2320*B^2*a^3*b^12*d^2 + 16896*B^2*a^5 *b^10*d^2 - 2304*B^2*a^7*b^8*d^2 - 2048*A*B*b^15*d^2 + 4764*A^2*a*b^14*...