Integrand size = 35, antiderivative size = 309 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(i a+b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (10 A b+7 a B) \sqrt {a+b \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (35 a^2 A-45 A b^2-77 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{105 a d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)} \]
(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))/d+(I*a+b)^(5/2)*(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/( a+b*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 )*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2)-2/35*a*(10*A*b+7*B*a)*(a+b*t an(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+2/105*(35*A*a^2-45*A*b^2-77*B*a*b)*(a+ b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)-2/7*a*A*(a+b*tan(d*x+c))^(3/2)/d/ta n(d*x+c)^(7/2)
Time = 6.80 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {420 \sqrt [4]{-1} a \left ((-a+i b)^{5/2} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+(a+i b)^{5/2} (-i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right ) \tan ^{\frac {7}{2}}(c+d x)-35 a b (4 A b+a B) \sqrt {a+b \tan (c+d x)}-5 a \left (24 a^2 A-28 A b^2-49 a b B\right ) \sqrt {a+b \tan (c+d x)}-6 a \left (60 a A b+28 a^2 B-35 b^2 B\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}+8 a \left (35 a^2 A-45 A b^2-77 a b B\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}+8 \left (245 a^2 A b-15 A b^3+105 a^3 B-161 a b^2 B\right ) \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}-210 a b B (a+b \tan (c+d x))^{3/2}}{420 a d \tan ^{\frac {7}{2}}(c+d x)} \]
(420*(-1)^(1/4)*a*((-a + I*b)^(5/2)*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + (a + I*b)^(5/2)*((-I )*A + B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T an[c + d*x]]])*Tan[c + d*x]^(7/2) - 35*a*b*(4*A*b + a*B)*Sqrt[a + b*Tan[c + d*x]] - 5*a*(24*a^2*A - 28*A*b^2 - 49*a*b*B)*Sqrt[a + b*Tan[c + d*x]] - 6*a*(60*a*A*b + 28*a^2*B - 35*b^2*B)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]] + 8*a*(35*a^2*A - 45*A*b^2 - 77*a*b*B)*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]] + 8*(245*a^2*A*b - 15*A*b^3 + 105*a^3*B - 161*a*b^2*B)*Tan[c + d*x]^ 3*Sqrt[a + b*Tan[c + d*x]] - 210*a*b*B*(a + b*Tan[c + d*x])^(3/2))/(420*a* d*Tan[c + d*x]^(7/2))
Time = 2.39 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.13, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{9/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan ^2(c+d x)-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{2 \tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan ^2(c+d x)-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (4 a A-7 b B) \tan (c+d x)^2-7 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (10 A b+7 a B)\right )}{\tan (c+d x)^{7/2}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int -\frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan ^2(c+d x)+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (-\frac {1}{5} \int \frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan ^2(c+d x)+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (-\frac {1}{5} \int \frac {b \left (28 B a^2+60 A b a-35 b^2 B\right ) \tan (c+d x)^2+35 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (35 A a^2-77 b B a-45 A b^2\right )}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \int -\frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}+\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 a b \left (35 A a^2-77 b B a-45 A b^2\right ) \tan (c+d x)^2-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+245 A b a^2-161 b^2 B a-15 A b^3\right )}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \int \frac {105 \left (\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {105 \int \frac {\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {105 \int \frac {\left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) a^2+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}\right )-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {1}{2} a^2 (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a^2 (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {1}{2} a^2 (a-i b)^3 (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a^2 (a+i b)^3 (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a^2 (a+i b)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}\right )\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a^2 (a+i b)^3 (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}\right )\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a-i b)^3 (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a^2 (a+i b)^3 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}\right )\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (-\frac {2 a (7 a B+10 A b) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (\frac {2 \left (35 a^2 A-77 a b B-45 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+245 a^2 A b-161 a b^2 B-15 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {105 \left (\frac {a^2 (a+i b)^3 (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a^2 (a-i b)^3 (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\right )}{a}}{3 a}\right )\right )\) |
(-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]*Sqrt[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 - 16 1*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/5)/7
3.5.48.3.1 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_))/((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]
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])
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])
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_)*((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]
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]
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])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.06 (sec) , antiderivative size = 2654465, normalized size of antiderivative = 8590.50
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 18856 vs. \(2 (257) = 514\).
Time = 4.85 (sec) , antiderivative size = 18856, normalized size of antiderivative = 61.02 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \]