Integrand size = 35, antiderivative size = 378 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (4 A b+3 a B) \sqrt {a+b \tan (c+d x)}}{21 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt {a+b \tan (c+d x)}}{105 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{315 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{315 a^2 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)} \]
(I*a-b)^(5/2)*(A+I*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)*(A-I*B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/( a+b*tan(d*x+c))^(1/2))/d-2/315*(315*A*a^4-483*A*a^2*b^2-10*A*b^4-735*B*a^3 *b+45*B*a*b^3)*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(1/2)-2/21*a*(4*A*b +3*B*a)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)+2/105*(21*A*a^2-25*A*b^2 -45*B*a*b)*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+2/315*(231*A*a^2*b-5* A*b^3+105*B*a^3-135*B*a*b^2)*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(3/2)-2 /9*a*A*(a+b*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(9/2)
Time = 7.15 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=-\frac {b B (a+b \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {3 b (2 A b+a B) \sqrt {a+b \tan (c+d x)}}{8 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{4} \left (-\frac {\left (16 a^2 A-18 A b^2-33 a b B\right ) \sqrt {a+b \tan (c+d x)}}{6 d \tan ^{\frac {9}{2}}(c+d x)}-\frac {2 \left (\frac {6 a \left (38 a A b+18 a^2 B-21 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {2 \left (\frac {18 a^2 \left (21 a^2 A-25 A b^2-45 a b B\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (-\frac {3 a^2 \left (231 a^2 A b-5 A b^3+105 a^3 B-135 a b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (-\frac {2835 a^4 \left (\sqrt [4]{-1} (-a+i b)^{5/2} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} (a+i b)^{5/2} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{4 d}-\frac {9 a^2 \left (315 a^4 A-483 a^2 A b^2-10 A b^4-735 a^3 b B+45 a b^3 B\right ) \sqrt {a+b \tan (c+d x)}}{2 d \sqrt {\tan (c+d x)}}\right )}{3 a}\right )}{5 a}\right )}{7 a}\right )}{9 a}\right )\right ) \]
-1/3*(b*B*(a + b*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(9/2)) + ((-3*b*(2*A *b + a*B)*Sqrt[a + b*Tan[c + d*x]])/(8*d*Tan[c + d*x]^(9/2)) + (-1/6*((16* a^2*A - 18*A*b^2 - 33*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Tan[c + d*x]^(9/ 2)) - (2*((6*a*(38*a*A*b + 18*a^2*B - 21*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/ (7*d*Tan[c + d*x]^(7/2)) - (2*((18*a^2*(21*a^2*A - 25*A*b^2 - 45*a*b*B)*Sq rt[a + b*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) - (2*((-3*a^2*(231*a^2*A* b - 5*A*b^3 + 105*a^3*B - 135*a*b^2*B)*Sqrt[a + b*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/2)) - (2*((-2835*a^4*((-1)^(1/4)*(-a + I*b)^(5/2)*(A - I*B)*ArcT an[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] ] + (-1)^(1/4)*(a + I*b)^(5/2)*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]* Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]))/(4*d) - (9*a^2*(315*a^4*A - 483*a^2*A*b^2 - 10*A*b^4 - 735*a^3*b*B + 45*a*b^3*B)*Sqrt[a + b*Tan[c + d*x]])/(2*d*Sqrt[Tan[c + d*x]])))/(3*a)))/(5*a)))/(7*a)))/(9*a))/4)/3
Time = 3.00 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.13, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.629, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4132, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{11/2}}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {2}{9} \int \frac {3 \sqrt {a+b \tan (c+d x)} \left (-b (2 a A-3 b B) \tan ^2(c+d x)-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)\right )}{2 \tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (2 a A-3 b B) \tan ^2(c+d x)-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)\right )}{\tan ^{\frac {9}{2}}(c+d x)}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-b (2 a A-3 b B) \tan (c+d x)^2-3 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (4 A b+3 a B)\right )}{\tan (c+d x)^{9/2}}dx-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4128 |
| \(\displaystyle \frac {1}{3} \left (\frac {2}{7} \int -\frac {b \left (18 B a^2+38 A b a-21 b^2 B\right ) \tan ^2(c+d x)+21 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (21 A a^2-45 b B a-25 A b^2\right )}{2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {1}{7} \int \frac {b \left (18 B a^2+38 A b a-21 b^2 B\right ) \tan ^2(c+d x)+21 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (21 A a^2-45 b B a-25 A b^2\right )}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (-\frac {1}{7} \int \frac {b \left (18 B a^2+38 A b a-21 b^2 B\right ) \tan (c+d x)^2+21 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)+a \left (21 A a^2-45 b B a-25 A b^2\right )}{\tan (c+d x)^{7/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \int -\frac {-4 a b \left (21 A a^2-45 b B a-25 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right )}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}+\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-4 a b \left (21 A a^2-45 b B a-25 A b^2\right ) \tan ^2(c+d x)-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right )}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-4 a b \left (21 A a^2-45 b B a-25 A b^2\right ) \tan (c+d x)^2-105 a \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)+a \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right )}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \int \frac {315 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2+2 b \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right ) \tan ^2(c+d x) a+\left (315 A a^4-735 b B a^3-483 A b^2 a^2+45 b^3 B a-10 A b^4\right ) a}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {\int \frac {315 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2+2 b \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right ) \tan ^2(c+d x) a+\left (315 A a^4-735 b B a^3-483 A b^2 a^2+45 b^3 B a-10 A b^4\right ) a}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {\int \frac {315 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x) a^2+2 b \left (105 B a^3+231 A b a^2-135 b^2 B a-5 A b^3\right ) \tan (c+d x)^2 a+\left (315 A a^4-735 b B a^3-483 A b^2 a^2+45 b^3 B a-10 A b^4\right ) a}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {-\frac {2 \int -\frac {315 \left (a^3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a^3 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {\frac {315 \int \frac {a^3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a^3 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {\frac {315 \int \frac {a^3 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right )-a^3 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )-\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {1}{2} a^3 (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {1}{2} a^3 (a-i b)^3 (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^3 (a+i b)^3 (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}}{5 a}\right )\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {a^3 (a-i b)^3 (B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^3 (a+i b)^3 (-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}}{5 a}\right )\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {a^3 (a-i b)^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (a+i b)^3 (-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}}{5 a}\right )\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {a^3 (a-i b)^3 (B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^3 (a+i b)^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\right )\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 a A (a+b \tan (c+d x))^{3/2}}{9 d \tan ^{\frac {9}{2}}(c+d x)}+\frac {1}{3} \left (-\frac {2 a (3 a B+4 A b) \sqrt {a+b \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 \left (21 a^2 A-45 a b B-25 A b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {2 \left (105 a^3 B+231 a^2 A b-135 a b^2 B-5 A b^3\right ) \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (315 a^4 A-735 a^3 b B-483 a^2 A b^2+45 a b^3 B-10 A b^4\right ) \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {315 \left (\frac {a^3 (a-i b)^3 (B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}-\frac {a^3 (a+i b)^3 (-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\right )}{a}}{3 a}}{5 a}\right )\right )\) |
(-2*a*A*(a + b*Tan[c + d*x])^(3/2))/(9*d*Tan[c + d*x]^(9/2)) + ((-2*a*(4*A *b + 3*a*B)*Sqrt[a + b*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) + ((2*(21*a ^2*A - 25*A*b^2 - 45*a*b*B)*Sqrt[a + b*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5 /2)) - ((-2*(231*a^2*A*b - 5*A*b^3 + 105*a^3*B - 135*a*b^2*B)*Sqrt[a + b*T an[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) - ((315*(-((a^3*(a + I*b)^3*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(S qrt[I*a - b]*d)) + (a^3*(a - I*b)^3*(I*A + 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*(315* a^4*A - 483*a^2*A*b^2 - 10*A*b^4 - 735*a^3*b*B + 45*a*b^3*B)*Sqrt[a + b*Ta n[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/(5*a))/7)/3
3.5.49.3.1 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 = 1.05 (sec) , antiderivative size = 2659448, normalized size of antiderivative = 7035.58
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 18922 vs. \(2 (320) = 640\).
Time = 4.90 (sec) , antiderivative size = 18922, normalized size of antiderivative = 50.06 \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {11}{2}}(c+d x)} \, dx=\text {Hanged} \]