Integrand size = 45, antiderivative size = 353 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(i a+b) (A-i B-C) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (A+i B-C) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (a \left (B c^2-2 c C d-B d^2\right )-b \left (c^2 C+2 B c d-C d^2\right )+A \left (2 a c d+b \left (c^2-d^2\right )\right )\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (A b c+a B c-b c C+a A d-b B d-a C d) (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 (A b+a B-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {2 (2 b c C-9 b B d-9 a C d) (c+d \tan (e+f x))^{7/2}}{63 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f} \]
-(I*a+b)*(A-I*B-C)*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1 /2))/f+(I*a-b)*(A+I*B-C)*(c+I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I *d)^(1/2))/f+2*(a*(B*c^2-B*d^2-2*C*c*d)-b*(2*B*c*d+C*c^2-C*d^2)+A*(2*a*c*d +b*(c^2-d^2)))*(c+d*tan(f*x+e))^(1/2)/f+2/3*(A*a*d+A*b*c+B*a*c-B*b*d-C*a*d -C*b*c)*(c+d*tan(f*x+e))^(3/2)/f+2/5*(A*b+B*a-C*b)*(c+d*tan(f*x+e))^(5/2)/ f-2/63*(-9*B*b*d-9*C*a*d+2*C*b*c)*(c+d*tan(f*x+e))^(7/2)/d^2/f+2/9*b*C*tan (f*x+e)*(c+d*tan(f*x+e))^(7/2)/d/f
Time = 5.51 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.92 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\frac {2 (-2 b c C+9 b B d+9 a C d) (c+d \tan (e+f x))^{7/2}}{d}+14 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}+\frac {63}{2} i (a-i b) (A-i B-C) d \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c-i d) \left (-3 (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )\right )-\frac {63}{2} i (a+i b) (A+i B-C) d \left (\frac {2}{5} (c+d \tan (e+f x))^{5/2}+\frac {2}{3} (c+i d) \left (-3 (c+i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )\right )}{63 d f} \]
Integrate[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f *x] + C*Tan[e + f*x]^2),x]
((2*(-2*b*c*C + 9*b*B*d + 9*a*C*d)*(c + d*Tan[e + f*x])^(7/2))/d + 14*b*C* Tan[e + f*x]*(c + d*Tan[e + f*x])^(7/2) + ((63*I)/2)*(a - I*b)*(A - I*B - C)*d*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c - I*d)*(-3*(c - I*d)^(3/2)* ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]] *(4*c - (3*I)*d + d*Tan[e + f*x])))/3) - ((63*I)/2)*(a + I*b)*(A + I*B - C )*d*((2*(c + d*Tan[e + f*x])^(5/2))/5 + (2*(c + I*d)*(-3*(c + I*d)^(3/2)*A rcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]* (4*c + (3*I)*d + d*Tan[e + f*x])))/3))/(63*d*f)
Time = 2.23 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4120, 27, 3042, 4113, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 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.
\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 4120 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {2 \int \frac {1}{2} (c+d \tan (e+f x))^{5/2} \left ((2 b c C-9 a d C-9 b B d) \tan ^2(e+f x)-9 (A b-C b+a B) d \tan (e+f x)+2 b c C-9 a A d\right )dx}{9 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left ((2 b c C-9 a d C-9 b B d) \tan ^2(e+f x)-9 (A b-C b+a B) d \tan (e+f x)+2 b c C-9 a A d\right )dx}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} \left ((2 b c C-9 a d C-9 b B d) \tan (e+f x)^2-9 (A b-C b+a B) d \tan (e+f x)+2 b c C-9 a A d\right )dx}{9 d}\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} (9 (b B-a (A-C)) d-9 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{5/2} (9 (b B-a (A-C)) d-9 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} (9 d (b B c+b (A-C) d-a (A c-C c-B d))-9 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x))dx-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int (c+d \tan (e+f x))^{3/2} (9 d (b B c+b (A-C) d-a (A c-C c-B d))-9 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x))dx-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (9 d \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )-9 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right ) \tan (e+f x)\right )dx-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (9 d \left (a \left (C c^2+2 B d c-C d^2-A \left (c^2-d^2\right )\right )+b \left (2 c (A-C) d+B \left (c^2-d^2\right )\right )\right )-9 d \left (2 a A c d-2 a c C d+A b \left (c^2-d^2\right )+a B \left (c^2-d^2\right )-b \left (C c^2+2 B d c-C d^2\right )\right ) \tan (e+f x)\right )dx-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \frac {9 d \left (a \left (C c^3+3 B d c^2-3 C d^2 c-B d^3-A \left (c^3-3 c d^2\right )\right )+b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )-9 d \left (A \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right )-b \left (C c^3+3 B d c^2-3 C d^2 c-B d^3\right )+a \left (B c^3-3 C d c^2-3 B d^2 c+C d^3\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\int \frac {9 d \left (a \left (C c^3+3 B d c^2-3 C d^2 c-B d^3-A \left (c^3-3 c d^2\right )\right )+b \left ((A-C) d \left (3 c^2-d^2\right )+B \left (c^3-3 c d^2\right )\right )\right )-9 d \left (A \left (b c^3+3 a d c^2-3 b d^2 c-a d^3\right )-b \left (C c^3+3 B d c^2-3 C d^2 c-B d^3\right )+a \left (B c^3-3 C d c^2-3 B d^2 c+C d^3\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9}{2} d (a+i b) (c+i d)^3 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {9}{2} d (a-i b) (c-i d)^3 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9}{2} d (a+i b) (c+i d)^3 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {9}{2} d (a-i b) (c-i d)^3 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9 i d (a-i b) (c-i d)^3 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {9 i d (a+i b) (c+i d)^3 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {\frac {9 i d (a-i b) (c-i d)^3 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {9 i d (a+i b) (c+i d)^3 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9 (a+i b) (c+i d)^3 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {9 (a-i b) (c-i d)^3 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{7/2}}{9 d f}-\frac {-\frac {9 d (a-i b) (c-i d)^{5/2} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {9 d (a+i b) (c+i d)^{5/2} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}-\frac {18 d \sqrt {c+d \tan (e+f x)} \left (2 a A c d+a B \left (c^2-d^2\right )-2 a c C d+A b \left (c^2-d^2\right )-b \left (2 B c d+c^2 C-C d^2\right )\right )}{f}-\frac {18 d (a B+A b-b C) (c+d \tan (e+f x))^{5/2}}{5 f}-\frac {6 d (c+d \tan (e+f x))^{3/2} (a A d+a B c-a C d+A b c-b B d-b c C)}{f}+\frac {2 (-9 a C d-9 b B d+2 b c C) (c+d \tan (e+f x))^{7/2}}{7 d f}}{9 d}\) |
Int[(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]
(2*b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])^(7/2))/(9*d*f) - ((-9*(a - I*b)*( A - I*B - C)*(c - I*d)^(5/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (9* (a + I*b)*(A + I*B - C)*(c + I*d)^(5/2)*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d ]])/f - (18*d*(2*a*A*c*d - 2*a*c*C*d + A*b*(c^2 - d^2) + a*B*(c^2 - d^2) - b*(c^2*C + 2*B*c*d - C*d^2))*Sqrt[c + d*Tan[e + f*x]])/f - (6*d*(A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d)*(c + d*Tan[e + f*x])^(3/2))/f - (18 *(A*b + a*B - b*C)*d*(c + d*Tan[e + f*x])^(5/2))/(5*f) + (2*(2*b*c*C - 9*b *B*d - 9*a*C*d)*(c + d*Tan[e + f*x])^(7/2))/(7*d*f))/(9*d)
3.2.5.3.1 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[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2)) Int[(c + d*Tan[e + f*x])^n*Si mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(7293\) vs. \(2(315)=630\).
Time = 0.21 (sec) , antiderivative size = 7294, normalized size of antiderivative = 20.66
method | result | size |
parts | \(\text {Expression too large to display}\) | \(7294\) |
derivativedivides | \(\text {Expression too large to display}\) | \(7402\) |
default | \(\text {Expression too large to display}\) | \(7402\) |
int((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2 ),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 48734 vs. \(2 (308) = 616\).
Time = 37.57 (sec) , antiderivative size = 48734, normalized size of antiderivative = 138.06 \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="fricas")
\[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right ) \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
Integral((a + b*tan(e + f*x))*(c + d*tan(e + f*x))**(5/2)*(A + B*tan(e + f *x) + C*tan(e + f*x)**2), x)
Timed out. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="maxima")
Timed out. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="giac")
Timed out. \[ \int (a+b \tan (e+f x)) (c+d \tan (e+f x))^{5/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Hanged} \]