Integrand size = 47, antiderivative size = 325 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(a-i b)^2 (B+i (A-C)) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(a+i b)^2 (B-i (A-C)) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 \left (a^2 B-b^2 B+2 a b (A-C)\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}}{105 d^3 f}-\frac {2 b (4 b c C-7 b B d-4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{35 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f} \]
-(a-I*b)^2*(B+I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I* d)^(1/2)/f-(a+I*b)^2*(B-I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1 /2))*(c+I*d)^(1/2)/f+2*(B*a^2-B*b^2+2*a*b*(A-C))*(c+d*tan(f*x+e))^(1/2)/f+ 2/105*(20*a^2*C*d^2-14*a*b*d*(-5*B*d+2*C*c)+b^2*(8*c^2*C-14*B*c*d+35*(A-C) *d^2))*(c+d*tan(f*x+e))^(3/2)/d^3/f-2/35*b*(-7*B*b*d-4*C*a*d+4*C*b*c)*tan( f*x+e)*(c+d*tan(f*x+e))^(3/2)/d^2/f+2/7*C*(a+b*tan(f*x+e))^2*(c+d*tan(f*x+ e))^(3/2)/d/f
Time = 5.24 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.97 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {2 \left (\left (20 a^2 C d^2+14 a b d (-2 c C+5 B d)+b^2 \left (8 c^2 C-14 B c d+35 (A-C) d^2\right )\right ) (c+d \tan (e+f x))^{3/2}+3 b d (-4 b c C+7 b B d+4 a C d) \tan (e+f x) (c+d \tan (e+f x))^{3/2}+15 C d^2 (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}+\frac {105}{2} (a-i b)^2 (i A+B-i C) d^3 \left (-\sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )+\frac {105}{2} (a+i b)^2 (-i A+B+i C) d^3 \left (-\sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )\right )}{105 d^3 f} \]
Integrate[(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]*(A + B*Tan[e + f *x] + C*Tan[e + f*x]^2),x]
(2*((20*a^2*C*d^2 + 14*a*b*d*(-2*c*C + 5*B*d) + b^2*(8*c^2*C - 14*B*c*d + 35*(A - C)*d^2))*(c + d*Tan[e + f*x])^(3/2) + 3*b*d*(-4*b*c*C + 7*b*B*d + 4*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^(3/2) + 15*C*d^2*(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2) + (105*(a - I*b)^2*(I*A + B - I*C)*d^3 *(-(Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]) + Sqrt[ c + d*Tan[e + f*x]]))/2 + (105*(a + I*b)^2*((-I)*A + B + I*C)*d^3*(-(Sqrt[ c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]) + Sqrt[c + d*Tan [e + f*x]]))/2))/(105*d^3*f)
Time = 2.31 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.362, Rules used = {3042, 4130, 27, 3042, 4120, 27, 3042, 4113, 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.
| \(\displaystyle \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {2 \int -\frac {1}{2} (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left ((4 b c C-4 a d C-7 b B d) \tan ^2(e+f x)-7 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (7 A-3 C) d\right )dx}{7 d}+\frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left ((4 b c C-4 a d C-7 b B d) \tan ^2(e+f x)-7 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (7 A-3 C) d\right )dx}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left ((4 b c C-4 a d C-7 b B d) \tan (e+f x)^2-7 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (7 A-3 C) d\right )dx}{7 d}\) |
| \(\Big \downarrow \) 4120 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 \int -\frac {1}{2} \sqrt {c+d \tan (e+f x)} \left (-2 c (4 c C-7 B d) b^2+28 a c C d b-5 a^2 (7 A-3 C) d^2-\left (\left (8 C c^2-14 B d c+35 (A-C) d^2\right ) b^2-14 a d (2 c C-5 B d) b+20 a^2 C d^2\right ) \tan ^2(e+f x)-35 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)\right )dx}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \sqrt {c+d \tan (e+f x)} \left (-2 c (4 c C-7 B d) b^2+28 a c C d b-5 a^2 (7 A-3 C) d^2-\left (\left (8 C c^2-14 B d c+35 (A-C) d^2\right ) b^2-14 a d (2 c C-5 B d) b+20 a^2 C d^2\right ) \tan ^2(e+f x)-35 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)\right )dx}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \sqrt {c+d \tan (e+f x)} \left (-2 c (4 c C-7 B d) b^2+28 a c C d b-5 a^2 (7 A-3 C) d^2-\left (\left (8 C c^2-14 B d c+35 (A-C) d^2\right ) b^2-14 a d (2 c C-5 B d) b+20 a^2 C d^2\right ) \tan (e+f x)^2-35 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)\right )dx}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \sqrt {c+d \tan (e+f x)} \left (35 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-35 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)\right )dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \sqrt {c+d \tan (e+f x)} \left (35 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-35 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)\right )dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \frac {-35 \left ((A c-C c-B d) a^2-2 b (B c+(A-C) d) a-b^2 (A c-C c-B d)\right ) d^2-35 \left ((B c+(A-C) d) a^2+2 b (A c-C c-B d) a-b^2 (B c+(A-C) d)\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {\int \frac {-35 \left ((A c-C c-B d) a^2-2 b (B c+(A-C) d) a-b^2 (A c-C c-B d)\right ) d^2-35 \left ((B c+(A-C) d) a^2+2 b (A c-C c-B d) a-b^2 (B c+(A-C) d)\right ) \tan (e+f x) d^2}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}+\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}}{7 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {-\frac {35}{2} d^2 (a+i b)^2 (c+i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {35}{2} d^2 (a-i b)^2 (c-i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {-\frac {35}{2} d^2 (a+i b)^2 (c+i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {35}{2} d^2 (a-i b)^2 (c-i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {-\frac {35 i d^2 (a-i b)^2 (c-i d) (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {35 i d^2 (a+i b)^2 (c+i d) (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {\frac {35 i d^2 (a-i b)^2 (c-i d) (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {35 i d^2 (a+i b)^2 (c+i d) (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {-\frac {35 d (a-i b)^2 (c-i d) (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {35 d (a+i b)^2 (c+i d) (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}}{5 d}}{7 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}}{7 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-7 b B d+4 b c C) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac {-\frac {2 (c+d \tan (e+f x))^{3/2} \left (20 a^2 C d^2-14 a b d (2 c C-5 B d)+b^2 \left (35 d^2 (A-C)-14 B c d+8 c^2 C\right )\right )}{3 d f}-\frac {70 d^2 \left (a^2 B+2 a b (A-C)-b^2 B\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {35 d^2 (a-i b)^2 \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {35 d^2 (a+i b)^2 \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}}{5 d}}{7 d}\) |
Int[(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]
(2*C*(a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2))/(7*d*f) - ((2*b*(4 *b*c*C - 7*b*B*d - 4*a*C*d)*Tan[e + f*x]*(c + d*Tan[e + f*x])^(3/2))/(5*d* f) + ((-35*(a - I*b)^2*(A - I*B - C)*Sqrt[c - I*d]*d^2*ArcTan[Tan[e + f*x] /Sqrt[c - I*d]])/f - (35*(a + I*b)^2*(A + I*B - C)*Sqrt[c + I*d]*d^2*ArcTa n[Tan[e + f*x]/Sqrt[c + I*d]])/f - (70*(a^2*B - b^2*B + 2*a*b*(A - C))*d^2 *Sqrt[c + d*Tan[e + f*x]])/f - (2*(20*a^2*C*d^2 - 14*a*b*d*(2*c*C - 5*B*d) + b^2*(8*c^2*C - 14*B*c*d + 35*(A - C)*d^2))*(c + d*Tan[e + f*x])^(3/2))/ (3*d*f))/(5*d))/(7*d)
3.1.91.3.1 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]
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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*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[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(3352\) vs. \(2(291)=582\).
Time = 0.16 (sec) , antiderivative size = 3353, normalized size of antiderivative = 10.32
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(3353\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(4775\) |
| default | \(\text {Expression too large to display}\) | \(4775\) |
int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e) ^2),x,method=_RETURNVERBOSE)
4/3/f/d*B*a*b*(c+d*tan(f*x+e))^(3/2)-1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a rctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2 )^(1/2)-2*c)^(1/2))*A*a^2+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*( c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c )^(1/2))*A*b^2+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1 /2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*C* a^2+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+( 2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*A*a^2-1/f*d/( 2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2) ^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*A*b^2-1/f*d/(2*(c^2+d^2) ^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c) ^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*C*a^2+1/2/f/d*ln(d*tan(f*x+e)+c+(c+ d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*B*(2*(c ^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b-1/2/f/d*ln(d*tan(f*x+e)+c+(c+ d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*B*(2*(c ^2+d^2)^(1/2)+2*c)^(1/2)*a*b*c-1/2/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d ^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*B*(2*(c^2+d^2)^(1/2)+ 2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b+1/2/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d ^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*B*(2*(c^2+d^2)^(1/2)+ 2*c)^(1/2)*a*b*c+2/f*b*(B*b+2*C*a)/d^2*(1/5*(c+d*tan(f*x+e))^(5/2)-1/3*...
Leaf count of result is larger than twice the leaf count of optimal. 23984 vs. \(2 (281) = 562\).
Time = 4.69 (sec) , antiderivative size = 23984, normalized size of antiderivative = 73.80 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan( f*x+e)^2),x, algorithm="fricas")
\[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
Integral((a + b*tan(e + f*x))**2*sqrt(c + d*tan(e + f*x))*(A + B*tan(e + f *x) + C*tan(e + f*x)**2), x)
\[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan( f*x+e)^2),x, algorithm="maxima")
integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^2*s qrt(d*tan(f*x + e) + c), x)
Timed out. \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan( f*x+e)^2),x, algorithm="giac")
Timed out. \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Hanged} \]