Integrand size = 32, antiderivative size = 315 \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=-\frac {\sqrt [4]{-1} a^{3/2} \left (3 i c^2+18 c d-11 i d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{4 \sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {a (3 i c+5 d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{4 f}+\frac {a^2 (c+i d) (c+d \tan (e+f x))^{3/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}} \]
-2*I*a^(3/2)*(c-I*d)^(3/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/ (c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))*2^(1/2)/f-1/4*(-1)^(1/4)*a^(3/2)*( 3*I*c^2+18*c*d-11*I*d^2)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e))^(1/ 2)/a^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/d^(1/2)+1/4*a*(3*I*c+5*d)*(a+I*a*tan( f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/f+1/2*a^2*(c+I*d)*(c+d*tan(f*x+e))^(3 /2)/d/f/(a+I*a*tan(f*x+e))^(1/2)-1/2*a^2*(c+d*tan(f*x+e))^(5/2)/d/f/(a+I*a *tan(f*x+e))^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(998\) vs. \(2(315)=630\).
Time = 6.81 (sec) , antiderivative size = 998, normalized size of antiderivative = 3.17 \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\frac {d (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}{2 f}-\frac {\frac {a \left (-\frac {1}{2} a^2 (5 c-3 i d) d-\frac {1}{2} i a^2 \left (4 c^2-3 i c d-d^2\right )\right ) \left (-\frac {2 \sqrt {2} \arctan \left (\frac {\sqrt {-a c+i a d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a c+i a d}}-\frac {2 (-1)^{3/4} \sqrt {c+i d} \sqrt {\frac {1}{\frac {c}{c+i d}+\frac {i d}{c+i d}}} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d} \sqrt {\frac {c}{c+i d}+\frac {i d}{c+i d}}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {a} \sqrt {d} \sqrt {c+d \tan (e+f x)}}\right )}{f}+\frac {(5 c-3 i d) (i a c-a d)^2 \sqrt {\frac {i a}{-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}} \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )^2 \sqrt {\frac {i a (c+d \tan (e+f x))}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}} \left (\frac {2 i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}-\frac {2 \sqrt [4]{-1} \sqrt {a} \sqrt {d} \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a c-a d} \sqrt {-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}} \sqrt {1+\frac {i a d (a+i a \tan (e+f x))}{(i a c-a d) \left (-\frac {a^2 c}{i a c-a d}-\frac {i a^2 d}{i a c-a d}\right )}}}\right )}{4 a d f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{2 a} \]
(d*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c + d*Tan[e + f*x]])/(2*f) - ((a*(-1/ 2*(a^2*(5*c - (3*I)*d)*d) - (I/2)*a^2*(4*c^2 - (3*I)*c*d - d^2))*((-2*Sqrt [2]*ArcTan[(Sqrt[-(a*c) + I*a*d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sq rt[c + d*Tan[e + f*x]])])/Sqrt[-(a*c) + I*a*d] - (2*(-1)^(3/4)*Sqrt[c + I* d]*Sqrt[(c/(c + I*d) + (I*d)/(c + I*d))^(-1)]*Sqrt[c/(c + I*d) + (I*d)/(c + I*d)]*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sq rt[c + I*d]*Sqrt[c/(c + I*d) + (I*d)/(c + I*d)])]*Sqrt[(c + d*Tan[e + f*x] )/(c + I*d)])/(Sqrt[a]*Sqrt[d]*Sqrt[c + d*Tan[e + f*x]])))/f + ((5*c - (3* I)*d)*(I*a*c - a*d)^2*Sqrt[(I*a)/(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I* a*c - a*d))]*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))^2*Sqrt[( I*a*(c + d*Tan[e + f*x]))/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)) )]*(((2*I)*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))) - (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[ ((-1)^(1/4)*Sqrt[a]*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d] *Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)])]*Sqrt[a + I*a*T an[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d) /(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-( (a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)))])))/(4*a*d*f*Sqrt[a + I *a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]))/(2*a)
Time = 1.98 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4039, 27, 3042, 4078, 3042, 4080, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 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.
| \(\displaystyle \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 4039 |
| \(\displaystyle \frac {a \int -\frac {(a (i c-9 d)+a (c-7 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{2 \sqrt {i \tan (e+f x) a+a}}dx}{2 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \int \frac {(a (i c-9 d)+a (c-7 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{\sqrt {i \tan (e+f x) a+a}}dx}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \int \frac {(a (i c-9 d)+a (c-7 i d) \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{\sqrt {i \tan (e+f x) a+a}}dx}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle -\frac {a \left (-\frac {\int \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)} \left ((5 c-3 i d) d a^2+d (3 i c+5 d) \tan (e+f x) a^2\right )dx}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \left (-\frac {\int \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)} \left ((5 c-3 i d) d a^2+d (3 i c+5 d) \tan (e+f x) a^2\right )dx}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (d \left (13 c^2-14 i d c-5 d^2\right ) a^3+d \left (3 i c^2+18 d c-11 i d^2\right ) \tan (e+f x) a^3\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (d \left (13 c^2-14 i d c-5 d^2\right ) a^3+d \left (3 i c^2+18 d c-11 i d^2\right ) \tan (e+f x) a^3\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (d \left (13 c^2-14 i d c-5 d^2\right ) a^3+d \left (3 i c^2+18 d c-11 i d^2\right ) \tan (e+f x) a^3\right )}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {16 a^3 d (c-i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-a^2 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {16 a^3 d (c-i d)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-a^2 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {-a^2 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {32 i a^5 d (c-i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {-a^2 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx-\frac {16 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {-\frac {a^4 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{f}-\frac {16 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle -\frac {a \left (-\frac {\frac {-\frac {2 a^4 d \left (3 c^2-18 i c d-11 d^2\right ) \int \frac {1}{i a-\frac {d (i \tan (e+f x) a+a)}{c+d \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}}{f}-\frac {16 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}-\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a^2 (c+d \tan (e+f x))^{5/2}}{2 d f \sqrt {a+i a \tan (e+f x)}}-\frac {a \left (-\frac {\frac {-\frac {2 (-1)^{3/4} a^{7/2} \sqrt {d} \left (3 c^2-18 i c d-11 d^2\right ) \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {16 i \sqrt {2} a^{7/2} d (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}}{2 a}+\frac {a^2 d (5 d+3 i c) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}}{a^2}-\frac {2 a (c+i d) (c+d \tan (e+f x))^{3/2}}{f \sqrt {a+i a \tan (e+f x)}}\right )}{4 d}\) |
-1/2*(a^2*(c + d*Tan[e + f*x])^(5/2))/(d*f*Sqrt[a + I*a*Tan[e + f*x]]) - ( a*((-2*a*(c + I*d)*(c + d*Tan[e + f*x])^(3/2))/(f*Sqrt[a + I*a*Tan[e + f*x ]]) - (((-2*(-1)^(3/4)*a^(7/2)*Sqrt[d]*(3*c^2 - (18*I)*c*d - 11*d^2)*ArcTa nh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan [e + f*x]])])/f - ((16*I)*Sqrt[2]*a^(7/2)*(c - I*d)^(3/2)*d*ArcTanh[(Sqrt[ 2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f *x]])])/f)/(2*a) + (a^2*d*((3*I)*c + 5*d)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[ c + d*Tan[e + f*x]])/f)/a^2))/(4*d)
3.12.44.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
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[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x ] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (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] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, 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] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*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] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1233 vs. \(2 (251 ) = 502\).
Time = 0.87 (sec) , antiderivative size = 1234, normalized size of antiderivative = 3.92
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1234\) |
| default | \(\text {Expression too large to display}\) | \(1234\) |
1/16/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a*(-3*2^(1/2)*(-a *(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+ d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^2+18*I*2^(1/2)* (-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))* (c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d-8*I*ln(1/2 *(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*( I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d*2^(1/2)*(-a*(I*d-c))^(1/2)-8*I*2^(1/2 )*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e) )*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c+11*ln(1/2* (2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I *a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d^2+10*(a*(1+ I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^( 1/2)*d+8*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a* (1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2)) *a*c-8*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(1 +I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a *d+8*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2 ^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*x+e)))^(1/2))/(ta n(f*x+e)+I))*a*c+8*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d* tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(1+I*tan(f*x+e))*(c+d*tan(f*...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1108 vs. \(2 (239) = 478\).
Time = 0.29 (sec) , antiderivative size = 1108, normalized size of antiderivative = 3.52 \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
1/8*(8*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) + f)*sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*log((sqrt(2)*f*sqrt(-(a^3*c^3 - 3*I*a^3*c^ 2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*e^(I*f*x + I*e) + sqrt(2)*(-I*a*c - a* d + (-I*a*c - a*d)*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e ) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))) *e^(-I*f*x - I*e)/(-I*a*c - a*d)) - 8*sqrt(2)*(f*e^(2*I*f*x + 2*I*e) + f)* sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*log(-(sqrt( 2)*f*sqrt(-(a^3*c^3 - 3*I*a^3*c^2*d - 3*a^3*c*d^2 + I*a^3*d^3)/f^2)*e^(I*f *x + I*e) - sqrt(2)*(-I*a*c - a*d + (-I*a*c - a*d)*e^(2*I*f*x + 2*I*e))*sq rt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sq rt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(-I*a*c - a*d)) + 2*sqrt (2)*((5*I*a*c + 7*a*d)*e^(3*I*f*x + 3*I*e) + (5*I*a*c + 3*a*d)*e^(I*f*x + I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) - (f*e^(2*I*f*x + 2*I*e) + f)*sqrt ((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I *a^3*d^4)/(d*f^2))*log((2*d*f*sqrt((-9*I*a^3*c^4 - 108*a^3*c^3*d + 390*I*a ^3*c^2*d^2 + 396*a^3*c*d^3 - 121*I*a^3*d^4)/(d*f^2))*e^(I*f*x + I*e) + sqr t(2)*(3*a*c^2 - 18*I*a*c*d - 11*a*d^2 + (3*a*c^2 - 18*I*a*c*d - 11*a*d^2)* e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2* I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e...
\[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(3*d-c>0)', see `assume?` for mor e details)
Exception generated. \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{18,[0,10,0,0,0]%%%}+%%%{-120,[0,9,0,0,1]%%%}+%%%{%%%{%%{[3 60,0]:[1,
Timed out. \[ \int (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2} \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \]