Integrand size = 36, antiderivative size = 310 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {((30+28 i) A-(7-5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((1+29 i) A-(6+i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{16}-\frac {i}{16}\right ) ((29+i) A+(1+6 i) B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {5 (6 A+i B)}{8 a^3 d \sqrt {\tan (c+d x)}}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}+\frac {5 A+2 i B}{12 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}+\frac {7 (4 A+i B)}{24 d \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
-1/32*((30+28*I)*A+(-7+5*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2) /a^3/d+(-1/32+1/32*I)*((1+29*I)*A-(6+I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/ 2))*2^(1/2)/a^3/d+(1/32-1/32*I)*((29+I)*A+(1+6*I)*B)*arctanh(2^(1/2)*tan(d *x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/a^3/d-5/8*(6*A+I*B)/a^3/d/tan(d*x+c)^( 1/2)+1/6*(A+I*B)/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^3+1/12*(5*A+2*I*B)/ a/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^2+7/24*(4*A+I*B)/d/tan(d*x+c)^(1/2 )/(a^3+I*a^3*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.95 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) \left (2 i \cos (c+d x) (7 A+4 i B+(35 A+11 i B) \cos (2 (c+d x))+(33 i A-9 B) \sin (2 (c+d x)))+6 (-29 i A+6 B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+6 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right ) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))\right )}{48 a^3 d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))^3} \] Input:
Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^ 3),x]
Output:
(Sec[c + d*x]^3*((2*I)*Cos[c + d*x]*(7*A + (4*I)*B + (35*A + (11*I)*B)*Cos [2*(c + d*x)] + ((33*I)*A - 9*B)*Sin[2*(c + d*x)]) + 6*((-29*I)*A + 6*B)*H ypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[c + d*x]]*(Cos[3*(c + d*x)] + I*Si n[3*(c + d*x)]) + 6*(A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, I*Tan[c + d* x]]*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)])))/(48*a^3*d*Sqrt[Tan[c + d* x]]*(-I + Tan[c + d*x])^3)
Time = 1.41 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.09, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.639, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4012, 25, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{3/2} (a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {a (13 A+i B)-7 a (i A-B) \tan (c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^2}dx}{6 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (13 A+i B)-7 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)^2}dx}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (13 A+i B)-7 a (i A-B) \tan (c+d x)}{\tan (c+d x)^{3/2} (i \tan (c+d x) a+a)^2}dx}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (a^2 (31 A+4 i B)-5 a^2 (5 i A-2 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)}dx}{4 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (31 A+4 i B)-5 a^2 (5 i A-2 B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (i \tan (c+d x) a+a)}dx}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (31 A+4 i B)-5 a^2 (5 i A-2 B) \tan (c+d x)}{\tan (c+d x)^{3/2} (i \tan (c+d x) a+a)}dx}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (6 A+i B)-7 a^3 (4 i A-B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (6 A+i B)-7 a^3 (4 i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (6 A+i B)-7 a^3 (4 i A-B) \tan (c+d x)}{\tan (c+d x)^{3/2}}dx}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {\frac {3 \left (\int -\frac {7 (4 i A-B) a^3+5 (6 A+i B) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)}}dx-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\int \frac {7 (4 i A-B) a^3+5 (6 A+i B) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)}}dx-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\int \frac {7 (4 i A-B) a^3+5 (6 A+i B) \tan (c+d x) a^3}{\sqrt {\tan (c+d x)}}dx-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 \int \frac {a^3 (7 (4 i A-B)+5 (6 A+i B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \int \frac {7 (4 i A-B)+5 (6 A+i B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}+\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {\frac {7 a^2 (4 A+i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}+\frac {3 \left (-\frac {2 a^3 \left (\frac {1}{2} ((30+28 i) A-(7-5 i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((29+i) A+(1+6 i) B) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}-\frac {10 a^3 (6 A+i B)}{d \sqrt {\tan (c+d x)}}\right )}{2 a^2}}{2 a^2}+\frac {a (5 A+2 i B)}{d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}}{12 a^2}+\frac {A+i B}{6 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^3}\) |
Input:
Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]
Output:
(A + I*B)/(6*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^3) + ((a*(5*A + ( 2*I)*B))/(d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^2) + ((3*((-2*a^3*(( ((30 + 28*I)*A - (7 - 5*I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sq rt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 - (1/2 - I/2)* ((29 + I)*A + (1 + 6*I)*B)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[ c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2* Sqrt[2]))))/d - (10*a^3*(6*A + I*B))/(d*Sqrt[Tan[c + d*x]])))/(2*a^2) + (7 *a^2*(4*A + I*B))/(d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])))/(2*a^2))/ (12*a^2)
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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^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] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.15 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.57
| method | result | size |
| derivativedivides | \(\frac {\frac {i \left (14 i A -5 B \right ) \tan \left (d x +c \right )^{\frac {5}{2}}+\left (\frac {98 i A}{3}-\frac {38 B}{3}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}+\left (9 i B +20 A \right ) \sqrt {\tan \left (d x +c \right )}}{8 \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {\left (6 i B +29 A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}}{d \,a^{3}}\) | \(177\) |
| default | \(\frac {\frac {i \left (14 i A -5 B \right ) \tan \left (d x +c \right )^{\frac {5}{2}}+\left (\frac {98 i A}{3}-\frac {38 B}{3}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}+\left (9 i B +20 A \right ) \sqrt {\tan \left (d x +c \right )}}{8 \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {\left (6 i B +29 A \right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{4 \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}}{d \,a^{3}}\) | \(177\) |
Input:
int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETUR NVERBOSE)
Output:
1/d/a^3*(1/8*(I*(14*I*A-5*B)*tan(d*x+c)^(5/2)+(98/3*I*A-38/3*B)*tan(d*x+c) ^(3/2)+(20*A+9*I*B)*tan(d*x+c)^(1/2))/(-I+tan(d*x+c))^3-1/4*(6*I*B+29*A)/( 2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))+4*(-1/16 *A+1/16*I*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1 /2)))-2*A/tan(d*x+c)^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 784 vs. \(2 (235) = 470\).
Time = 0.11 (sec) , antiderivative size = 784, normalized size of antiderivative = 2.53 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algori thm="fricas")
Output:
1/96*(3*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((I*A^ 2 + 2*A*B - I*B^2)/(a^6*d^2))*log(2*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*s qrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) + (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2 *I*c)/(I*A + B)) - 3*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6*I*d*x + 6*I*c ))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*log(-2*((a^3*d*e^(2*I*d*x + 2*I *c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))* sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e ^(-2*I*d*x - 2*I*c)/(I*A + B)) + 3*(a^3*d*e^(8*I*d*x + 8*I*c) - a^3*d*e^(6 *I*d*x + 6*I*c))*sqrt((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a^6*d^2))*log(1/8 *((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e ^(2*I*d*x + 2*I*c) + 1))*sqrt((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a^6*d^2)) + 29*A + 6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*(a^3*d*e^(8*I*d*x + 8*I *c) - a^3*d*e^(6*I*d*x + 6*I*c))*sqrt((-841*I*A^2 + 348*A*B + 36*I*B^2)/(a ^6*d^2))*log(-1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-841*I*A^2 + 348*A*B + 36* I*B^2)/(a^6*d^2)) - 29*A - 6*I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(73 *I*A - 10*B)*e^(8*I*d*x + 8*I*c) + 3*(35*I*A - 2*B)*e^(6*I*d*x + 6*I*c) - (49*I*A - 19*B)*e^(4*I*d*x + 4*I*c) + 3*(-3*I*A + 2*B)*e^(2*I*d*x + 2*I*c) - I*A + B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)...
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**3,x)
Output:
Exception raised: TypeError >> Invalid comparison of non-real -I
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algori thm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.69 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.49 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {3 \, \sqrt {2} {\left (-\left (29 i + 29\right ) \, A - \left (6 i - 6\right ) \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - 3 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {96 \, A}{\sqrt {\tan \left (d x + c\right )}} - \frac {2 \, {\left (42 \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 15 i \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 98 i \, A \tan \left (d x + c\right )^{\frac {3}{2}} + 38 \, B \tan \left (d x + c\right )^{\frac {3}{2}} - 60 \, A \sqrt {\tan \left (d x + c\right )} - 27 i \, B \sqrt {\tan \left (d x + c\right )}\right )}}{{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{48 \, a^{3} d} \] Input:
integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algori thm="giac")
Output:
1/48*(3*sqrt(2)*(-(29*I + 29)*A - (6*I - 6)*B)*arctan((1/2*I + 1/2)*sqrt(2 )*sqrt(tan(d*x + c))) - 3*sqrt(2)*(-(I - 1)*A - (I + 1)*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 96*A/sqrt(tan(d*x + c)) - 2*(42*A*tan (d*x + c)^(5/2) + 15*I*B*tan(d*x + c)^(5/2) - 98*I*A*tan(d*x + c)^(3/2) + 38*B*tan(d*x + c)^(3/2) - 60*A*sqrt(tan(d*x + c)) - 27*I*B*sqrt(tan(d*x + c)))/(tan(d*x + c) - I)^3)/(a^3*d)
Time = 3.85 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}}{29\,A}\right )\,\sqrt {-\frac {A^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}-\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a^3\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,17{}\mathrm {i}}{2\,a^3\,d}-\frac {121\,A\,{\mathrm {tan}\left (c+d\,x\right )}^2}{12\,a^3\,d}-\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^3\,15{}\mathrm {i}}{4\,a^3\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,3{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}-{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,1{}\mathrm {i}}+\frac {\frac {9\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {5\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1} \] Input:
int((A + B*tan(c + d*x))/(tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^3),x)
Output:
2*atanh((16*a^3*d*tan(c + d*x)^(1/2)*((A^2*1i)/(256*a^6*d^2))^(1/2))/A)*(( A^2*1i)/(256*a^6*d^2))^(1/2) + 2*atanh((16*a^3*d*tan(c + d*x)^(1/2)*(-(A^2 *841i)/(256*a^6*d^2))^(1/2))/(29*A))*(-(A^2*841i)/(256*a^6*d^2))^(1/2) - a tan((8*a^3*d*tan(c + d*x)^(1/2)*((B^2*9i)/(64*a^6*d^2))^(1/2))/(3*B))*((B^ 2*9i)/(64*a^6*d^2))^(1/2)*2i + atan((16*a^3*d*tan(c + d*x)^(1/2)*(-(B^2*1i )/(256*a^6*d^2))^(1/2))/B)*(-(B^2*1i)/(256*a^6*d^2))^(1/2)*2i - ((2*A)/(a^ 3*d) + (A*tan(c + d*x)*17i)/(2*a^3*d) - (121*A*tan(c + d*x)^2)/(12*a^3*d) - (A*tan(c + d*x)^3*15i)/(4*a^3*d))/(tan(c + d*x)^(1/2) + tan(c + d*x)^(3/ 2)*3i - 3*tan(c + d*x)^(5/2) - tan(c + d*x)^(7/2)*1i) + ((9*B*tan(c + d*x) ^(1/2))/(8*a^3*d) + (B*tan(c + d*x)^(3/2)*19i)/(12*a^3*d) - (5*B*tan(c + d *x)^(5/2))/(8*a^3*d))/(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3 *1i + 1)
\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 4*sqrt(tan(c + d*x))*tan(c + d*x)*b + 6*sqrt(tan(c + d*x))*b*i + 9*int (sqrt(tan(c + d*x))/(tan(c + d*x)**5 - 3*tan(c + d*x)**4*i - 3*tan(c + d*x )**3 + tan(c + d*x)**2*i),x)*tan(c + d*x)*a*d*i - 3*int(sqrt(tan(c + d*x)) /(tan(c + d*x)**5 - 3*tan(c + d*x)**4*i - 3*tan(c + d*x)**3 + tan(c + d*x) **2*i),x)*tan(c + d*x)*b*d - 2*int(sqrt(tan(c + d*x))/(tan(c + d*x)**4*i + 3*tan(c + d*x)**3 - 3*tan(c + d*x)**2*i - tan(c + d*x)),x)*tan(c + d*x)*b *d + 2*int((sqrt(tan(c + d*x))*tan(c + d*x)**4)/(tan(c + d*x)**3*i + 3*tan (c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)*b*d*i + 3*int((sqrt(t an(c + d*x))*tan(c + d*x)**3)/(tan(c + d*x)**3 - 3*tan(c + d*x)**2*i - 3*t an(c + d*x) + i),x)*tan(c + d*x)*b*d*i + 6*int((sqrt(tan(c + d*x))*tan(c + d*x)**3)/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x )*tan(c + d*x)*b*d + 9*int((sqrt(tan(c + d*x))*tan(c + d*x)**2)/(tan(c + d *x)**3 - 3*tan(c + d*x)**2*i - 3*tan(c + d*x) + i),x)*tan(c + d*x)*b*d - 4 *int((sqrt(tan(c + d*x))*tan(c + d*x)**2)/(tan(c + d*x)**3*i + 3*tan(c + d *x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)*b*d*i - 6*int((sqrt(tan(c + d*x))*tan(c + d*x))/(tan(c + d*x)**3 - 3*tan(c + d*x)**2*i - 3*tan(c + d* x) + i),x)*tan(c + d*x)*b*d*i + 4*int((sqrt(tan(c + d*x))*tan(c + d*x))/(t an(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d* x)*b*d)/(9*tan(c + d*x)*a**3*d)