Integrand size = 38, antiderivative size = 289 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+3 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {i A-7 B}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
2*(-1)^(1/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c ))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/a^(5/2)/d+(1/8+1/8*I)*(A-I*B)* arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c )^(1/2)*tan(d*x+c)^(1/2)/a^(5/2)/d+1/4*(-I*A+7*B)/a^2/d/cot(d*x+c)^(1/2)/( a+I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B)/d/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^ (5/2)+1/6*(A+3*I*B)/a/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)
Time = 9.97 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{120}+\frac {i}{120}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((-120-120 i) \sqrt [4]{-1} \sqrt {a} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-\sec ^2(c+d x) \sqrt {1+i \tan (c+d x)} \left ((-1-i) \sqrt {a} (-11 A-21 i B+2 (13 A+63 i B) \cos (2 (c+d x))+20 i (A+6 i B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}+15 (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )\right )}{a^{5/2} d \sqrt {1+i \tan (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
((1/120 + I/120)*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-120 - 120*I)*(-1 )^(1/4)*Sqrt[a]*B*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d*x]^3*(C os[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - Sec[c + d*x]^2*Sqrt[1 + I*Tan[c + d*x]]*((-1 - I)*Sqrt[a]*(-11*A - (21*I)*B + 2*(13*A + (63*I)*B)*Cos[2*(c + d*x)] + (20*I)*(A + (6*I)*B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]] + 15*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]]) ))/(a^(5/2)*d*Sqrt[1 + I*Tan[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a* Tan[c + d*x]])
Time = 1.82 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.99, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4729, 3042, 4078, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\cot (c+d x)^{5/2} (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(i \tan (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{5/2} (A+B \tan (c+d x))}{(i \tan (c+d x) a+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {5 \tan ^{\frac {3}{2}}(c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x)^{3/2} (a (i A-B)+2 i a B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {\int -\frac {3 \sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a^5 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {8 a^4 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {16 a^4 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {16 \sqrt [4]{-1} a^{7/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((I*A - B)*Tan[c + d*x]^(5/2))/(5*d *(a + I*a*Tan[c + d*x])^(5/2)) - (-1/3*(a*(A + (3*I)*B)*Tan[c + d*x]^(3/2) )/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (-1/2*((16*(-1)^(1/4)*a^(7/2)*B*ArcTa n[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((1 - I)*a^(7/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/S qrt[a + I*a*Tan[c + d*x]]])/d)/a^2 + (a^2*(I*A - 7*B)*Sqrt[Tan[c + d*x]])/ (d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a^2))/(2*a^2))
3.6.66.3.1 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[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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_)*((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*(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]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1551 vs. \(2 (230 ) = 460\).
Time = 0.55 (sec) , antiderivative size = 1552, normalized size of antiderivative = 5.37
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1552\) |
| default | \(\text {Expression too large to display}\) | \(1552\) |
-1/240/d/(1/tan(d*x+c))^(5/2)/tan(d*x+c)^2*(a*(1+I*tan(d*x+c)))^(1/2)/a^3* (240*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+ c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a-15*A*(I*a)^(1/2)*ln(-(-2*2^(1/2)* (-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(ta n(d*x+c)+I))*2^(1/2)*a+60*I*A*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^( 1/2)*a*tan(d*x+c)^3-420*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2 )*(-I*a)^(1/2)+240*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x +c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^4-1440*B *ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1 /2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2-220*A*(I*a)^(1/2)*(-I*a)^( 1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+148*A*(I*a)^(1/2)*(- I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+1548*B*(I*a) ^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-15* A*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c))) ^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^4+60*B*(I* a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 )+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^3+90*A*(I*a)^(1 /2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a -3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^2-60*B*(I*a)^(1/2...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (217) = 434\).
Time = 0.27 (sec) , antiderivative size = 799, normalized size of antiderivative = 2.76 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 30 \, a^{3} d \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) - 30 \, a^{3} d \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} - \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) + \sqrt {2} {\left ({\left (23 \, A + 123 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (17 \, A + 72 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (7 \, A + 12 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
-1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I* d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^ 3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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^5*d^2)) + (A - I* B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqr t((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*s qrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I *c) + 1))*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^5*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d *x - I*c)/(I*A + B)) + 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(5*I*d*x + 5*I* c)*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*(a^4*d*e^(3*I*d* x + 3*I*c) - a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt ((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-4*I*B^2/(a^5 *d^2)))*e^(-2*I*d*x - 2*I*c)/B) - 30*a^3*d*sqrt(-4*I*B^2/(a^5*d^2))*e^(5*I *d*x + 5*I*c)*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 - sqrt(2)*(a^4* d*e^(3*I*d*x + 3*I*c) - a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-4 *I*B^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/B) + sqrt(2)*((23*A + 123*I*B)*e^( 6*I*d*x + 6*I*c) - 2*(17*A + 72*I*B)*e^(4*I*d*x + 4*I*c) + 2*(7*A + 12*I*B )*e^(2*I*d*x + 2*I*c) - 3*A - 3*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*...
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]