Integrand size = 38, antiderivative size = 286 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {21 A+11 i B}{30 a d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {89 A+39 i B}{20 a^2 d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(707 i A-317 B) \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \]
(1/8+1/8*I)*(I*A+B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+ c))^(1/2))/a^(5/2)/d+1/60*(707*I*A-317*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/t an(d*x+c)^(1/2)+1/20*(89*A+39*I*B)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+ c)^(3/2)-1/60*(361*A+151*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(3 /2)+1/5*(A+I*B)/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2)+1/30*(21*A+11* I*B)/a/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)
Time = 7.77 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^2(c+d x) \left (\frac {15 \sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \tan ^2(c+d x)}{\sqrt {i a \tan (c+d x)}}+\frac {\sec ^2(c+d x) (-174 i A+84 B+(747 i A-317 B) \cos (2 (c+d x))+(-493 i A+233 B) \cos (4 (c+d x))-780 A \sin (2 (c+d x))-340 i B \sin (2 (c+d x))+490 A \sin (4 (c+d x))+230 i B \sin (4 (c+d x)))}{\sqrt {a+i a \tan (c+d x)}}\right )}{120 a^2 d \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))^2} \]
((-1/120*I)*Sec[c + d*x]^2*((15*Sqrt[2]*(I*A + B)*ArcTanh[(Sqrt[2]*Sqrt[I* a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*( c + d*x)])*Tan[c + d*x]^2)/Sqrt[I*a*Tan[c + d*x]] + (Sec[c + d*x]^2*((-174 *I)*A + 84*B + ((747*I)*A - 317*B)*Cos[2*(c + d*x)] + ((-493*I)*A + 233*B) *Cos[4*(c + d*x)] - 780*A*Sin[2*(c + d*x)] - (340*I)*B*Sin[2*(c + d*x)] + 490*A*Sin[4*(c + d*x)] + (230*I)*B*Sin[4*(c + d*x)]))/Sqrt[a + I*a*Tan[c + d*x]]))/(a^2*d*Tan[c + d*x]^(3/2)*(-I + Tan[c + d*x])^2)
Time = 1.75 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.07, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 218}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{5/2} (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {a (13 A+3 i B)-8 a (i A-B) \tan (c+d x)}{2 \tan ^{\frac {5}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (13 A+3 i B)-8 a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (13 A+3 i B)-8 a (i A-B) \tan (c+d x)}{\tan (c+d x)^{5/2} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (a^2 (47 A+17 i B)-2 a^2 (21 i A-11 B) \tan (c+d x)\right )}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (47 A+17 i B)-2 a^2 (21 i A-11 B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (47 A+17 i B)-2 a^2 (21 i A-11 B) \tan (c+d x)}{\tan (c+d x)^{5/2} \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (361 A+151 i B)-4 a^3 (89 i A-39 B) \tan (c+d x)\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx}{a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (361 A+151 i B)-4 a^3 (89 i A-39 B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (361 A+151 i B)-4 a^3 (89 i A-39 B) \tan (c+d x)\right )}{\tan (c+d x)^{5/2}}dx}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((707 i A-317 B) a^4+2 (361 A+151 i B) \tan (c+d x) a^4\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((707 i A-317 B) a^4+2 (361 A+151 i B) \tan (c+d x) a^4\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((707 i A-317 B) a^4+2 (361 A+151 i B) \tan (c+d x) a^4\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\frac {2 \int \frac {15 a^5 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^4 (-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {-\frac {15 a^4 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^4 (-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {-\frac {15 a^4 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^4 (-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {\frac {\frac {-\frac {-\frac {30 i a^6 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^4 (-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{2 a^2}+\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {a^2 (89 A+39 i B)}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {-\frac {2 a^3 (361 A+151 i B) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\frac {(15-15 i) a^{9/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^4 (-317 B+707 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}}{2 a^2}}{2 a^2}+\frac {a (21 A+11 i B)}{3 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {A+i B}{5 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}\) |
(A + I*B)/(5*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(21* A + (11*I)*B))/(3*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + ((a ^2*(89*A + (39*I)*B))/(d*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((-2*a^3*(361*A + (151*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x] ^(3/2)) - (((15 - 15*I)*a^(9/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Ta n[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^4*((707*I)*A - 317*B)*S qrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/(2*a^2))/(2*a^2) )/(10*a^2)
3.2.95.3.1 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_)^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*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]
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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1228 vs. \(2 (232 ) = 464\).
Time = 0.17 (sec) , antiderivative size = 1229, normalized size of antiderivative = 4.30
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1229\) |
| default | \(\text {Expression too large to display}\) | \(1229\) |
| parts | \(\text {Expression too large to display}\) | \(1273\) |
1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*A*2^(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))*a*tan(d*x+c)^6-12260*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^ (1/2)*tan(d*x+c)^3+15*B*2^(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))*a*tan(d*x+c)^6+ 15*I*A*2^(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))*a*tan(d*x+c)^2+60*I*B*2^(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))*a*tan(d*x+c)^3+60*A*2^(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))*a*tan(d*x+c)^5-2940*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c))) ^(1/2)*tan(d*x+c)^2-60*I*B*2^(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))*a*tan(d*x+c) ^5-90*B*2^(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))*a*tan(d*x+c)^4-1268*B*(-I*a)^(1 /2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^5+2828*I*A*(-I*a)^(1/ 2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^5-90*I*A*2^(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))*a*tan(d*x+c)^4-60*A*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*ta n(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^...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (218) = 436\).
Time = 0.29 (sec) , antiderivative size = 588, normalized size of antiderivative = 2.06 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (9 i \, d x + 9 i \, c\right )} - 2 \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + 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}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (9 i \, d x + 9 i \, c\right )} - 2 \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + 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}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (983 \, A + 463 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} - 2 \, {\left (272 \, A + 97 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, {\left (393 \, A + 163 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (381 \, A + 191 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (18 \, A + 13 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}}}{120 \, {\left (a^{3} d e^{\left (9 i \, d x + 9 i \, c\right )} - 2 \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]
-1/120*(15*sqrt(1/2)*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I *c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))* log((2*I*sqrt(1/2)*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 15*sqrt(1/2)*(a^3*d*e^(9*I*d*x + 9*I*c) - 2*a^3*d *e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))*sqrt((-I*A^2 - 2*A*B + I *B^2)/(a^5*d^2))*log((-2*I*sqrt(1/2)*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/( a^5*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + sqrt(2)*((983*A + 463*I*B)*e^(1 0*I*d*x + 10*I*c) - 2*(272*A + 97*I*B)*e^(8*I*d*x + 8*I*c) - 3*(393*A + 16 3*I*B)*e^(6*I*d*x + 6*I*c) + (381*A + 191*I*B)*e^(4*I*d*x + 4*I*c) + 2*(18 *A + 13*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))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(a^3 *d*e^(9*I*d*x + 9*I*c) - 2*a^3*d*e^(7*I*d*x + 7*I*c) + a^3*d*e^(5*I*d*x + 5*I*c))
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Not invertible Error: Bad Argument Value
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]