Integrand size = 38, antiderivative size = 249 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {(i A-B) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(i A-7 B) \sqrt {\tan (c+d x)}}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
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))/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))/a^(5/2)/d-1/4*(I*A-7*B)*tan(d*x+c)^(1/2)/a^ 2/d/(a+I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B)*tan(d*x+c)^(5/2)/d/(a+I*a*tan(d*x +c))^(5/2)+1/6*(A+3*I*B)*tan(d*x+c)^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 6.08 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {240 \sqrt [4]{-1} \sqrt {a} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sec ^3(c+d x) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))-(1+i) \sec ^2(c+d x) \sqrt {1+i \tan (c+d x)} \left ((-1-i) \sqrt {a} (-11 A-21 i B+2 (13 A+63 i B) \cos (2 (c+d x))+20 i (A+6 i B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}+15 (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{120 a^{5/2} d \sqrt {1+i \tan (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
(240*(-1)^(1/4)*Sqrt[a]*B*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sec[c + d *x]^3*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]) - (1 + I)*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[Ta n[c + d*x]] + 15*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sq rt[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]]))/(120*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 = 1.59 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.447, Rules used = {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.
| \(\displaystyle \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{5/2} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {5 \tan ^{\frac {3}{2}}(c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) (a (i A-B)+2 i a B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x)^{3/2} (a (i A-B)+2 i a B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {\int -\frac {3 \sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {\sqrt {\tan (c+d x)} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((i A-7 B) a^3+8 i B \tan (c+d x) a^3\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a^5 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-8 a^2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {8 a^4 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {16 a^4 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-7 B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {16 \sqrt [4]{-1} a^{7/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}}{2 a^2}-\frac {a (A+3 i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}\) |
((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*ArcTan[((-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]])/Sqrt[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)
3.2.90.3.1 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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1529 vs. \(2 (198 ) = 396\).
Time = 0.17 (sec) , antiderivative size = 1530, normalized size of antiderivative = 6.14
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1530\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1532\) |
| default | \(\text {Expression too large to display}\) | \(1532\) |
-1/240*A/d*(a*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^(1/2)/a^3*(148*(-I*a)^(1/ 2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+60*I*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))*tan(d*x+c)^3*a-15*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))*t an(d*x+c)^4*a-60*I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t an(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a-308*I*( -I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+90*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*t an(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+60*I*(-I*a)^(1/2)*(a*tan(d*x+c)* (1+I*tan(d*x+c)))^(1/2)-15*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-220*tan(d* x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+ I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(-I*a)^(1/2)-1/80*B/d*(a*(1+I*tan(d *x+c)))^(1/2)*tan(d*x+c)^(1/2)*(5*I*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))*(I*a)^(1/2) *2^(1/2)*tan(d*x+c)^4*a-30*I*(I*a)^(1/2)*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) )*tan(d*x+c)^2*a-320*I*(-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))*tan(d*x+c)^3*a+1...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 777 vs. \(2 (185) = 370\).
Time = 0.34 (sec) , antiderivative size = 777, normalized size of antiderivative = 3.12 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (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 {2 \, \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}} 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 {2 \, \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 ) - 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 {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B 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}} + {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\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 {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B 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}} - {\left (3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {-\frac {4 i \, B^{2}}{a^{5} d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) + \sqrt {2} {\left ({\left (-23 i \, A + 123 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (2 i \, A - 17 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-4 i \, A + 9 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, 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((2*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*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-(2*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)) - 30*a^3*d*sqr t(-4*I*B^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(52/605*(4*sqrt(2)*(B*e^(3*I* d*x + 3*I*c) + B*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)) + (3*a^3*d*e^(2*I*d *x + 2*I*c) - a^3*d)*sqrt(-4*I*B^2/(a^5*d^2)))/(B*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(52/605*(4*sq rt(2)*(B*e^(3*I*d*x + 3*I*c) + B*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)) - ( 3*a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(-4*I*B^2/(a^5*d^2)))/(B*e^(2*I*d *x + 2*I*c) + B)) + sqrt(2)*((-23*I*A + 123*B)*e^(6*I*d*x + 6*I*c) - 6*(2* I*A - 17*B)*e^(4*I*d*x + 4*I*c) - 2*(-4*I*A + 9*B)*e^(2*I*d*x + 2*I*c) - 3 *I*A + 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I...
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0] was discarded and replaced randomly by 0=[32]Warning, replacing 32 by 64, a substitutio n variabl
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]