Integrand size = 38, antiderivative size = 194 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(A+11 i B) \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(13 A-37 i B) \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
(-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+1/60*(13*A-37*I*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)^(3/2)/d/(a+I*a*tan(d*x+c))^(5/2)+1/3 0*(A+11*I*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 3.70 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\sec ^2(c+d x) \sqrt {\tan (c+d x)} \left (2 (A+11 i B+2 (7 A-13 i B) \cos (2 (c+d x))+20 (i A+B) \sin (2 (c+d x))) \sqrt {i a \tan (c+d x)}-15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{120 a^2 d \sqrt {i a \tan (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
-1/120*(Sec[c + d*x]^2*Sqrt[Tan[c + d*x]]*(2*(A + (11*I)*B + 2*(7*A - (13* I)*B)*Cos[2*(c + d*x)] + 20*(I*A + B)*Sin[2*(c + d*x)])*Sqrt[I*a*Tan[c + d *x]] - 15*Sqrt[2]*(A - I*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)])*Sqrt[a + I* a*Tan[c + d*x]]))/(a^2*d*Sqrt[I*a*Tan[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt [a + I*a*Tan[c + d*x]])
Time = 1.06 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4078, 27, 3042, 4078, 27, 3042, 4079, 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.
| \(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (3 a (i A-B)-2 a (A-4 i B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (3 a (i A-B)-2 a (A-4 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (3 a (i A-B)-2 a (A-4 i B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {-\frac {\int -\frac {(A+11 i B) a^2+2 (7 i A+13 B) \tan (c+d x) a^2}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {(A+11 i B) a^2+2 (7 i A+13 B) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {(A+11 i B) a^2+2 (7 i A+13 B) \tan (c+d x) a^2}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\int \frac {15 a^3 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {a^2 (13 A-37 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {a^2 (13 A-37 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {a^2 (13 A-37 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {-\frac {15 i a^3 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {a^2 (13 A-37 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\left (\frac {15}{2}-\frac {15 i}{2}\right ) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 (13 A-37 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (A+11 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\) |
((I*A - B)*Tan[c + d*x]^(3/2))/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) - (-1/3* (a*(A + (11*I)*B)*Sqrt[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + ( ((15/2 - (15*I)/2)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (a^2*(13*A - (37*I)*B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(6*a^2))/(10*a^2)
3.2.91.3.1 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*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[(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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1085 vs. \(2 (156 ) = 312\).
Time = 0.14 (sec) , antiderivative size = 1086, normalized size of antiderivative = 5.60
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1086\) |
| default | \(\text {Expression too large to display}\) | \(1086\) |
| parts | \(\text {Expression too large to display}\) | \(1143\) |
1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(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-148*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c )))^(1/2)*tan(d*x+c)^3+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)+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)^4-212*A*(-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*(-I*a)^(1/2) *(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-52*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*( 1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+60*A*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*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)^3-90*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)^2+308*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*( 1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^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)^4-60*A*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (144) = 288\).
Time = 0.27 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.37 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - \sqrt {2} {\left ({\left (17 \, A - 23 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (3 \, A - 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (A - 4 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
-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((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*sqrt((-I*A^2 - 2 *A*B + I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log((-2*I*sqrt(1/2)*a^3*d*sqr t((-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)) - sqr t(2)*((17*A - 23*I*B)*e^(6*I*d*x + 6*I*c) + 6*(3*A - 2*I*B)*e^(4*I*d*x + 4 *I*c) - 2*(A - 4*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)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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=[-83]Warning, replacing -83 by -8, a substitut ion varia
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]