Integrand size = 38, antiderivative size = 203 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 (-1)^{3/4} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A+3 i B) \sqrt {\tan (c+d x)}}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
2*(-1)^(3/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c ))^(1/2))/a^(3/2)/d+(-1/4+1/4*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^(3/2)/d+1/2*(A+3*I*B)*tan(d*x+c)^(1/2)/a /d/(a+I*a*tan(d*x+c))^(1/2)+1/3*(I*A-B)*tan(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+ c))^(3/2)
Time = 4.58 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {\sqrt {a+i a \tan (c+d x)} \left (12 \sqrt [4]{-1} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (1+i \tan (c+d x))^{3/2}+\sqrt {\tan (c+d x)} (-3 (A+3 i B)+(-5 i A+11 B) \tan (c+d x))\right )}{6 a^2 d (-i+\tan (c+d x))^2} \]
((1/4 + I/4)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + (Sqrt[a + I*a*Tan[c + d*x]]*(12*(-1)^ (1/4)*B*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*(1 + I*Tan[c + d*x])^(3/2) + Sqrt[Tan[c + d*x]]*(-3*(A + (3*I)*B) + ((-5*I)*A + 11*B)*Tan[c + d*x]))) /(6*a^2*d*(-I + Tan[c + d*x])^2)
Time = 1.21 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {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.
\(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {3 \sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}\) |
\(\Big \downarrow \) 4078 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 4084 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {4 i a^3 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {8 i a^3 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {8 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\) |
((I*A - B)*Tan[c + d*x]^(3/2))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) - (((-8* (-1)^(3/4)*a^(5/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((1 - I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqr t[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/(2*a^2) - (a*(A + (3*I)*B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a^2)
3.2.85.3.1 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. 1214 vs. \(2 (160 ) = 320\).
Time = 0.19 (sec) , antiderivative size = 1215, normalized size of antiderivative = 5.99
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1215\) |
default | \(\text {Expression too large to display}\) | \(1215\) |
parts | \(\text {Expression too large to display}\) | \(1229\) |
1/24*I/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(24*I*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+20*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan (d*x+c)*(1+I*tan(d*x+c)))^(1/2)-3*I*A*(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))*a-3*A*(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))*a*tan(d*x+ c)^3-9*I*B*(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))*a*tan(d*x+c)+9*I *A*(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))*a*tan(d*x+c)^2-36*I*B*(I *a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+9*B*2^(1/2)*l n((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*ta n(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*a*tan(d*x+c)^2+24*B*ln(1/2*(2*I*a*ta n(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)^3+44*I*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)^2-32*I*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)+9*A*(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))*a*tan(d*x+c)-72*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 758 vs. \(2 (149) = 298\).
Time = 0.35 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.73 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} 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 ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} 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 ) + 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 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 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) - 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 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 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) - \sqrt {2} {\left (2 \, {\left (2 \, A + 5 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - 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 (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]
-1/12*(3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d *x + 3*I*c)*log((2*I*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^ 2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sq rt(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)) - 3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A* B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((-2*I*sqrt(1/2)*a^2*d*sqrt(( -I*A^2 - 2*A*B + I*B^2)/(a^3*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)) + 3*a^2* d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*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))*sq rt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) - (3*I*a^2*d*e^ (2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(4*I*B^2/(a^3*d^2)))/(B*e^(2*I*d*x + 2*I* c) + B)) - 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*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*I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt(4*I*B^2/(a^3*d^2)))/(B *e^(2*I*d*x + 2*I*c) + B)) - sqrt(2)*(2*(2*A + 5*I*B)*e^(4*I*d*x + 4*I*c) + 3*(A + 3*I*B)*e^(2*I*d*x + 2*I*c) - A - 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^...
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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=[18]Warning, replacing 18 by 97, a substitutio n variabl
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]