Integrand size = 38, antiderivative size = 205 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(-1)^{3/4} (2 i A-B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d} \]
(-1)^(3/4)*(2*I*A-B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan (d*x+c))^(1/2))/d/a^(1/2)+(-1/2+1/2*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))/d/a^(1/2)-(A+2*I*B)*tan(d*x+c)^(1/2) *(a+I*a*tan(d*x+c))^(1/2)/a/d+(I*A-B)*tan(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c) )^(1/2)
Time = 2.76 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {(-1)^{3/4} (2 A+i B) \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} (-A-2 i B+B \tan (c+d x))}{d \sqrt {a+i a \tan (c+d x)}} \]
((1/2 + I/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) - ((-1)^(3/4)*(2*A + I*B)*ArcSinh[(-1)^ (1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[1 + I*Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (Sqrt[Tan[c + d*x]]*(-A - (2*I)*B + B*Tan[c + d*x]))/(d*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.23 (sec) , antiderivative size = 214, 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, 4080, 25, 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.
| \(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {1}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a (i A-B)+2 a (A+2 i B) \tan (c+d x))dx}{a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a (i A-B)+2 a (A+2 i B) \tan (c+d x))dx}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a (i A-B)+2 a (A+2 i B) \tan (c+d x))dx}{2 a^2}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (A+2 i B)-a^2 (2 i A-B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a}+\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (A+2 i B)-a^2 (2 i A-B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^2 (A+2 i B)-a^2 (2 i A-B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a (2 A+i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a (2 A+i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {2 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+a (2 A+i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a (2 A+i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {a^3 (2 A+i B) \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {2 a^3 (2 A+i B) \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}}{2 a^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {-\frac {2 \sqrt [4]{-1} a^{5/2} (2 A+i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}}{2 a^2}\) |
((I*A - B)*Tan[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (-(((-2*(- 1)^(1/4)*a^(5/2)*(2*A + I*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)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a) + (2*a *(A + (2*I)*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d)/(2*a^2)
3.2.79.3.1 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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 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. 1134 vs. \(2 (165 ) = 330\).
Time = 0.20 (sec) , antiderivative size = 1135, normalized size of antiderivative = 5.54
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1135\) |
| default | \(\text {Expression too large to display}\) | \(1135\) |
| parts | \(\text {Expression too large to display}\) | \(1199\) |
1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(-2*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-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))*(I*a)^(1/2)*a+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))*(I*a)^(1/2)*a*tan(d*x+c)^2-2*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))*(I*a)^(1/2)*a*tan(d*x+c)-8*I*A*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)+8*I*B*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+ c)*(1+I*tan(d*x+c)))^(1/2)+2*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)-4*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+2*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*tan(d*x+c )^2+4*A*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-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+4*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)+I*A*2^(1/2)*ln((2*2^(1/2)*(-I*a)^...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 807 vs. \(2 (155) = 310\).
Time = 0.34 (sec) , antiderivative size = 807, normalized size of antiderivative = 3.94 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {{\left (\sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a 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} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {-i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a 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 ) - a d \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {104 \, {\left (2 \, \sqrt {2} {\left ({\left (2 i \, A - B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 i \, A - B\right )} 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 d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left ({\left (2 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - B\right )}}\right ) + a d \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {104 \, {\left (2 \, \sqrt {2} {\left ({\left (2 i \, A - B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 i \, A - B\right )} 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 d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left ({\left (2 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - B\right )}}\right ) + 2 \, \sqrt {2} {\left ({\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 (-i \, d x - i \, c\right )}}{4 \, a d} \]
-1/4*(sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c)*l og((I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*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)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d* x + I*c)*log((-I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*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)) - a*d*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2))*e^ (I*d*x + I*c)*log(104/605*(2*sqrt(2)*((2*I*A - B)*e^(3*I*d*x + 3*I*c) + (2 *I*A - 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*d*e^(2*I*d*x + 2* I*c) - I*a*d)*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2)))/((2*I*A - B)*e^(2* I*d*x + 2*I*c) + 2*I*A - B)) + a*d*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2) )*e^(I*d*x + I*c)*log(104/605*(2*sqrt(2)*((2*I*A - B)*e^(3*I*d*x + 3*I*c) + (2*I*A - 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*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a*d^2)))/((2*I*A - B)* e^(2*I*d*x + 2*I*c) + 2*I*A - B)) + 2*sqrt(2)*((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 +...
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
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=[24]Warning, replacing 24 by 84, a substitutio n variabl
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]