Integrand size = 36, antiderivative size = 205 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d} \]
1/2*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2 )/a^(1/2)+4/5*(5*A+7*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d-1/5*(5*A+7*I*B)*(a+ I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/a/d+(I*A-B)*tan(d*x+c)^3/d/(a+I*a*tan(d *x+c))^(1/2)-1/15*(25*A+23*I*B)*(a+I*a*tan(d*x+c))^(3/2)/a^2/d
Time = 2.43 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.63 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {35 A+61 i B+(10 i A-38 B) \tan (c+d x)+2 (5 A+i B) \tan ^2(c+d x)+6 B \tan ^3(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}} \]
((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2] *Sqrt[a]*d) + (35*A + (61*I)*B + ((10*I)*A - 38*B)*Tan[c + d*x] + 2*(5*A + I*B)*Tan[c + d*x]^2 + 6*B*Tan[c + d*x]^3)/(15*d*Sqrt[a + I*a*Tan[c + d*x] ])
Time = 1.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4078, 27, 3042, 4080, 27, 3042, 4075, 3042, 4010, 3042, 3961, 219}
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.
| \(\displaystyle \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3 (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {1}{2} \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (6 a (i A-B)+a (5 A+7 i B) \tan (c+d x))dx}{a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (6 a (i A-B)+a (5 A+7 i B) \tan (c+d x))dx}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} (6 a (i A-B)+a (5 A+7 i B) \tan (c+d x))dx}{2 a^2}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 \int -\frac {1}{2} \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (5 A+7 i B)-a^2 (25 i A-23 B) \tan (c+d x)\right )dx}{5 a}+\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (5 A+7 i B)-a^2 (25 i A-23 B) \tan (c+d x)\right )dx}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 a^2 (5 A+7 i B)-a^2 (25 i A-23 B) \tan (c+d x)\right )dx}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left ((25 i A-23 B) a^2+4 (5 A+7 i B) \tan (c+d x) a^2\right )dx-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\int \sqrt {i \tan (c+d x) a+a} \left ((25 i A-23 B) a^2+4 (5 A+7 i B) \tan (c+d x) a^2\right )dx-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {5 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {8 a^2 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {5 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {8 a^2 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {10 i a^3 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {8 a^2 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {2 a (5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {-\frac {5 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 a^2 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{3 d}}{5 a}}{2 a^2}\) |
((I*A - B)*Tan[c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((2*a*(5*A + ( 7*I)*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (((-5*I)*Sqrt[2 ]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]) /d + (8*a^2*(5*A + (7*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*a*(25*A + ( 23*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d))/(5*a))/(2*a^2)
3.1.90.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
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]
Time = 0.17 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{3} \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{3} d}\) | \(167\) |
| default | \(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}+\frac {a^{3} \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}}{a^{3} d}\) | \(167\) |
| parts | \(\frac {2 A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{2}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}}+\frac {2 i B \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{3}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\) | \(207\) |
2/d/a^3*(1/5*I*B*(a+I*a*tan(d*x+c))^(5/2)-2/3*I*B*a*(a+I*a*tan(d*x+c))^(3/ 2)-1/3*A*a*(a+I*a*tan(d*x+c))^(3/2)+2*I*a^2*B*(a+I*a*tan(d*x+c))^(1/2)+A*a ^2*(a+I*a*tan(d*x+c))^(1/2)+1/4*a^(5/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a *tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/2*a^3*(A+I*B)/(a+I*a*tan(d*x+c))^(1/ 2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (163) = 326\).
Time = 0.28 (sec) , antiderivative size = 447, normalized size of antiderivative = 2.18 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, \sqrt {2} {\left ({\left (35 \, A + 103 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (25 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, {\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A + 15 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]
-1/60*(15*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) + 2*a*d*e^(3*I*d*x + 3*I*c) + a *d*e^(I*d*x + I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*((-I*A - B) *a*e^(I*d*x + I*c) + (I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d* x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I* A + B)) - 15*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) + 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) + (-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c) /(I*A + B)) - 2*sqrt(2)*((35*A + 103*I*B)*e^(6*I*d*x + 6*I*c) + 5*(25*A + 41*I*B)*e^(4*I*d*x + 4*I*c) + 15*(7*A + 11*I*B)*e^(2*I*d*x + 2*I*c) + 15*A + 15*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(a*d*e^(5*I*d*x + 5*I*c) + 2 *a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))
\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 24 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + 2 i \, B\right )} a^{2} - 120 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{3} - \frac {60 \, {\left (A + i \, B\right )} a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{60 \, a^{4} d} \]
-1/60*(15*sqrt(2)*(A - I*B)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d *x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - 24*I*(I*a* tan(d*x + c) + a)^(5/2)*B*a + 40*(I*a*tan(d*x + c) + a)^(3/2)*(A + 2*I*B)* a^2 - 120*sqrt(I*a*tan(d*x + c) + a)*(A + 2*I*B)*a^3 - 60*(A + I*B)*a^4/sq rt(I*a*tan(d*x + c) + a))/(a^4*d)
\[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Time = 8.70 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {A}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}+\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a^3\,d}+\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]
A/(d*(a + a*tan(c + d*x)*1i)^(1/2)) + (B*1i)/(d*(a + a*tan(c + d*x)*1i)^(1 /2)) + (2*A*(a + a*tan(c + d*x)*1i)^(1/2))/(a*d) - (2*A*(a + a*tan(c + d*x )*1i)^(3/2))/(3*a^2*d) + (B*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/(a*d) - (B*( a + a*tan(c + d*x)*1i)^(3/2)*4i)/(3*a^2*d) + (B*(a + a*tan(c + d*x)*1i)^(5 /2)*2i)/(5*a^3*d) + (2^(1/2)*B*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2) )/(2*(-a)^(1/2)))*1i)/(2*(-a)^(1/2)*d) - (2^(1/2)*A*atan((2^(1/2)*(a + a*t an(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*1i)/(2*a^(1/2)*d)