Integrand size = 36, antiderivative size = 197 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 a (7 i A+8 B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 a (7 i A+8 B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 (21 i A+19 B) (a+i a \tan (c+d x))^{3/2}}{105 d} \]
2*a^(3/2)*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^ (1/2)/d-8/35*a*(7*I*A+8*B)*(a+I*a*tan(d*x+c))^(1/2)/d+2/35*a*(7*I*A+8*B)*( a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d+2/7*I*a*B*(a+I*a*tan(d*x+c))^(1/2)* tan(d*x+c)^3/d-4/105*(21*I*A+19*B)*(a+I*a*tan(d*x+c))^(3/2)/d
Time = 1.95 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.68 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {210 \sqrt {2} a^{3/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a \sqrt {a+i a \tan (c+d x)} \left (-126 i A-134 B+(42 A-38 i B) \tan (c+d x)+3 (7 i A+8 B) \tan ^2(c+d x)+15 i B \tan ^3(c+d x)\right )}{105 d} \]
(210*Sqrt[2]*a^(3/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2] *Sqrt[a])] + 2*a*Sqrt[a + I*a*Tan[c + d*x]]*((-126*I)*A - 134*B + (42*A - (38*I)*B)*Tan[c + d*x] + 3*((7*I)*A + 8*B)*Tan[c + d*x]^2 + (15*I)*B*Tan[c + d*x]^3))/(105*d)
Time = 1.08 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4077, 27, 3042, 4080, 25, 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.
| \(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4077 |
| \(\displaystyle \frac {2}{7} \int \frac {1}{2} \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (a (7 A-6 i B)+a (7 i A+8 B) \tan (c+d x))dx+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \tan ^2(c+d x) \sqrt {i \tan (c+d x) a+a} (a (7 A-6 i B)+a (7 i A+8 B) \tan (c+d x))dx+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a} (a (7 A-6 i B)+a (7 i A+8 B) \tan (c+d x))dx+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 \int -\tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (2 a^2 (7 i A+8 B)-a^2 (21 A-19 i B) \tan (c+d x)\right )dx}{5 a}+\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (2 a^2 (7 i A+8 B)-a^2 (21 A-19 i B) \tan (c+d x)\right )dx}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \int \tan (c+d x) \sqrt {i \tan (c+d x) a+a} \left (2 a^2 (7 i A+8 B)-a^2 (21 A-19 i B) \tan (c+d x)\right )dx}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (\int \sqrt {i \tan (c+d x) a+a} \left ((21 A-19 i B) a^2+2 (7 i A+8 B) \tan (c+d x) a^2\right )dx+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (\int \sqrt {i \tan (c+d x) a+a} \left ((21 A-19 i B) a^2+2 (7 i A+8 B) \tan (c+d x) a^2\right )dx+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 4010 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (35 a^2 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {4 a^2 (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (35 a^2 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx+\frac {4 a^2 (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (-\frac {70 i a^3 (A-i B) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {4 a^2 (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a (8 B+7 i A) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \left (-\frac {35 i \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (8 B+7 i A) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (19 B+21 i A) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )}{5 a}\right )+\frac {2 i a B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\) |
(((2*I)/7)*a*B*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + ((2*a*((7*I) *A + 8*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*(((-35*I)* Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt [a])])/d + (4*a^2*((7*I)*A + 8*B)*Sqrt[a + I*a*Tan[c + d*x]])/d + (2*a*((2 1*I)*A + 19*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d)))/(5*a))/7
3.1.75.3.1 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[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan [e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && !LtQ[n, -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[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 = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.83
| method | result | size |
| derivativedivides | \(\frac {2 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+i a^{3} B \sqrt {a +i a \tan \left (d x +c \right )}-A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}}\) | \(164\) |
| default | \(\frac {2 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+i a^{3} B \sqrt {a +i a \tan \left (d x +c \right )}-A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}}\) | \(164\) |
| parts | \(\frac {2 i A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}+\frac {2 B \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5}-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{3}-a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d \,a^{2}}\) | \(189\) |
2*I/d/a^2*(1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)-1/5*I*B*a*(a+I*a*tan(d*x+c))^( 5/2)-1/5*A*a*(a+I*a*tan(d*x+c))^(5/2)+1/3*I*a^2*B*(a+I*a*tan(d*x+c))^(3/2) +I*a^3*B*(a+I*a*tan(d*x+c))^(1/2)-A*a^3*(a+I*a*tan(d*x+c))^(1/2)+a^(7/2)*( A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (152) = 304\).
Time = 0.27 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.37 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {105 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) - 105 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) + 2 \, \sqrt {2} {\left ({\left (189 i \, A + 211 \, B\right )} a e^{\left (7 i \, d x + 7 i \, c\right )} + 7 \, {\left (57 i \, A + 53 \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, {\left (9 i \, A + 11 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + 105 \, {\left (i \, A + B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-1/105*(105*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(6*I*d*x + 6 *I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((-I* A - B)*a^2*e^(I*d*x + I*c) + sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(2* I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(( -I*A - B)*a)) - 105*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(6*I *d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log (4*((-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt(-(A^2 - 2*I*A*B - B^2)*a^3/d^2)* (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a)) + 2*sqrt(2)*((189*I*A + 211*B)*a*e^(7*I*d*x + 7*I*c) + 7*(57*I*A + 53*B)*a*e^(5*I*d*x + 5*I*c) + 35*(9*I*A + 11*B)*a*e^(3*I*d* x + 3*I*c) + 105*(I*A + B)*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.78 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (105 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 30 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} B a + 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (A + i \, B\right )} a^{2} - 70 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a^{3} + 210 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{4}\right )}}{105 \, a^{3} d} \]
-1/105*I*(105*sqrt(2)*(A - I*B)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*t an(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - 30*I*( I*a*tan(d*x + c) + a)^(7/2)*B*a + 42*(I*a*tan(d*x + c) + a)^(5/2)*(A + I*B )*a^2 - 70*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a^3 + 210*sqrt(I*a*tan(d*x + c ) + a)*(A - I*B)*a^4)/(a^3*d)
\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{2} \,d x } \]
Time = 9.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.07 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {A\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}-\frac {2\,B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a\,d}+\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{7\,a^2\,d}-\frac {\sqrt {2}\,A\,{\left (-a\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,2{}\mathrm {i}}{d}-\frac {\sqrt {2}\,B\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,2{}\mathrm {i}}{d} \]
(2*B*(a + a*tan(c + d*x)*1i)^(5/2))/(5*a*d) - (A*a*(a + a*tan(c + d*x)*1i) ^(1/2)*2i)/d - (2*B*a*(a + a*tan(c + d*x)*1i)^(1/2))/d - (A*(a + a*tan(c + d*x)*1i)^(5/2)*2i)/(5*a*d) - (2*B*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) - (2*B*(a + a*tan(c + d*x)*1i)^(7/2))/(7*a^2*d) - (2^(1/2)*A*(-a)^(3/2)*atan ((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*2i)/d - (2^(1/2)* B*a^(3/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*2i) /d