Integrand size = 36, antiderivative size = 144 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B)}{15 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 (9 i A+5 B) \left (a^3+i a^3 \tan (c+d x)\right )}{15 d \tan ^{\frac {3}{2}}(c+d x)} \] Output:
8*(-1)^(1/4)*a^3*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d+16/15*a^3*( 6*A-5*I*B)/d/tan(d*x+c)^(1/2)-2/5*a*A*(a+I*a*tan(d*x+c))^2/d/tan(d*x+c)^(5 /2)-2/15*(9*I*A+5*B)*(a^3+I*a^3*tan(d*x+c))/d/tan(d*x+c)^(3/2)
Time = 1.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.69 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (-3 A+(-15 i A-5 B) \tan (c+d x)+15 (4 A-3 i B) \tan ^2(c+d x)+60 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {5}{2}}(c+d x)\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \] Input:
Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/ 2),x]
Output:
(2*a^3*(-3*A + ((-15*I)*A - 5*B)*Tan[c + d*x] + 15*(4*A - (3*I)*B)*Tan[c + d*x]^2 + 60*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Ta n[c + d*x]^(5/2)))/(15*d*Tan[c + d*x]^(5/2))
Time = 0.85 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4076, 27, 3042, 4076, 27, 3042, 4074, 27, 3042, 4016, 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.
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \frac {2}{5} \int \frac {(i \tan (c+d x) a+a)^2 (a (9 i A+5 B)-a (A-5 i B) \tan (c+d x))}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {(i \tan (c+d x) a+a)^2 (a (9 i A+5 B)-a (A-5 i B) \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(i \tan (c+d x) a+a)^2 (a (9 i A+5 B)-a (A-5 i B) \tan (c+d x))}{\tan (c+d x)^{5/2}}dx-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int -\frac {2 (i \tan (c+d x) a+a) \left (2 (6 A-5 i B) a^2+(3 i A+5 B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \int \frac {(i \tan (c+d x) a+a) \left (2 (6 A-5 i B) a^2+(3 i A+5 B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \int \frac {(i \tan (c+d x) a+a) \left (2 (6 A-5 i B) a^2+(3 i A+5 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^{3/2}}dx-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4074 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (\int \frac {15 \left (a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {4 a^3 (6 A-5 i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (15 \int \frac {a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {4 a^3 (6 A-5 i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (15 \int \frac {a^3 (i A+B)-a^3 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {4 a^3 (6 A-5 i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4016 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (\frac {30 a^6 (B+i A)^2 \int \frac {1}{(i A+B) a^3+(A-i B) \tan (c+d x) a^3}d\sqrt {\tan (c+d x)}}{d}-\frac {4 a^3 (6 A-5 i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (-\frac {30 \sqrt [4]{-1} a^3 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {4 a^3 (6 A-5 i B)}{d \sqrt {\tan (c+d x)}}\right )-\frac {2 (5 B+9 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A (a+i a \tan (c+d x))^2}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(7/2),x]
Output:
(-2*a*A*(a + I*a*Tan[c + d*x])^2)/(5*d*Tan[c + d*x]^(5/2)) + ((-4*((-30*(- 1)^(1/4)*a^3*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (4*a^3*( 6*A - (5*I)*B))/(d*Sqrt[Tan[c + d*x]])))/3 - (2*((9*I)*A + 5*B)*(a^3 + I*a ^3*Tan[c + d*x]))/(3*d*Tan[c + d*x]^(3/2)))/5
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[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2*(c^2/f) Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b *Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[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_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2 ))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1]
Time = 0.08 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.67
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (3 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (3 i B -4 A \right )}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\) | \(240\) |
| default | \(\frac {a^{3} \left (\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 A}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}-\frac {2 \left (3 i A +B \right )}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (3 i B -4 A \right )}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\) | \(240\) |
| parts | \(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {\left (3 i B \,a^{3}-3 A \,a^{3}\right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {A \,a^{3} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}-\frac {i B \,a^{3} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}\) | \(538\) |
Input:
int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x,method=_RETUR NVERBOSE)
Output:
1/d*a^3*(1/4*(-4*I*A-4*B)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2) +1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c) ^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-4*I*B+4*A)*2^(1/2)*(l n((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c)+2^(1/2)*tan(d*x+c)^( 1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+ c)^(1/2)))-2/5*A/tan(d*x+c)^(5/2)-2/3*(3*I*A+B)/tan(d*x+c)^(3/2)-2*(3*I*B- 4*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. 504 vs. \(2 (116) = 232\).
Time = 0.09 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.50 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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 {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{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 {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) + 2 \, {\left ({\left (-39 i \, A - 25 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (9 i \, A + 10 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (33 i \, A + 25 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (-6 i \, A - 5 \, B\right )} a^{3}\right )} \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 )}}{15 \, {\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 )}} \] Input:
integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori thm="fricas")
Output:
-2/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/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(-2*((A - I*B)* a^3*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(2* I*d*x + 2*I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^3)) - 15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/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(-2*((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt(- (-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-I*e^( 2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((- I*A - B)*a^3)) + 2*((-39*I*A - 25*B)*a^3*e^(6*I*d*x + 6*I*c) + 2*(9*I*A + 10*B)*a^3*e^(4*I*d*x + 4*I*c) + (33*I*A + 25*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(-6*I*A - 5*B)*a^3)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(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)
\[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=- i a^{3} \left (\int \left (- \frac {3 A}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {A}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 B}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx + \int B \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \frac {i A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 i A}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\right )\, dx + \int \frac {i B}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \left (- \frac {3 i B}{\sqrt {\tan {\left (c + d x \right )}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(7/2),x)
Output:
-I*a**3*(Integral(-3*A/tan(c + d*x)**(5/2), x) + Integral(A/sqrt(tan(c + d *x)), x) + Integral(-3*B/tan(c + d*x)**(3/2), x) + Integral(B*sqrt(tan(c + d*x)), x) + Integral(I*A/tan(c + d*x)**(7/2), x) + Integral(-3*I*A/tan(c + d*x)**(3/2), x) + Integral(I*B/tan(c + d*x)**(5/2), x) + Integral(-3*I*B /sqrt(tan(c + d*x)), x))
Time = 0.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.37 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} - \frac {2 \, {\left (15 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 5 \, {\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - 3 \, A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{15 \, d} \] Input:
integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori thm="maxima")
Output:
-1/15*(15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(-1/2*sq rt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B) *log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 - 2* (15*(4*A - 3*I*B)*a^3*tan(d*x + c)^2 + 5*(-3*I*A - B)*a^3*tan(d*x + c) - 3 *A*a^3)/tan(d*x + c)^(5/2))/d
Time = 0.59 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.74 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (30 \, \sqrt {2} {\left (\left (i - 1\right ) \, A a^{3} + \left (i + 1\right ) \, B a^{3}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {60 \, A a^{3} \tan \left (d x + c\right )^{2} - 45 i \, B a^{3} \tan \left (d x + c\right )^{2} - 15 i \, A a^{3} \tan \left (d x + c\right ) - 5 \, B a^{3} \tan \left (d x + c\right ) - 3 \, A a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}}\right )}}{15 \, d} \] Input:
integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x, algori thm="giac")
Output:
-2/15*(30*sqrt(2)*((I - 1)*A*a^3 + (I + 1)*B*a^3)*arctan(-(1/2*I - 1/2)*sq rt(2)*sqrt(tan(d*x + c))) - (60*A*a^3*tan(d*x + c)^2 - 45*I*B*a^3*tan(d*x + c)^2 - 15*I*A*a^3*tan(d*x + c) - 5*B*a^3*tan(d*x + c) - 3*A*a^3)/tan(d*x + c)^(5/2))/d
Time = 4.58 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.79 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {\frac {2\,A\,a^3}{5\,d}-\frac {8\,A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d}+\frac {A\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}-\frac {\frac {2\,B\,a^3}{3\,d}+\frac {B\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,6{}\mathrm {i}}{d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,A\,a^3\,\ln \left (8\,A\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,B\,a^3\,\ln \left (-B\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \] Input:
int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3)/tan(c + d*x)^(7/2),x)
Output:
(2^(1/2)*A*a^3*log(8*A*a^3*d - 2^(1/2)*A*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i) )*(2 + 2i))/d - ((2*B*a^3)/(3*d) + (B*a^3*tan(c + d*x)*6i)/d)/tan(c + d*x) ^(3/2) - ((2*A*a^3)/(5*d) + (A*a^3*tan(c + d*x)*2i)/d - (8*A*a^3*tan(c + d *x)^2)/d)/tan(c + d*x)^(5/2) - (16i^(1/2)*A*a^3*log(8*A*a^3*d + 2*16i^(1/2 )*A*a^3*d*tan(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^3*log(- B*a^3*d*8i - 2^(1/ 2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((-16i)^(1/2)*B*a^3* log(2*(-16i)^(1/2)*B*a^3*d*tan(c + d*x)^(1/2) - B*a^3*d*8i))/d
\[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=a^{3} \left (\left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{4}}d x \right ) a +3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{3}}d x \right ) a i +\left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{3}}d x \right ) b -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{2}}d x \right ) a +3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{2}}d x \right ) b i -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a i -3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b -\left (\int \sqrt {\tan \left (d x +c \right )}d x \right ) b i \right ) \] Input:
int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(7/2),x)
Output:
a**3*(int(sqrt(tan(c + d*x))/tan(c + d*x)**4,x)*a + 3*int(sqrt(tan(c + d*x ))/tan(c + d*x)**3,x)*a*i + int(sqrt(tan(c + d*x))/tan(c + d*x)**3,x)*b - 3*int(sqrt(tan(c + d*x))/tan(c + d*x)**2,x)*a + 3*int(sqrt(tan(c + d*x))/t an(c + d*x)**2,x)*b*i - int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a*i - 3*int (sqrt(tan(c + d*x))/tan(c + d*x),x)*b - int(sqrt(tan(c + d*x)),x)*b*i)