Integrand size = 34, antiderivative size = 225 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-8 a^4 (A-i B) x+\frac {8 a^4 (i A+B) \log (\cos (c+d x))}{d}+\frac {8 a^4 (A-i B) \tan (c+d x)}{d}+\frac {4 a^4 (i A+B) \tan ^2(c+d x)}{d}-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{60 d}+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{10 d}-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{20 d} \] Output:
-8*a^4*(A-I*B)*x+8*a^4*(I*A+B)*ln(cos(d*x+c))/d+8*a^4*(A-I*B)*tan(d*x+c)/d +4*a^4*(I*A+B)*tan(d*x+c)^2/d-1/60*a^4*(92*A-93*I*B)*tan(d*x+c)^3/d+1/6*I* a*B*tan(d*x+c)^3*(a+I*a*tan(d*x+c))^3/d-1/10*(2*A-3*I*B)*tan(d*x+c)^3*(a^2 +I*a^2*tan(d*x+c))^2/d-1/20*(12*A-13*I*B)*tan(d*x+c)^3*(a^4+I*a^4*tan(d*x+ c))/d
Time = 1.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.61 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {a^4 \left (-92 i A-93 B-480 i (A-i B) \log (i+\tan (c+d x))+480 (A-i B) \tan (c+d x)+240 (i A+B) \tan ^2(c+d x)-20 (7 A-8 i B) \tan ^3(c+d x)+(-60 i A-105 B) \tan ^4(c+d x)+12 (A-4 i B) \tan ^5(c+d x)+10 B \tan ^6(c+d x)\right )}{60 d} \] Input:
Integrate[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]
Output:
(a^4*((-92*I)*A - 93*B - (480*I)*(A - I*B)*Log[I + Tan[c + d*x]] + 480*(A - I*B)*Tan[c + d*x] + 240*(I*A + B)*Tan[c + d*x]^2 - 20*(7*A - (8*I)*B)*Ta n[c + d*x]^3 + ((-60*I)*A - 105*B)*Tan[c + d*x]^4 + 12*(A - (4*I)*B)*Tan[c + d*x]^5 + 10*B*Tan[c + d*x]^6))/(60*d)
Time = 1.46 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 4077, 27, 3042, 4077, 27, 3042, 4077, 3042, 4075, 3042, 4011, 3042, 4008, 3042, 3956}
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.
\(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^4 (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4077 |
\(\displaystyle \frac {1}{6} \int 3 \tan ^2(c+d x) (i \tan (c+d x) a+a)^3 (a (2 A-i B)+a (2 i A+3 B) \tan (c+d x))dx+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \tan ^2(c+d x) (i \tan (c+d x) a+a)^3 (a (2 A-i B)+a (2 i A+3 B) \tan (c+d x))dx+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \tan (c+d x)^2 (i \tan (c+d x) a+a)^3 (a (2 A-i B)+a (2 i A+3 B) \tan (c+d x))dx+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 4077 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{5} \int 2 \tan ^2(c+d x) (i \tan (c+d x) a+a)^2 \left ((8 A-7 i B) a^2+(12 i A+13 B) \tan (c+d x) a^2\right )dx-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \int \tan ^2(c+d x) (i \tan (c+d x) a+a)^2 \left ((8 A-7 i B) a^2+(12 i A+13 B) \tan (c+d x) a^2\right )dx-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \int \tan (c+d x)^2 (i \tan (c+d x) a+a)^2 \left ((8 A-7 i B) a^2+(12 i A+13 B) \tan (c+d x) a^2\right )dx-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 4077 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \int \tan ^2(c+d x) (i \tan (c+d x) a+a) \left ((68 A-67 i B) a^3+(92 i A+93 B) \tan (c+d x) a^3\right )dx-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \int \tan (c+d x)^2 (i \tan (c+d x) a+a) \left ((68 A-67 i B) a^3+(92 i A+93 B) \tan (c+d x) a^3\right )dx-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (\int \tan ^2(c+d x) \left (160 (A-i B) a^4+160 (i A+B) \tan (c+d x) a^4\right )dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (\int \tan (c+d x)^2 \left (160 (A-i B) a^4+160 (i A+B) \tan (c+d x) a^4\right )dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (\int \tan (c+d x) \left (160 a^4 (A-i B) \tan (c+d x)-160 a^4 (i A+B)\right )dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}+\frac {80 a^4 (B+i A) \tan ^2(c+d x)}{d}\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (\int \tan (c+d x) \left (160 a^4 (A-i B) \tan (c+d x)-160 a^4 (i A+B)\right )dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}+\frac {80 a^4 (B+i A) \tan ^2(c+d x)}{d}\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (-160 a^4 (B+i A) \int \tan (c+d x)dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}+\frac {80 a^4 (B+i A) \tan ^2(c+d x)}{d}+\frac {160 a^4 (A-i B) \tan (c+d x)}{d}-160 a^4 x (A-i B)\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (-160 a^4 (B+i A) \int \tan (c+d x)dx-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}+\frac {80 a^4 (B+i A) \tan ^2(c+d x)}{d}+\frac {160 a^4 (A-i B) \tan (c+d x)}{d}-160 a^4 x (A-i B)\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (\frac {1}{4} \left (-\frac {a^4 (92 A-93 i B) \tan ^3(c+d x)}{3 d}+\frac {80 a^4 (B+i A) \tan ^2(c+d x)}{d}+\frac {160 a^4 (A-i B) \tan (c+d x)}{d}+\frac {160 a^4 (B+i A) \log (\cos (c+d x))}{d}-160 a^4 x (A-i B)\right )-\frac {(12 A-13 i B) \tan ^3(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {(2 A-3 i B) \tan ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\right )+\frac {i a B \tan ^3(c+d x) (a+i a \tan (c+d x))^3}{6 d}\) |
Input:
Int[Tan[c + d*x]^2*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]
Output:
((I/6)*a*B*Tan[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + (-1/5*((2*A - (3*I )*B)*Tan[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (2*(-1/4*((12*A - (1 3*I)*B)*Tan[c + d*x]^3*(a^4 + I*a^4*Tan[c + d*x]))/d + (-160*a^4*(A - I*B) *x + (160*a^4*(I*A + B)*Log[Cos[c + d*x]])/d + (160*a^4*(A - I*B)*Tan[c + d*x])/d + (80*a^4*(I*A + B)*Tan[c + d*x]^2)/d - (a^4*(92*A - (93*I)*B)*Tan [c + d*x]^3)/(3*d))/4))/5)/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 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] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[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]
Time = 0.38 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {a^{4} \left (-\frac {4 i B \tan \left (d x +c \right )^{5}}{5}+\frac {B \tan \left (d x +c \right )^{6}}{6}-i A \tan \left (d x +c \right )^{4}+\frac {A \tan \left (d x +c \right )^{5}}{5}+\frac {8 i B \tan \left (d x +c \right )^{3}}{3}-\frac {7 B \tan \left (d x +c \right )^{4}}{4}+4 i A \tan \left (d x +c \right )^{2}-\frac {7 A \tan \left (d x +c \right )^{3}}{3}-8 i B \tan \left (d x +c \right )+4 B \tan \left (d x +c \right )^{2}+8 A \tan \left (d x +c \right )+\frac {\left (-8 i A -8 B \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (8 i B -8 A \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(168\) |
default | \(\frac {a^{4} \left (-\frac {4 i B \tan \left (d x +c \right )^{5}}{5}+\frac {B \tan \left (d x +c \right )^{6}}{6}-i A \tan \left (d x +c \right )^{4}+\frac {A \tan \left (d x +c \right )^{5}}{5}+\frac {8 i B \tan \left (d x +c \right )^{3}}{3}-\frac {7 B \tan \left (d x +c \right )^{4}}{4}+4 i A \tan \left (d x +c \right )^{2}-\frac {7 A \tan \left (d x +c \right )^{3}}{3}-8 i B \tan \left (d x +c \right )+4 B \tan \left (d x +c \right )^{2}+8 A \tan \left (d x +c \right )+\frac {\left (-8 i A -8 B \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (8 i B -8 A \right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(168\) |
norman | \(\left (8 i B \,a^{4}-8 A \,a^{4}\right ) x -\frac {\left (4 i A \,a^{4}+7 B \,a^{4}\right ) \tan \left (d x +c \right )^{4}}{4 d}-\frac {\left (-8 i B \,a^{4}+7 A \,a^{4}\right ) \tan \left (d x +c \right )^{3}}{3 d}+\frac {8 \left (-i B \,a^{4}+A \,a^{4}\right ) \tan \left (d x +c \right )}{d}+\frac {4 \left (i A \,a^{4}+B \,a^{4}\right ) \tan \left (d x +c \right )^{2}}{d}+\frac {\left (-4 i B \,a^{4}+A \,a^{4}\right ) \tan \left (d x +c \right )^{5}}{5 d}+\frac {B \,a^{4} \tan \left (d x +c \right )^{6}}{6 d}-\frac {4 \left (i A \,a^{4}+B \,a^{4}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d}\) | \(194\) |
parallelrisch | \(-\frac {480 i B \tan \left (d x +c \right ) a^{4}-10 B \tan \left (d x +c \right )^{6} a^{4}+48 i B \tan \left (d x +c \right )^{5} a^{4}-12 A \tan \left (d x +c \right )^{5} a^{4}-480 i B x \,a^{4} d +105 B \tan \left (d x +c \right )^{4} a^{4}-240 i A \tan \left (d x +c \right )^{2} a^{4}+140 A \tan \left (d x +c \right )^{3} a^{4}+240 i A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{4}-160 i B \tan \left (d x +c \right )^{3} a^{4}+480 A x \,a^{4} d +60 i A \tan \left (d x +c \right )^{4} a^{4}-240 B \tan \left (d x +c \right )^{2} a^{4}-480 A \tan \left (d x +c \right ) a^{4}+240 B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{4}}{60 d}\) | \(214\) |
risch | \(-\frac {16 i a^{4} B c}{d}+\frac {16 a^{4} A c}{d}+\frac {4 a^{4} \left (210 i A \,{\mathrm e}^{10 i \left (d x +c \right )}+270 B \,{\mathrm e}^{10 i \left (d x +c \right )}+765 i A \,{\mathrm e}^{8 i \left (d x +c \right )}+855 B \,{\mathrm e}^{8 i \left (d x +c \right )}+1210 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+1350 B \,{\mathrm e}^{6 i \left (d x +c \right )}+1020 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+1125 B \,{\mathrm e}^{4 i \left (d x +c \right )}+444 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+486 B \,{\mathrm e}^{2 i \left (d x +c \right )}+79 i A +86 B \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {8 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{d}+\frac {8 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}\) | \(220\) |
parts | \(\frac {\left (-4 i A \,a^{4}-6 B \,a^{4}\right ) \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {\left (-4 i B \,a^{4}+A \,a^{4}\right ) \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {\left (4 i A \,a^{4}+B \,a^{4}\right ) \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {\left (4 i B \,a^{4}-6 A \,a^{4}\right ) \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {A \,a^{4} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {B \,a^{4} \left (\frac {\tan \left (d x +c \right )^{6}}{6}-\frac {\tan \left (d x +c \right )^{4}}{4}+\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}\) | \(268\) |
Input:
int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x,method=_RETURNVER BOSE)
Output:
1/d*a^4*(-4/5*I*B*tan(d*x+c)^5+1/6*B*tan(d*x+c)^6-I*A*tan(d*x+c)^4+1/5*A*t an(d*x+c)^5+8/3*I*B*tan(d*x+c)^3-7/4*B*tan(d*x+c)^4+4*I*A*tan(d*x+c)^2-7/3 *A*tan(d*x+c)^3-8*I*B*tan(d*x+c)+4*B*tan(d*x+c)^2+8*A*tan(d*x+c)+1/2*(-8*I *A-8*B)*ln(1+tan(d*x+c)^2)+(-8*A+8*I*B)*arctan(tan(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.53 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {4 \, {\left (30 \, {\left (-7 i \, A - 9 \, B\right )} a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 45 \, {\left (-17 i \, A - 19 \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (-121 i \, A - 135 \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, {\left (-68 i \, A - 75 \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (-74 i \, A - 81 \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-79 i \, A - 86 \, B\right )} a^{4} + 30 \, {\left ({\left (-i \, A - B\right )} a^{4} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, {\left (-i \, A - B\right )} a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, {\left (-i \, A - B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, {\left (-i \, A - B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, {\left (-i \, A - B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (-i \, A - B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, A - B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm= "fricas")
Output:
-4/15*(30*(-7*I*A - 9*B)*a^4*e^(10*I*d*x + 10*I*c) + 45*(-17*I*A - 19*B)*a ^4*e^(8*I*d*x + 8*I*c) + 10*(-121*I*A - 135*B)*a^4*e^(6*I*d*x + 6*I*c) + 1 5*(-68*I*A - 75*B)*a^4*e^(4*I*d*x + 4*I*c) + 6*(-74*I*A - 81*B)*a^4*e^(2*I *d*x + 2*I*c) + (-79*I*A - 86*B)*a^4 + 30*((-I*A - B)*a^4*e^(12*I*d*x + 12 *I*c) + 6*(-I*A - B)*a^4*e^(10*I*d*x + 10*I*c) + 15*(-I*A - B)*a^4*e^(8*I* d*x + 8*I*c) + 20*(-I*A - B)*a^4*e^(6*I*d*x + 6*I*c) + 15*(-I*A - B)*a^4*e ^(4*I*d*x + 4*I*c) + 6*(-I*A - B)*a^4*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^4 )*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^ (4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)
Time = 0.88 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.55 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {8 i a^{4} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {316 i A a^{4} + 344 B a^{4} + \left (1776 i A a^{4} e^{2 i c} + 1944 B a^{4} e^{2 i c}\right ) e^{2 i d x} + \left (4080 i A a^{4} e^{4 i c} + 4500 B a^{4} e^{4 i c}\right ) e^{4 i d x} + \left (4840 i A a^{4} e^{6 i c} + 5400 B a^{4} e^{6 i c}\right ) e^{6 i d x} + \left (3060 i A a^{4} e^{8 i c} + 3420 B a^{4} e^{8 i c}\right ) e^{8 i d x} + \left (840 i A a^{4} e^{10 i c} + 1080 B a^{4} e^{10 i c}\right ) e^{10 i d x}}{15 d e^{12 i c} e^{12 i d x} + 90 d e^{10 i c} e^{10 i d x} + 225 d e^{8 i c} e^{8 i d x} + 300 d e^{6 i c} e^{6 i d x} + 225 d e^{4 i c} e^{4 i d x} + 90 d e^{2 i c} e^{2 i d x} + 15 d} \] Input:
integrate(tan(d*x+c)**2*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)
Output:
8*I*a**4*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (316*I*A*a**4 + 344 *B*a**4 + (1776*I*A*a**4*exp(2*I*c) + 1944*B*a**4*exp(2*I*c))*exp(2*I*d*x) + (4080*I*A*a**4*exp(4*I*c) + 4500*B*a**4*exp(4*I*c))*exp(4*I*d*x) + (484 0*I*A*a**4*exp(6*I*c) + 5400*B*a**4*exp(6*I*c))*exp(6*I*d*x) + (3060*I*A*a **4*exp(8*I*c) + 3420*B*a**4*exp(8*I*c))*exp(8*I*d*x) + (840*I*A*a**4*exp( 10*I*c) + 1080*B*a**4*exp(10*I*c))*exp(10*I*d*x))/(15*d*exp(12*I*c)*exp(12 *I*d*x) + 90*d*exp(10*I*c)*exp(10*I*d*x) + 225*d*exp(8*I*c)*exp(8*I*d*x) + 300*d*exp(6*I*c)*exp(6*I*d*x) + 225*d*exp(4*I*c)*exp(4*I*d*x) + 90*d*exp( 2*I*c)*exp(2*I*d*x) + 15*d)
Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.67 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {10 \, B a^{4} \tan \left (d x + c\right )^{6} + 12 \, {\left (A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right )^{5} - 15 \, {\left (4 i \, A + 7 \, B\right )} a^{4} \tan \left (d x + c\right )^{4} - 20 \, {\left (7 \, A - 8 i \, B\right )} a^{4} \tan \left (d x + c\right )^{3} - 240 \, {\left (-i \, A - B\right )} a^{4} \tan \left (d x + c\right )^{2} - 480 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{4} - 240 \, {\left (i \, A + B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 480 \, {\left (A - i \, B\right )} a^{4} \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm= "maxima")
Output:
1/60*(10*B*a^4*tan(d*x + c)^6 + 12*(A - 4*I*B)*a^4*tan(d*x + c)^5 - 15*(4* I*A + 7*B)*a^4*tan(d*x + c)^4 - 20*(7*A - 8*I*B)*a^4*tan(d*x + c)^3 - 240* (-I*A - B)*a^4*tan(d*x + c)^2 - 480*(d*x + c)*(A - I*B)*a^4 - 240*(I*A + B )*a^4*log(tan(d*x + c)^2 + 1) + 480*(A - I*B)*a^4*tan(d*x + c))/d
Time = 0.62 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.96 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {8 \, {\left (i \, A a^{4} + B a^{4}\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{d} + \frac {10 \, B a^{4} d^{5} \tan \left (d x + c\right )^{6} + 12 \, A a^{4} d^{5} \tan \left (d x + c\right )^{5} - 48 i \, B a^{4} d^{5} \tan \left (d x + c\right )^{5} - 60 i \, A a^{4} d^{5} \tan \left (d x + c\right )^{4} - 105 \, B a^{4} d^{5} \tan \left (d x + c\right )^{4} - 140 \, A a^{4} d^{5} \tan \left (d x + c\right )^{3} + 160 i \, B a^{4} d^{5} \tan \left (d x + c\right )^{3} + 240 i \, A a^{4} d^{5} \tan \left (d x + c\right )^{2} + 240 \, B a^{4} d^{5} \tan \left (d x + c\right )^{2} + 480 \, A a^{4} d^{5} \tan \left (d x + c\right ) - 480 i \, B a^{4} d^{5} \tan \left (d x + c\right )}{60 \, d^{6}} \] Input:
integrate(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm= "giac")
Output:
-8*(I*A*a^4 + B*a^4)*log(tan(d*x + c) + I)/d + 1/60*(10*B*a^4*d^5*tan(d*x + c)^6 + 12*A*a^4*d^5*tan(d*x + c)^5 - 48*I*B*a^4*d^5*tan(d*x + c)^5 - 60* I*A*a^4*d^5*tan(d*x + c)^4 - 105*B*a^4*d^5*tan(d*x + c)^4 - 140*A*a^4*d^5* tan(d*x + c)^3 + 160*I*B*a^4*d^5*tan(d*x + c)^3 + 240*I*A*a^4*d^5*tan(d*x + c)^2 + 240*B*a^4*d^5*tan(d*x + c)^2 + 480*A*a^4*d^5*tan(d*x + c) - 480*I *B*a^4*d^5*tan(d*x + c))/d^6
Time = 3.18 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.37 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-a^4\,\left (A-B\,1{}\mathrm {i}\right )+\frac {a^4\,\left (B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}+\frac {B\,a^4\,1{}\mathrm {i}}{3}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-A\,a^4-3\,a^4\,\left (A-B\,1{}\mathrm {i}\right )+a^4\,\left (B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}+B\,a^4\,1{}\mathrm {i}+a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {B\,a^4\,1{}\mathrm {i}}{5}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (8\,B\,a^4+A\,a^4\,8{}\mathrm {i}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {A\,a^4\,1{}\mathrm {i}}{2}+\frac {a^4\,\left (A-B\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2}+\frac {a^4\,\left (B+A\,3{}\mathrm {i}\right )}{2}+\frac {B\,a^4}{2}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )}{2}\right )}{d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {a^4\,\left (A-B\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4}+\frac {B\,a^4}{4}+\frac {a^4\,\left (3\,B+A\,1{}\mathrm {i}\right )}{4}\right )}{d}+\frac {B\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^6}{6\,d} \] Input:
int(tan(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)
Output:
(tan(c + d*x)^3*((a^4*(A*3i + B)*1i)/3 - a^4*(A - B*1i) + (B*a^4*1i)/3 + ( a^4*(A*1i + 3*B)*1i)/3))/d - (tan(c + d*x)*(a^4*(A*3i + B)*1i - 3*a^4*(A - B*1i) - A*a^4 + B*a^4*1i + a^4*(A*1i + 3*B)*1i))/d - (tan(c + d*x)^5*((B* a^4*1i)/5 + (a^4*(A*1i + 3*B)*1i)/5))/d - (log(tan(c + d*x) + 1i)*(A*a^4*8 i + 8*B*a^4))/d + (tan(c + d*x)^2*((A*a^4*1i)/2 + (a^4*(A - B*1i)*3i)/2 + (a^4*(A*3i + B))/2 + (B*a^4)/2 + (a^4*(A*1i + 3*B))/2))/d - (tan(c + d*x)^ 4*((a^4*(A - B*1i)*3i)/4 + (B*a^4)/4 + (a^4*(A*1i + 3*B))/4))/d + (B*a^4*t an(c + d*x)^6)/(6*d)
Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.76 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {a^{4} \left (-240 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a i -240 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) b +10 \tan \left (d x +c \right )^{6} b +12 \tan \left (d x +c \right )^{5} a -48 \tan \left (d x +c \right )^{5} b i -60 \tan \left (d x +c \right )^{4} a i -105 \tan \left (d x +c \right )^{4} b -140 \tan \left (d x +c \right )^{3} a +160 \tan \left (d x +c \right )^{3} b i +240 \tan \left (d x +c \right )^{2} a i +240 \tan \left (d x +c \right )^{2} b +480 \tan \left (d x +c \right ) a -480 \tan \left (d x +c \right ) b i -480 a d x +480 b d i x \right )}{60 d} \] Input:
int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)
Output:
(a**4*( - 240*log(tan(c + d*x)**2 + 1)*a*i - 240*log(tan(c + d*x)**2 + 1)* b + 10*tan(c + d*x)**6*b + 12*tan(c + d*x)**5*a - 48*tan(c + d*x)**5*b*i - 60*tan(c + d*x)**4*a*i - 105*tan(c + d*x)**4*b - 140*tan(c + d*x)**3*a + 160*tan(c + d*x)**3*b*i + 240*tan(c + d*x)**2*a*i + 240*tan(c + d*x)**2*b + 480*tan(c + d*x)*a - 480*tan(c + d*x)*b*i - 480*a*d*x + 480*b*d*i*x))/(6 0*d)