Integrand size = 24, antiderivative size = 116 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=-8 a^4 x+\frac {8 i a^4 \log (\cos (c+d x))}{d}+\frac {4 a^4 \tan (c+d x)}{d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]
-8*a^4*x+8*I*a^4*ln(cos(d*x+c))/d+4*a^4*tan(d*x+c)/d-1/3*I*a*(a+I*a*tan(d* x+c))^3/d-1/5*I*(a+I*a*tan(d*x+c))^5/a/d-I*(a^2+I*a^2*tan(d*x+c))^2/d
Time = 0.77 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \left (-i (23+120 \log (i+\tan (c+d x)))+120 \tan (c+d x)+60 i \tan ^2(c+d x)-35 \tan ^3(c+d x)-15 i \tan ^4(c+d x)+3 \tan ^5(c+d x)\right )}{15 d} \]
(a^4*((-I)*(23 + 120*Log[I + Tan[c + d*x]]) + 120*Tan[c + d*x] + (60*I)*Ta n[c + d*x]^2 - 35*Tan[c + d*x]^3 - (15*I)*Tan[c + d*x]^4 + 3*Tan[c + d*x]^ 5))/(15*d)
Time = 0.56 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 4026, 25, 3042, 3959, 3042, 3959, 3042, 3958, 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.
\(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^4dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle \int -(i \tan (c+d x) a+a)^4dx-\frac {i (a+i a \tan (c+d x))^5}{5 a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (i \tan (c+d x) a+a)^4dx-\frac {i (a+i a \tan (c+d x))^5}{5 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int (i \tan (c+d x) a+a)^4dx-\frac {i (a+i a \tan (c+d x))^5}{5 a d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle -2 a \int (i \tan (c+d x) a+a)^3dx-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a \int (i \tan (c+d x) a+a)^3dx-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle -2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3958 |
\(\displaystyle -2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -2 a \left (2 a \left (-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )-\frac {i (a+i a \tan (c+d x))^5}{5 a d}-\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
((-1/3*I)*a*(a + I*a*Tan[c + d*x])^3)/d - ((I/5)*(a + I*a*Tan[c + d*x])^5) /(a*d) - 2*a*(((I/2)*a*(a + I*a*Tan[c + d*x])^2)/d + 2*a*(2*a^2*x - ((2*I) *a^2*Log[Cos[c + d*x]])/d - (a^2*Tan[c + d*x])/d))
3.1.35.3.1 Defintions of rubi rules used
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[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2) *x, x] + (Simp[b^2*(Tan[c + d*x]/d), x] + Simp[2*a*b Int[Tan[c + d*x], x] , x]) /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a Int[(a + b*Tan[c + d* x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n , 1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {a^{4} \left (8 \tan \left (d x +c \right )+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-i \left (\tan ^{4}\left (d x +c \right )\right )-\frac {7 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+4 i \left (\tan ^{2}\left (d x +c \right )\right )-4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )-8 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
default | \(\frac {a^{4} \left (8 \tan \left (d x +c \right )+\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-i \left (\tan ^{4}\left (d x +c \right )\right )-\frac {7 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+4 i \left (\tan ^{2}\left (d x +c \right )\right )-4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )-8 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
parallelrisch | \(-\frac {15 i a^{4} \left (\tan ^{4}\left (d x +c \right )\right )-3 a^{4} \left (\tan ^{5}\left (d x +c \right )\right )-60 i a^{4} \left (\tan ^{2}\left (d x +c \right )\right )+35 a^{4} \left (\tan ^{3}\left (d x +c \right )\right )+60 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+120 a^{4} x d -120 a^{4} \tan \left (d x +c \right )}{15 d}\) | \(96\) |
risch | \(\frac {16 a^{4} c}{d}+\frac {4 i a^{4} \left (210 \,{\mathrm e}^{8 i \left (d x +c \right )}+555 \,{\mathrm e}^{6 i \left (d x +c \right )}+655 \,{\mathrm e}^{4 i \left (d x +c \right )}+365 \,{\mathrm e}^{2 i \left (d x +c \right )}+79\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {8 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(100\) |
norman | \(-8 a^{4} x +\frac {8 a^{4} \tan \left (d x +c \right )}{d}-\frac {7 a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {4 i a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {i a^{4} \left (\tan ^{4}\left (d x +c \right )\right )}{d}-\frac {4 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(108\) |
parts | \(\frac {a^{4} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a^{4} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {4 i a^{4} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {4 i a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}-\frac {6 a^{4} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(178\) |
1/d*a^4*(8*tan(d*x+c)+1/5*tan(d*x+c)^5-I*tan(d*x+c)^4-7/3*tan(d*x+c)^3+4*I *tan(d*x+c)^2-4*I*ln(1+tan(d*x+c)^2)-8*arctan(tan(d*x+c)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (98) = 196\).
Time = 0.24 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.87 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (-210 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 555 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 655 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 365 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 79 i \, a^{4} + 30 \, {\left (-i \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} - 5 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 10 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 10 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-4/15*(-210*I*a^4*e^(8*I*d*x + 8*I*c) - 555*I*a^4*e^(6*I*d*x + 6*I*c) - 65 5*I*a^4*e^(4*I*d*x + 4*I*c) - 365*I*a^4*e^(2*I*d*x + 2*I*c) - 79*I*a^4 + 3 0*(-I*a^4*e^(10*I*d*x + 10*I*c) - 5*I*a^4*e^(8*I*d*x + 8*I*c) - 10*I*a^4*e ^(6*I*d*x + 6*I*c) - 10*I*a^4*e^(4*I*d*x + 4*I*c) - 5*I*a^4*e^(2*I*d*x + 2 *I*c) - I*a^4)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(10*I*d*x + 10*I*c) + 5* d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c ) + 5*d*e^(2*I*d*x + 2*I*c) + d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (97) = 194\).
Time = 0.29 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.88 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 i a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {840 i a^{4} e^{8 i c} e^{8 i d x} + 2220 i a^{4} e^{6 i c} e^{6 i d x} + 2620 i a^{4} e^{4 i c} e^{4 i d x} + 1460 i a^{4} e^{2 i c} e^{2 i d x} + 316 i a^{4}}{15 d e^{10 i c} e^{10 i d x} + 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} + 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} + 15 d} \]
8*I*a**4*log(exp(2*I*d*x) + exp(-2*I*c))/d + (840*I*a**4*exp(8*I*c)*exp(8* I*d*x) + 2220*I*a**4*exp(6*I*c)*exp(6*I*d*x) + 2620*I*a**4*exp(4*I*c)*exp( 4*I*d*x) + 1460*I*a**4*exp(2*I*c)*exp(2*I*d*x) + 316*I*a**4)/(15*d*exp(10* I*c)*exp(10*I*d*x) + 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6 *I*d*x) + 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) + 1 5*d)
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {3 \, a^{4} \tan \left (d x + c\right )^{5} - 15 i \, a^{4} \tan \left (d x + c\right )^{4} - 35 \, a^{4} \tan \left (d x + c\right )^{3} + 60 i \, a^{4} \tan \left (d x + c\right )^{2} - 120 \, {\left (d x + c\right )} a^{4} - 60 i \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 120 \, a^{4} \tan \left (d x + c\right )}{15 \, d} \]
1/15*(3*a^4*tan(d*x + c)^5 - 15*I*a^4*tan(d*x + c)^4 - 35*a^4*tan(d*x + c) ^3 + 60*I*a^4*tan(d*x + c)^2 - 120*(d*x + c)*a^4 - 60*I*a^4*log(tan(d*x + c)^2 + 1) + 120*a^4*tan(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (98) = 196\).
Time = 0.70 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.36 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (-30 i \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 150 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 150 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 210 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 555 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 655 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 365 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 79 i \, a^{4}\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-4/15*(-30*I*a^4*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 150* I*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*a^4*e^(6*I* d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*a^4*e^(4*I*d*x + 4*I*c)* log(e^(2*I*d*x + 2*I*c) + 1) - 150*I*a^4*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d* x + 2*I*c) + 1) - 210*I*a^4*e^(8*I*d*x + 8*I*c) - 555*I*a^4*e^(6*I*d*x + 6 *I*c) - 655*I*a^4*e^(4*I*d*x + 4*I*c) - 365*I*a^4*e^(2*I*d*x + 2*I*c) - 30 *I*a^4*log(e^(2*I*d*x + 2*I*c) + 1) - 79*I*a^4)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4* I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)
Time = 4.53 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.75 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}-8\,a^4\,\mathrm {tan}\left (c+d\,x\right )-\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}+a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{d} \]