Integrand size = 24, antiderivative size = 93 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=2 i a^2 x+\frac {2 a^2 \log (\cos (c+d x))}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+\frac {a^2 \tan ^2(c+d x)}{d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan ^4(c+d x)}{4 d} \]
2*I*a^2*x+2*a^2*ln(cos(d*x+c))/d-2*I*a^2*tan(d*x+c)/d+a^2*tan(d*x+c)^2/d+2 /3*I*a^2*tan(d*x+c)^3/d-1/4*a^2*tan(d*x+c)^4/d
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \left (24 i \arctan (\tan (c+d x))+24 \log (\cos (c+d x))-24 i \tan (c+d x)+12 \tan ^2(c+d x)+8 i \tan ^3(c+d x)-3 \tan ^4(c+d x)\right )}{12 d} \]
(a^2*((24*I)*ArcTan[Tan[c + d*x]] + 24*Log[Cos[c + d*x]] - (24*I)*Tan[c + d*x] + 12*Tan[c + d*x]^2 + (8*I)*Tan[c + d*x]^3 - 3*Tan[c + d*x]^4))/(12*d )
Time = 0.56 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4026, 3042, 4011, 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.
\(\displaystyle \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a+i a \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle -\frac {a^2 \tan ^4(c+d x)}{4 d}+\int \tan ^3(c+d x) \left (2 i \tan (c+d x) a^2+2 a^2\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a^2 \tan ^4(c+d x)}{4 d}+\int \tan (c+d x)^3 \left (2 i \tan (c+d x) a^2+2 a^2\right )dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^2(c+d x) \left (2 a^2 \tan (c+d x)-2 i a^2\right )dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 \left (2 a^2 \tan (c+d x)-2 i a^2\right )dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan (c+d x) \left (-2 i \tan (c+d x) a^2-2 a^2\right )dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) \left (-2 i \tan (c+d x) a^2-2 a^2\right )dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle -2 a^2 \int \tan (c+d x)dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+2 i a^2 x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a^2 \int \tan (c+d x)dx-\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+2 i a^2 x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {a^2 \tan ^4(c+d x)}{4 d}+\frac {2 i a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan ^2(c+d x)}{d}-\frac {2 i a^2 \tan (c+d x)}{d}+\frac {2 a^2 \log (\cos (c+d x))}{d}+2 i a^2 x\) |
(2*I)*a^2*x + (2*a^2*Log[Cos[c + d*x]])/d - ((2*I)*a^2*Tan[c + d*x])/d + ( a^2*Tan[c + d*x]^2)/d + (((2*I)/3)*a^2*Tan[c + d*x]^3)/d - (a^2*Tan[c + d* x]^4)/(4*d)
3.1.14.3.1 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[(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_)*((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 = 0.17 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan ^{2}\left (d x +c \right )-\ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(70\) |
default | \(\frac {a^{2} \left (-2 i \tan \left (d x +c \right )-\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan ^{2}\left (d x +c \right )-\ln \left (1+\tan ^{2}\left (d x +c \right )\right )+2 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(70\) |
parallelrisch | \(\frac {8 i a^{2} \left (\tan ^{3}\left (d x +c \right )\right )-3 a^{2} \left (\tan ^{4}\left (d x +c \right )\right )+24 i a^{2} x d -24 i a^{2} \tan \left (d x +c \right )+12 \left (\tan ^{2}\left (d x +c \right )\right ) a^{2}-12 a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{12 d}\) | \(83\) |
risch | \(-\frac {4 i a^{2} c}{d}+\frac {2 a^{2} \left (21 \,{\mathrm e}^{6 i \left (d x +c \right )}+36 \,{\mathrm e}^{4 i \left (d x +c \right )}+29 \,{\mathrm e}^{2 i \left (d x +c \right )}+8\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(88\) |
norman | \(\frac {a^{2} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+2 i a^{2} x -\frac {a^{2} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {2 i a^{2} \tan \left (d x +c \right )}{d}+\frac {2 i a^{2} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(91\) |
parts | \(\frac {a^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 i a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {a^{2} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\) | \(110\) |
1/d*a^2*(-2*I*tan(d*x+c)-1/4*tan(d*x+c)^4+2/3*I*tan(d*x+c)^3+tan(d*x+c)^2- ln(1+tan(d*x+c)^2)+2*I*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. 174 vs. \(2 (83) = 166\).
Time = 0.24 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.87 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 \, {\left (21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{2} + 3 \, {\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
2/3*(21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2 *I*d*x + 2*I*c) + 8*a^2 + 3*(a^2*e^(8*I*d*x + 8*I*c) + 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) + 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*log( e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
Time = 0.33 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.81 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 a^{2} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {42 a^{2} e^{6 i c} e^{6 i d x} + 72 a^{2} e^{4 i c} e^{4 i d x} + 58 a^{2} e^{2 i c} e^{2 i d x} + 16 a^{2}}{3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} + 3 d} \]
2*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (42*a**2*exp(6*I*c)*exp(6*I*d*x ) + 72*a**2*exp(4*I*c)*exp(4*I*d*x) + 58*a**2*exp(2*I*c)*exp(2*I*d*x) + 16 *a**2)/(3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp(6*I*d*x) + 18*d* exp(4*I*c)*exp(4*I*d*x) + 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {3 \, a^{2} \tan \left (d x + c\right )^{4} - 8 i \, a^{2} \tan \left (d x + c\right )^{3} - 12 \, a^{2} \tan \left (d x + c\right )^{2} - 24 i \, {\left (d x + c\right )} a^{2} + 12 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 24 i \, a^{2} \tan \left (d x + c\right )}{12 \, d} \]
-1/12*(3*a^2*tan(d*x + c)^4 - 8*I*a^2*tan(d*x + c)^3 - 12*a^2*tan(d*x + c) ^2 - 24*I*(d*x + c)*a^2 + 12*a^2*log(tan(d*x + c)^2 + 1) + 24*I*a^2*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. 222 vs. \(2 (83) = 166\).
Time = 0.73 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.39 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2 \, {\left (3 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a^{2}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
2/3*(3*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(6* I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*a^2*e^(4*I*d*x + 4*I*c)*l og(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29 *a^2*e^(2*I*d*x + 2*I*c) + 3*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a^2)/(d* e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)
Time = 4.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )-a^2\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-\frac {a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3\,2{}\mathrm {i}}{3}}{d} \]