Integrand size = 15, antiderivative size = 89 \[ \int (a+i a \tan (c+d x))^4 \, dx=8 a^4 x-\frac {8 i a^4 \log (\cos (c+d x))}{d}-\frac {4 a^4 \tan (c+d x)}{d}+\frac {i a (a+i a \tan (c+d x))^3}{3 d}+\frac {i \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]
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+I*(a^2+I*a^2*tan(d*x+c))^2/d
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.65 \[ \int (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \left (4 i (1+6 \log (i+\tan (c+d x)))-21 \tan (c+d x)-6 i \tan ^2(c+d x)+\tan ^3(c+d x)\right )}{3 d} \]
(a^4*((4*I)*(1 + 6*Log[I + Tan[c + d*x]]) - 21*Tan[c + d*x] - (6*I)*Tan[c + d*x]^2 + Tan[c + d*x]^3))/(3*d)
Time = 0.43 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {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.
\(\displaystyle \int (a+i a \tan (c+d x))^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^4dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^3dx+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \int (i \tan (c+d x) a+a)^3dx+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \int (i \tan (c+d x) a+a)^2dx+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3958 |
\(\displaystyle 2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \left (2 a \left (2 i a^2 \int \tan (c+d x)dx-\frac {a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle 2 a \left (2 a \left (-\frac {a^2 \tan (c+d x)}{d}-\frac {2 i a^2 \log (\cos (c+d x))}{d}+2 a^2 x\right )+\frac {i a (a+i a \tan (c+d x))^2}{2 d}\right )+\frac {i a (a+i a \tan (c+d x))^3}{3 d}\) |
((I/3)*a*(a + I*a*Tan[c + d*x])^3)/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))
3.1.37.3.1 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]
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(\frac {a^{4} \left (-7 \tan \left (d x +c \right )+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-2 i \left (\tan ^{2}\left (d x +c \right )\right )+4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+8 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(61\) |
default | \(\frac {a^{4} \left (-7 \tan \left (d x +c \right )+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-2 i \left (\tan ^{2}\left (d x +c \right )\right )+4 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+8 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(61\) |
parallelrisch | \(\frac {-6 i a^{4} \left (\tan ^{2}\left (d x +c \right )\right )+a^{4} \left (\tan ^{3}\left (d x +c \right )\right )+12 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+24 a^{4} x d -21 a^{4} \tan \left (d x +c \right )}{3 d}\) | \(68\) |
norman | \(8 a^{4} x -\frac {7 a^{4} \tan \left (d x +c \right )}{d}+\frac {a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {2 i a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {4 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(75\) |
risch | \(-\frac {16 a^{4} c}{d}-\frac {4 i a^{4} \left (18 \,{\mathrm e}^{4 i \left (d x +c \right )}+27 \,{\mathrm e}^{2 i \left (d x +c \right )}+11\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {8 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(78\) |
parts | \(a^{4} x +\frac {a^{4} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {4 i a^{4} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {2 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {6 a^{4} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(117\) |
1/d*a^4*(-7*tan(d*x+c)+1/3*tan(d*x+c)^3-2*I*tan(d*x+c)^2+4*I*ln(1+tan(d*x+ c)^2)+8*arctan(tan(d*x+c)))
Time = 0.24 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.54 \[ \int (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (18 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 27 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 11 i \, a^{4} + 6 \, {\left (i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-4/3*(18*I*a^4*e^(4*I*d*x + 4*I*c) + 27*I*a^4*e^(2*I*d*x + 2*I*c) + 11*I*a ^4 + 6*(I*a^4*e^(6*I*d*x + 6*I*c) + 3*I*a^4*e^(4*I*d*x + 4*I*c) + 3*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^(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)
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.55 \[ \int (a+i a \tan (c+d x))^4 \, dx=- \frac {8 i a^{4} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 72 i a^{4} e^{4 i c} e^{4 i d x} - 108 i a^{4} e^{2 i c} e^{2 i d x} - 44 i a^{4}}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} \]
-8*I*a**4*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-72*I*a**4*exp(4*I*c)*exp(4 *I*d*x) - 108*I*a**4*exp(2*I*c)*exp(2*I*d*x) - 44*I*a**4)/(3*d*exp(6*I*c)* exp(6*I*d*x) + 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)
Time = 0.44 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.21 \[ \int (a+i a \tan (c+d x))^4 \, dx=a^{4} x + \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{4}}{3 \, d} + \frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{4}}{d} + \frac {2 i \, a^{4} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {4 i \, a^{4} \log \left (\sec \left (d x + c\right )\right )}{d} \]
a^4*x + 1/3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^4/d + 6*(d*x + c - tan(d*x + c))*a^4/d + 2*I*a^4*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1))/d + 4*I*a^4*log(sec(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (77) = 154\).
Time = 0.39 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.91 \[ \int (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (6 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 27 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 11 i \, a^{4}\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-4/3*(6*I*a^4*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*I*a^4* e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*I*a^4*e^(2*I*d*x + 2 *I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*I*a^4*e^(4*I*d*x + 4*I*c) + 27*I*a ^4*e^(2*I*d*x + 2*I*c) + 6*I*a^4*log(e^(2*I*d*x + 2*I*c) + 1) + 11*I*a^4)/ (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)
Time = 4.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int (a+i a \tan (c+d x))^4 \, dx=\frac {\frac {a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}-7\,a^4\,\mathrm {tan}\left (c+d\,x\right )+a^4\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2\,2{}\mathrm {i}}{d} \]