Integrand size = 22, antiderivative size = 64 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=a x+\frac {a \cot (c+d x)}{d}-\frac {i a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \log (\sin (c+d x))}{d} \] Output:
a*x+a*cot(d*x+c)/d-1/2*I*a*cot(d*x+c)^2/d-1/3*a*cot(d*x+c)^3/d-I*a*ln(sin( d*x+c))/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {i a \csc ^2(c+d x)}{2 d}-\frac {a \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )}{3 d}-\frac {i a \log (\sin (c+d x))}{d} \] Input:
Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x]),x]
Output:
((-1/2*I)*a*Csc[c + d*x]^2)/d - (a*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) - (I*a*Log[Sin[c + d*x]])/d
Time = 0.55 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4012, 3042, 4012, 25, 3042, 4012, 3042, 4014, 3042, 25, 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.
\(\displaystyle \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+i a \tan (c+d x)}{\tan (c+d x)^4}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {a \cot ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) (i a-a \tan (c+d x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {a \cot ^3(c+d x)}{3 d}+\int \frac {i a-a \tan (c+d x)}{\tan (c+d x)^3}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \int -\cot ^2(c+d x) (i \tan (c+d x) a+a)dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cot ^2(c+d x) (i \tan (c+d x) a+a)dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {i \tan (c+d x) a+a}{\tan (c+d x)^2}dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\int \cot (c+d x) (i a-a \tan (c+d x))dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {i a-a \tan (c+d x)}{\tan (c+d x)}dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle -i a \int \cot (c+d x)dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}+a x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i a \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}+a x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle i a \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}+a x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot (c+d x)}{d}-\frac {i a \log (-\sin (c+d x))}{d}+a x\) |
Input:
Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x]),x]
Output:
a*x + (a*Cot[c + d*x])/d - ((I/2)*a*Cot[c + d*x]^2)/d - (a*Cot[c + d*x]^3) /(3*d) - (I*a*Log[-Sin[c + d*x]])/d
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_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (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*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*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] && N eQ[a*c + b*d, 0]
Time = 0.96 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(-\frac {\left (i \cot \left (d x +c \right )^{2}+\frac {2 \cot \left (d x +c \right )^{3}}{3}+2 i \ln \left (\tan \left (d x +c \right )\right )-i \ln \left (\sec \left (d x +c \right )^{2}\right )-2 d x -2 \cot \left (d x +c \right )\right ) a}{2 d}\) | \(63\) |
derivativedivides | \(\frac {a \left (\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\arctan \left (\tan \left (d x +c \right )\right )-\frac {1}{3 \tan \left (d x +c \right )^{3}}-i \ln \left (\tan \left (d x +c \right )\right )-\frac {i}{2 \tan \left (d x +c \right )^{2}}+\frac {1}{\tan \left (d x +c \right )}\right )}{d}\) | \(67\) |
default | \(\frac {a \left (\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\arctan \left (\tan \left (d x +c \right )\right )-\frac {1}{3 \tan \left (d x +c \right )^{3}}-i \ln \left (\tan \left (d x +c \right )\right )-\frac {i}{2 \tan \left (d x +c \right )^{2}}+\frac {1}{\tan \left (d x +c \right )}\right )}{d}\) | \(67\) |
risch | \(-\frac {2 a c}{d}+\frac {2 i a \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}-9 \,{\mathrm e}^{2 i \left (d x +c \right )}+4\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {i a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(72\) |
norman | \(\frac {a x \tan \left (d x +c \right )^{3}+\frac {a \tan \left (d x +c \right )^{2}}{d}-\frac {a}{3 d}-\frac {i a \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {i a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {i a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(87\) |
Input:
int(cot(d*x+c)^4*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-1/2*(I*cot(d*x+c)^2+2/3*cot(d*x+c)^3+2*I*ln(tan(d*x+c))-I*ln(sec(d*x+c)^2 )-2*d*x-2*cot(d*x+c))*a/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (56) = 112\).
Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.95 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {18 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, {\left (i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 i \, a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
1/3*(18*I*a*e^(4*I*d*x + 4*I*c) - 18*I*a*e^(2*I*d*x + 2*I*c) - 3*(I*a*e^(6 *I*d*x + 6*I*c) - 3*I*a*e^(4*I*d*x + 4*I*c) + 3*I*a*e^(2*I*d*x + 2*I*c) - I*a)*log(e^(2*I*d*x + 2*I*c) - 1) + 8*I*a)/(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)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).
Time = 0.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.00 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=- \frac {i a \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {18 i a e^{4 i c} e^{4 i d x} - 18 i a e^{2 i c} e^{2 i d x} + 8 i a}{3 d e^{6 i c} e^{6 i d x} - 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} - 3 d} \] Input:
integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c)),x)
Output:
-I*a*log(exp(2*I*d*x) - exp(-2*I*c))/d + (18*I*a*exp(4*I*c)*exp(4*I*d*x) - 18*I*a*exp(2*I*c)*exp(2*I*d*x) + 8*I*a)/(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 = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.11 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {6 \, {\left (d x + c\right )} a + 3 i \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 i \, a \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, a \tan \left (d x + c\right )^{2} - 3 i \, a \tan \left (d x + c\right ) - 2 \, a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
1/6*(6*(d*x + c)*a + 3*I*a*log(tan(d*x + c)^2 + 1) - 6*I*a*log(tan(d*x + c )) + (6*a*tan(d*x + c)^2 - 3*I*a*tan(d*x + c) - 2*a)/tan(d*x + c)^3)/d
Time = 0.18 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {1}{6} i \, a {\left (\frac {6 \, \log \left (\tan \left (d x + c\right ) + i\right )}{d} - \frac {6 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{d} - \frac {6 i \, \tan \left (d x + c\right )^{2} + 3 \, \tan \left (d x + c\right ) - 2 i}{d \tan \left (d x + c\right )^{3}}\right )} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
1/6*I*a*(6*log(tan(d*x + c) + I)/d - 6*log(abs(tan(d*x + c)))/d - (6*I*tan (d*x + c)^2 + 3*tan(d*x + c) - 2*I)/(d*tan(d*x + c)^3))
Time = 0.94 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {2\,a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {-a\,{\mathrm {tan}\left (c+d\,x\right )}^2+\frac {1{}\mathrm {i}\,a\,\mathrm {tan}\left (c+d\,x\right )}{2}+\frac {a}{3}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^3} \] Input:
int(cot(c + d*x)^4*(a + a*tan(c + d*x)*1i),x)
Output:
(2*a*atan(2*tan(c + d*x) + 1i))/d - (a/3 + (a*tan(c + d*x)*1i)/2 - a*tan(c + d*x)^2)/(d*tan(c + d*x)^3)
Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.83 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \left (16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3} i -12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} i +12 \sin \left (d x +c \right )^{3} d x +3 \sin \left (d x +c \right )^{3} i -6 \sin \left (d x +c \right ) i \right )}{12 \sin \left (d x +c \right )^{3} d} \] Input:
int(cot(d*x+c)^4*(a+I*a*tan(d*x+c)),x)
Output:
(a*(16*cos(c + d*x)*sin(c + d*x)**2 - 4*cos(c + d*x) + 12*log(tan((c + d*x )/2)**2 + 1)*sin(c + d*x)**3*i - 12*log(tan((c + d*x)/2))*sin(c + d*x)**3* i + 12*sin(c + d*x)**3*d*x + 3*sin(c + d*x)**3*i - 6*sin(c + d*x)*i))/(12* sin(c + d*x)**3*d)