Integrand size = 24, antiderivative size = 142 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=-8 a^4 x-\frac {8 a^4 \cot (c+d x)}{d}+\frac {4 i a^4 \cot ^2(c+d x)}{d}+\frac {23 a^4 \cot ^3(c+d x)}{15 d}+\frac {8 i a^4 \log (\sin (c+d x))}{d}-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{5 d} \]
-8*a^4*x-8*a^4*cot(d*x+c)/d+4*I*a^4*cot(d*x+c)^2/d+23/15*a^4*cot(d*x+c)^3/ d+8*I*a^4*ln(sin(d*x+c))/d-1/5*cot(d*x+c)^5*(a^2+I*a^2*tan(d*x+c))^2/d-3/5 *I*cot(d*x+c)^4*(a^4+I*a^4*tan(d*x+c))/d
Time = 0.76 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.76 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=a^4 \left (-\frac {8 \cot (c+d x)}{d}+\frac {4 i \cot ^2(c+d x)}{d}+\frac {7 \cot ^3(c+d x)}{3 d}-\frac {i \cot ^4(c+d x)}{d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {8 i \log (\tan (c+d x))}{d}-\frac {8 i \log (i+\tan (c+d x))}{d}\right ) \]
a^4*((-8*Cot[c + d*x])/d + ((4*I)*Cot[c + d*x]^2)/d + (7*Cot[c + d*x]^3)/( 3*d) - (I*Cot[c + d*x]^4)/d - Cot[c + d*x]^5/(5*d) + ((8*I)*Log[Tan[c + d* x]])/d - ((8*I)*Log[I + Tan[c + d*x]])/d)
Time = 1.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.792, Rules used = {3042, 4036, 27, 3042, 4076, 25, 3042, 4074, 27, 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.
\(\displaystyle \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\tan (c+d x)^6}dx\) |
\(\Big \downarrow \) 4036 |
\(\displaystyle -\frac {1}{5} \int -4 \cot ^5(c+d x) (i \tan (c+d x) a+a)^2 \left (3 i a^2-2 a^2 \tan (c+d x)\right )dx-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{5} \int \cot ^5(c+d x) (i \tan (c+d x) a+a)^2 \left (3 i a^2-2 a^2 \tan (c+d x)\right )dx-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \int \frac {(i \tan (c+d x) a+a)^2 \left (3 i a^2-2 a^2 \tan (c+d x)\right )}{\tan (c+d x)^5}dx-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 4076 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \int -\cot ^4(c+d x) (i \tan (c+d x) a+a) \left (17 i \tan (c+d x) a^3+23 a^3\right )dx-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4}{5} \left (-\frac {1}{4} \int \cot ^4(c+d x) (i \tan (c+d x) a+a) \left (17 i \tan (c+d x) a^3+23 a^3\right )dx-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (-\frac {1}{4} \int \frac {(i \tan (c+d x) a+a) \left (17 i \tan (c+d x) a^3+23 a^3\right )}{\tan (c+d x)^4}dx-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 4074 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-\int 40 \cot ^3(c+d x) \left (i a^4-a^4 \tan (c+d x)\right )dx\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \int \cot ^3(c+d x) \left (i a^4-a^4 \tan (c+d x)\right )dx\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \int \frac {i a^4-a^4 \tan (c+d x)}{\tan (c+d x)^3}dx\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (\int -\cot ^2(c+d x) \left (i \tan (c+d x) a^4+a^4\right )dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-\int \cot ^2(c+d x) \left (i \tan (c+d x) a^4+a^4\right )dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-\int \frac {i \tan (c+d x) a^4+a^4}{\tan (c+d x)^2}dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-\int \cot (c+d x) \left (i a^4-a^4 \tan (c+d x)\right )dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-\int \frac {i a^4-a^4 \tan (c+d x)}{\tan (c+d x)}dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-i a^4 \int \cot (c+d x)dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}+a^4 x\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-i a^4 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}+a^4 x\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (i a^4 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}+a^4 x\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {4}{5} \left (\frac {1}{4} \left (\frac {23 a^4 \cot ^3(c+d x)}{3 d}-40 \left (-\frac {i a^4 \cot ^2(c+d x)}{2 d}+\frac {a^4 \cot (c+d x)}{d}-\frac {i a^4 \log (-\sin (c+d x))}{d}+a^4 x\right )\right )-\frac {3 i \cot ^4(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{4 d}\right )-\frac {\cot ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{5 d}\) |
-1/5*(Cot[c + d*x]^5*(a^2 + I*a^2*Tan[c + d*x])^2)/d + (4*(((23*a^4*Cot[c + d*x]^3)/(3*d) - 40*(a^4*x + (a^4*Cot[c + d*x])/d - ((I/2)*a^4*Cot[c + d* x]^2)/d - (I*a^4*Log[-Sin[c + d*x]])/d))/4 - (((3*I)/4)*Cot[c + d*x]^4*(a^ 4 + I*a^4*Tan[c + d*x]))/d))/5
3.1.43.3.1 Defintions of rubi rules used
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_)])^(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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] )^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si mp[a/(d*(b*c + a*d)*(n + 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*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*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^2, 0]
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[(-a^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Time = 0.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.60
method | result | size |
derivativedivides | \(\frac {a^{4} \left (-8 \cot \left (d x +c \right )-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-i \left (\cot ^{4}\left (d x +c \right )\right )+\frac {7 \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i \left (\cot ^{2}\left (d x +c \right )\right )-4 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )+4 \pi -8 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
default | \(\frac {a^{4} \left (-8 \cot \left (d x +c \right )-\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-i \left (\cot ^{4}\left (d x +c \right )\right )+\frac {7 \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i \left (\cot ^{2}\left (d x +c \right )\right )-4 i \ln \left (\cot ^{2}\left (d x +c \right )+1\right )+4 \pi -8 \,\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{d}\) | \(85\) |
parallelrisch | \(-\frac {\left (i \left (\cot ^{4}\left (d x +c \right )\right )+\frac {\left (\cot ^{5}\left (d x +c \right )\right )}{5}-4 i \left (\cot ^{2}\left (d x +c \right )\right )-\frac {7 \left (\cot ^{3}\left (d x +c \right )\right )}{3}-8 i \ln \left (\tan \left (d x +c \right )\right )+4 i \ln \left (\sec ^{2}\left (d x +c \right )\right )+8 d x +8 \cot \left (d x +c \right )\right ) a^{4}}{d}\) | \(86\) |
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 | \(\frac {-\frac {a^{4}}{5 d}-8 a^{4} x \left (\tan ^{5}\left (d x +c \right )\right )+\frac {7 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{3 d}-\frac {8 a^{4} \left (\tan ^{4}\left (d x +c \right )\right )}{d}-\frac {i a^{4} \tan \left (d x +c \right )}{d}+\frac {4 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{5}}+\frac {8 i a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 i a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(134\) |
1/d*a^4*(-8*cot(d*x+c)-1/5*cot(d*x+c)^5-I*cot(d*x+c)^4+7/3*cot(d*x+c)^3+4* I*cot(d*x+c)^2-4*I*ln(cot(d*x+c)^2+1)+4*Pi-8*arccot(cot(d*x+c)))
Time = 0.25 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.54 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {4 \, {\left (210 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 555 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 655 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 365 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 79 i \, a^{4} + 30 \, {\left (-i \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 10 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
-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) + 655 *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 + 30 *(-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)
Time = 0.35 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.54 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {8 i a^{4} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 840 i a^{4} e^{8 i c} e^{8 i d x} + 2220 i a^{4} e^{6 i c} e^{6 i d x} - 2620 i a^{4} e^{4 i c} e^{4 i d x} + 1460 i a^{4} e^{2 i c} e^{2 i d x} - 316 i a^{4}}{15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} - 15 d} \]
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) - 15*d)
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.77 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {120 \, {\left (d x + c\right )} a^{4} + 60 i \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 120 i \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac {120 \, a^{4} \tan \left (d x + c\right )^{4} - 60 i \, a^{4} \tan \left (d x + c\right )^{3} - 35 \, a^{4} \tan \left (d x + c\right )^{2} + 15 i \, a^{4} \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{5}}}{15 \, d} \]
-1/15*(120*(d*x + c)*a^4 + 60*I*a^4*log(tan(d*x + c)^2 + 1) - 120*I*a^4*lo g(tan(d*x + c)) + (120*a^4*tan(d*x + c)^4 - 60*I*a^4*tan(d*x + c)^3 - 35*a ^4*tan(d*x + c)^2 + 15*I*a^4*tan(d*x + c) + 3*a^4)/tan(d*x + c)^5)/d
Time = 0.75 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.49 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 155 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 600 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7680 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 3840 i \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 2370 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {-8768 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2370 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 600 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 155 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 30 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, d} \]
1/480*(3*a^4*tan(1/2*d*x + 1/2*c)^5 - 30*I*a^4*tan(1/2*d*x + 1/2*c)^4 - 15 5*a^4*tan(1/2*d*x + 1/2*c)^3 + 600*I*a^4*tan(1/2*d*x + 1/2*c)^2 - 7680*I*a ^4*log(tan(1/2*d*x + 1/2*c) + I) + 3840*I*a^4*log(tan(1/2*d*x + 1/2*c)) + 2370*a^4*tan(1/2*d*x + 1/2*c) + (-8768*I*a^4*tan(1/2*d*x + 1/2*c)^5 - 2370 *a^4*tan(1/2*d*x + 1/2*c)^4 + 600*I*a^4*tan(1/2*d*x + 1/2*c)^3 + 155*a^4*t an(1/2*d*x + 1/2*c)^2 - 30*I*a^4*tan(1/2*d*x + 1/2*c) - 3*a^4)/tan(1/2*d*x + 1/2*c)^5)/d
Time = 5.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.65 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {16\,a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {8\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^4-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3}+a^4\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}+\frac {a^4}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]