Integrand size = 24, antiderivative size = 143 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {25 i x}{8 a^3}+\frac {3 \log (\cos (c+d x))}{a^3 d}+\frac {25 i \tan (c+d x)}{8 a^3 d}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {11 i \tan ^3(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {3 \tan ^2(c+d x)}{2 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
-25/8*I*x/a^3+3*ln(cos(d*x+c))/a^3/d+25/8*I*tan(d*x+c)/a^3/d-1/6*tan(d*x+c )^4/d/(a+I*a*tan(d*x+c))^3+11/24*I*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^2+3 /2*tan(d*x+c)^2/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.50 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.30 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-i (50+147 \log (i-\tan (c+d x))-3 \log (i+\tan (c+d x)))+9 (49 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan (c+d x)+3 i (-76+147 \log (i-\tan (c+d x))-3 \log (i+\tan (c+d x))) \tan ^2(c+d x)+3 (78-49 \log (i-\tan (c+d x))+\log (i+\tan (c+d x))) \tan ^3(c+d x)+48 i \tan ^4(c+d x)}{48 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*(50 + 147*Log[I - Tan[c + d*x]] - 3*Log[I + Tan[c + d*x]]) + 9*(49*L og[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x] + (3*I)*(-76 + 147*Log[I - Tan[c + d*x]] - 3*Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + 3*(7 8 - 49*Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]])*Tan[c + d*x]^3 + (48 *I)*Tan[c + d*x]^4)/(48*a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.89 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.542, Rules used = {3042, 4041, 25, 3042, 4078, 27, 3042, 4078, 27, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a+i a \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^3(c+d x) (4 a-7 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\tan ^3(c+d x) (4 a-7 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^3 (4 a-7 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {3 \tan ^2(c+d x) \left (13 \tan (c+d x) a^2+11 i a^2\right )}{i \tan (c+d x) a+a}dx}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \int \frac {\tan ^2(c+d x) \left (13 \tan (c+d x) a^2+11 i a^2\right )}{i \tan (c+d x) a+a}dx}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \int \frac {\tan (c+d x)^2 \left (13 \tan (c+d x) a^2+11 i a^2\right )}{i \tan (c+d x) a+a}dx}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (-\frac {\int -2 \tan (c+d x) \left (24 a^3-25 i a^3 \tan (c+d x)\right )dx}{2 a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {\int \tan (c+d x) \left (24 a^3-25 i a^3 \tan (c+d x)\right )dx}{a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {\int \tan (c+d x) \left (24 a^3-25 i a^3 \tan (c+d x)\right )dx}{a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 4008 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {24 a^3 \int \tan (c+d x)dx-\frac {25 i a^3 \tan (c+d x)}{d}+25 i a^3 x}{a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {24 a^3 \int \tan (c+d x)dx-\frac {25 i a^3 \tan (c+d x)}{d}+25 i a^3 x}{a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {\frac {11 i a \tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3 \left (\frac {-\frac {25 i a^3 \tan (c+d x)}{d}-\frac {24 a^3 \log (\cos (c+d x))}{d}+25 i a^3 x}{a^2}-\frac {12 a^2 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}}{6 a^2}-\frac {\tan ^4(c+d x)}{6 d (a+i a \tan (c+d x))^3}\) |
Input:
Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^3,x]
Output:
-1/6*Tan[c + d*x]^4/(d*(a + I*a*Tan[c + d*x])^3) + ((((11*I)/4)*a*Tan[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^2) - (3*((-12*a^2*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])) + ((25*I)*a^3*x - (24*a^3*Log[Cos[c + d*x]])/d - ((25 *I)*a^3*Tan[c + d*x])/d)/a^2))/(4*a^2))/(6*a^2)
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_)])*((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_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[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] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[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_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*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] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.71 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {i \tan \left (d x +c \right )}{d \,a^{3}}+\frac {31 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {i}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {9}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}-\frac {25 i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}\) | \(111\) |
| default | \(\frac {i \tan \left (d x +c \right )}{d \,a^{3}}+\frac {31 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}-\frac {i}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {9}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {3 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}-\frac {25 i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}\) | \(111\) |
| risch | \(-\frac {49 i x}{8 a^{3}}-\frac {23 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {7 \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}-\frac {6 i c}{a^{3} d}-\frac {2}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\) | \(111\) |
| norman | \(\frac {\frac {i \tan \left (d x +c \right )^{7}}{a d}-\frac {5 \tan \left (d x +c \right )^{4}}{a d}-\frac {25 i x}{8 a}-\frac {35}{12 a d}-\frac {29 \tan \left (d x +c \right )^{2}}{4 a d}+\frac {25 i \tan \left (d x +c \right )}{8 d a}+\frac {25 i \tan \left (d x +c \right )^{3}}{3 d a}+\frac {55 i \tan \left (d x +c \right )^{5}}{8 a d}-\frac {75 i x \tan \left (d x +c \right )^{2}}{8 a}-\frac {75 i x \tan \left (d x +c \right )^{4}}{8 a}-\frac {25 i x \tan \left (d x +c \right )^{6}}{8 a}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{3}}\) | \(196\) |
Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
I/d/a^3*tan(d*x+c)+31/8*I/d/a^3/(-I+tan(d*x+c))-1/6*I/d/a^3/(-I+tan(d*x+c) )^3-9/8/d/a^3/(-I+tan(d*x+c))^2-3/2/d/a^3*ln(1+tan(d*x+c)^2)-25/8*I/d/a^3* arctan(tan(d*x+c))
Time = 0.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-588 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 6 \, {\left (98 i \, d x + 55\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 288 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 117 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 19 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2}{96 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/96*(-588*I*d*x*e^(8*I*d*x + 8*I*c) - 6*(98*I*d*x + 55)*e^(6*I*d*x + 6*I* c) + 288*(e^(8*I*d*x + 8*I*c) + e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I* c) + 1) - 117*e^(4*I*d*x + 4*I*c) + 19*e^(2*I*d*x + 2*I*c) - 2)/(a^3*d*e^( 8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))
Time = 0.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.50 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 35328 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 5376 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 49 i e^{6 i c} + 23 i e^{4 i c} - 7 i e^{2 i c} + i\right ) e^{- 6 i c}}{8 a^{3}} + \frac {49 i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {2}{a^{3} d e^{2 i c} e^{2 i d x} + a^{3} d} - \frac {49 i x}{8 a^{3}} + \frac {3 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \] Input:
integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((-35328*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 5376*a**6*d**2*ex p(8*I*c)*exp(-4*I*d*x) - 512*a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I *c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((-49*I*exp(6*I*c ) + 23*I*exp(4*I*c) - 7*I*exp(2*I*c) + I)*exp(-6*I*c)/(8*a**3) + 49*I/(8*a **3)), True)) - 2/(a**3*d*exp(2*I*c)*exp(2*I*d*x) + a**3*d) - 49*I*x/(8*a* *3) + 3*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)
Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} - \frac {49 \, \log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} + \frac {i \, \tan \left (d x + c\right )}{a^{3} d} + \frac {i \, {\left (93 \, \tan \left (d x + c\right )^{2} - 159 i \, \tan \left (d x + c\right ) - 70\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:
integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/16*log(tan(d*x + c) + I)/(a^3*d) - 49/16*log(tan(d*x + c) - I)/(a^3*d) + I*tan(d*x + c)/(a^3*d) + 1/24*I*(93*tan(d*x + c)^2 - 159*I*tan(d*x + c) - 70)/(a^3*d*(tan(d*x + c) - I)^3)
Time = 1.01 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.85 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {35}{12\,a^3}-\frac {31\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8\,a^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,53{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {49\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{a^3\,d} \] Input:
int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^3,x)
Output:
log(tan(c + d*x) + 1i)/(16*a^3*d) - (49*log(tan(c + d*x) - 1i))/(16*a^3*d) - ((tan(c + d*x)*53i)/(8*a^3) + 35/(12*a^3) - (31*tan(c + d*x)^2)/(8*a^3) )/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1)) + (tan( c + d*x)*1i)/(a^3*d)
\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x}{a^{3}} \] Input:
int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^3,x)
Output:
( - int(tan(c + d*x)**5/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x))/a**3