\(\int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [57]

Optimal result

Integrand size = 24, antiderivative size = 124 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {15 i x}{4 a^2}-\frac {4 \log (\cos (c+d x))}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:

15/4*I*x/a^2-4*ln(cos(d*x+c))/a^2/d-15/4*I*tan(d*x+c)/a^2/d-2*tan(d*x+c)^2 
/a^2/d+5/4*I*tan(d*x+c)^3/a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^4/d/(a+I*a 
*tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {30+31 \log (i-\tan (c+d x))+\log (i+\tan (c+d x))+2 i (15+31 \log (i-\tan (c+d x))+\log (i+\tan (c+d x))) \tan (c+d x)-(-14+31 \log (i-\tan (c+d x))+\log (i+\tan (c+d x))) \tan ^2(c+d x)+8 i \tan ^3(c+d x)+4 \tan ^4(c+d x)}{8 a^2 d (-i+\tan (c+d x))^2} \] Input:

Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/8*(30 + 31*Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]] + (2*I)*(15 + 
31*Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]])*Tan[c + d*x] - (-14 + 31 
*Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + (8*I)*Tan 
[c + d*x]^3 + 4*Tan[c + d*x]^4)/(a^2*d*(-I + Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 4041, 27, 3042, 4078, 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.

\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^5}{(a+i a \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {2 \tan ^3(c+d x) (2 a-3 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^3(c+d x) (2 a-3 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^3 (2 a-3 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {\int \tan ^2(c+d x) \left (16 \tan (c+d x) a^2+15 i a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x)^2 \left (16 \tan (c+d x) a^2+15 i a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {\frac {8 a^2 \tan ^2(c+d x)}{d}+\int \tan (c+d x) \left (15 i a^2 \tan (c+d x)-16 a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {\frac {8 a^2 \tan ^2(c+d x)}{d}+\int \tan (c+d x) \left (15 i a^2 \tan (c+d x)-16 a^2\right )dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4008

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {-16 a^2 \int \tan (c+d x)dx+\frac {8 a^2 \tan ^2(c+d x)}{d}+\frac {15 i a^2 \tan (c+d x)}{d}-15 i a^2 x}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {-16 a^2 \int \tan (c+d x)dx+\frac {8 a^2 \tan ^2(c+d x)}{d}+\frac {15 i a^2 \tan (c+d x)}{d}-15 i a^2 x}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {5 i \tan ^3(c+d x)}{2 d (1+i \tan (c+d x))}-\frac {\frac {8 a^2 \tan ^2(c+d x)}{d}+\frac {15 i a^2 \tan (c+d x)}{d}+\frac {16 a^2 \log (\cos (c+d x))}{d}-15 i a^2 x}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/4*Tan[c + d*x]^4/(d*(a + I*a*Tan[c + d*x])^2) + ((((5*I)/2)*Tan[c + d*x 
]^3)/(d*(1 + I*Tan[c + d*x])) - ((-15*I)*a^2*x + (16*a^2*Log[Cos[c + d*x]] 
)/d + ((15*I)*a^2*Tan[c + d*x])/d + (8*a^2*Tan[c + d*x]^2)/d)/(2*a^2))/(2* 
a^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4008

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]
 

rule 4011

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]
 

rule 4041

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])
 

rule 4078

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]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {31 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {8 i c}{a^{2} d}+\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}+4}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(104\)
derivativedivides	\(-\frac {2 i \tan \left (d x +c \right )}{d \,a^{2}}-\frac {\tan \left (d x +c \right )^{2}}{2 a^{2} d}-\frac {9 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}+\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\)	\(107\)
default	\(-\frac {2 i \tan \left (d x +c \right )}{d \,a^{2}}-\frac {\tan \left (d x +c \right )^{2}}{2 a^{2} d}-\frac {9 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}+\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\)	\(107\)
norman	\(\frac {\frac {3}{a d}-\frac {\tan \left (d x +c \right )^{6}}{2 a d}+\frac {15 i x}{4 a}-\frac {15 i \tan \left (d x +c \right )}{4 d a}-\frac {25 i \tan \left (d x +c \right )^{3}}{4 d a}-\frac {2 i \tan \left (d x +c \right )^{5}}{a d}+\frac {15 i x \tan \left (d x +c \right )^{2}}{2 a}+\frac {15 i x \tan \left (d x +c \right )^{4}}{4 a}+\frac {4 \tan \left (d x +c \right )^{2}}{a d}}{a \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {2 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}\)	\(164\)

Input:

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

31/4*I*x/a^2+1/a^2/d*exp(-2*I*(d*x+c))-1/16/a^2/d*exp(-4*I*(d*x+c))+8*I/a^ 
2/d*c+2*(exp(2*I*(d*x+c))+2)/d/a^2/(exp(2*I*(d*x+c))+1)^2-4/a^2/d*ln(exp(2 
*I*(d*x+c))+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.22 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \, {\left (-31 i \, d x - 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

1/16*(124*I*d*x*e^(8*I*d*x + 8*I*c) - 8*(-31*I*d*x - 6)*e^(6*I*d*x + 6*I*c 
) + (124*I*d*x + 95)*e^(4*I*d*x + 4*I*c) - 64*(e^(8*I*d*x + 8*I*c) + 2*e^( 
6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) + 14* 
e^(2*I*d*x + 2*I*c) - 1)/(a^2*d*e^(8*I*d*x + 8*I*c) + 2*a^2*d*e^(6*I*d*x + 
 6*I*c) + a^2*d*e^(4*I*d*x + 4*I*c))
 

Sympy [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.73 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {2 e^{2 i c} e^{2 i d x} + 4}{a^{2} d e^{4 i c} e^{4 i d x} + 2 a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (31 i e^{4 i c} - 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac {31 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{4 a^{2}} - \frac {4 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \] Input:

integrate(tan(d*x+c)**5/(a+I*a*tan(d*x+c))**2,x)
 

Output:

(2*exp(2*I*c)*exp(2*I*d*x) + 4)/(a**2*d*exp(4*I*c)*exp(4*I*d*x) + 2*a**2*d 
*exp(2*I*c)*exp(2*I*d*x) + a**2*d) + Piecewise(((16*a**2*d*exp(4*I*c)*exp( 
-2*I*d*x) - a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(16*a**4*d**2), N 
e(a**4*d**2*exp(6*I*c), 0)), (x*((31*I*exp(4*I*c) - 8*I*exp(2*I*c) + I)*ex 
p(-4*I*c)/(4*a**2) - 31*I/(4*a**2)), True)) + 31*I*x/(4*a**2) - 4*log(exp( 
2*I*d*x) + exp(-2*I*c))/(a**2*d)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{8 \, a^{2} d} + \frac {31 \, \log \left (\tan \left (d x + c\right ) - i\right )}{8 \, a^{2} d} + \frac {-9 i \, \tan \left (d x + c\right ) - 8}{4 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} - \frac {a^{2} d \tan \left (d x + c\right )^{2} + 4 i \, a^{2} d \tan \left (d x + c\right )}{2 \, a^{4} d^{2}} \] Input:

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

1/8*log(tan(d*x + c) + I)/(a^2*d) + 31/8*log(tan(d*x + c) - I)/(a^2*d) + 1 
/4*(-9*I*tan(d*x + c) - 8)/(a^2*d*(tan(d*x + c) - I)^2) - 1/2*(a^2*d*tan(d 
*x + c)^2 + 4*I*a^2*d*tan(d*x + c))/(a^4*d^2)
 

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{a^2\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,a^2\,d}+\frac {\frac {9\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {2{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \] Input:

int(tan(c + d*x)^5/(a + a*tan(c + d*x)*1i)^2,x)
                                                                                    
                                                                                    
 

Output:

(31*log(tan(c + d*x) - 1i))/(8*a^2*d) + log(tan(c + d*x) + 1i)/(8*a^2*d) - 
 (tan(c + d*x)*2i)/(a^2*d) - tan(c + d*x)^2/(2*a^2*d) + ((9*tan(c + d*x))/ 
(4*a^2) - 2i/a^2)/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))
 

Reduce [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{5}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - int(tan(c + d*x)**5/(tan(c + d*x)**2 - 2*tan(c + d*x)*i - 1),x))/a**2