\(\int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [58]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {9 x}{4 a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:

9/4*x/a^2+2*I*ln(cos(d*x+c))/a^2/d-9/4*tan(d*x+c)/a^2/d+I*tan(d*x+c)^2/a^2 
/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c)^3/d/(a+I*a*tan(d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {-i (18+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x)))+2 (9+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan (c+d x)+i (-10+17 \log (i-\tan (c+d x))-\log (i+\tan (c+d x))) \tan ^2(c+d x)+8 \tan ^3(c+d x)}{8 a^2 d (-i+\tan (c+d x))^2} \] Input:

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/8*((-I)*(18 + 17*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]]) + 2*(9 
+ 17*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x] + I*(-10 
+ 17*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]])*Tan[c + d*x]^2 + 8*Tan 
[c + d*x]^3)/(a^2*d*(-I + Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4041, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int 2 \tan (c+d x) \left (9 \tan (c+d x) a^2+8 i a^2\right )dx}{2 a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x) \left (9 \tan (c+d x) a^2+8 i a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {\int \tan (c+d x) \left (9 \tan (c+d x) a^2+8 i a^2\right )dx}{a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 4008

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {8 i a^2 \int \tan (c+d x)dx+\frac {9 a^2 \tan (c+d x)}{d}-9 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {8 i a^2 \int \tan (c+d x)dx+\frac {9 a^2 \tan (c+d x)}{d}-9 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\frac {4 i \tan ^2(c+d x)}{d (1+i \tan (c+d x))}-\frac {\frac {9 a^2 \tan (c+d x)}{d}-\frac {8 i a^2 \log (\cos (c+d x))}{d}-9 a^2 x}{a^2}}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\)

Input:

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]
 

Output:

-1/4*Tan[c + d*x]^3/(d*(a + I*a*Tan[c + d*x])^2) + (((4*I)*Tan[c + d*x]^2) 
/(d*(1 + I*Tan[c + d*x])) - (-9*a^2*x - ((8*I)*a^2*Log[Cos[c + d*x]])/d + 
(9*a^2*Tan[c + d*x])/d)/a^2)/(4*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 = 0.73 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(-\frac {\tan \left (d x +c \right )}{a^{2} d}-\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}+\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {7}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}\)	\(90\)
default	\(-\frac {\tan \left (d x +c \right )}{a^{2} d}-\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}+\frac {9 \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {7}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}\)	\(90\)
risch	\(\frac {17 x}{4 a^{2}}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {4 c}{a^{2} d}-\frac {2 i}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\)	\(96\)
norman	\(\frac {\frac {9 x}{4 a}-\frac {\tan \left (d x +c \right )^{5}}{a d}+\frac {9 x \tan \left (d x +c \right )^{2}}{2 a}+\frac {9 x \tan \left (d x +c \right )^{4}}{4 a}-\frac {3 i}{2 a d}-\frac {9 \tan \left (d x +c \right )}{4 a d}-\frac {15 \tan \left (d x +c \right )^{3}}{4 a d}-\frac {2 i \tan \left (d x +c \right )^{2}}{a d}}{a \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}-\frac {i \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{d \,a^{2}}\)	\(145\)

Input:

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

-tan(d*x+c)/a^2/d-I/d/a^2*ln(1+tan(d*x+c)^2)+9/4/d/a^2*arctan(tan(d*x+c))- 
1/4*I/d/a^2/(-I+tan(d*x+c))^2-7/4/d/a^2/(-I+tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 4 \, {\left (17 \, d x - 11 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 32 \, {\left (-i \, e^{\left (6 i \, d x + 6 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

1/16*(68*d*x*e^(6*I*d*x + 6*I*c) + 4*(17*d*x - 11*I)*e^(4*I*d*x + 4*I*c) - 
 32*(-I*e^(6*I*d*x + 6*I*c) - I*e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I* 
c) + 1) - 11*I*e^(2*I*d*x + 2*I*c) + I)/(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 = 0.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 48 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (17 e^{4 i c} - 6 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {17}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{2} d e^{2 i c} e^{2 i d x} + a^{2} d} + \frac {17 x}{4 a^{2}} + \frac {2 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \] Input:

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**2,x)
 

Output:

Piecewise(((-48*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**2*d*exp(2*I*c)* 
exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d**2), Ne(a**4*d**2*exp(6*I*c), 0)), ( 
x*((17*exp(4*I*c) - 6*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) - 17/(4*a**2)), 
 True)) - 2*I/(a**2*d*exp(2*I*c)*exp(2*I*d*x) + a**2*d) + 17*x/(4*a**2) + 
2*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^4/(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 = 77, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{8 \, a^{2} d} - \frac {17 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{8 \, a^{2} d} - \frac {\tan \left (d x + c\right )}{a^{2} d} - \frac {7 \, \tan \left (d x + c\right ) - 6 i}{4 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

1/8*I*log(tan(d*x + c) + I)/(a^2*d) - 17/8*I*log(tan(d*x + c) - I)/(a^2*d) 
 - tan(d*x + c)/(a^2*d) - 1/4*(7*tan(d*x + c) - 6*I)/(a^2*d*(tan(d*x + c) 
- I)^2)
 

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,17{}\mathrm {i}}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}-\frac {\frac {3}{2\,a^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,7{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \] Input:

int(tan(c + d*x)^4/(a + a*tan(c + d*x)*1i)^2,x)
                                                                                    
                                                                                    
 

Output:

(log(tan(c + d*x) + 1i)*1i)/(8*a^2*d) - (log(tan(c + d*x) - 1i)*17i)/(8*a^ 
2*d) - tan(c + d*x)/(a^2*d) - ((tan(c + d*x)*7i)/(4*a^2) + 3/(2*a^2))/(d*( 
2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))
 

Reduce [F]

\[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^2,x)
 

Output:

( - int(tan(c + d*x)**4/(tan(c + d*x)**2 - 2*tan(c + d*x)*i - 1),x))/a**2