\(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [78]

Optimal result

Integrand size = 24, antiderivative size = 126 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i x}{16 a^4}+\frac {3}{16 a^4 d (1+i \tan (c+d x))}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2} \] Output:

1/16*I*x/a^4+3/16/a^4/d/(1+I*tan(d*x+c))+1/8*tan(d*x+c)^4/d/(a+I*a*tan(d*x 
+c))^4+1/12*I*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^3-1/16/d/(a^2+I*a^2*tan( 
d*x+c))^2
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (16 \cos (2 (c+d x))+3 (1+8 i d x) \cos (4 (c+d x))+32 i \sin (2 (c+d x))-3 i \sin (4 (c+d x))-24 d x \sin (4 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \] Input:

Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

(Sec[c + d*x]^4*(16*Cos[2*(c + d*x)] + 3*(1 + (8*I)*d*x)*Cos[4*(c + d*x)] 
+ (32*I)*Sin[2*(c + d*x)] - (3*I)*Sin[4*(c + d*x)] - 24*d*x*Sin[4*(c + d*x 
)]))/(384*a^4*d*(-I + Tan[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4030, 3042, 4029, 3042, 4023, 3042, 4009, 24}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {\tan ^2(c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {\tan (c+d x)^2}{(i \tan (c+d x) a+a)^2}dx}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4023

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {\int \frac {a-2 i a \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {\int \frac {a-2 i a \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 4009

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {-\frac {\int 1dx}{2}+\frac {3 i}{2 d (1+i \tan (c+d x))}}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {-\frac {x}{2}+\frac {3 i}{2 d (1+i \tan (c+d x))}}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}}{2 a}+\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\)

Input:

Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^4,x]
 

Output:

Tan[c + d*x]^4/(8*d*(a + I*a*Tan[c + d*x])^4) + (((I/6)*Tan[c + d*x]^3)/(d 
*(a + I*a*Tan[c + d*x])^3) - ((I/2)*((-1/2*x + ((3*I)/2)/(d*(1 + I*Tan[c + 
 d*x])))/(2*a^2) - (I/4)/(d*(a + I*a*Tan[c + d*x])^2)))/a)/(2*a)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 4009

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/(2*a 
*f*m)), x] + Simp[(b*c + a*d)/(2*a*b)   Int[(a + b*Tan[e + f*x])^(m + 1), x 
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 
, 0] && LtQ[m, 0]
 

rule 4023

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ 
m/(2*a^3*f*m)), x] + Simp[1/(2*a^2)   Int[(a + b*Tan[e + f*x])^(m + 1)*Simp 
[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, 
 c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
 

rule 4029

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^n/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 

rule 4030

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a)   Int[(a + b*Tan[e 
 + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, 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] && EqQ[ 
m + n + 1, 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.48

method	result	size
risch	\(\frac {i x}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{4} d}+\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\)	\(60\)
derivativedivides	\(\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}-\frac {5 i}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}-\frac {7}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\)	\(96\)
default	\(\frac {i}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )}-\frac {5 i}{12 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{3}}+\frac {1}{8 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{4}}-\frac {7}{16 d \,a^{4} \left (-i+\tan \left (d x +c \right )\right )^{2}}+\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\)	\(96\)
norman	\(\frac {\frac {i x}{16 a}+\frac {1}{12 a d}-\frac {\tan \left (d x +c \right )^{6}}{2 a d}-\frac {i \tan \left (d x +c \right )}{16 d a}-\frac {11 i \tan \left (d x +c \right )^{3}}{48 d a}-\frac {53 i \tan \left (d x +c \right )^{5}}{48 a d}+\frac {i \tan \left (d x +c \right )^{7}}{16 a d}+\frac {i x \tan \left (d x +c \right )^{2}}{4 a}+\frac {3 i x \tan \left (d x +c \right )^{4}}{8 a}+\frac {i x \tan \left (d x +c \right )^{6}}{4 a}+\frac {i x \tan \left (d x +c \right )^{8}}{16 a}+\frac {3 \tan \left (d x +c \right )^{4}}{4 a d}+\frac {\tan \left (d x +c \right )^{2}}{3 a d}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{4} a^{3}}\)	\(207\)

Input:

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

1/16*I*x/a^4+1/16/a^4/d*exp(-2*I*(d*x+c))-1/48/a^4/d*exp(-6*I*(d*x+c))+1/1 
28/a^4/d*exp(-8*I*(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.43 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \] Input:

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/384*(24*I*d*x*e^(8*I*d*x + 8*I*c) + 24*e^(6*I*d*x + 6*I*c) - 8*e^(2*I*d* 
x + 2*I*c) + 3)*e^(-8*I*d*x - 8*I*c)/(a^4*d)
 

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.24 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (6144 a^{8} d^{2} e^{14 i c} e^{- 2 i d x} - 2048 a^{8} d^{2} e^{10 i c} e^{- 6 i d x} + 768 a^{8} d^{2} e^{8 i c} e^{- 8 i d x}\right ) e^{- 16 i c}}{98304 a^{12} d^{3}} & \text {for}\: a^{12} d^{3} e^{16 i c} \neq 0 \\x \left (\frac {\left (i e^{8 i c} - 2 i e^{6 i c} + 2 i e^{2 i c} - i\right ) e^{- 8 i c}}{16 a^{4}} - \frac {i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {i x}{16 a^{4}} \] Input:

integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**4,x)
 

Output:

Piecewise(((6144*a**8*d**2*exp(14*I*c)*exp(-2*I*d*x) - 2048*a**8*d**2*exp( 
10*I*c)*exp(-6*I*d*x) + 768*a**8*d**2*exp(8*I*c)*exp(-8*I*d*x))*exp(-16*I* 
c)/(98304*a**12*d**3), Ne(a**12*d**3*exp(16*I*c), 0)), (x*((I*exp(8*I*c) - 
 2*I*exp(6*I*c) + 2*I*exp(2*I*c) - I)*exp(-8*I*c)/(16*a**4) - I/(16*a**4)) 
, True)) + I*x/(16*a**4)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{32 \, a^{4} d} + \frac {\log \left (\tan \left (d x + c\right ) - i\right )}{32 \, a^{4} d} - \frac {-3 i \, \tan \left (d x + c\right )^{3} + 12 \, \tan \left (d x + c\right )^{2} - 13 i \, \tan \left (d x + c\right ) - 4}{48 \, a^{4} d {\left (\tan \left (d x + c\right ) - i\right )}^{4}} \] Input:

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
 

Output:

-1/32*log(tan(d*x + c) + I)/(a^4*d) + 1/32*log(tan(d*x + c) - I)/(a^4*d) - 
 1/48*(-3*I*tan(d*x + c)^3 + 12*tan(d*x + c)^2 - 13*I*tan(d*x + c) - 4)/(a 
^4*d*(tan(d*x + c) - I)^4)
 

Mupad [B] (verification not implemented)

Time = 1.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.48 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {x\,1{}\mathrm {i}}{16\,a^4}+\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{16}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,13{}\mathrm {i}}{48}+\frac {1}{12}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \] Input:

int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^4,x)
 

Output:

(x*1i)/(16*a^4) + ((tan(c + d*x)*13i)/48 - tan(c + d*x)^2/4 + (tan(c + d*x 
)^3*1i)/16 + 1/12)/(a^4*d*(tan(c + d*x)*1i + 1)^4)
 

Reduce [F]

\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i -6 \tan \left (d x +c \right )^{2}+4 \tan \left (d x +c \right ) i +1}d x}{a^{4}} \] Input:

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^4,x)
 

Output:

int(tan(c + d*x)**3/(tan(c + d*x)**4 - 4*tan(c + d*x)**3*i - 6*tan(c + d*x 
)**2 + 4*tan(c + d*x)*i + 1),x)/a**4