Integrand size = 24, antiderivative size = 92 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i x}{8 a^3}+\frac {3}{8 a^3 d (1+i \tan (c+d x))}+\frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {1}{8 a d (a+i a \tan (c+d x))^2} \] Output:
1/8*I*x/a^3+3/8/a^3/d/(1+I*tan(d*x+c))+1/6*I*tan(d*x+c)^3/d/(a+I*a*tan(d*x +c))^3-1/8/a/d/(a+I*a*tan(d*x+c))^2
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {3 \arctan (\tan (c+d x)) (-1-i \tan (c+d x))^3+\tan (c+d x) \left (3+9 i \tan (c+d x)-10 \tan ^2(c+d x)\right )}{24 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^3,x]
Output:
(3*ArcTan[Tan[c + d*x]]*(-1 - I*Tan[c + d*x])^3 + Tan[c + d*x]*(3 + (9*I)* Tan[c + d*x] - 10*Tan[c + d*x]^2))/(24*a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 4029 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {\tan ^2(c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {\tan (c+d x)^2}{(i \tan (c+d x) a+a)^2}dx}{2 a}\) |
\(\Big \downarrow \) 4023 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {\int \frac {a-2 i a \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {\int \frac {a-2 i a \tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {-\frac {\int 1dx}{2}+\frac {3 i}{2 d (1+i \tan (c+d x))}}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {i \left (\frac {-\frac {x}{2}+\frac {3 i}{2 d (1+i \tan (c+d x))}}{2 a^2}-\frac {i}{4 d (a+i a \tan (c+d x))^2}\right )}{2 a}\) |
Input:
Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^3,x]
Output:
((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
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]
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]
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)]
Time = 0.64 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.65
method | result | size |
risch | \(\frac {i x}{8 a^{3}}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\) | \(60\) |
derivativedivides | \(\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {7 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}+\frac {5}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\) | \(77\) |
default | \(\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}+\frac {i}{6 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{3}}-\frac {7 i}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )}+\frac {5}{8 d \,a^{3} \left (-i+\tan \left (d x +c \right )\right )^{2}}\) | \(77\) |
norman | \(\frac {\frac {5}{12 a d}+\frac {i x}{8 a}+\frac {5 \tan \left (d x +c \right )^{2}}{4 a d}-\frac {i \tan \left (d x +c \right )}{8 d a}-\frac {i \tan \left (d x +c \right )^{3}}{3 d a}-\frac {7 i \tan \left (d x +c \right )^{5}}{8 a d}+\frac {3 i x \tan \left (d x +c \right )^{2}}{8 a}+\frac {3 i x \tan \left (d x +c \right )^{4}}{8 a}+\frac {i x \tan \left (d x +c \right )^{6}}{8 a}+\frac {3 \tan \left (d x +c \right )^{4}}{2 a d}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}\) | \(159\) |
Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*I*x/a^3+3/16/a^3/d*exp(-2*I*(d*x+c))-3/32/a^3/d*exp(-4*I*(d*x+c))+1/48 /a^3/d*exp(-6*I*(d*x+c))
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.59 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (12 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/96*(12*I*d*x*e^(6*I*d*x + 6*I*c) + 18*e^(4*I*d*x + 4*I*c) - 9*e^(2*I*d*x + 2*I*c) + 2)*e^(-6*I*d*x - 6*I*c)/(a^3*d)
Time = 0.24 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (4608 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 2304 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (i e^{6 i c} - 3 i e^{4 i c} + 3 i e^{2 i c} - i\right ) e^{- 6 i c}}{8 a^{3}} - \frac {i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {i x}{8 a^{3}} \] Input:
integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((4608*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) - 2304*a**6*d**2*exp( 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*((I*exp(6*I*c) - 3* I*exp(4*I*c) + 3*I*exp(2*I*c) - I)*exp(-6*I*c)/(8*a**3) - I/(8*a**3)), Tru e)) + I*x/(8*a**3)
Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^3/(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 = 73, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{16 \, a^{3} d} + \frac {\log \left (\tan \left (d x + c\right ) - i\right )}{16 \, a^{3} d} - \frac {i \, {\left (21 \, \tan \left (d x + c\right )^{2} - 27 i \, \tan \left (d x + c\right ) - 10\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/16*log(tan(d*x + c) + I)/(a^3*d) + 1/16*log(tan(d*x + c) - I)/(a^3*d) - 1/24*I*(21*tan(d*x + c)^2 - 27*I*tan(d*x + c) - 10)/(a^3*d*(tan(d*x + c) - I)^3)
Time = 0.95 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.53 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {x\,1{}\mathrm {i}}{8\,a^3}+\frac {-\frac {7\,{\mathrm {tan}\left (c+d\,x\right )}^2}{8}+\frac {\mathrm {tan}\left (c+d\,x\right )\,9{}\mathrm {i}}{8}+\frac {5}{12}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \] Input:
int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^3,x)
Output:
(x*1i)/(8*a^3) + ((tan(c + d*x)*9i)/8 - (7*tan(c + d*x)^2)/8 + 5/12)/(a^3* d*(tan(c + d*x)*1i + 1)^3)
\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x}{a^{3}} \] Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x)
Output:
( - int(tan(c + d*x)**3/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x))/a**3