Integrand size = 24, antiderivative size = 79 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {3}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \] Output:
-3/4*I*x/a^2+ln(cos(d*x+c))/a^2/d-3/4/a^2/d/(1+I*tan(d*x+c))-1/4*tan(d*x+c )^2/d/(a+I*a*tan(d*x+c))^2
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {4 \log (\cos (c+d x))+i (3+8 \log (\cos (c+d x))) \tan (c+d x)-4 (1+\log (\cos (c+d x))) \tan ^2(c+d x)+3 i \arctan (\tan (c+d x)) (-i+\tan (c+d x))^2}{4 a^2 d (-i+\tan (c+d x))^2} \] Input:
Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]
Output:
-1/4*(4*Log[Cos[c + d*x]] + I*(3 + 8*Log[Cos[c + d*x]])*Tan[c + d*x] - 4*( 1 + Log[Cos[c + d*x]])*Tan[c + d*x]^2 + (3*I)*ArcTan[Tan[c + d*x]]*(-I + T an[c + d*x])^2)/(a^2*d*(-I + Tan[c + d*x])^2)
Time = 0.53 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4041, 27, 3042, 4072, 27, 3042, 3956, 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.
\(\displaystyle \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^3}{(a+i a \tan (c+d x))^2}dx\) |
\(\Big \downarrow \) 4041 |
\(\displaystyle -\frac {\int -\frac {2 \tan (c+d x) (a-2 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\tan (c+d x) (a-2 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\tan (c+d x) (a-2 i a \tan (c+d x))}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4072 |
\(\displaystyle \frac {-2 \int \tan (c+d x)dx-\frac {i \int \frac {3 i a^2 \tan (c+d x)}{i \tan (c+d x) a+a}dx}{a}}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-2 \int \tan (c+d x)dx+3 a \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-2 \int \tan (c+d x)dx+3 a \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\frac {2 \log (\cos (c+d x))}{d}+3 a \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle \frac {\frac {2 \log (\cos (c+d x))}{d}+3 a \left (-\frac {i \int 1dx}{2 a}-\frac {1}{2 d (a+i a \tan (c+d x))}\right )}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {2 \log (\cos (c+d x))}{d}+3 a \left (-\frac {1}{2 d (a+i a \tan (c+d x))}-\frac {i x}{2 a}\right )}{2 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}\) |
Input:
Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]
Output:
-1/4*Tan[c + d*x]^2/(d*(a + I*a*Tan[c + d*x])^2) + ((2*Log[Cos[c + d*x]])/ d + 3*a*(((-1/2*I)*x)/a - 1/(2*d*(a + I*a*Tan[c + d*x]))))/(2*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_)])^(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_)])^(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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[1/b Int[Simp[A*b*c + (A*b*d + B*(b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d , e, f, A, B}, x] && NeQ[b*c - a*d, 0]
Time = 0.69 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {7 i x}{4 a^{2}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {2 i c}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{2} d}\) | \(72\) |
derivativedivides | \(\frac {5 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\) | \(76\) |
default | \(\frac {5 i}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )}-\frac {1}{4 d \,a^{2} \left (-i+\tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\) | \(76\) |
norman | \(\frac {-\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {3 \tan \left (d x +c \right )^{2}}{2 a d}+\frac {3 i \tan \left (d x +c \right )}{4 d a}+\frac {5 i \tan \left (d x +c \right )^{3}}{4 d a}-\frac {3 i x \tan \left (d x +c \right )^{2}}{2 a}-\frac {3 i x \tan \left (d x +c \right )^{4}}{4 a}}{a \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \,a^{2}}\) | \(131\) |
Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-7/4*I*x/a^2-1/2/a^2/d*exp(-2*I*(d*x+c))+1/16/a^2/d*exp(-4*I*(d*x+c))-2*I/ a^2/d*c+1/a^2/d*ln(exp(2*I*(d*x+c))+1)
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d* x + 2*I*c) + 1) - 8*e^(2*I*d*x + 2*I*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)
Time = 0.31 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.87 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} + 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \] Input:
integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**2,x)
Output:
Piecewise(((-16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp( -4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2), Ne(a**4*d**2*exp(6*I*c), 0)), (x*(( -7*I*exp(4*I*c) + 4*I*exp(2*I*c) - I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a**2)) , True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)
Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^3/(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.
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{8 \, a^{2} d} - \frac {7 \, \log \left (\tan \left (d x + c\right ) - i\right )}{8 \, a^{2} d} + \frac {5 i \, \tan \left (d x + c\right ) + 4}{4 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \] Input:
integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
-1/8*log(tan(d*x + c) + I)/(a^2*d) - 7/8*log(tan(d*x + c) - I)/(a^2*d) + 1 /4*(5*I*tan(d*x + c) + 4)/(a^2*d*(tan(d*x + c) - I)^2)
Time = 1.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {\frac {5\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \] Input:
int(tan(c + d*x)^3/(a + a*tan(c + d*x)*1i)^2,x)
Output:
- (7*log(tan(c + d*x) - 1i))/(8*a^2*d) - log(tan(c + d*x) + 1i)/(8*a^2*d) - ((5*tan(c + d*x))/(4*a^2) - 1i/a^2)/(d*(2*tan(c + d*x) + tan(c + d*x)^2* 1i - 1i))
\[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\int \frac {\tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{2}-2 \tan \left (d x +c \right ) i -1}d x}{a^{2}} \] Input:
int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x)
Output:
( - int(tan(c + d*x)**3/(tan(c + d*x)**2 - 2*tan(c + d*x)*i - 1),x))/a**2