\(\int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [59]

Optimal result

Integrand size = 24, antiderivative size = 79 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {3}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3639, 3670, 3556, 12, 3607, 8} \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {3 i x}{4 a^2}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[In]

Int[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3556

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

Rule 3607

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] + Dist[(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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[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))*Tan[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] && (IntegerQ[m] || IntegersQ[2*m
, 2*n])

Rule 3670

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[B*(d/b), Int[Tan[e + f*x], x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan (c+d x) (-2 a+4 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {i \int -\frac {6 i a^2 \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^3}-\frac {\int \tan (c+d x) \, dx}{a^2} \\ & = \frac {\log (\cos (c+d x))}{a^2 d}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {3 \int \frac {\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{2 a} \\ & = \frac {\log (\cos (c+d x))}{a^2 d}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {(3 i) \int 1 \, dx}{4 a^2} \\ & = -\frac {3 i x}{4 a^2}+\frac {\log (\cos (c+d x))}{a^2 d}-\frac {\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {3}{4 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {4 \log (\cos (c+d x))+i (3+8 \log (\cos (c+d x))) \tan (c+d x)-4 (1+\log (\cos (c+d x))) \tan ^2(c+d x)+3 i \arctan (\tan (c+d x)) (-i+\tan (c+d x))^2}{4 a^2 d (-i+\tan (c+d x))^2} \]

[In]

Integrate[Tan[c + d*x]^3/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-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 + Tan[c + d*x])^2)/(a^2*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.23 (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	\(\frac {5 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\)	\(76\)
default	\(\frac {5 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}\)	\(76\)
norman	\(\frac {-\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2 a d}+\frac {3 i \tan \left (d x +c \right )}{4 d a}+\frac {5 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}-\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}\)	\(131\)

[In]

int(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

[In]

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.87 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} + 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \]

[In]

integrate(tan(d*x+c)**3/(a+I*a*tan(d*x+c))**2,x)

[Out]

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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.75 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {21 \, \tan \left (d x + c\right )^{2} - 22 i \, \tan \left (d x + c\right ) - 5}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

[In]

integrate(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*log(tan(d*x + c) + I)/a^2 + 14*log(tan(d*x + c) - I)/a^2 - (21*tan(d*x + c)^2 - 22*I*tan(d*x + c) - 5
)/(a^2*(tan(d*x + c) - I)^2))/d

Mupad [B] (verification not implemented)

Time = 4.95 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {\frac {5\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

[In]

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

[Out]

- (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))