\(\int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx\) [48]

Optimal result

Integrand size = 24, antiderivative size = 74 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {3 i x}{2 a}-\frac {\log (\cos (c+d x))}{a d}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3631, 3606, 3556} \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 i \tan (c+d x)}{2 a d}-\frac {\log (\cos (c+d x))}{a d}+\frac {3 i x}{2 a} \]

[In]

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

[Out]

(((3*I)/2)*x)/a - Log[Cos[c + d*x]]/(a*d) - (((3*I)/2)*Tan[c + d*x])/(a*d) - Tan[c + d*x]^2/(2*d*(a + I*a*Tan[
c + d*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 3606

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

Rule 3631

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

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {2 i \log (\cos (c+d x))+\arctan (\tan (c+d x)) (3+3 i \tan (c+d x))-(3+2 \log (\cos (c+d x))) \tan (c+d x)-2 i \tan ^2(c+d x)}{2 a d (-i+\tan (c+d x))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {i \tan \left (d x +c \right )}{d a}-\frac {i}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}+\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}\)	\(72\)
default	\(-\frac {i \tan \left (d x +c \right )}{d a}-\frac {i}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}+\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{2 d a}\)	\(72\)
risch	\(\frac {5 i x}{2 a}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{4 a d}+\frac {2 i c}{a d}+\frac {2}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a d}\)	\(77\)
norman	\(\frac {\frac {1}{2 a d}+\frac {3 i x}{2 a}-\frac {3 i \tan \left (d x +c \right )}{2 d a}-\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{d a}+\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}}{1+\tan ^{2}\left (d x +c \right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {10 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (10 i \, d x + 9\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left (e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 1}{4 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

[In]

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

[Out]

1/4*(10*I*d*x*e^(4*I*d*x + 4*I*c) + (10*I*d*x + 9)*e^(2*I*d*x + 2*I*c) - 4*(e^(4*I*d*x + 4*I*c) + e^(2*I*d*x +
2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) + 1)/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=\begin {cases} \frac {e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (\frac {\left (5 i e^{2 i c} - i\right ) e^{- 2 i c}}{2 a} - \frac {5 i}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {2}{a d e^{2 i c} e^{2 i d x} + a d} + \frac {5 i x}{2 a} - \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \]

[In]

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

[Out]

Piecewise((exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2*I*c), 0)), (x*((5*I*exp(2*I*c) - I)*exp(-2*I*c)/(2*
a) - 5*I/(2*a)), True)) + 2/(a*d*exp(2*I*c)*exp(2*I*d*x) + a*d) + 5*I*x/(2*a) - log(exp(2*I*d*x) + exp(-2*I*c)
)/(a*d)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=-\frac {\frac {\log \left (\tan \left (d x + c\right ) + i\right )}{a} - \frac {5 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {4 i \, \tan \left (d x + c\right )}{a} + \frac {5 \, \tan \left (d x + c\right ) - 3 i}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]

[In]

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

[Out]

-1/4*(log(tan(d*x + c) + I)/a - 5*log(tan(d*x + c) - I)/a + 4*I*tan(d*x + c)/a + (5*tan(d*x + c) - 3*I)/(a*(ta
n(d*x + c) - I)))/d

Mupad [B] (verification not implemented)

Time = 4.62 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99 \[ \int \frac {\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx=\frac {5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{4\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{4\,a\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{a\,d}+\frac {1}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

(5*log(tan(c + d*x) - 1i))/(4*a*d) - log(tan(c + d*x) + 1i)/(4*a*d) - (tan(c + d*x)*1i)/(a*d) + 1/(2*a*d*(tan(
c + d*x)*1i + 1))