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

Optimal result

Integrand size = 22, antiderivative size = 67 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=i a x+\frac {a \log (\cos (c+d x))}{d}-\frac {i a \tan (c+d x)}{d}+\frac {a \tan ^2(c+d x)}{2 d}+\frac {i a \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3609, 3606, 3556} \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \tan ^3(c+d x)}{3 d}+\frac {a \tan ^2(c+d x)}{2 d}-\frac {i a \tan (c+d x)}{d}+\frac {a \log (\cos (c+d x))}{d}+i a x \]

[In]

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

[Out]

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

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 3609

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \arctan (\tan (c+d x))}{d}-\frac {i a \tan (c+d x)}{d}+\frac {i a \tan ^3(c+d x)}{3 d}+\frac {a \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {a \left (-i \tan \left (d x +c \right )+\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(60\)
default	\(\frac {a \left (-i \tan \left (d x +c \right )+\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(60\)
parallelrisch	\(-\frac {-2 i a \left (\tan ^{3}\left (d x +c \right )\right )-6 i a x d +6 i a \tan \left (d x +c \right )-3 \left (\tan ^{2}\left (d x +c \right )\right ) a +3 a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{6 d}\)	\(60\)
parts	\(\frac {a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {i a \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(64\)
norman	\(i a x +\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {i a \tan \left (d x +c \right )}{d}+\frac {i a \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(66\)
risch	\(-\frac {2 i a c}{d}+\frac {2 a \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}+4\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(70\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (57) = 114\).

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.79 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, {\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

1/3*(18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c) + 3*(a*e^(6*I*d*x + 6*I*c) + 3*a*e^(4*I*d*x + 4*I*c)
+ 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4
*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).

Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.81 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {18 a e^{4 i c} e^{4 i d x} + 18 a e^{2 i c} e^{2 i d x} + 8 a}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

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

[Out]

a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (18*a*exp(4*I*c)*exp(4*I*d*x) + 18*a*exp(2*I*c)*exp(2*I*d*x) + 8*a)/(3*d
*exp(6*I*c)*exp(6*I*d*x) + 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {-2 i \, a \tan \left (d x + c\right )^{3} - 3 \, a \tan \left (d x + c\right )^{2} - 6 i \, {\left (d x + c\right )} a + 3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 i \, a \tan \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

-1/6*(-2*I*a*tan(d*x + c)^3 - 3*a*tan(d*x + c)^2 - 6*I*(d*x + c)*a + 3*a*log(tan(d*x + c)^2 + 1) + 6*I*a*tan(d
*x + c))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (57) = 114\).

Time = 0.65 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.33 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {3 \, a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

1/3*(3*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) +
1) + 9*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c
) + 3*a*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x
+ 2*I*c) + d)

Mupad [B] (verification not implemented)

Time = 4.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{3}+a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d} \]

[In]

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

[Out]

-(a*tan(c + d*x)*1i - (a*tan(c + d*x)^2)/2 - (a*tan(c + d*x)^3*1i)/3 + a*log(tan(c + d*x) + 1i))/d