3.25 \(\int \tan ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=90 \[ \frac{2 a^3 \tan (c+d x)}{d}+\frac{4 i a^3 \log (\cos (c+d x))}{d}-4 a^3 x-\frac{i (a+i a \tan (c+d x))^4}{4 a d}-\frac{i a (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0698095, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3543, 3478, 3477, 3475} \[ \frac{2 a^3 \tan (c+d x)}{d}+\frac{4 i a^3 \log (\cos (c+d x))}{d}-4 a^3 x-\frac{i (a+i a \tan (c+d x))^4}{4 a d}-\frac{i a (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

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

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3475

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

Rubi steps

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

Mathematica [B]  time = 1.5869, size = 228, normalized size = 2.53 \[ -\frac{a^3 \sec (c) \sec ^4(c+d x) \left (-13 \sin (c+2 d x)+7 \sin (3 c+2 d x)-5 \sin (3 c+4 d x)+8 d x \cos (3 c+2 d x)-5 i \cos (3 c+2 d x)+2 d x \cos (3 c+4 d x)+2 d x \cos (5 c+4 d x)-4 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+2 \cos (c) \left (-3 i \log \left (\cos ^2(c+d x)\right )+6 d x-4 i\right )+\cos (c+2 d x) \left (-4 i \log \left (\cos ^2(c+d x)\right )+8 d x-5 i\right )-i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+15 \sin (c)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sec[c]*Sec[c + d*x]^4*((-5*I)*Cos[3*c + 2*d*x] + 8*d*x*Cos[3*c + 2*d*x] + 2*d*x*Cos[3*c + 4*d*x] + 2*d*x
*Cos[5*c + 4*d*x] + 2*Cos[c]*(-4*I + 6*d*x - (3*I)*Log[Cos[c + d*x]^2]) + Cos[c + 2*d*x]*(-5*I + 8*d*x - (4*I)
*Log[Cos[c + d*x]^2]) - (4*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - I*Cos[3*c + 4*d*x]*Log[Cos[c + d*x]^2] -
I*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 15*Sin[c] - 13*Sin[c + 2*d*x] + 7*Sin[3*c + 2*d*x] - 5*Sin[3*c + 4*d*
x]))/(8*d)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 101, normalized size = 1.1 \begin{align*} 4\,{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}-{\frac{{\frac{i}{4}}{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{2\,i{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^3,x)

[Out]

4*a^3*tan(d*x+c)/d-1/4*I/d*a^3*tan(d*x+c)^4-1/d*a^3*tan(d*x+c)^3+2*I/d*a^3*tan(d*x+c)^2-2*I/d*a^3*ln(1+tan(d*x
+c)^2)-4/d*a^3*arctan(tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 2.41642, size = 111, normalized size = 1.23 \begin{align*} -\frac{i \, a^{3} \tan \left (d x + c\right )^{4} + 4 \, a^{3} \tan \left (d x + c\right )^{3} - 8 i \, a^{3} \tan \left (d x + c\right )^{2} + 16 \,{\left (d x + c\right )} a^{3} + 8 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 16 \, a^{3} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(I*a^3*tan(d*x + c)^4 + 4*a^3*tan(d*x + c)^3 - 8*I*a^3*tan(d*x + c)^2 + 16*(d*x + c)*a^3 + 8*I*a^3*log(ta
n(d*x + c)^2 + 1) - 16*a^3*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [B]  time = 2.22138, size = 506, normalized size = 5.62 \begin{align*} \frac{24 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 46 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 36 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i \, a^{3} +{\left (4 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 16 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(24*I*a^3*e^(6*I*d*x + 6*I*c) + 46*I*a^3*e^(4*I*d*x + 4*I*c) + 36*I*a^3*e^(2*I*d*x + 2*I*c) + 10*I*a^3 + (4*I*
a^3*e^(8*I*d*x + 8*I*c) + 16*I*a^3*e^(6*I*d*x + 6*I*c) + 24*I*a^3*e^(4*I*d*x + 4*I*c) + 16*I*a^3*e^(2*I*d*x +
2*I*c) + 4*I*a^3)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*
d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

Sympy [B]  time = 7.05857, size = 180, normalized size = 2. \begin{align*} \frac{4 i a^{3} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{24 i a^{3} e^{- 2 i c} e^{6 i d x}}{d} + \frac{46 i a^{3} e^{- 4 i c} e^{4 i d x}}{d} + \frac{36 i a^{3} e^{- 6 i c} e^{2 i d x}}{d} + \frac{10 i a^{3} e^{- 8 i c}}{d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*I*a**3*log(exp(2*I*d*x) + exp(-2*I*c))/d + (24*I*a**3*exp(-2*I*c)*exp(6*I*d*x)/d + 46*I*a**3*exp(-4*I*c)*exp
(4*I*d*x)/d + 36*I*a**3*exp(-6*I*c)*exp(2*I*d*x)/d + 10*I*a**3*exp(-8*I*c)/d)/(exp(8*I*d*x) + 4*exp(-2*I*c)*ex
p(6*I*d*x) + 6*exp(-4*I*c)*exp(4*I*d*x) + 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

________________________________________________________________________________________

Giac [B]  time = 1.6925, size = 298, normalized size = 3.31 \begin{align*} \frac{4 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 16 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 16 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 46 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 36 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 10 i \, a^{3}}{d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*I*a^3*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 16*I*a^3*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*
c) + 1) + 24*I*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 16*I*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*
d*x + 2*I*c) + 1) + 24*I*a^3*e^(6*I*d*x + 6*I*c) + 46*I*a^3*e^(4*I*d*x + 4*I*c) + 36*I*a^3*e^(2*I*d*x + 2*I*c)
+ 4*I*a^3*log(e^(2*I*d*x + 2*I*c) + 1) + 10*I*a^3)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(
4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)