3.36 \(\int \tan (c+d x) (a+i a \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=108 \[ \frac{4 i a^4 \tan (c+d x)}{d}+\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac{8 a^4 \log (\cos (c+d x))}{d}-8 i a^4 x+\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d} \]

[Out]

(-8*I)*a^4*x - (8*a^4*Log[Cos[c + d*x]])/d + ((4*I)*a^4*Tan[c + d*x])/d + (a*(a + I*a*Tan[c + d*x])^3)/(3*d) +
(a + I*a*Tan[c + d*x])^4/(4*d) + (a^2 + I*a^2*Tan[c + d*x])^2/d

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Rubi [A]  time = 0.0799437, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3527, 3478, 3477, 3475} \[ \frac{4 i a^4 \tan (c+d x)}{d}+\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac{8 a^4 \log (\cos (c+d x))}{d}-8 i a^4 x+\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-8*I)*a^4*x - (8*a^4*Log[Cos[c + d*x]])/d + ((4*I)*a^4*Tan[c + d*x])/d + (a*(a + I*a*Tan[c + d*x])^3)/(3*d) +
(a + I*a*Tan[c + d*x])^4/(4*d) + (a^2 + I*a^2*Tan[c + d*x])^2/d

Rule 3527

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 (c+d x) (a+i a \tan (c+d x))^4 \, dx &=\frac{(a+i a \tan (c+d x))^4}{4 d}-i \int (a+i a \tan (c+d x))^4 \, dx\\ &=\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d}-(2 i a) \int (a+i a \tan (c+d x))^3 \, dx\\ &=\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d}+\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\left (4 i a^2\right ) \int (a+i a \tan (c+d x))^2 \, dx\\ &=-8 i a^4 x+\frac{4 i a^4 \tan (c+d x)}{d}+\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d}+\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (8 a^4\right ) \int \tan (c+d x) \, dx\\ &=-8 i a^4 x-\frac{8 a^4 \log (\cos (c+d x))}{d}+\frac{4 i a^4 \tan (c+d x)}{d}+\frac{a (a+i a \tan (c+d x))^3}{3 d}+\frac{(a+i a \tan (c+d x))^4}{4 d}+\frac{\left (a^2+i a^2 \tan (c+d x)\right )^2}{d}\\ \end{align*}

Mathematica [B]  time = 1.20427, size = 231, normalized size = 2.14 \[ -\frac{i a^4 \sec (c) \sec ^4(c+d x) \left (-38 \sin (c+2 d x)+18 \sin (3 c+2 d x)-14 \sin (3 c+4 d x)+24 d x \cos (3 c+2 d x)-12 i \cos (3 c+2 d x)+6 d x \cos (3 c+4 d x)+6 d x \cos (5 c+4 d x)-12 i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+12 \cos (c+2 d x) \left (-i \log \left (\cos ^2(c+d x)\right )+2 d x-i\right )+3 \cos (c) \left (-6 i \log \left (\cos ^2(c+d x)\right )+12 d x-7 i\right )-3 i \cos (3 c+4 d x) \log \left (\cos ^2(c+d x)\right )-3 i \cos (5 c+4 d x) \log \left (\cos ^2(c+d x)\right )+42 \sin (c)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^4,x]

[Out]

((-I/12)*a^4*Sec[c]*Sec[c + d*x]^4*((-12*I)*Cos[3*c + 2*d*x] + 24*d*x*Cos[3*c + 2*d*x] + 6*d*x*Cos[3*c + 4*d*x
] + 6*d*x*Cos[5*c + 4*d*x] + 12*Cos[c + 2*d*x]*(-I + 2*d*x - I*Log[Cos[c + d*x]^2]) + 3*Cos[c]*(-7*I + 12*d*x
- (6*I)*Log[Cos[c + d*x]^2]) - (12*I)*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] - (3*I)*Cos[3*c + 4*d*x]*Log[Cos[c
+ d*x]^2] - (3*I)*Cos[5*c + 4*d*x]*Log[Cos[c + d*x]^2] + 42*Sin[c] - 38*Sin[c + 2*d*x] + 18*Sin[3*c + 2*d*x] -
14*Sin[3*c + 4*d*x]))/d

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Maple [A]  time = 0.005, size = 101, normalized size = 0.9 \begin{align*}{\frac{8\,i{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{\frac{4\,i}{3}}{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{7\,{a}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+4\,{\frac{{a}^{4}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{8\,i{a}^{4}\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)*(a+I*a*tan(d*x+c))^4,x)

[Out]

8*I/d*a^4*tan(d*x+c)+1/4*a^4*tan(d*x+c)^4/d-4/3*I/d*a^4*tan(d*x+c)^3-7/2*a^4*tan(d*x+c)^2/d+4/d*a^4*ln(1+tan(d
*x+c)^2)-8*I/d*a^4*arctan(tan(d*x+c))

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Maxima [A]  time = 1.61168, size = 111, normalized size = 1.03 \begin{align*} \frac{3 \, a^{4} \tan \left (d x + c\right )^{4} - 16 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} - 96 i \,{\left (d x + c\right )} a^{4} + 48 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 96 i \, a^{4} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/12*(3*a^4*tan(d*x + c)^4 - 16*I*a^4*tan(d*x + c)^3 - 42*a^4*tan(d*x + c)^2 - 96*I*(d*x + c)*a^4 + 48*a^4*log
(tan(d*x + c)^2 + 1) + 96*I*a^4*tan(d*x + c))/d

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Fricas [A]  time = 2.20163, size = 482, normalized size = 4.46 \begin{align*} -\frac{4 \,{\left (30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 14 \, a^{4} + 6 \,{\left (a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \,{\left (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\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-4/3*(30*a^4*e^(6*I*d*x + 6*I*c) + 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) + 14*a^4 + 6*(a^4*e
^(8*I*d*x + 8*I*c) + 4*a^4*e^(6*I*d*x + 6*I*c) + 6*a^4*e^(4*I*d*x + 4*I*c) + 4*a^4*e^(2*I*d*x + 2*I*c) + a^4)*
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)

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Sympy [A]  time = 6.35128, size = 177, normalized size = 1.64 \begin{align*} - \frac{8 a^{4} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{40 a^{4} e^{- 2 i c} e^{6 i d x}}{d} - \frac{84 a^{4} e^{- 4 i c} e^{4 i d x}}{d} - \frac{200 a^{4} e^{- 6 i c} e^{2 i d x}}{3 d} - \frac{56 a^{4} e^{- 8 i c}}{3 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)*(a+I*a*tan(d*x+c))**4,x)

[Out]

-8*a**4*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-40*a**4*exp(-2*I*c)*exp(6*I*d*x)/d - 84*a**4*exp(-4*I*c)*exp(4*I
*d*x)/d - 200*a**4*exp(-6*I*c)*exp(2*I*d*x)/(3*d) - 56*a**4*exp(-8*I*c)/(3*d))/(exp(8*I*d*x) + 4*exp(-2*I*c)*e
xp(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))

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Giac [B]  time = 1.47225, size = 300, normalized size = 2.78 \begin{align*} -\frac{4 \,{\left (6 \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 36 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 24 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, a^{4} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 \, a^{4}\right )}}{3 \,{\left (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\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-4/3*(6*a^4*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*a^4*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) + 36*a^4*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 24*a^4*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 30*a^4*e^(6*I*d*x + 6*I*c) + 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) + 6*a^4*
log(e^(2*I*d*x + 2*I*c) + 1) + 14*a^4)/(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)