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

Optimal. Leaf size=97 \[ -\frac{15 i a^4 \sec (c+d x)}{2 d}-\frac{15 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d} \]

[Out]

(-15*a^4*ArcTanh[Sin[c + d*x]])/(2*d) - (((15*I)/2)*a^4*Sec[c + d*x])/d - ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c
+ d*x])^3)/d - (((5*I)/2)*Sec[c + d*x]*(a^4 + I*a^4*Tan[c + d*x]))/d

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Rubi [A]  time = 0.0744401, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3496, 3498, 3486, 3770} \[ -\frac{15 i a^4 \sec (c+d x)}{2 d}-\frac{15 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^4*ArcTanh[Sin[c + d*x]])/(2*d) - (((15*I)/2)*a^4*Sec[c + d*x])/d - ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c
+ d*x])^3)/d - (((5*I)/2)*Sec[c + d*x]*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3498

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3770

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

Rubi steps

\begin{align*} \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}-\left (5 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}-\frac{5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d}-\frac{1}{2} \left (15 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac{15 i a^4 \sec (c+d x)}{2 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}-\frac{5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d}-\frac{1}{2} \left (15 a^4\right ) \int \sec (c+d x) \, dx\\ &=-\frac{15 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{15 i a^4 \sec (c+d x)}{2 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}-\frac{5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d}\\ \end{align*}

Mathematica [B]  time = 6.43177, size = 906, normalized size = 9.34 \[ \frac{\cos ^4(c+d x) (8 \cos (3 c)-8 i \sin (3 c)) \sin (d x) (i \tan (c+d x) a+a)^4}{d (\cos (d x)+i \sin (d x))^4}-\frac{i \cos ^4(c+d x) (4 \cos (4 c)-4 i \sin (4 c)) \sin \left (\frac{d x}{2}\right ) (i \tan (c+d x) a+a)^4}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{i \cos ^4(c+d x) (4 \cos (4 c)-4 i \sin (4 c)) \sin \left (\frac{d x}{2}\right ) (i \tan (c+d x) a+a)^4}{d \left (\cos \left (\frac{c}{2}\right )+\sin \left (\frac{c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\cos ^4(c+d x) \left (\frac{1}{4} \cos (4 c)-\frac{1}{4} i \sin (4 c)\right ) (i \tan (c+d x) a+a)^4}{d (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{\cos ^4(c+d x) \left (\frac{1}{4} i \sin (4 c)-\frac{1}{4} \cos (4 c)\right ) (i \tan (c+d x) a+a)^4}{d (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{15 \cos (4 c) \cos ^4(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) (i \tan (c+d x) a+a)^4}{2 d (\cos (d x)+i \sin (d x))^4}-\frac{15 \cos (4 c) \cos ^4(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) (i \tan (c+d x) a+a)^4}{2 d (\cos (d x)+i \sin (d x))^4}+\frac{\cos (d x) \cos ^4(c+d x) (-8 i \cos (3 c)-8 \sin (3 c)) (i \tan (c+d x) a+a)^4}{d (\cos (d x)+i \sin (d x))^4}+\frac{\cos ^4(c+d x) \sec (c) (-4 i \cos (4 c)-4 \sin (4 c)) (i \tan (c+d x) a+a)^4}{d (\cos (d x)+i \sin (d x))^4}-\frac{15 i \cos ^4(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sin (4 c) (i \tan (c+d x) a+a)^4}{2 d (\cos (d x)+i \sin (d x))^4}+\frac{15 i \cos ^4(c+d x) \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )+\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \sin (4 c) (i \tan (c+d x) a+a)^4}{2 d (\cos (d x)+i \sin (d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Cos[4*c]*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(a + I*a*Tan[c + d*x])^4)/(2*d*(Cos[d
*x] + I*Sin[d*x])^4) - (15*Cos[4*c]*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(a + I*a*Tan[c
+ d*x])^4)/(2*d*(Cos[d*x] + I*Sin[d*x])^4) + (Cos[d*x]*Cos[c + d*x]^4*((-8*I)*Cos[3*c] - 8*Sin[3*c])*(a + I*a
*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) + (Cos[c + d*x]^4*Sec[c]*((-4*I)*Cos[4*c] - 4*Sin[4*c])*(a + I
*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) - (((15*I)/2)*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/
2 + (d*x)/2]]*Sin[4*c]*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) + (((15*I)/2)*Cos[c + d*x]^4*Lo
g[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sin[4*c]*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) +
(Cos[c + d*x]^4*(8*Cos[3*c] - (8*I)*Sin[3*c])*Sin[d*x]*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4)
+ (Cos[c + d*x]^4*(Cos[4*c]/4 - (I/4)*Sin[4*c])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4*(Cos[c
/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) - (I*Cos[c + d*x]^4*(4*Cos[4*c] - (4*I)*Sin[4*c])*Sin[(d*x)/2]*(a + I*a
*Tan[c + d*x])^4)/(d*(Cos[c/2] - Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^4*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])
) + (Cos[c + d*x]^4*(-Cos[4*c]/4 + (I/4)*Sin[4*c])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4*(Cos
[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (I*Cos[c + d*x]^4*(4*Cos[4*c] - (4*I)*Sin[4*c])*Sin[(d*x)/2]*(a + I
*a*Tan[c + d*x])^4)/(d*(Cos[c/2] + Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^4*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2
]))

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Maple [A]  time = 0.052, size = 141, normalized size = 1.5 \begin{align*}{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{17\,{a}^{4}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{15\,{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}-{\frac{4\,i{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}-{\frac{4\,i{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }{d}}-{\frac{12\,i{a}^{4}\cos \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+I*a*tan(d*x+c))^4,x)

[Out]

1/2/d*a^4*sin(d*x+c)^5/cos(d*x+c)^2+1/2*a^4*sin(d*x+c)^3/d+17/2*a^4*sin(d*x+c)/d-15/2/d*a^4*ln(sec(d*x+c)+tan(
d*x+c))-4*I/d*a^4*sin(d*x+c)^4/cos(d*x+c)-4*I/d*a^4*sin(d*x+c)^2*cos(d*x+c)-12*I/d*a^4*cos(d*x+c)

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Maxima [A]  time = 1.12994, size = 185, normalized size = 1.91 \begin{align*} -\frac{a^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} + 16 i \, a^{4}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 12 \, a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 16 i \, a^{4} \cos \left (d x + c\right ) - 4 \, a^{4} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin(d*x
+ c)) + 16*I*a^4*(1/cos(d*x + c) + cos(d*x + c)) + 12*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*
sin(d*x + c)) + 16*I*a^4*cos(d*x + c) - 4*a^4*sin(d*x + c))/d

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Fricas [B]  time = 1.20279, size = 444, normalized size = 4.58 \begin{align*} \frac{-16 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 50 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 30 i \, a^{4} e^{\left (i \, d x + i \, c\right )} - 15 \,{\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 15 \,{\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{2 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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(cos(d*x+c)*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/2*(-16*I*a^4*e^(5*I*d*x + 5*I*c) - 50*I*a^4*e^(3*I*d*x + 3*I*c) - 30*I*a^4*e^(I*d*x + I*c) - 15*(a^4*e^(4*I*
d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(I*d*x + I*c) + I) + 15*(a^4*e^(4*I*d*x + 4*I*c) + 2*a^4
*e^(2*I*d*x + 2*I*c) + a^4)*log(e^(I*d*x + I*c) - I))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 1.34373, size = 153, normalized size = 1.58 \begin{align*} \frac{15 a^{4} \left (\frac{\log{\left (e^{i d x} - i e^{- i c} \right )}}{2} - \frac{\log{\left (e^{i d x} + i e^{- i c} \right )}}{2}\right )}{d} + \frac{- \frac{9 i a^{4} e^{- i c} e^{3 i d x}}{d} - \frac{7 i a^{4} e^{- 3 i c} e^{i d x}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \begin{cases} - \frac{8 i a^{4} e^{i c} e^{i d x}}{d} & \text{for}\: d \neq 0 \\8 a^{4} x e^{i c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+I*a*tan(d*x+c))**4,x)

[Out]

15*a**4*(log(exp(I*d*x) - I*exp(-I*c))/2 - log(exp(I*d*x) + I*exp(-I*c))/2)/d + (-9*I*a**4*exp(-I*c)*exp(3*I*d
*x)/d - 7*I*a**4*exp(-3*I*c)*exp(I*d*x)/d)/(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c)) + Piecewi
se((-8*I*a**4*exp(I*c)*exp(I*d*x)/d, Ne(d, 0)), (8*a**4*x*exp(I*c), True))

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Giac [B]  time = 1.41703, size = 502, normalized size = 5.18 \begin{align*} \frac{235 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 470 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 5 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 235 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 470 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 5 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 256 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 800 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 480 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 235 \, a^{4} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 5 \, a^{4} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 235 \, a^{4} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 5 \, a^{4} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right )}{32 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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(cos(d*x+c)*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/32*(235*a^4*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 470*a^4*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*
c) + 1) - 5*a^4*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) - 10*a^4*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I
*c) - 1) - 235*a^4*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 470*a^4*e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d
*x + I*c) + 1) + 5*a^4*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 10*a^4*e^(2*I*d*x + 2*I*c)*log(-I*e^(
I*d*x + I*c) - 1) - 256*I*a^4*e^(5*I*d*x + 5*I*c) - 800*I*a^4*e^(3*I*d*x + 3*I*c) - 480*I*a^4*e^(I*d*x + I*c)
+ 235*a^4*log(I*e^(I*d*x + I*c) + 1) - 5*a^4*log(I*e^(I*d*x + I*c) - 1) - 235*a^4*log(-I*e^(I*d*x + I*c) + 1)
+ 5*a^4*log(-I*e^(I*d*x + I*c) - 1))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)