\(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [55]

Optimal result

Integrand size = 24, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d} \]

[Out]

a^4*arctanh(sin(d*x+c))/d-2/3*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^3/d+2*I*cos(d*x+c)*(a^4+I*a^4*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 3855} \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]

[Out]

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

Rule 3577

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 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-a^2 \int \cos (c+d x) (a+i a \tan (c+d x))^2 \, dx \\ & = -\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+a^4 \int \sec (c+d x) \, dx \\ & = \frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(246\) vs. \(2(78)=156\).

Time = 0.94 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.15 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \left (-3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \cos (3 d x) \sin (c)+6 \cos (d x) \sin (3 c)+3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)-3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)+\cos (3 c) (6 i \cos (d x)-6 \sin (d x))+6 i \sin (3 c) \sin (d x)-2 i \sin (c) \sin (3 d x)+2 \cos (c) (-i \cos (3 d x)+\sin (3 d x))\right ) (\cos (c+d x)+i \sin (c+d x))^4}{3 d (\cos (d x)+i \sin (d x))^4} \]

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*(-3*Cos[4*c]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[4*c]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]] - 2*Cos[3*d*x]*Sin[c] + 6*Cos[d*x]*Sin[3*c] + (3*I)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[4*c] - (3*
I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*c] + Cos[3*c]*((6*I)*Cos[d*x] - 6*Sin[d*x]) + (6*I)*Sin[3*c]
*Sin[d*x] - (2*I)*Sin[c]*Sin[3*d*x] + 2*Cos[c]*((-I)*Cos[3*d*x] + Sin[3*d*x]))*(Cos[c + d*x] + I*Sin[c + d*x])
^4)/(3*d*(Cos[d*x] + I*Sin[d*x])^4)

Maple [A] (verified)

Time = 13.92 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01

method	result	size
risch	\(-\frac {2 i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{3 d}+\frac {2 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(79\)
derivativedivides	\(\frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {4 i a^{4} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 a^{4} \left (\sin ^{3}\left (d x +c \right )\right )-\frac {4 i a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(113\)
default	\(\frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {4 i a^{4} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}-2 a^{4} \left (\sin ^{3}\left (d x +c \right )\right )-\frac {4 i a^{4} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(113\)

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {-2 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (- \log {\left (e^{i d x} - i e^{- i c} \right )} + \log {\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \begin {cases} \frac {- 2 i a^{4} d e^{3 i c} e^{3 i d x} + 6 i a^{4} d e^{i c} e^{i d x}}{3 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (2 a^{4} e^{3 i c} - 2 a^{4} e^{i c}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**4,x)

[Out]

a**4*(-log(exp(I*d*x) - I*exp(-I*c)) + log(exp(I*d*x) + I*exp(-I*c)))/d + Piecewise(((-2*I*a**4*d*exp(3*I*c)*e
xp(3*I*d*x) + 6*I*a**4*d*exp(I*c)*exp(I*d*x))/(3*d**2), Ne(d**2, 0)), (x*(2*a**4*exp(3*I*c) - 2*a**4*exp(I*c))
, True))

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 i \, a^{4} \cos \left (d x + c\right )^{3} + 12 \, a^{4} \sin \left (d x + c\right )^{3} + 8 i \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} + {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{4} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/6*(8*I*a^4*cos(d*x + c)^3 + 12*a^4*sin(d*x + c)^3 + 8*I*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^4 + (2*sin(d*x
+ c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a^4 + 2*(sin(d*x + c)^3 - 3*sin(d
*x + c))*a^4)/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. 1299 vs. \(2 (68) = 136\).

Time = 0.97 (sec) , antiderivative size = 1299, normalized size of antiderivative = 16.65 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/768*(1110*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 6660*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 16650*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 16650*a^4*e^(4*I*d*x - 2*I*c)*log(I*
e^(I*d*x + I*c) + 1) + 6660*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 22200*a^4*e^(6*I*d*x)*log(I*e
^(I*d*x + I*c) + 1) + 1110*a^4*e^(-6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1875*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(
I*d*x + I*c) - 1) + 11250*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^(8*I*d*x + 2*I*c)*
log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 11250*a^4*e^(2*I*d*x -
4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 37500*a^4*e^(6*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1875*a^4*e^(-6*I*c)*lo
g(I*e^(I*d*x + I*c) - 1) - 1110*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(10*I*d*x +
4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(
4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2220
0*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 1110*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1875*a^4*e^(
12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 11250*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 2
8125*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 28125*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c
) - 1) - 11250*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 37500*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I
*c) - 1) - 1875*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 3*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) - 18*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 45*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x) +
e^(-I*c)) - 45*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 18*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*
x) + e^(-I*c)) - 60*a^4*e^(6*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) - 3*a^4*e^(-6*I*c)*log(I*e^(I*d*x) + e^(-I*c))
+ 3*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 18*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^
(-I*c)) + 45*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 45*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x
) + e^(-I*c)) + 18*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 60*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x)
+ e^(-I*c)) + 3*a^4*e^(-6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 512*I*a^4*e^(15*I*d*x + 9*I*c) - 1536*I*a^4*e^(1
3*I*d*x + 7*I*c) + 1536*I*a^4*e^(11*I*d*x + 5*I*c) + 12800*I*a^4*e^(9*I*d*x + 3*I*c) + 23040*I*a^4*e^(7*I*d*x
+ I*c) + 19968*I*a^4*e^(5*I*d*x - I*c) + 8704*I*a^4*e^(3*I*d*x - 3*I*c) + 1536*I*a^4*e^(I*d*x - 5*I*c))/(d*e^(
12*I*d*x + 6*I*c) + 6*d*e^(10*I*d*x + 4*I*c) + 15*d*e^(8*I*d*x + 2*I*c) + 15*d*e^(4*I*d*x - 2*I*c) + 6*d*e^(2*
I*d*x - 4*I*c) + 20*d*e^(6*I*d*x) + d*e^(-6*I*c))

Mupad [B] (verification not implemented)

Time = 4.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {2\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\frac {8\,a^4}{3}-a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,8{}\mathrm {i}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(2*a^4*atanh(tan(c/2 + (d*x)/2)))/d - ((8*a^4)/3 - a^4*tan(c/2 + (d*x)/2)*8i)/(d*(3*tan(c/2 + (d*x)/2) - tan(c
/2 + (d*x)/2)^2*3i - tan(c/2 + (d*x)/2)^3 + 1i))