\(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx\) [83]

Optimal result

Integrand size = 24, antiderivative size = 124 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=80 a^8 x-\frac {80 i a^8 \log (\cos (c+d x))}{d}-\frac {31 a^8 \tan (c+d x)}{d}-\frac {4 i a^8 \tan ^2(c+d x)}{d}+\frac {a^8 \tan ^3(c+d x)}{3 d}-\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}+\frac {80 i a^9}{d (a-i a \tan (c+d x))} \]

[Out]

80*a^8*x-80*I*a^8*ln(cos(d*x+c))/d-31*a^8*tan(d*x+c)/d-4*I*a^8*tan(d*x+c)^2/d+1/3*a^8*tan(d*x+c)^3/d-16*I*a^10
/d/(a-I*a*tan(d*x+c))^2+80*I*a^9/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 124, 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 = {3568, 45} \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}+\frac {80 i a^9}{d (a-i a \tan (c+d x))}+\frac {a^8 \tan ^3(c+d x)}{3 d}-\frac {4 i a^8 \tan ^2(c+d x)}{d}-\frac {31 a^8 \tan (c+d x)}{d}-\frac {80 i a^8 \log (\cos (c+d x))}{d}+80 a^8 x \]

[In]

Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]

[Out]

80*a^8*x - ((80*I)*a^8*Log[Cos[c + d*x]])/d - (31*a^8*Tan[c + d*x])/d - ((4*I)*a^8*Tan[c + d*x]^2)/d + (a^8*Ta
n[c + d*x]^3)/(3*d) - ((16*I)*a^10)/(d*(a - I*a*Tan[c + d*x])^2) + ((80*I)*a^9)/(d*(a - I*a*Tan[c + d*x]))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {(a+x)^5}{(a-x)^3} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \left (-31 a^2+\frac {32 a^5}{(a-x)^3}-\frac {80 a^4}{(a-x)^2}+\frac {80 a^3}{a-x}-8 a x-x^2\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = 80 a^8 x-\frac {80 i a^8 \log (\cos (c+d x))}{d}-\frac {31 a^8 \tan (c+d x)}{d}-\frac {4 i a^8 \tan ^2(c+d x)}{d}+\frac {a^8 \tan ^3(c+d x)}{3 d}-\frac {16 i a^{10}}{d (a-i a \tan (c+d x))^2}+\frac {80 i a^9}{d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.69 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {i a^8 \left (-93 i \tan (c+d x)+12 \tan ^2(c+d x)+i \tan ^3(c+d x)+48 \left (-5 \log (i+\tan (c+d x))+\frac {4-5 i \tan (c+d x)}{(i+\tan (c+d x))^2}\right )\right )}{3 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^8,x]

[Out]

((-1/3*I)*a^8*((-93*I)*Tan[c + d*x] + 12*Tan[c + d*x]^2 + I*Tan[c + d*x]^3 + 48*(-5*Log[I + Tan[c + d*x]] + (4
- (5*I)*Tan[c + d*x])/(I + Tan[c + d*x])^2)))/d

Maple [A] (verified)

Time = 219.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {4 i a^{8} {\mathrm e}^{4 i \left (d x +c \right )}}{d}+\frac {32 i a^{8} {\mathrm e}^{2 i \left (d x +c \right )}}{d}-\frac {160 a^{8} c}{d}-\frac {4 i a^{8} \left (60 \,{\mathrm e}^{4 i \left (d x +c \right )}+105 \,{\mathrm e}^{2 i \left (d x +c \right )}+47\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {80 i a^{8} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(114\)
derivativedivides	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{\cos \left (d x +c \right )}-2 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )+\frac {35 d x}{8}+\frac {35 c}{8}\right )-14 i a^{8} \left (\sin ^{4}\left (d x +c \right )\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+56 i a^{8} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-2 i a^{8} \left (\cos ^{4}\left (d x +c \right )\right )-28 a^{8} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+a^{8} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(409\)
default	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{\cos \left (d x +c \right )}-2 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )+\frac {35 d x}{8}+\frac {35 c}{8}\right )-14 i a^{8} \left (\sin ^{4}\left (d x +c \right )\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+56 i a^{8} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+70 a^{8} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-2 i a^{8} \left (\cos ^{4}\left (d x +c \right )\right )-28 a^{8} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}+3 \ln \left (\cos \left (d x +c \right )\right )\right )+a^{8} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(409\)

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

-4*I/d*a^8*exp(4*I*(d*x+c))+32*I/d*a^8*exp(2*I*(d*x+c))-160/d*a^8*c-4/3*I*a^8*(60*exp(4*I*(d*x+c))+105*exp(2*I
*(d*x+c))+47)/d/(exp(2*I*(d*x+c))+1)^3-80*I/d*a^8*ln(exp(2*I*(d*x+c))+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.44 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {4 \, {\left (3 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 15 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 63 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 81 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + 47 i \, a^{8} + 60 \, {\left (i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-4/3*(3*I*a^8*e^(10*I*d*x + 10*I*c) - 15*I*a^8*e^(8*I*d*x + 8*I*c) - 63*I*a^8*e^(6*I*d*x + 6*I*c) - 9*I*a^8*e^
(4*I*d*x + 4*I*c) + 81*I*a^8*e^(2*I*d*x + 2*I*c) + 47*I*a^8 + 60*(I*a^8*e^(6*I*d*x + 6*I*c) + 3*I*a^8*e^(4*I*d
*x + 4*I*c) + 3*I*a^8*e^(2*I*d*x + 2*I*c) + I*a^8)*log(e^(2*I*d*x + 2*I*c) + 1))/(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 [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.74 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=- \frac {80 i a^{8} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 240 i a^{8} e^{4 i c} e^{4 i d x} - 420 i a^{8} e^{2 i c} e^{2 i d x} - 188 i a^{8}}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} + \begin {cases} \frac {- 4 i a^{8} d e^{4 i c} e^{4 i d x} + 32 i a^{8} d e^{2 i c} e^{2 i d x}}{d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (16 a^{8} e^{4 i c} - 64 a^{8} e^{2 i c}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**8,x)

[Out]

-80*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-240*I*a**8*exp(4*I*c)*exp(4*I*d*x) - 420*I*a**8*exp(2*I*c)*ex
p(2*I*d*x) - 188*I*a**8)/(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) + Piecewise(((-4*I*a**8*d*exp(4*I*c)*exp(4*I*d*x) + 32*I*a**8*d*exp(2*I*c)*exp(2*I*d*x))/d**2, Ne(d
**2, 0)), (x*(16*a**8*exp(4*I*c) - 64*a**8*exp(2*I*c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^{8} \tan \left (d x + c\right )^{3} - 12 i \, a^{8} \tan \left (d x + c\right )^{2} + 240 \, {\left (d x + c\right )} a^{8} + 120 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 93 \, a^{8} \tan \left (d x + c\right ) - \frac {48 \, {\left (5 \, a^{8} \tan \left (d x + c\right )^{3} - 6 i \, a^{8} \tan \left (d x + c\right )^{2} + 3 \, a^{8} \tan \left (d x + c\right ) - 4 i \, a^{8}\right )}}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/3*(a^8*tan(d*x + c)^3 - 12*I*a^8*tan(d*x + c)^2 + 240*(d*x + c)*a^8 + 120*I*a^8*log(tan(d*x + c)^2 + 1) - 93
*a^8*tan(d*x + c) - 48*(5*a^8*tan(d*x + c)^3 - 6*I*a^8*tan(d*x + c)^2 + 3*a^8*tan(d*x + c) - 4*I*a^8)/(tan(d*x
+ c)^4 + 2*tan(d*x + c)^2 + 1))/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. 785 vs. \(2 (110) = 220\).

Time = 1.04 (sec) , antiderivative size = 785, normalized size of antiderivative = 6.33 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-4/3*(60*I*a^8*e^(28*I*d*x + 14*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 840*I*a^8*e^(26*I*d*x + 12*I*c)*log(e^(2*I
*d*x + 2*I*c) + 1) + 5460*I*a^8*e^(24*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 21840*I*a^8*e^(22*I*d*x +
8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 60060*I*a^8*e^(20*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 120120*
I*a^8*e^(18*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 180180*I*a^8*e^(16*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2
*I*c) + 1) + 180180*I*a^8*e^(12*I*d*x - 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 120120*I*a^8*e^(10*I*d*x - 4*I*c
)*log(e^(2*I*d*x + 2*I*c) + 1) + 60060*I*a^8*e^(8*I*d*x - 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 21840*I*a^8*e^
(6*I*d*x - 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 5460*I*a^8*e^(4*I*d*x - 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1)
+ 840*I*a^8*e^(2*I*d*x - 12*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 205920*I*a^8*e^(14*I*d*x)*log(e^(2*I*d*x + 2*I
*c) + 1) + 60*I*a^8*e^(-14*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 3*I*a^8*e^(32*I*d*x + 18*I*c) + 18*I*a^8*e^(30*
I*d*x + 16*I*c) - 63*I*a^8*e^(28*I*d*x + 14*I*c) - 1032*I*a^8*e^(26*I*d*x + 12*I*c) - 4968*I*a^8*e^(24*I*d*x +
10*I*c) - 13516*I*a^8*e^(22*I*d*x + 8*I*c) - 22847*I*a^8*e^(20*I*d*x + 6*I*c) - 22066*I*a^8*e^(18*I*d*x + 4*I
*c) - 3234*I*a^8*e^(16*I*d*x + 2*I*c) + 44979*I*a^8*e^(12*I*d*x - 2*I*c) + 43332*I*a^8*e^(10*I*d*x - 4*I*c) +
27672*I*a^8*e^(8*I*d*x - 6*I*c) + 12048*I*a^8*e^(6*I*d*x - 8*I*c) + 3467*I*a^8*e^(4*I*d*x - 10*I*c) + 598*I*a^
8*e^(2*I*d*x - 12*I*c) + 25674*I*a^8*e^(14*I*d*x) + 47*I*a^8*e^(-14*I*c))/(d*e^(28*I*d*x + 14*I*c) + 14*d*e^(2
6*I*d*x + 12*I*c) + 91*d*e^(24*I*d*x + 10*I*c) + 364*d*e^(22*I*d*x + 8*I*c) + 1001*d*e^(20*I*d*x + 6*I*c) + 20
02*d*e^(18*I*d*x + 4*I*c) + 3003*d*e^(16*I*d*x + 2*I*c) + 3003*d*e^(12*I*d*x - 2*I*c) + 2002*d*e^(10*I*d*x - 4
*I*c) + 1001*d*e^(8*I*d*x - 6*I*c) + 364*d*e^(6*I*d*x - 8*I*c) + 91*d*e^(4*I*d*x - 10*I*c) + 14*d*e^(2*I*d*x -
12*I*c) + 3432*d*e^(14*I*d*x) + d*e^(-14*I*c))

Mupad [B] (verification not implemented)

Time = 4.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {80\,a^8\,\mathrm {tan}\left (c+d\,x\right )+a^8\,64{}\mathrm {i}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}-1\right )}-\frac {31\,a^8\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {a^8\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,80{}\mathrm {i}}{d}-\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}}{d} \]

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^8,x)

[Out]

(a^8*log(tan(c + d*x) + 1i)*80i)/d - (31*a^8*tan(c + d*x))/d - (80*a^8*tan(c + d*x) + a^8*64i)/(d*(tan(c + d*x
)*2i + tan(c + d*x)^2 - 1)) - (a^8*tan(c + d*x)^2*4i)/d + (a^8*tan(c + d*x)^3)/(3*d)