\(\int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [75]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {65 x}{16 a^4}-\frac {4 i \log (\cos (c+d x))}{a^4 d}+\frac {65 \tan (c+d x)}{16 a^4 d}-\frac {2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[Out]

-65/16*x/a^4-4*I*ln(cos(d*x+c))/a^4/d+65/16*tan(d*x+c)/a^4/d-2*I*tan(d*x+c)^2/a^4/d/(1+I*tan(d*x+c))+31/48*tan
(d*x+c)^3/a^4/d/(1+I*tan(d*x+c))^2-1/8*tan(d*x+c)^5/d/(a+I*a*tan(d*x+c))^4+7/24*I*tan(d*x+c)^4/a/d/(a+I*a*tan(
d*x+c))^3

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3639, 3676, 3606, 3556} \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {65 \tan (c+d x)}{16 a^4 d}-\frac {4 i \log (\cos (c+d x))}{a^4 d}-\frac {65 x}{16 a^4}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[In]

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-65*x)/(16*a^4) - ((4*I)*Log[Cos[c + d*x]])/(a^4*d) + (65*Tan[c + d*x])/(16*a^4*d) - ((2*I)*Tan[c + d*x]^2)/(
a^4*d*(1 + I*Tan[c + d*x])) + (31*Tan[c + d*x]^3)/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - Tan[c + d*x]^5/(8*d*(a +
I*a*Tan[c + d*x])^4) + (((7*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d*x])^3)

Rule 3556

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

Rule 3606

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

Rule 3639

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

Rule 3676

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac {\int \frac {\tan ^4(c+d x) (-5 a+9 i a \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2} \\ & = -\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\tan ^3(c+d x) \left (-56 i a^2-68 a^2 \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4} \\ & = \frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^2(c+d x) \left (372 a^3-396 i a^3 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6} \\ & = \frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \tan (c+d x) \left (1536 i a^4+1560 a^4 \tan (c+d x)\right ) \, dx}{384 a^8} \\ & = -\frac {65 x}{16 a^4}+\frac {65 \tan (c+d x)}{16 a^4 d}+\frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {(4 i) \int \tan (c+d x) \, dx}{a^4} \\ & = -\frac {65 x}{16 a^4}-\frac {4 i \log (\cos (c+d x))}{a^4 d}+\frac {65 \tan (c+d x)}{16 a^4 d}+\frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac {2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^5(c+d x) (268 \cos (c+d x)-550 \cos (5 (c+d x))+774 \cos (5 (c+d x)) \log (i-\tan (c+d x))+6 \cos (3 (c+d x)) (47+129 \log (i-\tan (c+d x))-\log (i+\tan (c+d x)))-6 \cos (5 (c+d x)) \log (i+\tan (c+d x))+416 i \sin (c+d x)+253 i \sin (3 (c+d x))+774 i \log (i-\tan (c+d x)) \sin (3 (c+d x))-6 i \log (i+\tan (c+d x)) \sin (3 (c+d x))-547 i \sin (5 (c+d x))+774 i \log (i-\tan (c+d x)) \sin (5 (c+d x))-6 i \log (i+\tan (c+d x)) \sin (5 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \]

[In]

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/384)*Sec[c + d*x]^5*(268*Cos[c + d*x] - 550*Cos[5*(c + d*x)] + 774*Cos[5*(c + d*x)]*Log[I - Tan[c + d*x]]
+ 6*Cos[3*(c + d*x)]*(47 + 129*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]]) - 6*Cos[5*(c + d*x)]*Log[I + Tan
[c + d*x]] + (416*I)*Sin[c + d*x] + (253*I)*Sin[3*(c + d*x)] + (774*I)*Log[I - Tan[c + d*x]]*Sin[3*(c + d*x)]
- (6*I)*Log[I + Tan[c + d*x]]*Sin[3*(c + d*x)] - (547*I)*Sin[5*(c + d*x)] + (774*I)*Log[I - Tan[c + d*x]]*Sin[
5*(c + d*x)] - (6*I)*Log[I + Tan[c + d*x]]*Sin[5*(c + d*x)]))/(a^4*d*(-I + Tan[c + d*x])^4)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )}{a^{4} d}+\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {11}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {111}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\)	\(128\)
default	\(\frac {\tan \left (d x +c \right )}{a^{4} d}+\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {11}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {111}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\)	\(128\)
risch	\(-\frac {129 x}{16 a^{4}}+\frac {9 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{4} d}-\frac {15 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{12 a^{4} d}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}-\frac {8 c}{a^{4} d}+\frac {2 i}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{4} d}\)	\(132\)
norman	\(\frac {\frac {\tan ^{9}\left (d x +c \right )}{a d}-\frac {65 x}{16 a}+\frac {949 \left (\tan ^{5}\left (d x +c \right )\right )}{48 a d}+\frac {175 \left (\tan ^{7}\left (d x +c \right )\right )}{16 a d}-\frac {65 x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}-\frac {195 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {65 x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}-\frac {65 x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {14 i}{3 a d}+\frac {65 \tan \left (d x +c \right )}{16 a d}+\frac {715 \left (\tan ^{3}\left (d x +c \right )\right )}{48 a d}+\frac {10 i \left (\tan ^{6}\left (d x +c \right )\right )}{a d}+\frac {21 i \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {50 i \left (\tan ^{2}\left (d x +c \right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}\)	\(238\)

[In]

int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

tan(d*x+c)/a^4/d+49/16*I/d/a^4/(tan(d*x+c)-I)^2-1/8*I/d/a^4/(tan(d*x+c)-I)^4+2*I/d/a^4*ln(1+tan(d*x+c)^2)-65/1
6/d/a^4*arctan(tan(d*x+c))-11/12/d/a^4/(tan(d*x+c)-I)^3+111/16/d/a^4/(tan(d*x+c)-I)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 24 \, {\left (129 \, d x - 68 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 1536 \, {\left (i \, e^{\left (10 i \, d x + 10 i \, c\right )} + i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{384 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/384*(3096*d*x*e^(10*I*d*x + 10*I*c) + 24*(129*d*x - 68*I)*e^(8*I*d*x + 8*I*c) + 1536*(I*e^(10*I*d*x + 10*I*
c) + I*e^(8*I*d*x + 8*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 684*I*e^(6*I*d*x + 6*I*c) + 148*I*e^(4*I*d*x + 4*I*
c) - 29*I*e^(2*I*d*x + 2*I*c) + 3*I)/(a^4*d*e^(10*I*d*x + 10*I*c) + a^4*d*e^(8*I*d*x + 8*I*c))

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (442368 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 92160 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 16384 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1536 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{196608 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (- 129 e^{8 i c} + 72 e^{6 i c} - 30 e^{4 i c} + 8 e^{2 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {129}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {2 i}{a^{4} d e^{2 i c} e^{2 i d x} + a^{4} d} - \frac {129 x}{16 a^{4}} - \frac {4 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} \]

[In]

integrate(tan(d*x+c)**6/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((442368*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 92160*I*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 163
84*I*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) - 1536*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(196608*
a**16*d**4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((-129*exp(8*I*c) + 72*exp(6*I*c) - 30*exp(4*I*c) + 8*exp(2*I*
c) - 1)*exp(-8*I*c)/(16*a**4) + 129/(16*a**4)), True)) + 2*I/(a**4*d*exp(2*I*c)*exp(2*I*d*x) + a**4*d) - 129*x
/(16*a**4) - 4*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 2.70 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, \tan \left (d x + c\right )}{a^{4}} - \frac {-3225 i \, \tan \left (d x + c\right )^{4} - 10236 \, \tan \left (d x + c\right )^{3} + 12534 i \, \tan \left (d x + c\right )^{2} + 6908 \, \tan \left (d x + c\right ) - 1433 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/384*(12*I*log(tan(d*x + c) + I)/a^4 - 1548*I*log(tan(d*x + c) - I)/a^4 - 384*tan(d*x + c)/a^4 - (-3225*I*ta
n(d*x + c)^4 - 10236*tan(d*x + c)^3 + 12534*I*tan(d*x + c)^2 + 6908*tan(d*x + c) - 1433*I)/(a^4*(tan(d*x + c)
- I)^4))/d

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a^4\,d}-\frac {65\,x}{16\,a^4}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,2{}\mathrm {i}}{a^4\,d}-\frac {\frac {749\,\mathrm {tan}\left (c+d\,x\right )}{48\,a^4}-\frac {111\,{\mathrm {tan}\left (c+d\,x\right )}^3}{16\,a^4}-\frac {14{}\mathrm {i}}{3\,a^4}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,71{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]

[In]

int(tan(c + d*x)^6/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

tan(c + d*x)/(a^4*d) - (65*x)/(16*a^4) + (log(tan(c + d*x)^2 + 1)*2i)/(a^4*d) - ((749*tan(c + d*x))/(48*a^4) -
14i/(3*a^4) + (tan(c + d*x)^2*71i)/(4*a^4) - (111*tan(c + d*x)^3)/(16*a^4))/(d*(tan(c + d*x)*4i - 6*tan(c + d
*x)^2 - tan(c + d*x)^3*4i + tan(c + d*x)^4 + 1))