3.1.66 \(\int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [66]

Optimal. Leaf size=161 \[ \frac {55 x}{8 a^3}+\frac {7 i \log (\cos (c+d x))}{a^3 d}-\frac {55 \tan (c+d x)}{8 a^3 d}+\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

55/8*x/a^3+7*I*ln(cos(d*x+c))/a^3/d-55/8*tan(d*x+c)/a^3/d+7/2*I*tan(d*x+c)^2/a^3/d-1/6*tan(d*x+c)^5/d/(a+I*a*t
an(d*x+c))^3+13/24*I*tan(d*x+c)^4/a/d/(a+I*a*tan(d*x+c))^2+55/24*tan(d*x+c)^3/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]
time = 0.21, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3639, 3676, 3609, 3606, 3556} \begin {gather*} \frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {55 \tan (c+d x)}{8 a^3 d}+\frac {7 i \log (\cos (c+d x))}{a^3 d}+\frac {55 x}{8 a^3}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(55*x)/(8*a^3) + ((7*I)*Log[Cos[c + d*x]])/(a^3*d) - (55*Tan[c + d*x])/(8*a^3*d) + (((7*I)/2)*Tan[c + d*x]^2)/
(a^3*d) - Tan[c + d*x]^5/(6*d*(a + I*a*Tan[c + d*x])^3) + (((13*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d
*x])^2) + (55*Tan[c + d*x]^3)/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

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 3609

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 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*} \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^4(c+d x) (-5 a+8 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^3(c+d x) \left (-52 i a^2-58 a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \tan ^2(c+d x) \left (330 a^3-336 i a^3 \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \tan (c+d x) \left (336 i a^3+330 a^3 \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac {55 x}{8 a^3}-\frac {55 \tan (c+d x)}{8 a^3 d}+\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(7 i) \int \tan (c+d x) \, dx}{a^3}\\ &=\frac {55 x}{8 a^3}+\frac {7 i \log (\cos (c+d x))}{a^3 d}-\frac {55 \tan (c+d x)}{8 a^3 d}+\frac {7 i \tan ^2(c+d x)}{2 a^3 d}-\frac {\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 4.26, size = 264, normalized size = 1.64 \begin {gather*} \frac {\sec ^3(c+d x) (\cos (d x)+i \sin (d x))^3 \left (234 \cos (2 d x) \sin (c)+27 \cos (4 d x) \sin (c)+660 i d x \sin (3 c)-2 \cos (6 d x) \sin (3 c)-672 \log (\cos (c+d x)) \sin (3 c)-48 \sec ^2(c+d x) \sin (3 c)-288 i \sec (c) \sec (c+d x) \sin (3 c) \sin (d x)-234 i \sin (c) \sin (2 d x)+9 \cos (c) (-23 i \cos (d x)+29 \sin (d x)) (\cos (3 d x)-i \sin (3 d x))-27 i \sin (c) \sin (4 d x)+\cos (3 c) \left (660 d x-2 i \cos (6 d x)+672 i \log (\cos (c+d x))+48 i \sec ^2(c+d x)-288 \sec (c) \sec (c+d x) \sin (d x)-2 \sin (6 d x)\right )+2 i \sin (3 c) \sin (6 d x)\right )}{96 d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^3*(234*Cos[2*d*x]*Sin[c] + 27*Cos[4*d*x]*Sin[c] + (660*I)*d*x*Sin[3*c]
- 2*Cos[6*d*x]*Sin[3*c] - 672*Log[Cos[c + d*x]]*Sin[3*c] - 48*Sec[c + d*x]^2*Sin[3*c] - (288*I)*Sec[c]*Sec[c
+ d*x]*Sin[3*c]*Sin[d*x] - (234*I)*Sin[c]*Sin[2*d*x] + 9*Cos[c]*((-23*I)*Cos[d*x] + 29*Sin[d*x])*(Cos[3*d*x] -
I*Sin[3*d*x]) - (27*I)*Sin[c]*Sin[4*d*x] + Cos[3*c]*(660*d*x - (2*I)*Cos[6*d*x] + (672*I)*Log[Cos[c + d*x]] +
(48*I)*Sec[c + d*x]^2 - 288*Sec[c]*Sec[c + d*x]*Sin[d*x] - 2*Sin[6*d*x]) + (2*I)*Sin[3*c]*Sin[6*d*x]))/(96*d*
(a + I*a*Tan[c + d*x])^3)

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Maple [A]
time = 0.19, size = 94, normalized size = 0.58

method	result	size
derivativedivides	\(\frac {-3 \tan \left (d x +c \right )+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {111 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {11 i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{8 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(94\)
default	\(\frac {-3 \tan \left (d x +c \right )+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {111 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {11 i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{8 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\)	\(94\)
risch	\(\frac {111 x}{8 a^{3}}-\frac {39 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {9 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}+\frac {14 c}{a^{3} d}-\frac {2 i \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}+3\right )}{d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\)	\(127\)
norman	\(\frac {\frac {55 x}{8 a}-\frac {121 \left (\tan ^{5}\left (d x +c \right )\right )}{8 d a}-\frac {3 \left (\tan ^{7}\left (d x +c \right )\right )}{d a}+\frac {165 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}+\frac {165 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {55 x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}-\frac {77 i}{12 d a}-\frac {55 \tan \left (d x +c \right )}{8 d a}-\frac {55 \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}-\frac {63 i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}-\frac {21 i \left (\tan ^{4}\left (d x +c \right )\right )}{2 d a}+\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{2 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}-\frac {7 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 a^{3} d}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.38, size = 167, normalized size = 1.04 \begin {gather*} \frac {1332 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 6 \, {\left (444 \, d x - 103 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 9 \, {\left (148 \, d x - 113 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 672 \, {\left (-i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 2 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 182 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i}{96 \, {\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 2 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(1332*d*x*e^(10*I*d*x + 10*I*c) + 6*(444*d*x - 103*I)*e^(8*I*d*x + 8*I*c) + 9*(148*d*x - 113*I)*e^(6*I*d*
x + 6*I*c) - 672*(-I*e^(10*I*d*x + 10*I*c) - 2*I*e^(8*I*d*x + 8*I*c) - I*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x +
2*I*c) + 1) - 182*I*e^(4*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^3*d*e^(10*I*d*x + 10*I*c) + 2*a^
3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))

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Sympy [A]
time = 0.42, size = 257, normalized size = 1.60 \begin {gather*} \frac {- 4 i e^{2 i c} e^{2 i d x} - 6 i}{a^{3} d e^{4 i c} e^{4 i d x} + 2 a^{3} d e^{2 i c} e^{2 i d x} + a^{3} d} + \begin {cases} \frac {\left (- 59904 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 6912 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (111 e^{6 i c} - 39 e^{4 i c} + 9 e^{2 i c} - 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {111}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {111 x}{8 a^{3}} + \frac {7 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+I*a*tan(d*x+c))**3,x)

[Out]

(-4*I*exp(2*I*c)*exp(2*I*d*x) - 6*I)/(a**3*d*exp(4*I*c)*exp(4*I*d*x) + 2*a**3*d*exp(2*I*c)*exp(2*I*d*x) + a**3
*d) + Piecewise(((-59904*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 6912*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 5
12*I*a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((1
11*exp(6*I*c) - 39*exp(4*I*c) + 9*exp(2*I*c) - 1)*exp(-6*I*c)/(8*a**3) - 111/(8*a**3)), True)) + 111*x/(8*a**3
) + 7*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)

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Giac [A]
time = 2.66, size = 111, normalized size = 0.69 \begin {gather*} -\frac {\frac {666 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {48 \, {\left (-i \, a^{3} \tan \left (d x + c\right )^{2} + 6 \, a^{3} \tan \left (d x + c\right )\right )}}{a^{6}} - \frac {1221 i \, \tan \left (d x + c\right )^{3} + 3075 \, \tan \left (d x + c\right )^{2} - 2619 i \, \tan \left (d x + c\right ) - 749}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(666*I*log(tan(d*x + c) - I)/a^3 + 6*I*log(I*tan(d*x + c) - 1)/a^3 + 48*(-I*a^3*tan(d*x + c)^2 + 6*a^3*t
an(d*x + c))/a^6 - (1221*I*tan(d*x + c)^3 + 3075*tan(d*x + c)^2 - 2619*I*tan(d*x + c) - 749)/(a^3*(tan(d*x + c
) - I)^3))/d

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Mupad [B]
time = 3.95, size = 140, normalized size = 0.87 \begin {gather*} \frac {\frac {87\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^3}-\frac {59{}\mathrm {i}}{12\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,49{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,111{}\mathrm {i}}{16\,a^3\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d}-\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{a^3\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((87*tan(c + d*x))/(8*a^3) - 59i/(12*a^3) + (tan(c + d*x)^2*49i)/(8*a^3))/(d*(tan(c + d*x)*3i - 3*tan(c + d*x)
^2 - tan(c + d*x)^3*1i + 1)) - (log(tan(c + d*x) - 1i)*111i)/(16*a^3*d) - (log(tan(c + d*x) + 1i)*1i)/(16*a^3*
d) - (3*tan(c + d*x))/(a^3*d) + (tan(c + d*x)^2*1i)/(2*a^3*d)

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