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

Optimal. Leaf size=171 \[ \frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {65 \tan (c+d x)}{16 a^4 d}-\frac {4 i \log (\cos (c+d x))}{a^4 d}-\frac {65 x}{16 a^4}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[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

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Rubi [A]  time = 0.39, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3558, 3595, 3525, 3475} \[ \frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {65 \tan (c+d x)}{16 a^4 d}-\frac {4 i \log (\cos (c+d x))}{a^4 d}-\frac {65 x}{16 a^4}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[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 3475

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

Rule 3525

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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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))^4} \, dx &=-\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*}

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Mathematica [B]  time = 0.82, size = 429, normalized size = 2.51 \[ -\frac {\sec (c) \sec ^5(c+d x) (832 \sin (2 c+d x)+1560 i d x \sin (2 c+3 d x)+835 \sin (2 c+3 d x)+1560 i d x \sin (4 c+3 d x)+1603 \sin (4 c+3 d x)+1560 i d x \sin (4 c+5 d x)-765 \sin (4 c+5 d x)+1560 i d x \sin (6 c+5 d x)+3 \sin (6 c+5 d x)-536 i \cos (2 c+d x)+1560 d x \cos (2 c+3 d x)-893 i \cos (2 c+3 d x)+1560 d x \cos (4 c+3 d x)-1661 i \cos (4 c+3 d x)+1560 d x \cos (4 c+5 d x)+771 i \cos (4 c+5 d x)+1560 d x \cos (6 c+5 d x)+3 i \cos (6 c+5 d x)+1536 i \cos (2 c+3 d x) \log (\cos (c+d x))+1536 i \cos (4 c+3 d x) \log (\cos (c+d x))+1536 i \cos (4 c+5 d x) \log (\cos (c+d x))+1536 i \cos (6 c+5 d x) \log (\cos (c+d x))-1536 \sin (2 c+3 d x) \log (\cos (c+d x))-1536 \sin (4 c+3 d x) \log (\cos (c+d x))-1536 \sin (4 c+5 d x) \log (\cos (c+d x))-1536 \sin (6 c+5 d x) \log (\cos (c+d x))+832 \sin (d x)-536 i \cos (d x))}{1536 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1536*(Sec[c]*Sec[c + d*x]^5*((-536*I)*Cos[d*x] - (536*I)*Cos[2*c + d*x] - (893*I)*Cos[2*c + 3*d*x] + 1560*d
*x*Cos[2*c + 3*d*x] - (1661*I)*Cos[4*c + 3*d*x] + 1560*d*x*Cos[4*c + 3*d*x] + (771*I)*Cos[4*c + 5*d*x] + 1560*
d*x*Cos[4*c + 5*d*x] + (3*I)*Cos[6*c + 5*d*x] + 1560*d*x*Cos[6*c + 5*d*x] + (1536*I)*Cos[2*c + 3*d*x]*Log[Cos[
c + d*x]] + (1536*I)*Cos[4*c + 3*d*x]*Log[Cos[c + d*x]] + (1536*I)*Cos[4*c + 5*d*x]*Log[Cos[c + d*x]] + (1536*
I)*Cos[6*c + 5*d*x]*Log[Cos[c + d*x]] + 832*Sin[d*x] + 832*Sin[2*c + d*x] + 835*Sin[2*c + 3*d*x] + (1560*I)*d*
x*Sin[2*c + 3*d*x] - 1536*Log[Cos[c + d*x]]*Sin[2*c + 3*d*x] + 1603*Sin[4*c + 3*d*x] + (1560*I)*d*x*Sin[4*c +
3*d*x] - 1536*Log[Cos[c + d*x]]*Sin[4*c + 3*d*x] - 765*Sin[4*c + 5*d*x] + (1560*I)*d*x*Sin[4*c + 5*d*x] - 1536
*Log[Cos[c + d*x]]*Sin[4*c + 5*d*x] + 3*Sin[6*c + 5*d*x] + (1560*I)*d*x*Sin[6*c + 5*d*x] - 1536*Log[Cos[c + d*
x]]*Sin[6*c + 5*d*x]))/(a^4*d*(-I + Tan[c + d*x])^4)

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fricas [A]  time = 0.44, size = 134, normalized size = 0.78 \[ -\frac {3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (3096 \, d x - 1632 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - {\left (-1536 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1536 i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{384 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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) + (3096*d*x - 1632*I)*e^(8*I*d*x + 8*I*c) - (-1536*I*e^(10*I*d*x + 10*I
*c) - 1536*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))

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giac [A]  time = 22.10, size = 100, normalized size = 0.58 \[ -\frac {\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, \tan \left (d x + c\right )}{a^{4}} - \frac {-3225 i \, \tan \left (d x + c\right )^{4} - 10236 \, \tan \left (d x + c\right )^{3} + 12534 i \, \tan \left (d x + c\right )^{2} + 6908 \, \tan \left (d x + c\right ) - 1433 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.16, size = 131, normalized size = 0.77 \[ \frac {\tan \left (d x +c \right )}{a^{4} d}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}+\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {129 i \ln \left (\tan \left (d x +c \right )-i\right )}{32 d \,a^{4}}-\frac {11}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {111}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(d*x+c)/a^4/d-1/32*I/d/a^4*ln(tan(d*x+c)+I)+49/16*I/d/a^4/(tan(d*x+c)-I)^2-1/8*I/d/a^4/(tan(d*x+c)-I)^4+129
/32*I/d/a^4*ln(tan(d*x+c)-I)-11/12/d/a^4/(tan(d*x+c)-I)^3+111/16/d/a^4/(tan(d*x+c)-I)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+I*a*tan(d*x+c))^4,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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mupad [B]  time = 3.91, size = 132, normalized size = 0.77 \[ \frac {\mathrm {tan}\left (c+d\,x\right )}{a^4\,d}-\frac {65\,x}{16\,a^4}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,2{}\mathrm {i}}{a^4\,d}-\frac {\frac {749\,\mathrm {tan}\left (c+d\,x\right )}{48\,a^4}-\frac {111\,{\mathrm {tan}\left (c+d\,x\right )}^3}{16\,a^4}-\frac {14{}\mathrm {i}}{3\,a^4}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,71{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.86, size = 250, normalized size = 1.46 \[ \begin {cases} \frac {\left (442368 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 92160 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 16384 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1536 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{196608 a^{16} d^{4}} & \text {for}\: 196608 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (- 129 e^{8 i c} + 72 e^{6 i c} - 30 e^{4 i c} + 8 e^{2 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {129}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {2 i}{a^{4} d e^{2 i c} e^{2 i d x} + a^{4} d} - \frac {129 x}{16 a^{4}} - \frac {4 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(196608*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*e
xp(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)

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