3.31 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=101 \[ \frac {2 a^3 \cot (c+d x)}{d}-\frac {4 i a^3 \log (\sin (c+d x))}{d}+4 a^3 x-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

4*a^3*x+2*a^3*cot(d*x+c)/d-4*I*a^3*ln(sin(d*x+c))/d-1/2*I*a*cot(d*x+c)^2*(a+I*a*tan(d*x+c))^2/d-1/3*cot(d*x+c)
^3*(a+I*a*tan(d*x+c))^3/d

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Rubi [A]  time = 0.15, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3548, 3545, 3542, 3531, 3475} \[ \frac {2 a^3 \cot (c+d x)}{d}-\frac {4 i a^3 \log (\sin (c+d x))}{d}+4 a^3 x-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3475

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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] && NeQ[a*c + b*d, 0]

Rule 3542

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

Rule 3545

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

Rule 3548

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

Rubi steps

\begin {align*} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+i \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-(2 a) \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {2 a^3 \cot (c+d x)}{d}-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-(2 a) \int \cot (c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=4 a^3 x+\frac {2 a^3 \cot (c+d x)}{d}-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\left (4 i a^3\right ) \int \cot (c+d x) \, dx\\ &=4 a^3 x+\frac {2 a^3 \cot (c+d x)}{d}-\frac {4 i a^3 \log (\sin (c+d x))}{d}-\frac {i a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac {\cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\\ \end {align*}

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Mathematica [B]  time = 1.33, size = 251, normalized size = 2.49 \[ \frac {a^3 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \csc ^3(c+d x) (\cos (3 d x)+i \sin (3 d x)) \left (-15 \sin (2 c+d x)+13 \sin (2 c+3 d x)-36 d x \cos (2 c+d x)+9 i \cos (2 c+d x)-12 d x \cos (2 c+3 d x)+12 d x \cos (4 c+3 d x)-48 \sin (c) \sin ^3(c+d x) \tan ^{-1}(\tan (4 c+d x))+9 \cos (d x) \left (-i \log \left (\sin ^2(c+d x)\right )+4 d x-i\right )+9 i \cos (2 c+d x) \log \left (\sin ^2(c+d x)\right )+3 i \cos (2 c+3 d x) \log \left (\sin ^2(c+d x)\right )-3 i \cos (4 c+3 d x) \log \left (\sin ^2(c+d x)\right )-24 \sin (d x)\right )}{24 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Csc[c/2]*Csc[c + d*x]^3*Sec[c/2]*(Cos[3*d*x] + I*Sin[3*d*x])*((9*I)*Cos[2*c + d*x] - 36*d*x*Cos[2*c + d*x
] - 12*d*x*Cos[2*c + 3*d*x] + 12*d*x*Cos[4*c + 3*d*x] + 9*Cos[d*x]*(-I + 4*d*x - I*Log[Sin[c + d*x]^2]) + (9*I
)*Cos[2*c + d*x]*Log[Sin[c + d*x]^2] + (3*I)*Cos[2*c + 3*d*x]*Log[Sin[c + d*x]^2] - (3*I)*Cos[4*c + 3*d*x]*Log
[Sin[c + d*x]^2] - 24*Sin[d*x] - 48*ArcTan[Tan[4*c + d*x]]*Sin[c]*Sin[c + d*x]^3 - 15*Sin[2*c + d*x] + 13*Sin[
2*c + 3*d*x]))/(24*d*(Cos[d*x] + I*Sin[d*x])^3)

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fricas [A]  time = 0.42, size = 138, normalized size = 1.37 \[ \frac {48 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 66 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 26 i \, a^{3} + {\left (-12 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(48*I*a^3*e^(4*I*d*x + 4*I*c) - 66*I*a^3*e^(2*I*d*x + 2*I*c) + 26*I*a^3 + (-12*I*a^3*e^(6*I*d*x + 6*I*c) +
36*I*a^3*e^(4*I*d*x + 4*I*c) - 36*I*a^3*e^(2*I*d*x + 2*I*c) + 12*I*a^3)*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)

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giac [A]  time = 2.18, size = 146, normalized size = 1.45 \[ \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 192 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 96 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 51 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {-176 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 51 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(a^3*tan(1/2*d*x + 1/2*c)^3 - 9*I*a^3*tan(1/2*d*x + 1/2*c)^2 + 192*I*a^3*log(tan(1/2*d*x + 1/2*c) + I) -
96*I*a^3*log(tan(1/2*d*x + 1/2*c)) - 51*a^3*tan(1/2*d*x + 1/2*c) - (-176*I*a^3*tan(1/2*d*x + 1/2*c)^3 - 51*a^3
*tan(1/2*d*x + 1/2*c)^2 + 9*I*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c)^3)/d

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maple [A]  time = 0.32, size = 80, normalized size = 0.79 \[ -\frac {4 i a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}+4 a^{3} x +\frac {4 a^{3} \cot \left (d x +c \right )}{d}+\frac {4 a^{3} c}{d}-\frac {3 i a^{3} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{3} \left (\cot ^{3}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x)

[Out]

-4*I*a^3*ln(sin(d*x+c))/d+4*a^3*x+4*a^3*cot(d*x+c)/d+4/d*a^3*c-3/2*I/d*a^3*cot(d*x+c)^2-1/3*a^3*cot(d*x+c)^3/d

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maxima [A]  time = 0.51, size = 83, normalized size = 0.82 \[ \frac {24 \, {\left (d x + c\right )} a^{3} + 12 i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 i \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {24 \, a^{3} \tan \left (d x + c\right )^{2} - 9 i \, a^{3} \tan \left (d x + c\right ) - 2 \, a^{3}}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(24*(d*x + c)*a^3 + 12*I*a^3*log(tan(d*x + c)^2 + 1) - 24*I*a^3*log(tan(d*x + c)) + (24*a^3*tan(d*x + c)^2
- 9*I*a^3*tan(d*x + c) - 2*a^3)/tan(d*x + c)^3)/d

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mupad [B]  time = 3.79, size = 68, normalized size = 0.67 \[ \frac {4\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {8\,a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {a^3\,{\mathrm {cot}\left (c+d\,x\right )}^3}{3\,d}-\frac {a^3\,{\mathrm {cot}\left (c+d\,x\right )}^2\,3{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.51, size = 136, normalized size = 1.35 \[ - \frac {4 i a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 48 i a^{3} e^{4 i c} e^{4 i d x} + 66 i a^{3} e^{2 i c} e^{2 i d x} - 26 i a^{3}}{- 3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} - 9 d e^{2 i c} e^{2 i d x} + 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*I*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-48*I*a**3*exp(4*I*c)*exp(4*I*d*x) + 66*I*a**3*exp(2*I*c)*exp(2
*I*d*x) - 26*I*a**3)/(-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)

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