3.1.30 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [30]

Optimal. Leaf size=71 \[ -4 i a^3 x-\frac {2 i a^3 \cot (c+d x)}{d}-\frac {4 a^3 \log (\sin (c+d x))}{d}-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3626, 3623, 3612, 3556} \begin {gather*} -\frac {2 i a^3 \cot (c+d x)}{d}-\frac {4 a^3 \log (\sin (c+d x))}{d}-4 i a^3 x-\frac {a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3556

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

Rule 3612

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 3623

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 3626

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]

Rubi steps

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

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Mathematica [A]
time = 0.78, size = 126, normalized size = 1.77 \begin {gather*} \frac {a^3 \csc \left (\frac {c}{2}\right ) \csc ^2(c+d x) \sec \left (\frac {c}{2}\right ) \left (3 i \cos (c)-3 i \cos (c+2 d x)+\left (-1-4 i d x-2 \log \left (\sin ^2(c+d x)\right )+2 \cos (2 (c+d x)) \left (2 i d x+\log \left (\sin ^2(c+d x)\right )\right )\right ) \sin (c)\right ) (\cos (3 d x)+i \sin (3 d x))}{4 d (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Csc[c/2]*Csc[c + d*x]^2*Sec[c/2]*((3*I)*Cos[c] - (3*I)*Cos[c + 2*d*x] + (-1 - (4*I)*d*x - 2*Log[Sin[c + d
*x]^2] + 2*Cos[2*(c + d*x)]*((2*I)*d*x + Log[Sin[c + d*x]^2]))*Sin[c])*(Cos[3*d*x] + I*Sin[3*d*x]))/(4*d*(Cos[
d*x] + I*Sin[d*x])^3)

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Maple [A]
time = 0.22, size = 75, normalized size = 1.06

method	result	size
risch	\(\frac {8 i a^{3} c}{d}+\frac {2 a^{3} \left (4 \,{\mathrm e}^{2 i \left (d x +c \right )}-3\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(66\)
derivativedivides	\(\frac {-i a^{3} \left (d x +c \right )-3 a^{3} \ln \left (\sin \left (d x +c \right )\right )+3 i a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(75\)
default	\(\frac {-i a^{3} \left (d x +c \right )-3 a^{3} \ln \left (\sin \left (d x +c \right )\right )+3 i a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(75\)
norman	\(\frac {-\frac {a^{3}}{2 d}-\frac {3 i a^{3} \tan \left (d x +c \right )}{d}-4 i a^{3} x \left (\tan ^{2}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{2}}-\frac {4 a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 68, normalized size = 0.96 \begin {gather*} -\frac {8 i \, {\left (d x + c\right )} a^{3} - 4 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 8 \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 i \, a^{3} \tan \left (d x + c\right ) + a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(8*I*(d*x + c)*a^3 - 4*a^3*log(tan(d*x + c)^2 + 1) + 8*a^3*log(tan(d*x + c)) + (6*I*a^3*tan(d*x + c) + a^
3)/tan(d*x + c)^2)/d

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Fricas [A]
time = 0.49, size = 94, normalized size = 1.32 \begin {gather*} \frac {2 \, {\left (4 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, a^{3} - 2 \, {\left (a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(4*a^3*e^(2*I*d*x + 2*I*c) - 3*a^3 - 2*(a^3*e^(4*I*d*x + 4*I*c) - 2*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*
I*d*x + 2*I*c) - 1))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]
time = 0.22, size = 87, normalized size = 1.23 \begin {gather*} - \frac {4 a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {8 a^{3} e^{2 i c} e^{2 i d x} - 6 a^{3}}{d e^{4 i c} e^{4 i d x} - 2 d e^{2 i c} e^{2 i d x} + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (8*a**3*exp(2*I*c)*exp(2*I*d*x) - 6*a**3)/(d*exp(4*I*c)*exp(4*I*d*
x) - 2*d*exp(2*I*c)*exp(2*I*d*x) + d)

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Giac [A]
time = 1.02, size = 116, normalized size = 1.63 \begin {gather*} -\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 64 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 32 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 12 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {48 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 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 )^{2}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 64*a^3*log(tan(1/2*d*x + 1/2*c) + I) + 32*a^3*log(tan(1/2*d*x + 1/2*c)) - 1
2*I*a^3*tan(1/2*d*x + 1/2*c) - (48*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*I*a^3*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d
*x + 1/2*c)^2)/d

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Mupad [B]
time = 3.78, size = 53, normalized size = 0.75 \begin {gather*} -\frac {\frac {a^3}{2}+a^3\,\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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