3.1.31 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [31]

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

[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 = 0.11, 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 = {3629, 3626, 3623, 3612, 3556} \begin {gather*} \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} \end {gather*}

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 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]

Rule 3629

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(251\) vs. \(2(101)=202\).
time = 0.93, size = 251, normalized size = 2.49 \begin {gather*} \frac {a^3 \csc \left (\frac {c}{2}\right ) \csc ^3(c+d x) \sec \left (\frac {c}{2}\right ) (\cos (3 d x)+i \sin (3 d x)) \left (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) \left (-i+4 d x-i \log \left (\sin ^2(c+d x)\right )\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)-48 \text {ArcTan}(\tan (4 c+d x)) \sin (c) \sin ^3(c+d x)-15 \sin (2 c+d x)+13 \sin (2 c+3 d x)\right )}{24 d (\cos (d x)+i \sin (d x))^3} \end {gather*}

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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Maple [A]
time = 0.22, size = 91, normalized size = 0.90

method	result	size
risch	\(-\frac {8 a^{3} c}{d}+\frac {2 i a^{3} \left (24 \,{\mathrm e}^{4 i \left (d x +c \right )}-33 \,{\mathrm e}^{2 i \left (d x +c \right )}+13\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(78\)
derivativedivides	\(\frac {-i a^{3} \ln \left (\sin \left (d x +c \right )\right )-3 a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 i a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(91\)
default	\(\frac {-i a^{3} \ln \left (\sin \left (d x +c \right )\right )-3 a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 i a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(91\)
norman	\(\frac {-\frac {a^{3}}{3 d}+4 a^{3} x \left (\tan ^{3}\left (d x +c \right )\right )+\frac {4 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {3 i a^{3} \tan \left (d x +c \right )}{2 d}}{\tan \left (d x +c \right )^{3}}-\frac {4 i a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {2 i a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(101\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.58, size = 83, normalized size = 0.82 \begin {gather*} \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} \end {gather*}

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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Fricas [A]
time = 0.46, size = 139, normalized size = 1.38 \begin {gather*} -\frac {2 \, {\left (-24 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 33 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 13 i \, a^{3} + 6 \, {\left (i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\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 )}} \end {gather*}

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]

-2/3*(-24*I*a^3*e^(4*I*d*x + 4*I*c) + 33*I*a^3*e^(2*I*d*x + 2*I*c) - 13*I*a^3 + 6*(I*a^3*e^(6*I*d*x + 6*I*c) -
3*I*a^3*e^(4*I*d*x + 4*I*c) + 3*I*a^3*e^(2*I*d*x + 2*I*c) - 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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Sympy [A]
time = 0.28, size = 136, normalized size = 1.35 \begin {gather*} - \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} \end {gather*}

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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Giac [A]
time = 1.08, size = 146, normalized size = 1.45 \begin {gather*} \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} \end {gather*}

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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Mupad [B]
time = 3.79, size = 68, normalized size = 0.67 \begin {gather*} \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} \end {gather*}

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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