3.1.32 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [32]

Optimal. Leaf size=108 \[ 4 i a^3 x+\frac {4 i a^3 \cot (c+d x)}{d}+\frac {2 a^3 \cot ^2(c+d x)}{d}-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}+\frac {4 a^3 \log (\sin (c+d x))}{d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 108, 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 = {3634, 3672, 3610, 3612, 3556} \begin {gather*} -\frac {3 i a^3 \cot ^3(c+d x)}{4 d}+\frac {2 a^3 \cot ^2(c+d x)}{d}+\frac {4 i a^3 \cot (c+d x)}{d}+\frac {4 a^3 \log (\sin (c+d x))}{d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}+4 i a^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(4*I)*a^3*x + ((4*I)*a^3*Cot[c + d*x])/d + (2*a^3*Cot[c + d*x]^2)/d - (((3*I)/4)*a^3*Cot[c + d*x]^3)/d + (4*a^
3*Log[Sin[c + d*x]])/d - (Cot[c + d*x]^4*(a^3 + I*a^3*Tan[c + d*x]))/(4*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 3610

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

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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(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] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3672

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

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}-\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \left (-9 i a^2+7 a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}-\frac {1}{4} \int \cot ^3(c+d x) \left (16 a^3+16 i a^3 \tan (c+d x)\right ) \, dx\\ &=\frac {2 a^3 \cot ^2(c+d x)}{d}-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}-\frac {1}{4} \int \cot ^2(c+d x) \left (16 i a^3-16 a^3 \tan (c+d x)\right ) \, dx\\ &=\frac {4 i a^3 \cot (c+d x)}{d}+\frac {2 a^3 \cot ^2(c+d x)}{d}-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}-\frac {1}{4} \int \cot (c+d x) \left (-16 a^3-16 i a^3 \tan (c+d x)\right ) \, dx\\ &=4 i a^3 x+\frac {4 i a^3 \cot (c+d x)}{d}+\frac {2 a^3 \cot ^2(c+d x)}{d}-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 d}+\left (4 a^3\right ) \int \cot (c+d x) \, dx\\ &=4 i a^3 x+\frac {4 i a^3 \cot (c+d x)}{d}+\frac {2 a^3 \cot ^2(c+d x)}{d}-\frac {3 i a^3 \cot ^3(c+d x)}{4 d}+\frac {4 a^3 \log (\sin (c+d x))}{d}-\frac {\cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{4 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. \(254\) vs. \(2(108)=216\).
time = 0.96, size = 254, normalized size = 2.35 \begin {gather*} \frac {a^3 \csc \left (\frac {c}{2}\right ) \csc ^4(c+d x) \sec \left (\frac {c}{2}\right ) \left (-15 i \cos (c)+13 i \cos (c+2 d x)+7 i \cos (3 c+2 d x)-5 i \cos (3 c+4 d x)+8 \sin (c)+12 i d x \sin (c)+6 \log \left (\sin ^2(c+d x)\right ) \sin (c)+5 \sin (c+2 d x)+8 i d x \sin (c+2 d x)+4 \log \left (\sin ^2(c+d x)\right ) \sin (c+2 d x)-5 \sin (3 c+2 d x)-8 i d x \sin (3 c+2 d x)-4 \log \left (\sin ^2(c+d x)\right ) \sin (3 c+2 d x)-2 i d x \sin (3 c+4 d x)-\log \left (\sin ^2(c+d x)\right ) \sin (3 c+4 d x)+2 i d x \sin (5 c+4 d x)+\log \left (\sin ^2(c+d x)\right ) \sin (5 c+4 d x)\right )}{16 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Csc[c/2]*Csc[c + d*x]^4*Sec[c/2]*((-15*I)*Cos[c] + (13*I)*Cos[c + 2*d*x] + (7*I)*Cos[3*c + 2*d*x] - (5*I)
*Cos[3*c + 4*d*x] + 8*Sin[c] + (12*I)*d*x*Sin[c] + 6*Log[Sin[c + d*x]^2]*Sin[c] + 5*Sin[c + 2*d*x] + (8*I)*d*x
*Sin[c + 2*d*x] + 4*Log[Sin[c + d*x]^2]*Sin[c + 2*d*x] - 5*Sin[3*c + 2*d*x] - (8*I)*d*x*Sin[3*c + 2*d*x] - 4*L
og[Sin[c + d*x]^2]*Sin[3*c + 2*d*x] - (2*I)*d*x*Sin[3*c + 4*d*x] - Log[Sin[c + d*x]^2]*Sin[3*c + 4*d*x] + (2*I
)*d*x*Sin[5*c + 4*d*x] + Log[Sin[c + d*x]^2]*Sin[5*c + 4*d*x]))/(16*d)

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Maple [A]
time = 0.20, size = 112, normalized size = 1.04

method	result	size
risch	\(-\frac {8 i a^{3} c}{d}-\frac {2 a^{3} \left (12 \,{\mathrm e}^{6 i \left (d x +c \right )}-23 \,{\mathrm e}^{4 i \left (d x +c \right )}+18 \,{\mathrm e}^{2 i \left (d x +c \right )}-5\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(88\)
derivativedivides	\(\frac {-i a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )-3 a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+3 i a^{3} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{3} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(112\)
default	\(\frac {-i a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )-3 a^{3} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+3 i a^{3} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{3} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(112\)
norman	\(\frac {-\frac {a^{3}}{4 d}+\frac {2 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {i a^{3} \tan \left (d x +c \right )}{d}+\frac {4 i a^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{d}+4 i a^{3} x \left (\tan ^{4}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{4}}+\frac {4 a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.55, size = 94, normalized size = 0.87 \begin {gather*} -\frac {-16 i \, {\left (d x + c\right )} a^{3} + 8 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 16 \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {-16 i \, a^{3} \tan \left (d x + c\right )^{3} - 8 \, a^{3} \tan \left (d x + c\right )^{2} + 4 i \, a^{3} \tan \left (d x + c\right ) + a^{3}}{\tan \left (d x + c\right )^{4}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/4*(-16*I*(d*x + c)*a^3 + 8*a^3*log(tan(d*x + c)^2 + 1) - 16*a^3*log(tan(d*x + c)) + (-16*I*a^3*tan(d*x + c)
^3 - 8*a^3*tan(d*x + c)^2 + 4*I*a^3*tan(d*x + c) + a^3)/tan(d*x + c)^4)/d

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Fricas [A]
time = 0.46, size = 174, normalized size = 1.61 \begin {gather*} -\frac {2 \, {\left (12 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 23 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 \, a^{3} - 2 \, {\left (a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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 (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-2*(12*a^3*e^(6*I*d*x + 6*I*c) - 23*a^3*e^(4*I*d*x + 4*I*c) + 18*a^3*e^(2*I*d*x + 2*I*c) - 5*a^3 - 2*(a^3*e^(8
*I*d*x + 8*I*c) - 4*a^3*e^(6*I*d*x + 6*I*c) + 6*a^3*e^(4*I*d*x + 4*I*c) - 4*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log
(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^
(2*I*d*x + 2*I*c) + d)

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Sympy [A]
time = 0.42, size = 165, normalized size = 1.53 \begin {gather*} \frac {4 a^{3} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 24 a^{3} e^{6 i c} e^{6 i d x} + 46 a^{3} e^{4 i c} e^{4 i d x} - 36 a^{3} e^{2 i c} e^{2 i d x} + 10 a^{3}}{d e^{8 i c} e^{8 i d x} - 4 d e^{6 i c} e^{6 i d x} + 6 d e^{4 i c} e^{4 i d x} - 4 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)**5*(a+I*a*tan(d*x+c))**3,x)

[Out]

4*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-24*a**3*exp(6*I*c)*exp(6*I*d*x) + 46*a**3*exp(4*I*c)*exp(4*I*d*x)
- 36*a**3*exp(2*I*c)*exp(2*I*d*x) + 10*a**3)/(d*exp(8*I*c)*exp(8*I*d*x) - 4*d*exp(6*I*c)*exp(6*I*d*x) + 6*d*e
xp(4*I*c)*exp(4*I*d*x) - 4*d*exp(2*I*c)*exp(2*I*d*x) + d)

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Giac [A]
time = 1.29, size = 180, normalized size = 1.67 \begin {gather*} -\frac {3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1536 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 768 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 456 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {1600 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 456 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/192*(3*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*I*a^3*tan(1/2*d*x + 1/2*c)^3 - 108*a^3*tan(1/2*d*x + 1/2*c)^2 + 1536
*a^3*log(tan(1/2*d*x + 1/2*c) + I) - 768*a^3*log(tan(1/2*d*x + 1/2*c)) + 456*I*a^3*tan(1/2*d*x + 1/2*c) + (160
0*a^3*tan(1/2*d*x + 1/2*c)^4 - 456*I*a^3*tan(1/2*d*x + 1/2*c)^3 - 108*a^3*tan(1/2*d*x + 1/2*c)^2 + 24*I*a^3*ta
n(1/2*d*x + 1/2*c) + 3*a^3)/tan(1/2*d*x + 1/2*c)^4)/d

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Mupad [B]
time = 3.91, size = 80, normalized size = 0.74 \begin {gather*} \frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d}-\frac {-a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-2\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2+a^3\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}+\frac {a^3}{4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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