3.1.42 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [42]

Optimal. Leaf size=134 \[ 8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]

[Out]

8*I*a^4*x+4*I*a^4*cot(d*x+c)/d+8*a^4*ln(sin(d*x+c))/d-1/3*I*a*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3/d-1/4*cot(d*x+
c)^4*(a+I*a*tan(d*x+c))^4/d+cot(d*x+c)^2*(a^2+I*a^2*tan(d*x+c))^2/d

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Rubi [A]
time = 0.14, antiderivative size = 134, 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 = {3629, 3626, 3623, 3612, 3556} \begin {gather*} \frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}+8 i a^4 x+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(8*I)*a^4*x + ((4*I)*a^4*Cot[c + d*x])/d + (8*a^4*Log[Sin[c + d*x]])/d - ((I/3)*a*Cot[c + d*x]^3*(a + I*a*Tan[
c + d*x])^3)/d - (Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^4)/(4*d) + (Cot[c + d*x]^2*(a^2 + I*a^2*Tan[c + d*x])^
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]

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 ^5(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+i \int \cot ^4(c+d x) (a+i a \tan (c+d x))^4 \, dx\\ &=-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}-(2 a) \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\left (4 i a^2\right ) \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=\frac {4 i a^4 \cot (c+d x)}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\left (4 i a^2\right ) \int \cot (c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (8 a^4\right ) \int \cot (c+d x) \, dx\\ &=8 i a^4 x+\frac {4 i a^4 \cot (c+d x)}{d}+\frac {8 a^4 \log (\sin (c+d x))}{d}-\frac {i a \cot ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-\frac {\cot ^4(c+d x) (a+i a \tan (c+d x))^4}{4 d}+\frac {\cot ^2(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 245, normalized size = 1.83 \begin {gather*} \frac {a^4 \csc (c) \csc ^4(c+d x) \left (-42 i \cos (c)+38 i \cos (c+2 d x)+18 i \cos (3 c+2 d x)-14 i \cos (3 c+4 d x)+21 \sin (c)+36 i d x \sin (c)+18 \log \left (\sin ^2(c+d x)\right ) \sin (c)+12 \sin (c+2 d x)+24 i d x \sin (c+2 d x)+12 \log \left (\sin ^2(c+d x)\right ) \sin (c+2 d x)-12 \sin (3 c+2 d x)-24 i d x \sin (3 c+2 d x)-12 \log \left (\sin ^2(c+d x)\right ) \sin (3 c+2 d x)-6 i d x \sin (3 c+4 d x)-3 \log \left (\sin ^2(c+d x)\right ) \sin (3 c+4 d x)+6 i d x \sin (5 c+4 d x)+3 \log \left (\sin ^2(c+d x)\right ) \sin (5 c+4 d x)\right )}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*Csc[c]*Csc[c + d*x]^4*((-42*I)*Cos[c] + (38*I)*Cos[c + 2*d*x] + (18*I)*Cos[3*c + 2*d*x] - (14*I)*Cos[3*c
+ 4*d*x] + 21*Sin[c] + (36*I)*d*x*Sin[c] + 18*Log[Sin[c + d*x]^2]*Sin[c] + 12*Sin[c + 2*d*x] + (24*I)*d*x*Sin[
c + 2*d*x] + 12*Log[Sin[c + d*x]^2]*Sin[c + 2*d*x] - 12*Sin[3*c + 2*d*x] - (24*I)*d*x*Sin[3*c + 2*d*x] - 12*Lo
g[Sin[c + d*x]^2]*Sin[3*c + 2*d*x] - (6*I)*d*x*Sin[3*c + 4*d*x] - 3*Log[Sin[c + d*x]^2]*Sin[3*c + 4*d*x] + (6*
I)*d*x*Sin[5*c + 4*d*x] + 3*Log[Sin[c + d*x]^2]*Sin[5*c + 4*d*x]))/(12*d)

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Maple [A]
time = 0.20, size = 123, normalized size = 0.92

method	result	size
risch	\(-\frac {16 i a^{4} c}{d}-\frac {4 a^{4} \left (30 \,{\mathrm e}^{6 i \left (d x +c \right )}-63 \,{\mathrm e}^{4 i \left (d x +c \right )}+50 \,{\mathrm e}^{2 i \left (d x +c \right )}-14\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {8 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(88\)
norman	\(\frac {-\frac {a^{4}}{4 d}+\frac {7 a^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {4 i a^{4} \tan \left (d x +c \right )}{3 d}+\frac {8 i a^{4} \left (\tan ^{3}\left (d x +c \right )\right )}{d}+8 i a^{4} x \left (\tan ^{4}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{4}}+\frac {8 a^{4} \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 a^{4} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(117\)
derivativedivides	\(\frac {a^{4} \ln \left (\sin \left (d x +c \right )\right )-4 i a^{4} \left (-\cot \left (d x +c \right )-d x -c \right )-6 a^{4} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+4 i a^{4} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{4} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(123\)
default	\(\frac {a^{4} \ln \left (\sin \left (d x +c \right )\right )-4 i a^{4} \left (-\cot \left (d x +c \right )-d x -c \right )-6 a^{4} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+4 i a^{4} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+a^{4} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^4*ln(sin(d*x+c))-4*I*a^4*(-cot(d*x+c)-d*x-c)-6*a^4*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+4*I*a^4*(-1/3*cot
(d*x+c)^3+cot(d*x+c)+d*x+c)+a^4*(-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 = 0.52, size = 96, normalized size = 0.72 \begin {gather*} -\frac {-96 i \, {\left (d x + c\right )} a^{4} + 48 \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 96 \, a^{4} \log \left (\tan \left (d x + c\right )\right ) + \frac {-96 i \, a^{4} \tan \left (d x + c\right )^{3} - 42 \, a^{4} \tan \left (d x + c\right )^{2} + 16 i \, a^{4} \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{4}}}{12 \, 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))^4,x, algorithm="maxima")

[Out]

-1/12*(-96*I*(d*x + c)*a^4 + 48*a^4*log(tan(d*x + c)^2 + 1) - 96*a^4*log(tan(d*x + c)) + (-96*I*a^4*tan(d*x +
c)^3 - 42*a^4*tan(d*x + c)^2 + 16*I*a^4*tan(d*x + c) + 3*a^4)/tan(d*x + c)^4)/d

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Fricas [A]
time = 0.36, size = 174, normalized size = 1.30 \begin {gather*} -\frac {4 \, {\left (30 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 63 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 50 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - 14 \, a^{4} - 6 \, {\left (a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (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\right )}} \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))^4,x, algorithm="fricas")

[Out]

-4/3*(30*a^4*e^(6*I*d*x + 6*I*c) - 63*a^4*e^(4*I*d*x + 4*I*c) + 50*a^4*e^(2*I*d*x + 2*I*c) - 14*a^4 - 6*(a^4*e
^(8*I*d*x + 8*I*c) - 4*a^4*e^(6*I*d*x + 6*I*c) + 6*a^4*e^(4*I*d*x + 4*I*c) - 4*a^4*e^(2*I*d*x + 2*I*c) + a^4)*
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 = 1.73, size = 168, normalized size = 1.25 \begin {gather*} \frac {8 a^{4} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 120 a^{4} e^{6 i c} e^{6 i d x} + 252 a^{4} e^{4 i c} e^{4 i d x} - 200 a^{4} e^{2 i c} e^{2 i d x} + 56 a^{4}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 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)**5*(a+I*a*tan(d*x+c))**4,x)

[Out]

8*a**4*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-120*a**4*exp(6*I*c)*exp(6*I*d*x) + 252*a**4*exp(4*I*c)*exp(4*I*d*
x) - 200*a**4*exp(2*I*c)*exp(2*I*d*x) + 56*a**4)/(3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*d*x) +
18*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

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Giac [A]
time = 1.47, size = 180, normalized size = 1.34 \begin {gather*} -\frac {3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 32 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3072 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 1536 \, a^{4} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 864 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3200 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 864 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 32 i \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4}}{\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))^4,x, algorithm="giac")

[Out]

-1/192*(3*a^4*tan(1/2*d*x + 1/2*c)^4 - 32*I*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*a^4*tan(1/2*d*x + 1/2*c)^2 + 3072
*a^4*log(tan(1/2*d*x + 1/2*c) + I) - 1536*a^4*log(tan(1/2*d*x + 1/2*c)) + 864*I*a^4*tan(1/2*d*x + 1/2*c) + (32
00*a^4*tan(1/2*d*x + 1/2*c)^4 - 864*I*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*a^4*tan(1/2*d*x + 1/2*c)^2 + 32*I*a^4*t
an(1/2*d*x + 1/2*c) + 3*a^4)/tan(1/2*d*x + 1/2*c)^4)/d

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Mupad [B]
time = 4.01, size = 80, normalized size = 0.60 \begin {gather*} \frac {a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}}{d}-\frac {-a^4\,{\mathrm {tan}\left (c+d\,x\right )}^3\,8{}\mathrm {i}-\frac {7\,a^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a^4\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3}+\frac {a^4}{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)^4,x)

[Out]

(a^4*atan(2*tan(c + d*x) + 1i)*16i)/d - ((a^4*tan(c + d*x)*4i)/3 + a^4/4 - (7*a^4*tan(c + d*x)^2)/2 - a^4*tan(
c + d*x)^3*8i)/(d*tan(c + d*x)^4)

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