3.1.9 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx\) [9]

Optimal. Leaf size=50 \[ -i a x-\frac {i a \cot (c+d x)}{d}-\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3610, 3612, 3556} \begin {gather*} -\frac {a \cot ^2(c+d x)}{2 d}-\frac {i a \cot (c+d x)}{d}-\frac {a \log (\sin (c+d x))}{d}-i a x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.19, size = 68, normalized size = 1.36 \begin {gather*} -\frac {i a \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )}{d}-\frac {a \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (a*(Cot[c + d*x]^2 + 2*Log[Cos[c +
d*x]] + 2*Log[Tan[c + d*x]]))/(2*d)

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Maple [A]
time = 0.21, size = 48, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {i a \left (-\cot \left (d x +c \right )-d x -c \right )+a \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(48\)
default	\(\frac {i a \left (-\cot \left (d x +c \right )-d x -c \right )+a \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(48\)
risch	\(\frac {2 i a c}{d}+\frac {2 a \left (2 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(60\)
norman	\(\frac {-\frac {a}{2 d}-\frac {i a \tan \left (d x +c \right )}{d}-i a x \left (\tan ^{2}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{2}}-\frac {a \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 58, normalized size = 1.16 \begin {gather*} -\frac {2 i \, {\left (d x + c\right )} a - a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a \log \left (\tan \left (d x + c\right )\right ) + \frac {2 i \, a \tan \left (d x + c\right ) + a}{\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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 83, normalized size = 1.66 \begin {gather*} \frac {4 \, a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (a e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 2 \, a}{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)),x, algorithm="fricas")

[Out]

(4*a*e^(2*I*d*x + 2*I*c) - (a*e^(4*I*d*x + 4*I*c) - 2*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) - 1)
- 2*a)/(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 = 0.16, size = 80, normalized size = 1.60 \begin {gather*} - \frac {a \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {4 a e^{2 i c} e^{2 i d x} - 2 a}{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)),x)

[Out]

-a*log(exp(2*I*d*x) - exp(-2*I*c))/d + (4*a*exp(2*I*c)*exp(2*I*d*x) - 2*a)/(d*exp(4*I*c)*exp(4*I*d*x) - 2*d*ex
p(2*I*c)*exp(2*I*d*x) + d)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (44) = 88\).
time = 0.55, size = 102, normalized size = 2.04 \begin {gather*} -\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 8 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 4 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a}{\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)),x, algorithm="giac")

[Out]

-1/8*(a*tan(1/2*d*x + 1/2*c)^2 - 16*a*log(tan(1/2*d*x + 1/2*c) + I) + 8*a*log(tan(1/2*d*x + 1/2*c)) - 4*I*a*ta
n(1/2*d*x + 1/2*c) - (12*a*tan(1/2*d*x + 1/2*c)^2 - 4*I*a*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^2)/d

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Mupad [B]
time = 3.79, size = 47, normalized size = 0.94 \begin {gather*} -\frac {a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d}-\frac {\frac {a}{2}+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2} \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),x)

[Out]

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

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