3.1.20 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [20]

Optimal. Leaf size=58 \[ -2 i a^2 x-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a^2 \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 58, 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 = {3623, 3610, 3612, 3556} \begin {gather*} -\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {2 a^2 \log (\sin (c+d x))}{d}-2 i a^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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]

Rubi steps

\begin {align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac {a^2 \cot ^2(c+d x)}{2 d}+\int \cot ^2(c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}+\int \cot (c+d x) \left (-2 a^2-2 i a^2 \tan (c+d x)\right ) \, dx\\ &=-2 i a^2 x-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\left (2 a^2\right ) \int \cot (c+d x) \, dx\\ &=-2 i a^2 x-\frac {2 i a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {2 a^2 \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 = 0.22, size = 64, normalized size = 1.10 \begin {gather*} -\frac {a^2 \left (\cot ^2(c+d x)+4 i \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )+4 (\log (\cos (c+d x))+\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])^2,x]

[Out]

-1/2*(a^2*(Cot[c + d*x]^2 + (4*I)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 4*(Log[Cos[c
+ d*x]] + Log[Tan[c + d*x]])))/d

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Maple [A]
time = 0.21, size = 64, normalized size = 1.10

method	result	size
derivativedivides	\(\frac {-a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(64\)
default	\(\frac {-a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 i a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(64\)
risch	\(\frac {4 i a^{2} c}{d}+\frac {2 a^{2} \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(66\)
norman	\(\frac {-\frac {a^{2}}{2 d}-\frac {2 i a^{2} \tan \left (d x +c \right )}{d}-2 i a^{2} x \left (\tan ^{2}\left (d x +c \right )\right )}{\tan \left (d x +c \right )^{2}}+\frac {a^{2} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 68, normalized size = 1.17 \begin {gather*} -\frac {4 i \, {\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {4 i \, a^{2} \tan \left (d x + c\right ) + a^{2}}{\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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 94, normalized size = 1.62 \begin {gather*} \frac {2 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, a^{2} - {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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))^2,x, algorithm="fricas")

[Out]

2*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*a^2 - (a^2*e^(4*I*d*x + 4*I*c) - 2*a^2*e^(2*I*d*x + 2*I*c) + a^2)*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 = 0.26, size = 87, normalized size = 1.50 \begin {gather*} - \frac {2 a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {6 a^{2} e^{2 i c} e^{2 i d x} - 4 a^{2}}{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))**2,x)

[Out]

-2*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (6*a**2*exp(2*I*c)*exp(2*I*d*x) - 4*a**2)/(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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (52) = 104\).
time = 0.82, size = 116, normalized size = 2.00 \begin {gather*} -\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 32 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 16 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\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))^2,x, algorithm="giac")

[Out]

-1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 32*a^2*log(tan(1/2*d*x + 1/2*c) + I) + 16*a^2*log(tan(1/2*d*x + 1/2*c)) - 8
*I*a^2*tan(1/2*d*x + 1/2*c) - (24*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x
+ 1/2*c)^2)/d

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Mupad [B]
time = 3.84, size = 53, normalized size = 0.91 \begin {gather*} -\frac {\frac {a^2}{2}+a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\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)^2,x)

[Out]

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

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