3.365 \(\int (\frac {1}{2}-3 \cot (x)) (3-2 \cot (x))^3 \, dx\)

Optimal. Leaf size=33 \[ -\frac {285 x}{2}+(3-2 \cot (x))^3+5 (3-2 \cot (x))^2-42 \cot (x)+4 \log (\sin (x)) \]

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Rubi [A]  time = 0.06, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3528, 3525, 3475} \[ -\frac {285 x}{2}+(3-2 \cot (x))^3+5 (3-2 \cot (x))^2-42 \cot (x)+4 \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1/2 - 3*Cot[x])*(3 - 2*Cot[x])^3,x]

[Out]

(-285*x)/2 + 5*(3 - 2*Cot[x])^2 + (3 - 2*Cot[x])^3 - 42*Cot[x] + 4*Log[Sin[x]]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3525

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

Rule 3528

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

Rubi steps

\begin {align*} \int \left (\frac {1}{2}-3 \cot (x)\right ) (3-2 \cot (x))^3 \, dx &=(3-2 \cot (x))^3+\int \left (-\frac {9}{2}-10 \cot (x)\right ) (3-2 \cot (x))^2 \, dx\\ &=5 (3-2 \cot (x))^2+(3-2 \cot (x))^3+\int \left (-\frac {67}{2}-21 \cot (x)\right ) (3-2 \cot (x)) \, dx\\ &=-\frac {285 x}{2}+5 (3-2 \cot (x))^2+(3-2 \cot (x))^3-42 \cot (x)+4 \int \cot (x) \, dx\\ &=-\frac {285 x}{2}+5 (3-2 \cot (x))^2+(3-2 \cot (x))^3-42 \cot (x)+4 \log (\sin (x))\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 0.88 \[ -\frac {285 x}{2}-148 \cot (x)+56 \csc ^2(x)+4 \log (\sin (x))-8 \cot (x) \csc ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1/2 - 3*Cot[x])*(3 - 2*Cot[x])^3,x]

[Out]

(-285*x)/2 - 148*Cot[x] + 56*Csc[x]^2 - 8*Cot[x]*Csc[x]^2 + 4*Log[Sin[x]]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {1}{2}-3 \cot (x)\right ) (3-2 \cot (x))^3 \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[(1/2 - 3*Cot[x])*(3 - 2*Cot[x])^3,x]

[Out]

Could not integrate

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fricas [B]  time = 0.86, size = 71, normalized size = 2.15 \[ \frac {4 \, {\left (\cos \left (2 \, x\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, x\right ) + \frac {1}{2}\right ) \sin \left (2 \, x\right ) - 296 \, \cos \left (2 \, x\right )^{2} - {\left (285 \, x \cos \left (2 \, x\right ) - 285 \, x + 224\right )} \sin \left (2 \, x\right ) + 32 \, \cos \left (2 \, x\right ) + 328}{2 \, {\left (\cos \left (2 \, x\right ) - 1\right )} \sin \left (2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2-3*cot(x))*(3-2*cot(x))^3,x, algorithm="fricas")

[Out]

1/2*(4*(cos(2*x) - 1)*log(-1/2*cos(2*x) + 1/2)*sin(2*x) - 296*cos(2*x)^2 - (285*x*cos(2*x) - 285*x + 224)*sin(
2*x) + 32*cos(2*x) + 328)/((cos(2*x) - 1)*sin(2*x))

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giac [B]  time = 0.67, size = 75, normalized size = 2.27 \[ \tan \left (\frac {1}{2} \, x\right )^{3} + 14 \, \tan \left (\frac {1}{2} \, x\right )^{2} - \frac {285}{2} \, x - \frac {22 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 225 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 42 \, \tan \left (\frac {1}{2} \, x\right ) + 3}{3 \, \tan \left (\frac {1}{2} \, x\right )^{3}} - 4 \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 4 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right ) + 75 \, \tan \left (\frac {1}{2} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2-3*cot(x))*(3-2*cot(x))^3,x, algorithm="giac")

[Out]

tan(1/2*x)^3 + 14*tan(1/2*x)^2 - 285/2*x - 1/3*(22*tan(1/2*x)^3 + 225*tan(1/2*x)^2 - 42*tan(1/2*x) + 3)/tan(1/
2*x)^3 - 4*log(tan(1/2*x)^2 + 1) + 4*log(abs(tan(1/2*x))) + 75*tan(1/2*x)

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maple [A]  time = 0.06, size = 35, normalized size = 1.06

method	result	size
derivativedivides	\(-8 \left (\cot ^{3}\relax (x )\right )+56 \left (\cot ^{2}\relax (x )\right )-156 \cot \relax (x )-2 \ln \left (\cot ^{2}\relax (x )+1\right )+\frac {285 \pi }{4}-\frac {285 \,\mathrm {arccot}\left (\cot \relax (x )\right )}{2}\)	\(35\)
default	\(-8 \left (\cot ^{3}\relax (x )\right )+56 \left (\cot ^{2}\relax (x )\right )-156 \cot \relax (x )-2 \ln \left (\cot ^{2}\relax (x )+1\right )+\frac {285 \pi }{4}-\frac {285 \,\mathrm {arccot}\left (\cot \relax (x )\right )}{2}\)	\(35\)
norman	\(\frac {-8-156 \left (\tan ^{2}\relax (x )\right )-\frac {285 x \left (\tan ^{3}\relax (x )\right )}{2}+56 \tan \relax (x )}{\tan \relax (x )^{3}}+4 \ln \left (\tan \relax (x )\right )-2 \ln \left (1+\tan ^{2}\relax (x )\right )\)	\(40\)
risch	\(-\frac {285 x}{2}-4 i x +\frac {\left (-\frac {224}{1873}-\frac {264 i}{1873}\right ) \left (1873 \,{\mathrm e}^{4 i x}-1260 i {\mathrm e}^{2 i x}-3358 \,{\mathrm e}^{2 i x}+1221+1036 i\right )}{\left ({\mathrm e}^{2 i x}-1\right )^{3}}+4 \ln \left ({\mathrm e}^{2 i x}-1\right )\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/2-3*cot(x))*(3-2*cot(x))^3,x,method=_RETURNVERBOSE)

[Out]

-8*cot(x)^3+56*cot(x)^2-156*cot(x)-2*ln(cot(x)^2+1)+285/4*Pi-285/2*arccot(cot(x))

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maxima [A]  time = 1.27, size = 36, normalized size = 1.09 \[ -\frac {285}{2} \, x - \frac {4 \, {\left (39 \, \tan \relax (x)^{2} - 14 \, \tan \relax (x) + 2\right )}}{\tan \relax (x)^{3}} - 2 \, \log \left (\tan \relax (x)^{2} + 1\right ) + 4 \, \log \left (\tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2-3*cot(x))*(3-2*cot(x))^3,x, algorithm="maxima")

[Out]

-285/2*x - 4*(39*tan(x)^2 - 14*tan(x) + 2)/tan(x)^3 - 2*log(tan(x)^2 + 1) + 4*log(tan(x))

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mupad [B]  time = 0.52, size = 75, normalized size = 2.27 \[ x\,\left (-\frac {285}{2}-4{}\mathrm {i}\right )+4\,\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}-1\right )+\frac {64{}\mathrm {i}}{3\,{\mathrm {e}}^{x\,2{}\mathrm {i}}-3\,{\mathrm {e}}^{x\,4{}\mathrm {i}}+{\mathrm {e}}^{x\,6{}\mathrm {i}}-1}+\frac {-224+96{}\mathrm {i}}{1+{\mathrm {e}}^{x\,4{}\mathrm {i}}-2\,{\mathrm {e}}^{x\,2{}\mathrm {i}}}+\frac {-224-264{}\mathrm {i}}{{\mathrm {e}}^{x\,2{}\mathrm {i}}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*cot(x) - 3)^3*(3*cot(x) - 1/2),x)

[Out]

4*log(exp(x*2i) - 1) - x*(285/2 + 4i) + 64i/(3*exp(x*2i) - 3*exp(x*4i) + exp(x*6i) - 1) - (224 - 96i)/(exp(x*4
i) - 2*exp(x*2i) + 1) - (224 + 264i)/(exp(x*2i) - 1)

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sympy [A]  time = 0.58, size = 39, normalized size = 1.18 \[ - \frac {285 x}{2} - 2 \log {\left (\tan ^{2}{\relax (x )} + 1 \right )} + 4 \log {\left (\tan {\relax (x )} \right )} - \frac {156}{\tan {\relax (x )}} + \frac {56}{\tan ^{2}{\relax (x )}} - \frac {8}{\tan ^{3}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2-3*cot(x))*(3-2*cot(x))**3,x)

[Out]

-285*x/2 - 2*log(tan(x)**2 + 1) + 4*log(tan(x)) - 156/tan(x) + 56/tan(x)**2 - 8/tan(x)**3

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