3.107 \(\int (c+d x)^2 \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=97 \[ -\frac {i d^2 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d} \]

[Out]

-I*(d*x+c)^2/b-1/3*(d*x+c)^3/d-(d*x+c)^2*cot(b*x+a)/b+2*d*(d*x+c)*ln(1-exp(2*I*(b*x+a)))/b^2-I*d^2*polylog(2,e
xp(2*I*(b*x+a)))/b^3

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Rubi [A]  time = 0.13, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3720, 3717, 2190, 2279, 2391, 32} \[ -\frac {i d^2 \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cot[a + b*x]^2,x]

[Out]

((-I)*(c + d*x)^2)/b - (c + d*x)^3/(3*d) - ((c + d*x)^2*Cot[a + b*x])/b + (2*d*(c + d*x)*Log[1 - E^((2*I)*(a +
b*x))])/b^2 - (I*d^2*PolyLog[2, E^((2*I)*(a + b*x))])/b^3

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3720

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

Rubi steps

\begin {align*} \int (c+d x)^2 \cot ^2(a+b x) \, dx &=-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {(2 d) \int (c+d x) \cot (a+b x) \, dx}{b}-\int (c+d x)^2 \, dx\\ &=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {(4 i d) \int \frac {e^{2 i (a+b x)} (c+d x)}{1-e^{2 i (a+b x)}} \, dx}{b}\\ &=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {\left (2 d^2\right ) \int \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}+\frac {\left (i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{b^3}\\ &=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d^2 \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^3}\\ \end {align*}

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Mathematica [B]  time = 6.03, size = 198, normalized size = 2.04 \[ -\frac {2 c d (b x \cot (a)-\log (\sin (a+b x)))}{b^2}+\frac {d^2 \left (-b^2 x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt {\sec ^2(a)}-i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (\pi -2 \tan ^{-1}(\tan (a))\right )+2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+\pi \log \left (1+e^{-2 i b x}\right )-\pi \log (\cos (b x))\right )}{b^3}+\frac {\csc (a) \sin (b x) (c+d x)^2 \csc (a+b x)}{b}-\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Cot[a + b*x]^2,x]

[Out]

-1/3*(x*(3*c^2 + 3*c*d*x + d^2*x^2)) - (2*c*d*(b*x*Cot[a] - Log[Sin[a + b*x]]))/b^2 + (d^2*(I*b*x*(Pi - 2*ArcT
an[Tan[a]]) + Pi*Log[1 + E^((-2*I)*b*x)] + 2*(b*x + ArcTan[Tan[a]])*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))]
- Pi*Log[Cos[b*x]] - 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] - I*PolyLog[2, E^((2*I)*(b*x + ArcTan[Tan
[a]]))] - b^2*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2]))/b^3 + ((c + d*x)^2*Csc[a]*Csc[a + b*x]*Sin[b*x]
)/b

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fricas [B]  time = 0.72, size = 384, normalized size = 3.96 \[ -\frac {6 \, b^{2} d^{2} x^{2} + 12 \, b^{2} c d x + 6 \, b^{2} c^{2} + 3 i \, d^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 i \, d^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) + 6 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (b^{3} d^{2} x^{3} + 3 \, b^{3} c d x^{2} + 3 \, b^{3} c^{2} x\right )} \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3} \sin \left (2 \, b x + 2 \, a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/6*(6*b^2*d^2*x^2 + 12*b^2*c*d*x + 6*b^2*c^2 + 3*I*d^2*dilog(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a))*sin(2*b*
x + 2*a) - 3*I*d^2*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a))*sin(2*b*x + 2*a) - 6*(b*c*d - a*d^2)*log(-1/2*
cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) - 6*(b*c*d - a*d^2)*log(-1/2*cos(2*b*x + 2*a
) - 1/2*I*sin(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) - 6*(b*d^2*x + a*d^2)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x +
2*a) + 1)*sin(2*b*x + 2*a) - 6*(b*d^2*x + a*d^2)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1)*sin(2*b*x +
2*a) + 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(2*b*x + 2*a) + 2*(b^3*d^2*x^3 + 3*b^3*c*d*x^2 + 3*b^3*c^2*x
)*sin(2*b*x + 2*a))/(b^3*sin(2*b*x + 2*a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \cot \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*cot(b*x + a)^2, x)

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maple [B]  time = 0.09, size = 297, normalized size = 3.06 \[ -\frac {d^{2} x^{3}}{3}-c d \,x^{2}-c^{2} x -\frac {4 i d^{2} a x}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} a^{2}}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} x^{2}}{b}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {4 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*cot(b*x+a)^2,x)

[Out]

-1/3*d^2*x^3-c*d*x^2-c^2*x-4*I/b^2*d^2*a*x+2/b^2*d*c*ln(exp(I*(b*x+a))-1)+2/b^2*d*c*ln(exp(I*(b*x+a))+1)-4/b^2
*d*c*ln(exp(I*(b*x+a)))-2*I/b^3*d^2*polylog(2,-exp(I*(b*x+a)))-2*I/b^3*d^2*a^2-2*I*d^2*polylog(2,exp(I*(b*x+a)
))/b^3+2/b^2*d^2*ln(exp(I*(b*x+a))+1)*x-2*I*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I*(b*x+a))-1)+2/b^2*d^2*ln(1-exp(I*
(b*x+a)))*x+2/b^3*d^2*ln(1-exp(I*(b*x+a)))*a-2*I/b*d^2*x^2-2/b^3*d^2*a*ln(exp(I*(b*x+a))-1)+4/b^3*d^2*a*ln(exp
(I*(b*x+a)))

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maxima [B]  time = 0.74, size = 646, normalized size = 6.66 \[ \frac {-i \, b^{3} d^{2} x^{3} - 3 i \, b^{3} c d x^{2} - 3 i \, b^{3} c^{2} x - 6 \, b^{2} c^{2} - {\left (6 \, b d^{2} x + 6 \, b c d - 6 \, {\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (6 i \, b d^{2} x + 6 i \, b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (6 \, b c d \cos \left (2 \, b x + 2 \, a\right ) + 6 i \, b c d \sin \left (2 \, b x + 2 \, a\right ) - 6 \, b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) - {\left (6 \, b d^{2} x \cos \left (2 \, b x + 2 \, a\right ) + 6 i \, b d^{2} x \sin \left (2 \, b x + 2 \, a\right ) - 6 \, b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + {\left (i \, b^{3} d^{2} x^{3} + {\left (3 i \, b^{3} c d - 6 \, b^{2} d^{2}\right )} x^{2} - 3 \, {\left (-i \, b^{3} c^{2} + 4 \, b^{2} c d\right )} x\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left (3 i \, b d^{2} x + 3 i \, b c d + {\left (-3 i \, b d^{2} x - 3 i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + {\left (3 i \, b d^{2} x + 3 i \, b c d + {\left (-3 i \, b d^{2} x - 3 i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (b^{3} d^{2} x^{3} + 3 \, {\left (b^{3} c d + 2 i \, b^{2} d^{2}\right )} x^{2} + {\left (3 \, b^{3} c^{2} + 12 i \, b^{2} c d\right )} x\right )} \sin \left (2 \, b x + 2 \, a\right )}{-3 i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + 3 \, b^{3} \sin \left (2 \, b x + 2 \, a\right ) + 3 i \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cot(b*x+a)^2,x, algorithm="maxima")

[Out]

(-I*b^3*d^2*x^3 - 3*I*b^3*c*d*x^2 - 3*I*b^3*c^2*x - 6*b^2*c^2 - (6*b*d^2*x + 6*b*c*d - 6*(b*d^2*x + b*c*d)*cos
(2*b*x + 2*a) - (6*I*b*d^2*x + 6*I*b*c*d)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*b*c*d
*cos(2*b*x + 2*a) + 6*I*b*c*d*sin(2*b*x + 2*a) - 6*b*c*d)*arctan2(sin(b*x + a), cos(b*x + a) - 1) - (6*b*d^2*x
*cos(2*b*x + 2*a) + 6*I*b*d^2*x*sin(2*b*x + 2*a) - 6*b*d^2*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (I*b^
3*d^2*x^3 + (3*I*b^3*c*d - 6*b^2*d^2)*x^2 - 3*(-I*b^3*c^2 + 4*b^2*c*d)*x)*cos(2*b*x + 2*a) - 6*(d^2*cos(2*b*x
+ 2*a) + I*d^2*sin(2*b*x + 2*a) - d^2)*dilog(-e^(I*b*x + I*a)) - 6*(d^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x + 2
*a) - d^2)*dilog(e^(I*b*x + I*a)) + (3*I*b*d^2*x + 3*I*b*c*d + (-3*I*b*d^2*x - 3*I*b*c*d)*cos(2*b*x + 2*a) + 3
*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (3*I*b*d^2*x
+ 3*I*b*c*d + (-3*I*b*d^2*x - 3*I*b*c*d)*cos(2*b*x + 2*a) + 3*(b*d^2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x
+ a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (b^3*d^2*x^3 + 3*(b^3*c*d + 2*I*b^2*d^2)*x^2 + (3*b^3*c^2 + 12
*I*b^2*c*d)*x)*sin(2*b*x + 2*a))/(-3*I*b^3*cos(2*b*x + 2*a) + 3*b^3*sin(2*b*x + 2*a) + 3*I*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(a + b*x)^2*(c + d*x)^2,x)

[Out]

int(cot(a + b*x)^2*(c + d*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \cot ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*cot(b*x+a)**2,x)

[Out]

Integral((c + d*x)**2*cot(a + b*x)**2, x)

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