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

Optimal. Leaf size=171 \[ \frac{2 i d (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac{2 d (c+d x) \sin (a+b x)}{b^2}-\frac{2 d^2 \cos (a+b x)}{b^3}+\frac{(c+d x)^2 \cos (a+b x)}{b}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

(-2*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b - (2*d^2*Cos[a + b*x])/b^3 + ((c + d*x)^2*Cos[a + b*x])/b + ((2*I)
*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 - (2*d^2*
PolyLog[3, -E^(I*(a + b*x))])/b^3 + (2*d^2*PolyLog[3, E^(I*(a + b*x))])/b^3 - (2*d*(c + d*x)*Sin[a + b*x])/b^2

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Rubi [A]  time = 0.138333, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {4408, 3296, 2638, 4183, 2531, 2282, 6589} \[ \frac{2 i d (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac{2 d (c+d x) \sin (a+b x)}{b^2}-\frac{2 d^2 \cos (a+b x)}{b^3}+\frac{(c+d x)^2 \cos (a+b x)}{b}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b - (2*d^2*Cos[a + b*x])/b^3 + ((c + d*x)^2*Cos[a + b*x])/b + ((2*I)
*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((2*I)*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 - (2*d^2*
PolyLog[3, -E^(I*(a + b*x))])/b^3 + (2*d^2*PolyLog[3, E^(I*(a + b*x))])/b^3 - (2*d*(c + d*x)*Sin[a + b*x])/b^2

Rule 4408

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

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

Rule 2638

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

Rule 4183

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

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (c+d x)^2 \cos (a+b x) \cot (a+b x) \, dx &=\int (c+d x)^2 \csc (a+b x) \, dx-\int (c+d x)^2 \sin (a+b x) \, dx\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{(c+d x)^2 \cos (a+b x)}{b}-\frac{(2 d) \int (c+d x) \cos (a+b x) \, dx}{b}-\frac{(2 d) \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac{(2 d) \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{(c+d x)^2 \cos (a+b x)}{b}+\frac{2 i d (c+d x) \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{2 d (c+d x) \sin (a+b x)}{b^2}-\frac{\left (2 i d^2\right ) \int \text{Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (2 i d^2\right ) \int \text{Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (2 d^2\right ) \int \sin (a+b x) \, dx}{b^2}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{2 d^2 \cos (a+b x)}{b^3}+\frac{(c+d x)^2 \cos (a+b x)}{b}+\frac{2 i d (c+d x) \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{2 d (c+d x) \sin (a+b x)}{b^2}-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}+\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{2 d^2 \cos (a+b x)}{b^3}+\frac{(c+d x)^2 \cos (a+b x)}{b}+\frac{2 i d (c+d x) \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac{2 d (c+d x) \sin (a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.874117, size = 221, normalized size = 1.29 \[ \frac{2 i b d (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )-2 i b d (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )-2 d^2 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )+2 d^2 \text{PolyLog}\left (3,e^{i (a+b x)}\right )+\cos (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )-2 b d \sin (a) (c+d x)\right )-\sin (b x) \left (\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )+2 b d \cos (a) (c+d x)\right )+b^2 (c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-b^2 (c+d x)^2 \log \left (1+e^{i (a+b x)}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.184, size = 479, normalized size = 2.8 \begin{align*}{\frac{ \left ({d}^{2}{x}^{2}{b}^{2}+2\,{b}^{2}cdx+{b}^{2}{c}^{2}+2\,ib{d}^{2}x-2\,{d}^{2}+2\,ibcd \right ){{\rm e}^{i \left ( bx+a \right ) }}}{2\,{b}^{3}}}+{\frac{ \left ({d}^{2}{x}^{2}{b}^{2}+2\,{b}^{2}cdx+{b}^{2}{c}^{2}-2\,ib{d}^{2}x-2\,{d}^{2}-2\,ibcd \right ){{\rm e}^{-i \left ( bx+a \right ) }}}{2\,{b}^{3}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}{d}^{2}{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) a}{{b}^{2}}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+2\,{\frac{cd\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-2\,{\frac{{c}^{2}{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}-{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){a}^{2}}{{b}^{3}}}-{\frac{2\,icd{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){a}^{2}}{{b}^{3}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+4\,{\frac{cda{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{2\,icd{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.47035, size = 684, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 0.638458, size = 1462, normalized size = 8.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} \cos{\left (a + b x \right )} \cot{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2} \cos \left (b x + a\right ) \cot \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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