3.33 \(\int (c+d x)^3 \cot (a+b x) \, dx\)

Optimal. Leaf size=127 \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i (c+d x)^4}{4 d} \]

[Out]

((-I/4)*(c + d*x)^4)/d + ((c + d*x)^3*Log[1 - E^((2*I)*(a + b*x))])/b - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, E^
((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4
, E^((2*I)*(a + b*x))])/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.192486, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3717, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 i d^3 \text{PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Cot[a + b*x],x]

[Out]

((-I/4)*(c + d*x)^4)/d + ((c + d*x)^3*Log[1 - E^((2*I)*(a + b*x))])/b - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, E^
((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, E^((2*I)*(a + b*x))])/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4
, E^((2*I)*(a + b*x))])/b^4

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 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 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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)^3 \cot (a+b x) \, dx &=-\frac{i (c+d x)^4}{4 d}-2 i \int \frac{e^{2 i (a+b x)} (c+d x)^3}{1-e^{2 i (a+b x)}} \, dx\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{(3 d) \int (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{\left (3 i d^2\right ) \int (c+d x) \text{Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}-\frac{\left (3 d^3\right ) \int \text{Li}_3\left (e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac{\left (3 i d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=-\frac{i (c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 d^2 (c+d x) \text{Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i d^3 \text{Li}_4\left (e^{2 i (a+b x)}\right )}{4 b^4}\\ \end{align*}

Mathematica [B]  time = 2.75977, size = 560, normalized size = 4.41 \[ \frac{-6 i b^2 c^2 d \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+12 i b^2 d^2 x (2 c+d x) \text{PolyLog}\left (2,-e^{-i (a+b x)}\right )+12 i b^2 d^2 x (2 c+d x) \text{PolyLog}\left (2,e^{-i (a+b x)}\right )+24 b c d^2 \text{PolyLog}\left (3,-e^{-i (a+b x)}\right )+24 b c d^2 \text{PolyLog}\left (3,e^{-i (a+b x)}\right )+24 b d^3 x \text{PolyLog}\left (3,-e^{-i (a+b x)}\right )+24 b d^3 x \text{PolyLog}\left (3,e^{-i (a+b x)}\right )-24 i d^3 \text{PolyLog}\left (4,-e^{-i (a+b x)}\right )-24 i d^3 \text{PolyLog}\left (4,e^{-i (a+b x)}\right )+6 b^4 c^2 d x^2 \cot (a)-6 b^4 c^2 d x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt{\sec ^2(a)}-12 i b^3 c^2 d x \tan ^{-1}(\tan (a))+12 b^3 c^2 d x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+12 b^2 c^2 d \tan ^{-1}(\tan (a)) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-12 b^2 c^2 d \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+4 b^3 c^3 \log (\sin (a+b x))+12 b^3 c d^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+12 b^3 c d^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+4 b^3 d^3 x^3 \log \left (1-e^{-i (a+b x)}\right )+4 b^3 d^3 x^3 \log \left (1+e^{-i (a+b x)}\right )+6 i \pi b^3 c^2 d x+6 \pi b^2 c^2 d \log \left (1+e^{-2 i b x}\right )-6 \pi b^2 c^2 d \log (\cos (b x))+4 i b^4 c d^2 x^3+i b^4 d^3 x^4}{4 b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Cot[a + b*x],x]

[Out]

((6*I)*b^3*c^2*d*Pi*x + (4*I)*b^4*c*d^2*x^3 + I*b^4*d^3*x^4 - (12*I)*b^3*c^2*d*x*ArcTan[Tan[a]] + 6*b^4*c^2*d*
x^2*Cot[a] + 6*b^2*c^2*d*Pi*Log[1 + E^((-2*I)*b*x)] + 12*b^3*c*d^2*x^2*Log[1 - E^((-I)*(a + b*x))] + 4*b^3*d^3
*x^3*Log[1 - E^((-I)*(a + b*x))] + 12*b^3*c*d^2*x^2*Log[1 + E^((-I)*(a + b*x))] + 4*b^3*d^3*x^3*Log[1 + E^((-I
)*(a + b*x))] + 12*b^3*c^2*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 12*b^2*c^2*d*ArcTan[Tan[a]]*Log[1 -
E^((2*I)*(b*x + ArcTan[Tan[a]]))] - 6*b^2*c^2*d*Pi*Log[Cos[b*x]] + 4*b^3*c^3*Log[Sin[a + b*x]] - 12*b^2*c^2*d
*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + (12*I)*b^2*d^2*x*(2*c + d*x)*PolyLog[2, -E^((-I)*(a + b*x))]
+ (12*I)*b^2*d^2*x*(2*c + d*x)*PolyLog[2, E^((-I)*(a + b*x))] - (6*I)*b^2*c^2*d*PolyLog[2, E^((2*I)*(b*x + Arc
Tan[Tan[a]]))] + 24*b*c*d^2*PolyLog[3, -E^((-I)*(a + b*x))] + 24*b*d^3*x*PolyLog[3, -E^((-I)*(a + b*x))] + 24*
b*c*d^2*PolyLog[3, E^((-I)*(a + b*x))] + 24*b*d^3*x*PolyLog[3, E^((-I)*(a + b*x))] - (24*I)*d^3*PolyLog[4, -E^
((-I)*(a + b*x))] - (24*I)*d^3*PolyLog[4, E^((-I)*(a + b*x))] - 6*b^4*c^2*d*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sq
rt[Sec[a]^2])/(4*b^4)

________________________________________________________________________________________

Maple [B]  time = 0.296, size = 783, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*cos(b*x+a)*csc(b*x+a),x)

[Out]

6/b^3*d^3*polylog(3,exp(I*(b*x+a)))*x-1/b^4*d^3*a^3*ln(exp(I*(b*x+a))-1)+6/b^3*d^3*polylog(3,-exp(I*(b*x+a)))*
x-3/2*I/b^4*d^3*a^4+6*I/b^4*d^3*polylog(4,-exp(I*(b*x+a)))+2/b^4*d^3*a^3*ln(exp(I*(b*x+a)))+6/b^3*c*d^2*polylo
g(3,-exp(I*(b*x+a)))+6/b^3*c*d^2*polylog(3,exp(I*(b*x+a)))-1/4*I*d^3*x^4+6*I/b^2*c*d^2*a^2*x-6*I/b^2*c*d^2*pol
ylog(2,-exp(I*(b*x+a)))*x-6*I/b^2*c*d^2*polylog(2,exp(I*(b*x+a)))*x-6*I/b*c^2*d*a*x-I*c*d^2*x^3-3/2*I*c^2*d*x^
2+6*I*d^3*polylog(4,exp(I*(b*x+a)))/b^4+3/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2+3/b*c^2*d*ln(1-exp(I*(b*x+a)))*x+3/
b^2*c^2*d*ln(1-exp(I*(b*x+a)))*a+3/b*c*d^2*ln(1-exp(I*(b*x+a)))*x^2-2/b*c^3*ln(exp(I*(b*x+a)))+1/b*c^3*ln(exp(
I*(b*x+a))-1)+1/b*c^3*ln(exp(I*(b*x+a))+1)-3/b^3*c*d^2*ln(1-exp(I*(b*x+a)))*a^2+3/b*c^2*d*ln(exp(I*(b*x+a))+1)
*x+1/b*d^3*ln(1-exp(I*(b*x+a)))*x^3+1/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^3+1/b*d^3*ln(exp(I*(b*x+a))+1)*x^3+3/b^3*
c*d^2*a^2*ln(exp(I*(b*x+a))-1)+6/b^2*c^2*d*a*ln(exp(I*(b*x+a)))-3/b^2*c^2*d*a*ln(exp(I*(b*x+a))-1)-6/b^3*c*d^2
*a^2*ln(exp(I*(b*x+a)))-3*I/b^2*d^3*polylog(2,-exp(I*(b*x+a)))*x^2-3*I/b^2*d^3*polylog(2,exp(I*(b*x+a)))*x^2-3
*I/b^2*c^2*d*polylog(2,exp(I*(b*x+a)))-3*I/b^2*c^2*d*polylog(2,-exp(I*(b*x+a)))+4*I/b^3*c*d^2*a^3-2*I/b^3*d^3*
a^3*x-3*I/b^2*c^2*d*a^2+I*c^3*x

________________________________________________________________________________________

Maxima [B]  time = 1.70128, size = 1008, normalized size = 7.94 \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)^3*cos(b*x+a)*csc(b*x+a),x, algorithm="maxima")

[Out]

1/4*(4*c^3*log(sin(b*x + a)) - 12*a*c^2*d*log(sin(b*x + a))/b + 12*a^2*c*d^2*log(sin(b*x + a))/b^2 - 4*a^3*d^3
*log(sin(b*x + a))/b^3 + (-I*(b*x + a)^4*d^3 + (-4*I*b*c*d^2 + 4*I*a*d^3)*(b*x + a)^3 + 24*I*d^3*polylog(4, -e
^(I*b*x + I*a)) + 24*I*d^3*polylog(4, e^(I*b*x + I*a)) + (-6*I*b^2*c^2*d + 12*I*a*b*c*d^2 - 6*I*a^2*d^3)*(b*x
+ a)^2 + (4*I*(b*x + a)^3*d^3 + (12*I*b*c*d^2 - 12*I*a*d^3)*(b*x + a)^2 + (12*I*b^2*c^2*d - 24*I*a*b*c*d^2 + 1
2*I*a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (-4*I*(b*x + a)^3*d^3 + (-12*I*b*c*d^2 + 12*
I*a*d^3)*(b*x + a)^2 + (-12*I*b^2*c^2*d + 24*I*a*b*c*d^2 - 12*I*a^2*d^3)*(b*x + a))*arctan2(sin(b*x + a), -cos
(b*x + a) + 1) + (-12*I*b^2*c^2*d + 24*I*a*b*c*d^2 - 12*I*(b*x + a)^2*d^3 - 12*I*a^2*d^3 + (-24*I*b*c*d^2 + 24
*I*a*d^3)*(b*x + a))*dilog(-e^(I*b*x + I*a)) + (-12*I*b^2*c^2*d + 24*I*a*b*c*d^2 - 12*I*(b*x + a)^2*d^3 - 12*I
*a^2*d^3 + (-24*I*b*c*d^2 + 24*I*a*d^3)*(b*x + a))*dilog(e^(I*b*x + I*a)) + 2*((b*x + a)^3*d^3 + 3*(b*c*d^2 -
a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*
cos(b*x + a) + 1) + 2*((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^
3)*(b*x + a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 24*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)
*polylog(3, -e^(I*b*x + I*a)) + 24*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*polylog(3, e^(I*b*x + I*a)))/b^3)/b

________________________________________________________________________________________

Fricas [C]  time = 0.626955, size = 2056, normalized size = 16.19 \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)^3*cos(b*x+a)*csc(b*x+a),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*cos(b*x+a)*csc(b*x+a),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \cos \left (b x + a\right ) \csc \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*cos(b*x+a)*csc(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*cos(b*x + a)*csc(b*x + a), x)