3.23 \(\int (c+d x)^3 \csc (a+b x) \, dx\)

Optimal. Leaf size=185 \[ -\frac{6 d^2 (c+d x) \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{6 d^2 (c+d x) \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{6 i d^3 \text{PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac{6 i d^3 \text{PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

(-2*(c + d*x)^3*ArcTanh[E^(I*(a + b*x))])/b + ((3*I)*d*(c + d*x)^2*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((3*I)*
d*(c + d*x)^2*PolyLog[2, E^(I*(a + b*x))])/b^2 - (6*d^2*(c + d*x)*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (6*d^2*(
c + d*x)*PolyLog[3, E^(I*(a + b*x))])/b^3 - ((6*I)*d^3*PolyLog[4, -E^(I*(a + b*x))])/b^4 + ((6*I)*d^3*PolyLog[
4, E^(I*(a + b*x))])/b^4

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Rubi [A]  time = 0.136988, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4183, 2531, 6609, 2282, 6589} \[ -\frac{6 d^2 (c+d x) \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{6 d^2 (c+d x) \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{6 i d^3 \text{PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac{6 i d^3 \text{PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csc[a + b*x],x]

[Out]

(-2*(c + d*x)^3*ArcTanh[E^(I*(a + b*x))])/b + ((3*I)*d*(c + d*x)^2*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((3*I)*
d*(c + d*x)^2*PolyLog[2, E^(I*(a + b*x))])/b^2 - (6*d^2*(c + d*x)*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (6*d^2*(
c + d*x)*PolyLog[3, E^(I*(a + b*x))])/b^3 - ((6*I)*d^3*PolyLog[4, -E^(I*(a + b*x))])/b^4 + ((6*I)*d^3*PolyLog[
4, E^(I*(a + b*x))])/b^4

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 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 \csc (a+b x) \, dx &=-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{(3 d) \int (c+d x)^2 \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac{(3 d) \int (c+d x)^2 \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{\left (6 i d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (6 i d^2\right ) \int (c+d x) \text{Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{6 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{6 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{\left (6 d^3\right ) \int \text{Li}_3\left (-e^{i (a+b x)}\right ) \, dx}{b^3}-\frac{\left (6 d^3\right ) \int \text{Li}_3\left (e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{6 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{6 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac{\left (6 i d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac{\left (6 i d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}\\ &=-\frac{2 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{3 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{6 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{6 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac{6 i d^3 \text{Li}_4\left (-e^{i (a+b x)}\right )}{b^4}+\frac{6 i d^3 \text{Li}_4\left (e^{i (a+b x)}\right )}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.457009, size = 221, normalized size = 1.19 \[ \frac{3 i d \left (b^2 (c+d x)^2 \text{PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))+2 i b d (c+d x) \text{PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))-2 d^2 \text{PolyLog}(4,-\cos (a+b x)-i \sin (a+b x))\right )-3 i d \left (b^2 (c+d x)^2 \text{PolyLog}(2,\cos (a+b x)+i \sin (a+b x))+2 i b d (c+d x) \text{PolyLog}(3,\cos (a+b x)+i \sin (a+b x))-2 d^2 \text{PolyLog}(4,\cos (a+b x)+i \sin (a+b x))\right )-2 b^3 (c+d x)^3 \tanh ^{-1}(\cos (a+b x)+i \sin (a+b x))}{b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Csc[a + b*x],x]

[Out]

(-2*b^3*(c + d*x)^3*ArcTanh[Cos[a + b*x] + I*Sin[a + b*x]] + (3*I)*d*(b^2*(c + d*x)^2*PolyLog[2, -Cos[a + b*x]
- I*Sin[a + b*x]] + (2*I)*b*d*(c + d*x)*PolyLog[3, -Cos[a + b*x] - I*Sin[a + b*x]] - 2*d^2*PolyLog[4, -Cos[a
+ b*x] - I*Sin[a + b*x]]) - (3*I)*d*(b^2*(c + d*x)^2*PolyLog[2, Cos[a + b*x] + I*Sin[a + b*x]] + (2*I)*b*d*(c
+ d*x)*PolyLog[3, Cos[a + b*x] + I*Sin[a + b*x]] - 2*d^2*PolyLog[4, Cos[a + b*x] + I*Sin[a + b*x]]))/b^4

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Maple [B]  time = 0.095, size = 633, normalized size = 3.4 \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*csc(b*x+a),x)

[Out]

-6/b^3*c*d^2*polylog(3,-exp(I*(b*x+a)))+2/b^4*d^3*a^3*arctanh(exp(I*(b*x+a)))+6/b^3*c*d^2*polylog(3,exp(I*(b*x
+a)))+6/b^3*d^3*polylog(3,exp(I*(b*x+a)))*x-6/b^3*d^3*polylog(3,-exp(I*(b*x+a)))*x+3/b*c*d^2*ln(1-exp(I*(b*x+a
)))*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^2*d*ln(exp(I*(b*x+a))+1)*x-3
/b^3*c*d^2*a^2*ln(1-exp(I*(b*x+a)))-3/b^2*c^2*d*ln(exp(I*(b*x+a))+1)*a+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)))-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^2-1/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)-3/b*c*d^2*ln(exp(I*(b*x
+a))+1)*x^2-1/b^4*d^3*ln(exp(I*(b*x+a))+1)*a^3-6/b^3*c*d^2*a^2*arctanh(exp(I*(b*x+a)))+6/b^2*c^2*d*a*arctanh(e
xp(I*(b*x+a)))+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^3-2/b*c^3*arctanh(exp(I*(b*x+
a)))+6*I/b^2*c*d^2*polylog(2,-exp(I*(b*x+a)))*x-6*I/b^2*c*d^2*polylog(2,exp(I*(b*x+a)))*x+6*I*d^3*polylog(4,ex
p(I*(b*x+a)))/b^4-6*I*d^3*polylog(4,-exp(I*(b*x+a)))/b^4

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Maxima [B]  time = 1.5116, size = 953, normalized size = 5.15 \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*csc(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.1118, size = 2056, normalized size = 11.11 \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*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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**3*csc(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 )}^{3} \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*csc(b*x+a),x, algorithm="giac")

[Out]

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