3.376 \(\int (c+d x)^3 \csc ^2(a+b x) \sin (3 a+3 b x) \, dx\)

Optimal. Leaf size=255 \[ -\frac{18 d^2 (c+d x) \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{18 d^2 (c+d x) \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{9 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{18 i d^3 \text{PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac{18 i d^3 \text{PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}-\frac{24 d^2 (c+d x) \cos (a+b x)}{b^3}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}+\frac{24 d^3 \sin (a+b x)}{b^4}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

(-6*(c + d*x)^3*ArcTanh[E^(I*(a + b*x))])/b - (24*d^2*(c + d*x)*Cos[a + b*x])/b^3 + (4*(c + d*x)^3*Cos[a + b*x
])/b + ((9*I)*d*(c + d*x)^2*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((9*I)*d*(c + d*x)^2*PolyLog[2, E^(I*(a + b*x)
)])/b^2 - (18*d^2*(c + d*x)*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (18*d^2*(c + d*x)*PolyLog[3, E^(I*(a + b*x))])
/b^3 - ((18*I)*d^3*PolyLog[4, -E^(I*(a + b*x))])/b^4 + ((18*I)*d^3*PolyLog[4, E^(I*(a + b*x))])/b^4 + (24*d^3*
Sin[a + b*x])/b^4 - (12*d*(c + d*x)^2*Sin[a + b*x])/b^2

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Rubi [A]  time = 0.34706, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {4431, 4408, 3296, 2637, 4183, 2531, 6609, 2282, 6589} \[ -\frac{18 d^2 (c+d x) \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{18 d^2 (c+d x) \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}+\frac{9 i d (c+d x)^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{18 i d^3 \text{PolyLog}\left (4,-e^{i (a+b x)}\right )}{b^4}+\frac{18 i d^3 \text{PolyLog}\left (4,e^{i (a+b x)}\right )}{b^4}-\frac{24 d^2 (c+d x) \cos (a+b x)}{b^3}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}+\frac{24 d^3 \sin (a+b x)}{b^4}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csc[a + b*x]^2*Sin[3*a + 3*b*x],x]

[Out]

(-6*(c + d*x)^3*ArcTanh[E^(I*(a + b*x))])/b - (24*d^2*(c + d*x)*Cos[a + b*x])/b^3 + (4*(c + d*x)^3*Cos[a + b*x
])/b + ((9*I)*d*(c + d*x)^2*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((9*I)*d*(c + d*x)^2*PolyLog[2, E^(I*(a + b*x)
)])/b^2 - (18*d^2*(c + d*x)*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (18*d^2*(c + d*x)*PolyLog[3, E^(I*(a + b*x))])
/b^3 - ((18*I)*d^3*PolyLog[4, -E^(I*(a + b*x))])/b^4 + ((18*I)*d^3*PolyLog[4, E^(I*(a + b*x))])/b^4 + (24*d^3*
Sin[a + b*x])/b^4 - (12*d*(c + d*x)^2*Sin[a + b*x])/b^2

Rule 4431

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int
[ExpandTrigExpand[(e + f*x)^m*G[c + d*x]^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 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 ^2(a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^3 \cos (a+b x) \cot (a+b x)-(c+d x)^3 \sin (a+b x)\right ) \, dx\\ &=3 \int (c+d x)^3 \cos (a+b x) \cot (a+b x) \, dx-\int (c+d x)^3 \sin (a+b x) \, dx\\ &=\frac{(c+d x)^3 \cos (a+b x)}{b}+3 \int (c+d x)^3 \csc (a+b x) \, dx-3 \int (c+d x)^3 \sin (a+b x) \, dx-\frac{(3 d) \int (c+d x)^2 \cos (a+b x) \, dx}{b}\\ &=-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}-\frac{3 d (c+d x)^2 \sin (a+b x)}{b^2}-\frac{(9 d) \int (c+d x)^2 \cos (a+b x) \, dx}{b}-\frac{(9 d) \int (c+d x)^2 \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac{(9 d) \int (c+d x)^2 \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}+\frac{\left (6 d^2\right ) \int (c+d x) \sin (a+b x) \, dx}{b^2}\\ &=-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{6 d^2 (c+d x) \cos (a+b x)}{b^3}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}+\frac{9 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}-\frac{\left (18 i d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (18 i d^2\right ) \int (c+d x) \text{Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}+\frac{\left (18 d^2\right ) \int (c+d x) \sin (a+b x) \, dx}{b^2}+\frac{\left (6 d^3\right ) \int \cos (a+b x) \, dx}{b^3}\\ &=-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{24 d^2 (c+d x) \cos (a+b x)}{b^3}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}+\frac{9 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{18 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{18 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{6 d^3 \sin (a+b x)}{b^4}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}+\frac{\left (18 d^3\right ) \int \cos (a+b x) \, dx}{b^3}+\frac{\left (18 d^3\right ) \int \text{Li}_3\left (-e^{i (a+b x)}\right ) \, dx}{b^3}-\frac{\left (18 d^3\right ) \int \text{Li}_3\left (e^{i (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{24 d^2 (c+d x) \cos (a+b x)}{b^3}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}+\frac{9 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{18 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{18 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}+\frac{24 d^3 \sin (a+b x)}{b^4}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}-\frac{\left (18 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 (18 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{6 (c+d x)^3 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{24 d^2 (c+d x) \cos (a+b x)}{b^3}+\frac{4 (c+d x)^3 \cos (a+b x)}{b}+\frac{9 i d (c+d x)^2 \text{Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac{9 i d (c+d x)^2 \text{Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac{18 d^2 (c+d x) \text{Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac{18 d^2 (c+d x) \text{Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac{18 i d^3 \text{Li}_4\left (-e^{i (a+b x)}\right )}{b^4}+\frac{18 i d^3 \text{Li}_4\left (e^{i (a+b x)}\right )}{b^4}+\frac{24 d^3 \sin (a+b x)}{b^4}-\frac{12 d (c+d x)^2 \sin (a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 1.57214, size = 459, normalized size = 1.8 \[ \frac{3 \left (3 i d \left (b^2 (c+d x)^2 \text{PolyLog}(2,-\cos (a+b x)-i \sin (a+b x))+2 i b d (c+d x) \text{PolyLog}(3,-\cos (a+b x)-i \sin (a+b x))-2 d^2 \text{PolyLog}(4,-\cos (a+b x)-i \sin (a+b x))\right )-3 i d \left (b^2 (c+d x)^2 \text{PolyLog}(2,\cos (a+b x)+i \sin (a+b x))+2 i b d (c+d x) \text{PolyLog}(3,\cos (a+b x)+i \sin (a+b x))-2 d^2 \text{PolyLog}(4,\cos (a+b x)+i \sin (a+b x))\right )-2 b^3 (c+d x)^3 \tanh ^{-1}(\cos (a+b x)+i \sin (a+b x))\right )}{b^4}+\frac{4 \cos (b x) \left (3 b^3 c^2 d x \cos (a)-3 b^2 c^2 d \sin (a)+b^3 c^3 \cos (a)+3 b^3 c d^2 x^2 \cos (a)-6 b^2 c d^2 x \sin (a)-3 b^2 d^3 x^2 \sin (a)+b^3 d^3 x^3 \cos (a)-6 b c d^2 \cos (a)-6 b d^3 x \cos (a)+6 d^3 \sin (a)\right )}{b^4}-\frac{4 \sin (b x) \left (3 b^3 c^2 d x \sin (a)+3 b^2 c^2 d \cos (a)+b^3 c^3 \sin (a)+3 b^3 c d^2 x^2 \sin (a)+6 b^2 c d^2 x \cos (a)+b^3 d^3 x^3 \sin (a)+3 b^2 d^3 x^2 \cos (a)-6 b c d^2 \sin (a)-6 b d^3 x \sin (a)-6 d^3 \cos (a)\right )}{b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Csc[a + b*x]^2*Sin[3*a + 3*b*x],x]

[Out]

(3*(-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 +
(4*Cos[b*x]*(b^3*c^3*Cos[a] - 6*b*c*d^2*Cos[a] + 3*b^3*c^2*d*x*Cos[a] - 6*b*d^3*x*Cos[a] + 3*b^3*c*d^2*x^2*Co
s[a] + b^3*d^3*x^3*Cos[a] - 3*b^2*c^2*d*Sin[a] + 6*d^3*Sin[a] - 6*b^2*c*d^2*x*Sin[a] - 3*b^2*d^3*x^2*Sin[a]))/
b^4 - (4*(3*b^2*c^2*d*Cos[a] - 6*d^3*Cos[a] + 6*b^2*c*d^2*x*Cos[a] + 3*b^2*d^3*x^2*Cos[a] + b^3*c^3*Sin[a] - 6
*b*c*d^2*Sin[a] + 3*b^3*c^2*d*x*Sin[a] - 6*b*d^3*x*Sin[a] + 3*b^3*c*d^2*x^2*Sin[a] + b^3*d^3*x^3*Sin[a])*Sin[b
*x])/b^4

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Maple [B]  time = 0.196, size = 849, normalized size = 3.3 \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)^2*sin(3*b*x+3*a),x)

[Out]

2*(d^3*x^3*b^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3-3*I*b^2*d^3*x^2-6*b*d^3*x-6*I*b^2*c*d^2*x-6*c*d^2*b-3*I*b
^2*c^2*d+6*I*d^3)/b^4*exp(-I*(b*x+a))-9/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2+9/b*c*d^2*ln(1-exp(I*(b*x+a)))*x^2+18
*I*d^3*polylog(4,exp(I*(b*x+a)))/b^4-9*I/b^2*d^3*polylog(2,exp(I*(b*x+a)))*x^2+9*I/b^2*d^3*polylog(2,-exp(I*(b
*x+a)))*x^2-9*I/b^2*c^2*d*polylog(2,exp(I*(b*x+a)))+6/b^4*d^3*a^3*arctanh(exp(I*(b*x+a)))-3/b^4*d^3*ln(exp(I*(
b*x+a))+1)*a^3+9/b^3*c*d^2*a^2*ln(exp(I*(b*x+a))+1)+9*I/b^2*c^2*d*polylog(2,-exp(I*(b*x+a)))+2*(d^3*x^3*b^3+3*
b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3+3*I*b^2*d^3*x^2-6*b*d^3*x+6*I*b^2*c*d^2*x-6*c*d^2*b+3*I*b^2*c^2*d-6*I*d^3)
/b^4*exp(I*(b*x+a))+18/b^2*c^2*d*a*arctanh(exp(I*(b*x+a)))-18/b^3*c*d^2*a^2*arctanh(exp(I*(b*x+a)))-9/b^2*c^2*
d*ln(exp(I*(b*x+a))+1)*a-18*I/b^2*c*d^2*polylog(2,exp(I*(b*x+a)))*x+18*I/b^2*c*d^2*polylog(2,-exp(I*(b*x+a)))*
x+3/b*d^3*ln(1-exp(I*(b*x+a)))*x^3+3/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^3-3/b*d^3*ln(exp(I*(b*x+a))+1)*x^3-9/b^3*c
*d^2*a^2*ln(1-exp(I*(b*x+a)))-9/b*c^2*d*ln(exp(I*(b*x+a))+1)*x+9/b*c^2*d*ln(1-exp(I*(b*x+a)))*x+9/b^2*c^2*d*ln
(1-exp(I*(b*x+a)))*a-6/b*c^3*arctanh(exp(I*(b*x+a)))-18*I*d^3*polylog(4,-exp(I*(b*x+a)))/b^4+18/b^3*d^3*polylo
g(3,exp(I*(b*x+a)))*x-18/b^3*d^3*polylog(3,-exp(I*(b*x+a)))*x+18/b^3*c*d^2*polylog(3,exp(I*(b*x+a)))-18/b^3*c*
d^2*polylog(3,-exp(I*(b*x+a)))

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Maxima [B]  time = 1.80947, size = 813, normalized size = 3.19 \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*csc(b*x+a)^2*sin(3*b*x+3*a),x, algorithm="maxima")

[Out]

1/2*c^3*(8*cos(b*x + a) - 3*log(cos(b*x)^2 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + s
in(a)^2) + 3*log(cos(b*x)^2 - 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2))/b - 1
/2*(36*I*d^3*polylog(4, -e^(I*b*x + I*a)) - 36*I*d^3*polylog(4, e^(I*b*x + I*a)) + (6*I*b^3*d^3*x^3 + 18*I*b^3
*c*d^2*x^2 + 18*I*b^3*c^2*d*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*I*b^3*d^3*x^3 + 18*I*b^3*c*d^2*x^2
+ 18*I*b^3*c^2*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 8*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 - 6*b*c*d^2 +
3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a) + (-18*I*b^2*d^3*x^2 - 36*I*b^2*c*d^2*x - 18*I*b^2*c^2*d)*dilog(-e^(I
*b*x + I*a)) + (18*I*b^2*d^3*x^2 + 36*I*b^2*c*d^2*x + 18*I*b^2*c^2*d)*dilog(e^(I*b*x + I*a)) + 3*(b^3*d^3*x^3
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - 3*(b^3*d^3*x^3
+ 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 36*(b*d^3*x + b
*c*d^2)*polylog(3, -e^(I*b*x + I*a)) - 36*(b*d^3*x + b*c*d^2)*polylog(3, e^(I*b*x + I*a)) + 24*(b^2*d^3*x^2 +
2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*sin(b*x + a))/b^4

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Fricas [C]  time = 0.75591, size = 2314, normalized size = 9.07 \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)^2*sin(3*b*x+3*a),x, algorithm="fricas")

[Out]

1/2*(18*I*d^3*polylog(4, cos(b*x + a) + I*sin(b*x + a)) - 18*I*d^3*polylog(4, cos(b*x + a) - I*sin(b*x + a)) +
18*I*d^3*polylog(4, -cos(b*x + a) + I*sin(b*x + a)) - 18*I*d^3*polylog(4, -cos(b*x + a) - I*sin(b*x + a)) + 8
*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a) + (-9*I*b^2*d^
3*x^2 - 18*I*b^2*c*d^2*x - 9*I*b^2*c^2*d)*dilog(cos(b*x + a) + I*sin(b*x + a)) + (9*I*b^2*d^3*x^2 + 18*I*b^2*c
*d^2*x + 9*I*b^2*c^2*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-9*I*b^2*d^3*x^2 - 18*I*b^2*c*d^2*x - 9*I*b^2*
c^2*d)*dilog(-cos(b*x + a) + I*sin(b*x + a)) + (9*I*b^2*d^3*x^2 + 18*I*b^2*c*d^2*x + 9*I*b^2*c^2*d)*dilog(-cos
(b*x + a) - I*sin(b*x + a)) - 3*(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) - 3*(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) + 3*(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) + 3*(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
) + 3*(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) + 3*(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) + 18*(b*d^3*x + b*c*d^2)*polylog(3, cos(b*x + a) + I*sin(b*x
+ a)) + 18*(b*d^3*x + b*c*d^2)*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 18*(b*d^3*x + b*c*d^2)*polylog(3,
-cos(b*x + a) + I*sin(b*x + a)) - 18*(b*d^3*x + b*c*d^2)*polylog(3, -cos(b*x + a) - I*sin(b*x + a)) - 24*(b^2*
d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*sin(b*x + a))/b^4

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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*csc(b*x+a)**2*sin(3*b*x+3*a),x)

[Out]

Timed out

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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 )^{2} \sin \left (3 \, b x + 3 \, 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)^2*sin(3*b*x+3*a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*csc(b*x + a)^2*sin(3*b*x + 3*a), x)