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

Optimal. Leaf size=172 \[ -\frac {6 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac {8 d^2 \cos (a+b x)}{b^3}+\frac {6 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {6 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {8 d (c+d x) \sin (a+b x)}{b^2}+\frac {4 (c+d x)^2 \cos (a+b x)}{b}-\frac {6 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

-6*(d*x+c)^2*arctanh(exp(I*(b*x+a)))/b-8*d^2*cos(b*x+a)/b^3+4*(d*x+c)^2*cos(b*x+a)/b+6*I*d*(d*x+c)*polylog(2,-
exp(I*(b*x+a)))/b^2-6*I*d*(d*x+c)*polylog(2,exp(I*(b*x+a)))/b^2-6*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+6*d^2*pol
ylog(3,exp(I*(b*x+a)))/b^3-8*d*(d*x+c)*sin(b*x+a)/b^2

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Rubi [A]  time = 0.23, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4431, 4408, 3296, 2638, 4183, 2531, 2282, 6589} \[ \frac {6 i d (c+d x) \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {6 i d (c+d x) \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 \text {PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 \text {PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac {8 d (c+d x) \sin (a+b x)}{b^2}-\frac {8 d^2 \cos (a+b x)}{b^3}+\frac {4 (c+d x)^2 \cos (a+b x)}{b}-\frac {6 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(c + d*x)^2*ArcTanh[E^(I*(a + b*x))])/b - (8*d^2*Cos[a + b*x])/b^3 + (4*(c + d*x)^2*Cos[a + b*x])/b + ((6*
I)*d*(c + d*x)*PolyLog[2, -E^(I*(a + b*x))])/b^2 - ((6*I)*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))])/b^2 - (6*d^
2*PolyLog[3, -E^(I*(a + b*x))])/b^3 + (6*d^2*PolyLog[3, E^(I*(a + b*x))])/b^3 - (8*d*(c + d*x)*Sin[a + b*x])/b
^2

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 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 2638

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

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 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 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 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 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 \csc ^2(a+b x) \sin (3 a+3 b x) \, dx &=\int \left (3 (c+d x)^2 \cos (a+b x) \cot (a+b x)-(c+d x)^2 \sin (a+b x)\right ) \, dx\\ &=3 \int (c+d x)^2 \cos (a+b x) \cot (a+b x) \, dx-\int (c+d x)^2 \sin (a+b x) \, dx\\ &=\frac {(c+d x)^2 \cos (a+b x)}{b}+3 \int (c+d x)^2 \csc (a+b x) \, dx-3 \int (c+d x)^2 \sin (a+b x) \, dx-\frac {(2 d) \int (c+d x) \cos (a+b x) \, dx}{b}\\ &=-\frac {6 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {4 (c+d x)^2 \cos (a+b x)}{b}-\frac {2 d (c+d x) \sin (a+b x)}{b^2}-\frac {(6 d) \int (c+d x) \cos (a+b x) \, dx}{b}-\frac {(6 d) \int (c+d x) \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac {(6 d) \int (c+d x) \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}+\frac {\left (2 d^2\right ) \int \sin (a+b x) \, dx}{b^2}\\ &=-\frac {6 (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 {4 (c+d x)^2 \cos (a+b x)}{b}+\frac {6 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {6 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {8 d (c+d x) \sin (a+b x)}{b^2}-\frac {\left (6 i d^2\right ) \int \text {Li}_2\left (-e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (6 i d^2\right ) \int \text {Li}_2\left (e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (6 d^2\right ) \int \sin (a+b x) \, dx}{b^2}\\ &=-\frac {6 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {8 d^2 \cos (a+b x)}{b^3}+\frac {4 (c+d x)^2 \cos (a+b x)}{b}+\frac {6 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {6 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {8 d (c+d x) \sin (a+b x)}{b^2}-\frac {\left (6 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 (6 d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^3}\\ &=-\frac {6 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {8 d^2 \cos (a+b x)}{b^3}+\frac {4 (c+d x)^2 \cos (a+b x)}{b}+\frac {6 i d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {6 i d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {6 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )}{b^3}+\frac {6 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3}-\frac {8 d (c+d x) \sin (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 1.10, size = 223, normalized size = 1.30 \[ \frac {4 \cos (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )-2 b d \sin (a) (c+d x)\right )-4 \sin (b x) \left (\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )+2 b d \cos (a) (c+d x)\right )+3 b^2 (c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-3 b^2 (c+d x)^2 \log \left (1+e^{i (a+b x)}\right )+6 i b d (c+d x) \text {Li}_2\left (-e^{i (a+b x)}\right )-6 i b d (c+d x) \text {Li}_2\left (e^{i (a+b x)}\right )-6 d^2 \text {Li}_3\left (-e^{i (a+b x)}\right )+6 d^2 \text {Li}_3\left (e^{i (a+b x)}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*(c + d*x)^2*Log[1 - E^(I*(a + b*x))] - 3*b^2*(c + d*x)^2*Log[1 + E^(I*(a + b*x))] + (6*I)*b*d*(c + d*x)
*PolyLog[2, -E^(I*(a + b*x))] - (6*I)*b*d*(c + d*x)*PolyLog[2, E^(I*(a + b*x))] - 6*d^2*PolyLog[3, -E^(I*(a +
b*x))] + 6*d^2*PolyLog[3, E^(I*(a + b*x))] + 4*Cos[b*x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] - 2*b*d*(c + d*x)*S
in[a]) - 4*(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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fricas [C]  time = 0.54, size = 562, normalized size = 3.27 \[ \frac {6 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 6 \, d^{2} {\rm polylog}\left (3, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 6 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 6 \, d^{2} {\rm polylog}\left (3, -\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right ) + {\left (-6 i \, b d^{2} x - 6 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{2} x + 6 i \, b c d\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (-6 i \, b d^{2} x - 6 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + {\left (6 i \, b d^{2} x + 6 i \, b c d\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + 2 \, a b c d - a^{2} d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 16 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*d^2*polylog(3, cos(b*x + a) + I*sin(b*x + a)) + 6*d^2*polylog(3, cos(b*x + a) - I*sin(b*x + a)) - 6*d^2
*polylog(3, -cos(b*x + a) + I*sin(b*x + a)) - 6*d^2*polylog(3, -cos(b*x + a) - I*sin(b*x + a)) + 8*(b^2*d^2*x^
2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a) + (-6*I*b*d^2*x - 6*I*b*c*d)*dilog(cos(b*x + a) + I*sin(b*x +
a)) + (6*I*b*d^2*x + 6*I*b*c*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (-6*I*b*d^2*x - 6*I*b*c*d)*dilog(-cos(b
*x + a) + I*sin(b*x + a)) + (6*I*b*d^2*x + 6*I*b*c*d)*dilog(-cos(b*x + a) - I*sin(b*x + a)) - 3*(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) - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(c
os(b*x + a) - I*sin(b*x + a) + 1) + 3*(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) + 3*(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) + 3*(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) + 3*(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) - 16*(b*d^2*x + b*c*d)*sin(b*x + a))/b^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \csc \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.20, size = 481, normalized size = 2.80 \[ \frac {2 \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b^{3}}+\frac {2 \left (d^{2} x^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{b^{3}}+\frac {12 c d a \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 i c d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {6 i c d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {6 c^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {6 d^{2} a^{2} \arctanh \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a^{2}}{b^{3}}+\frac {6 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {3 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {3 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}-\frac {6 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {6 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {6 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {6 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {6 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}-\frac {6 d^{2} \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 d^{2} \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*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))+12/b^2*c*d*a*arctanh(exp(I*(b*x+a)))-6*I/b^2*c*d*poly
log(2,exp(I*(b*x+a)))-6*I/b^2*polylog(2,exp(I*(b*x+a)))*d^2*x-6/b*c^2*arctanh(exp(I*(b*x+a)))-6/b^3*d^2*a^2*ar
ctanh(exp(I*(b*x+a)))-3/b*d^2*ln(exp(I*(b*x+a))+1)*x^2+3/b^3*d^2*ln(exp(I*(b*x+a))+1)*a^2+6*I/b^2*c*d*polylog(
2,-exp(I*(b*x+a)))+3/b*d^2*ln(1-exp(I*(b*x+a)))*x^2-3/b^3*d^2*ln(1-exp(I*(b*x+a)))*a^2+6*I/b^2*d^2*polylog(2,-
exp(I*(b*x+a)))*x+6/b*c*d*ln(1-exp(I*(b*x+a)))*x+6/b^2*c*d*ln(1-exp(I*(b*x+a)))*a-6/b*c*d*ln(exp(I*(b*x+a))+1)
*x-6/b^2*c*d*ln(exp(I*(b*x+a))+1)*a-6*d^2*polylog(3,-exp(I*(b*x+a)))/b^3+6*d^2*polylog(3,exp(I*(b*x+a)))/b^3

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maxima [B]  time = 0.49, size = 409, normalized size = 2.38 \[ \frac {c^{2} {\left (8 \, \cos \left (b x + a\right ) - 3 \, \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) + 3 \, \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right )\right )}}{2 \, b} - \frac {12 \, d^{2} {\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 12 \, d^{2} {\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) + {\left (6 i \, b^{2} d^{2} x^{2} + 12 i \, b^{2} c d x\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + {\left (6 i \, b^{2} d^{2} x^{2} + 12 i \, b^{2} c d x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x - 2 \, d^{2}\right )} \cos \left (b x + a\right ) + {\left (-12 i \, b d^{2} x - 12 i \, b c d\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + {\left (12 i \, b d^{2} x + 12 i \, b c d\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - 3 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) + 16 \, {\left (b d^{2} x + b c d\right )} \sin \left (b x + a\right )}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^2*(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*(12*d^2*polylog(3, -e^(I*b*x + I*a)) - 12*d^2*polylog(3, e^(I*b*x + I*a)) + (6*I*b^2*d^2*x^2 + 12*I*b^2*c*d
*x)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*I*b^2*d^2*x^2 + 12*I*b^2*c*d*x)*arctan2(sin(b*x + a), -cos(b*
x + a) + 1) - 8*(b^2*d^2*x^2 + 2*b^2*c*d*x - 2*d^2)*cos(b*x + a) + (-12*I*b*d^2*x - 12*I*b*c*d)*dilog(-e^(I*b*
x + I*a)) + (12*I*b*d^2*x + 12*I*b*c*d)*dilog(e^(I*b*x + I*a)) + 3*(b^2*d^2*x^2 + 2*b^2*c*d*x)*log(cos(b*x + a
)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x)*log(cos(b*x + a)^2 + sin(b*x + a)^2
- 2*cos(b*x + a) + 1) + 16*(b*d^2*x + b*c*d)*sin(b*x + a))/b^3

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(3*a + 3*b*x)*(c + d*x)^2)/sin(a + b*x)^2,x)

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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