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

Optimal. Leaf size=93 \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac{(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.166433, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3717, 2190, 2531, 2282, 6589} \[ -\frac{i d (c+d x) \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}+\frac{d^2 \text{PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}+\frac{(c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac{i (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Cot[a + b*x],x]

[Out]

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

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

Mathematica [B]  time = 1.42455, size = 356, normalized size = 3.83 \[ \frac{-3 i b c d \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+6 i b d^2 x \text{PolyLog}\left (2,-e^{-i (a+b x)}\right )+6 i b d^2 x \text{PolyLog}\left (2,e^{-i (a+b x)}\right )+6 d^2 \text{PolyLog}\left (3,-e^{-i (a+b x)}\right )+6 d^2 \text{PolyLog}\left (3,e^{-i (a+b x)}\right )+3 b^2 c^2 \log (\sin (a+b x))+3 b^3 c d x^2 \cot (a)-3 b^3 c d x^2 e^{i \tan ^{-1}(\tan (a))} \cot (a) \sqrt{\sec ^2(a)}-6 i b^2 c d x \tan ^{-1}(\tan (a))+6 b^2 c d x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+3 b^2 d^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 b^2 d^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+6 b c d \tan ^{-1}(\tan (a)) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-6 b c d \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )+3 i \pi b^2 c d x+i b^3 d^2 x^3+3 \pi b c d \log \left (1+e^{-2 i b x}\right )-3 \pi b c d \log (\cos (b x))}{3 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^2*Cot[a + b*x],x]

[Out]

((3*I)*b^2*c*d*Pi*x + I*b^3*d^2*x^3 - (6*I)*b^2*c*d*x*ArcTan[Tan[a]] + 3*b^3*c*d*x^2*Cot[a] + 3*b*c*d*Pi*Log[1
+ E^((-2*I)*b*x)] + 3*b^2*d^2*x^2*Log[1 - E^((-I)*(a + b*x))] + 3*b^2*d^2*x^2*Log[1 + E^((-I)*(a + b*x))] + 6
*b^2*c*d*x*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 6*b*c*d*ArcTan[Tan[a]]*Log[1 - E^((2*I)*(b*x + ArcTan[T
an[a]]))] - 3*b*c*d*Pi*Log[Cos[b*x]] + 3*b^2*c^2*Log[Sin[a + b*x]] - 6*b*c*d*ArcTan[Tan[a]]*Log[Sin[b*x + ArcT
an[Tan[a]]]] + (6*I)*b*d^2*x*PolyLog[2, -E^((-I)*(a + b*x))] + (6*I)*b*d^2*x*PolyLog[2, E^((-I)*(a + b*x))] -
(3*I)*b*c*d*PolyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] + 6*d^2*PolyLog[3, -E^((-I)*(a + b*x))] + 6*d^2*PolyL
og[3, E^((-I)*(a + b*x))] - 3*b^3*c*d*E^(I*ArcTan[Tan[a]])*x^2*Cot[a]*Sqrt[Sec[a]^2])/(3*b^3)

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Maple [B]  time = 0.27, size = 468, normalized size = 5. \begin{align*} -2\,{\frac{{a}^{2}{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{a}^{2}{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}-{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){a}^{2}}{{b}^{3}}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}+{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}+{\frac{{\frac{4\,i}{3}}{d}^{2}{a}^{3}}{{b}^{3}}}-icd{x}^{2}-{\frac{4\,icdax}{b}}+{\frac{2\,i{d}^{2}{a}^{2}x}{{b}^{2}}}-{\frac{2\,icd{\it polylog} \left ( 2,{{\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}}}-{\frac{2\,icd{a}^{2}}{{b}^{2}}}-{\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}}}+i{c}^{2}x+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}+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{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}}+4\,{\frac{cda\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{{d}^{2}{\it polylog} \left ( 3,-{{\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}}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{b}}-2\,{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{c}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{b}}-{\frac{i}{3}}{d}^{2}{x}^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.52594, size = 545, normalized size = 5.86 \begin{align*} \frac{6 \, c^{2} \log \left (\sin \left (b x + a\right )\right ) - \frac{12 \, a c d \log \left (\sin \left (b x + a\right )\right )}{b} + \frac{6 \, a^{2} d^{2} \log \left (\sin \left (b x + a\right )\right )}{b^{2}} + \frac{-2 i \,{\left (b x + a\right )}^{3} d^{2} +{\left (-6 i \, b c d + 6 i \, a d^{2}\right )}{\left (b x + a\right )}^{2} + 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 \,{\left (b x + a\right )}^{2} d^{2} +{\left (12 i \, b c d - 12 i \, a d^{2}\right )}{\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) +{\left (-6 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (-12 i \, b c d + 12 i \, a d^{2}\right )}{\left (b x + a\right )}\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) +{\left (-12 i \, b c d - 12 i \,{\left (b x + a\right )} d^{2} + 12 i \, a d^{2}\right )}{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) +{\left (-12 i \, b c d - 12 i \,{\left (b x + a\right )} d^{2} + 12 i \, a d^{2}\right )}{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 3 \,{\left ({\left (b x + a\right )}^{2} d^{2} + 2 \,{\left (b c d - a d^{2}\right )}{\left (b x + a\right )}\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 ({\left (b x + a\right )}^{2} d^{2} + 2 \,{\left (b c d - a d^{2}\right )}{\left (b x + a\right )}\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 )}{b^{2}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 0.566539, size = 1330, normalized size = 14.3 \begin{align*} \frac{2 \, d^{2}{\rm polylog}\left (3, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 \, d^{2}{\rm polylog}\left (3, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + 2 \, d^{2}{\rm polylog}\left (3, -\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 2 \, d^{2}{\rm polylog}\left (3, -\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) +{\left (-2 i \, b d^{2} x - 2 i \, b c d\right )}{\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) +{\left (2 i \, b d^{2} x + 2 i \, b c d\right )}{\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) +{\left (2 i \, b d^{2} x + 2 i \, b c d\right )}{\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) +{\left (-2 i \, b d^{2} x - 2 i \, b c d\right )}{\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) +{\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 ) +{\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 ) +{\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 ) +{\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 ) +{\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 ) +{\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 )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*cos(b*x+a)*csc(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*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(-co
s(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))/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 )} \csc{\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)*csc(b*x+a),x)

[Out]

Integral((c + d*x)**2*cos(a + b*x)*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 )}^{2} \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)^2*cos(b*x+a)*csc(b*x+a),x, algorithm="giac")

[Out]

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