3.24 \(\int (c+d x)^2 \csc (a+b x) \, dx\)

Optimal. Leaf size=123 \[ \frac{2 i d (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0882635, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4183, 2531, 2282, 6589} \[ \frac{2 i d (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{2 i d (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{2 d^2 \text{PolyLog}\left (3,-e^{i (a+b x)}\right )}{b^3}+\frac{2 d^2 \text{PolyLog}\left (3,e^{i (a+b x)}\right )}{b^3}-\frac{2 (c+d x)^2 \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csc[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.319925, size = 148, normalized size = 1.2 \[ \frac{\frac{2 i d \left (b (c+d x) \text{PolyLog}\left (2,-e^{i (a+b x)}\right )+i d \text{PolyLog}\left (3,-e^{i (a+b x)}\right )\right )}{b^2}+\frac{2 d \left (d \text{PolyLog}\left (3,e^{i (a+b x)}\right )-i b (c+d x) \text{PolyLog}\left (2,e^{i (a+b x)}\right )\right )}{b^2}+(c+d x)^2 \log \left (1-e^{i (a+b x)}\right )-(c+d x)^2 \log \left (1+e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Csc[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.048, size = 361, normalized size = 2.9 \begin{align*} 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}}}-2\,{\frac{{c}^{2}{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-2\,{\frac{{a}^{2}{d}^{2}{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-{\frac{2\,icd{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+4\,{\frac{cda{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){x}^{2}}{b}}-{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ){a}^{2}}{{b}^{3}}}+{\frac{2\,icd{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){x}^{2}}{b}}+{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ){a}^{2}}{{b}^{3}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}-2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) a}{{b}^{2}}}+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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.32668, size = 529, normalized size = 4.3 \begin{align*} -\frac{2 \, c^{2} \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right ) - \frac{4 \, a c d \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right )}{b} + \frac{2 \, a^{2} d^{2} \log \left (\cot \left (b x + a\right ) + \csc \left (b x + a\right )\right )}{b^{2}} + \frac{4 \, d^{2}{\rm Li}_{3}(-e^{\left (i \, b x + i \, a\right )}) - 4 \, d^{2}{\rm Li}_{3}(e^{\left (i \, b x + i \, a\right )}) +{\left (2 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (4 i \, b c d - 4 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 (2 i \,{\left (b x + a\right )}^{2} d^{2} +{\left (4 i \, b c d - 4 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 (-4 i \, b c d - 4 i \,{\left (b x + a\right )} d^{2} + 4 i \, a d^{2}\right )}{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) +{\left (4 i \, b c d + 4 i \,{\left (b x + a\right )} d^{2} - 4 i \, a d^{2}\right )}{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) +{\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 ) -{\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}}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.9378, size = 1330, normalized size = 10.81 \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*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} \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*csc(b*x+a),x)

[Out]

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

[Out]

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