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

Optimal. Leaf size=90 \[ \frac{2 i d^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac{4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \csc (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0622306, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4410, 4183, 2279, 2391} \[ \frac{2 i d^2 \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}-\frac{2 i d^2 \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac{4 d (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4410

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp
[((c + d*x)^m*Csc[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Fr
eeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [B]  time = 2.0513, size = 234, normalized size = 2.6 \[ \frac{4 d^2 \left (2 \tan ^{-1}(\tan (a)) \tanh ^{-1}\left (\cos (a)-\sin (a) \tan \left (\frac{b x}{2}\right )\right )+\frac{\sec (a) \left (i \text{PolyLog}\left (2,-e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-i \text{PolyLog}\left (2,e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+\left (\tan ^{-1}(\tan (a))+b x\right ) \left (\log \left (1-e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-\log \left (1+e^{i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )\right )\right )}{\sqrt{\sec ^2(a)}}\right )-2 b^2 \csc (a) (c+d x)^2+b^2 \csc \left (\frac{a}{2}\right ) \sin \left (\frac{b x}{2}\right ) (c+d x)^2 \csc \left (\frac{1}{2} (a+b x)\right )-b^2 \sec \left (\frac{a}{2}\right ) \sin \left (\frac{b x}{2}\right ) (c+d x)^2 \sec \left (\frac{1}{2} (a+b x)\right )-8 b c d \tanh ^{-1}\left (\cos (a)-\sin (a) \tan \left (\frac{b x}{2}\right )\right )}{2 b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*b*c*d*ArcTanh[Cos[a] - Sin[a]*Tan[(b*x)/2]] - 2*b^2*(c + d*x)^2*Csc[a] + 4*d^2*(2*ArcTan[Tan[a]]*ArcTanh[C
os[a] - Sin[a]*Tan[(b*x)/2]] + (((b*x + ArcTan[Tan[a]])*(Log[1 - E^(I*(b*x + ArcTan[Tan[a]]))] - Log[1 + E^(I*
(b*x + ArcTan[Tan[a]]))]) + I*PolyLog[2, -E^(I*(b*x + ArcTan[Tan[a]]))] - I*PolyLog[2, E^(I*(b*x + ArcTan[Tan[
a]]))])*Sec[a])/Sqrt[Sec[a]^2]) + b^2*(c + d*x)^2*Csc[a/2]*Csc[(a + b*x)/2]*Sin[(b*x)/2] - b^2*(c + d*x)^2*Sec
[a/2]*Sec[(a + b*x)/2]*Sin[(b*x)/2])/(2*b^3)

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Maple [B]  time = 0.205, size = 212, normalized size = 2.4 \begin{align*}{\frac{-2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ){{\rm e}^{i \left ( bx+a \right ) }}}{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) }}-4\,{\frac{cd{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{2}}}+2\,{\frac{{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{3}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{2}}}-2\,{\frac{{d}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) a}{{b}^{3}}}+{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+4\,{\frac{{d}^{2}a{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{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)^2,x)

[Out]

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

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Maxima [B]  time = 1.63384, size = 751, normalized size = 8.34 \begin{align*} \frac{{\left (2 \, b d^{2} x + 2 \, b c d - 2 \,{\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) -{\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) +{\left (2 \, b c d \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b c d \sin \left (2 \, b x + 2 \, a\right ) - 2 \, b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) -{\left (2 \, b d^{2} x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b d^{2} x \sin \left (2 \, b x + 2 \, a\right ) - 2 \, b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (b x + a\right ) + 2 \,{\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )}{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 2 \,{\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )}{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) -{\left (i \, b d^{2} x + i \, b c d +{\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, 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 (-i \, b d^{2} x - i \, b c d +{\left (i \, b d^{2} x + i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) -{\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, 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 (2 i \, b^{2} d^{2} x^{2} + 4 i \, b^{2} c d x + 2 i \, b^{2} c^{2}\right )} \sin \left (b x + a\right )}{-i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + b^{3} \sin \left (2 \, b x + 2 \, a\right ) + i \, 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)^2,x, algorithm="maxima")

[Out]

((2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(2*b*x + 2*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(2*b*x + 2*a))*arc
tan2(sin(b*x + a), cos(b*x + a) + 1) + (2*b*c*d*cos(2*b*x + 2*a) + 2*I*b*c*d*sin(2*b*x + 2*a) - 2*b*c*d)*arcta
n2(sin(b*x + a), cos(b*x + a) - 1) - (2*b*d^2*x*cos(2*b*x + 2*a) + 2*I*b*d^2*x*sin(2*b*x + 2*a) - 2*b*d^2*x)*a
rctan2(sin(b*x + a), -cos(b*x + a) + 1) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(b*x + a) + 2*(d^2*cos(2*
b*x + 2*a) + I*d^2*sin(2*b*x + 2*a) - d^2)*dilog(-e^(I*b*x + I*a)) - 2*(d^2*cos(2*b*x + 2*a) + I*d^2*sin(2*b*x
+ 2*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(2*b*x + 2*a) + (b*d^
2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (-I*b*d^2*x - I*b*c
*d + (I*b*d^2*x + I*b*c*d)*cos(2*b*x + 2*a) - (b*d^2*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x
+ a)^2 - 2*cos(b*x + a) + 1) - (2*I*b^2*d^2*x^2 + 4*I*b^2*c*d*x + 2*I*b^2*c^2)*sin(b*x + a))/(-I*b^3*cos(2*b*
x + 2*a) + b^3*sin(2*b*x + 2*a) + I*b^3)

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Fricas [B]  time = 0.558857, size = 1006, normalized size = 11.18 \begin{align*} -\frac{b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2}{\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2}{\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d^{2}{\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2}{\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) +{\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) +{\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) -{\left (b c d - a 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 ) \sin \left (b x + a\right ) -{\left (b c d - a 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 ) \sin \left (b x + a\right ) -{\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) -{\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right )}{b^{3} \sin \left (b x + a\right )} \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)^2,x, algorithm="fricas")

[Out]

-(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + I*d^2*dilog(cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2*dilog(
cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + I*d^2*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2
*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) + I*sin(b*x + a) + 1)
*sin(b*x + a) + (b*d^2*x + b*c*d)*log(cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) - (b*c*d - a*d^2)*log(-1
/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) - (b*c*d - a*d^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b
*x + a) + 1/2)*sin(b*x + a) - (b*d^2*x + a*d^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1)*sin(b*x + a) - (b*d^2*
x + a*d^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a))/(b^3*sin(b*x + a))

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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)**2*cos(b*x+a)*csc(b*x+a)**2,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 )}^{2} \cos \left (b x + a\right ) \csc \left (b x + a\right )^{2}\,{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)^2,x, algorithm="giac")

[Out]

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