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

Optimal. Leaf size=108 \[ -\frac{i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{(c+d x) \cot (a+b x) \csc (a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.107197, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4415, 4183, 2279, 2391, 4185} \[ -\frac{i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{2 b^2}+\frac{i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{2 b^2}-\frac{d \csc (a+b x)}{2 b^2}+\frac{(c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac{(c+d x) \cot (a+b x) \csc (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4415

Int[Cot[(a_.) + (b_.)*(x_)]^(p_)*Csc[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[(c + d*
x)^m*Csc[a + b*x]*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Csc[a + b*x]^3*Cot[a + b*x]^(p - 2), x] /; FreeQ[
{a, b, c, d, m}, x] && IGtQ[p/2, 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]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

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

Mathematica [B]  time = 1.83628, size = 260, normalized size = 2.41 \[ -\frac{d \left (i \left (\text{PolyLog}\left (2,-e^{i (a+b x)}\right )-\text{PolyLog}\left (2,e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )\right )}{2 b^2}-\frac{d \tan \left (\frac{1}{2} (a+b x)\right )}{4 b^2}-\frac{d \cot \left (\frac{1}{2} (a+b x)\right )}{4 b^2}+\frac{a d \log \left (\tan \left (\frac{1}{2} (a+b x)\right )\right )}{2 b^2}-\frac{c \csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{c \sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{c \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}+\frac{c \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}-\frac{d x \csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{d x \sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.139, size = 246, normalized size = 2.3 \begin{align*}{\frac{dxb{{\rm e}^{3\,i \left ( bx+a \right ) }}+bc{{\rm e}^{3\,i \left ( bx+a \right ) }}+dxb{{\rm e}^{i \left ( bx+a \right ) }}+bc{{\rm e}^{i \left ( bx+a \right ) }}-id{{\rm e}^{3\,i \left ( bx+a \right ) }}+id{{\rm e}^{i \left ( bx+a \right ) }}}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}}}+{\frac{c{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{2\,b}}-{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{2\,{b}^{2}}}+{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{2\,b}}+{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) a}{2\,{b}^{2}}}-{\frac{{\frac{i}{2}}d{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{ad{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.90929, size = 1037, normalized size = 9.6 \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)*cot(b*x+a)^2*csc(b*x+a),x, algorithm="maxima")

[Out]

((2*b*d*x + 2*b*c + 2*(b*d*x + b*c)*cos(4*b*x + 4*a) - 4*(b*d*x + b*c)*cos(2*b*x + 2*a) + (2*I*b*d*x + 2*I*b*c
)*sin(4*b*x + 4*a) + (-4*I*b*d*x - 4*I*b*c)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b*c
*cos(4*b*x + 4*a) - 4*b*c*cos(2*b*x + 2*a) + 2*I*b*c*sin(4*b*x + 4*a) - 4*I*b*c*sin(2*b*x + 2*a) + 2*b*c)*arct
an2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d*x*cos(4*b*x + 4*a) - 4*b*d*x*cos(2*b*x + 2*a) + 2*I*b*d*x*sin(4*b
*x + 4*a) - 4*I*b*d*x*sin(2*b*x + 2*a) + 2*b*d*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + (-4*I*b*d*x - 4*I
*b*c - 4*d)*cos(3*b*x + 3*a) + (-4*I*b*d*x - 4*I*b*c + 4*d)*cos(b*x + a) - (2*d*cos(4*b*x + 4*a) - 4*d*cos(2*b
*x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(-e^(I*b*x + I*a)) + (2*d*cos(4*b*x +
4*a) - 4*d*cos(2*b*x + 2*a) + 2*I*d*sin(4*b*x + 4*a) - 4*I*d*sin(2*b*x + 2*a) + 2*d)*dilog(e^(I*b*x + I*a)) +
(-I*b*d*x - I*b*c + (-I*b*d*x - I*b*c)*cos(4*b*x + 4*a) + (2*I*b*d*x + 2*I*b*c)*cos(2*b*x + 2*a) + (b*d*x + b*
c)*sin(4*b*x + 4*a) - 2*(b*d*x + b*c)*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*x + I*b*c + (I*b*d*x + I*b*c)*cos(4*b*x + 4*a) + (-2*I*b*d*x - 2*I*b*c)*cos(2*b*x + 2*a) - (b*d*x
+ b*c)*sin(4*b*x + 4*a) + 2*(b*d*x + b*c)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x +
a) + 1) + 4*(b*d*x + b*c - I*d)*sin(3*b*x + 3*a) + 4*(b*d*x + b*c + I*d)*sin(b*x + a))/(-4*I*b^2*cos(4*b*x +
4*a) + 8*I*b^2*cos(2*b*x + 2*a) + 4*b^2*sin(4*b*x + 4*a) - 8*b^2*sin(2*b*x + 2*a) - 4*I*b^2)

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Fricas [B]  time = 0.576427, size = 1191, normalized size = 11.03 \begin{align*} \frac{2 \,{\left (b d x + b c\right )} \cos \left (b x + a\right ) +{\left (i \, d \cos \left (b x + a\right )^{2} - i \, d\right )}{\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) +{\left (-i \, d \cos \left (b x + a\right )^{2} + i \, d\right )}{\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) +{\left (i \, d \cos \left (b x + a\right )^{2} - i \, d\right )}{\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) +{\left (-i \, d \cos \left (b x + a\right )^{2} + i \, d\right )}{\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) -{\left (b d x -{\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) -{\left (b d x -{\left (b d x + b c\right )} \cos \left (b x + a\right )^{2} + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) -{\left ({\left (b c - a d\right )} \cos \left (b x + a\right )^{2} - b c + a d\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 ({\left (b c - a d\right )} \cos \left (b x + a\right )^{2} - b c + a d\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 d x -{\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) +{\left (b d x -{\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + 2 \, d \sin \left (b x + a\right )}{4 \,{\left (b^{2} \cos \left (b x + a\right )^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(b*d*x + b*c)*cos(b*x + a) + (I*d*cos(b*x + a)^2 - I*d)*dilog(cos(b*x + a) + I*sin(b*x + a)) + (-I*d*co
s(b*x + a)^2 + I*d)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (I*d*cos(b*x + a)^2 - I*d)*dilog(-cos(b*x + a) + I*
sin(b*x + a)) + (-I*d*cos(b*x + a)^2 + I*d)*dilog(-cos(b*x + a) - I*sin(b*x + a)) - (b*d*x - (b*d*x + b*c)*cos
(b*x + a)^2 + b*c)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - (b*d*x - (b*d*x + b*c)*cos(b*x + a)^2 + b*c)*log(c
os(b*x + a) - I*sin(b*x + a) + 1) - ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos(b*x + a) + 1/2*I*sin
(b*x + a) + 1/2) - ((b*c - a*d)*cos(b*x + a)^2 - b*c + a*d)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2)
+ (b*d*x - (b*d*x + a*d)*cos(b*x + a)^2 + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) + (b*d*x - (b*d*x + a*d
)*cos(b*x + a)^2 + a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + 1) + 2*d*sin(b*x + a))/(b^2*cos(b*x + a)^2 - b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right ) \cot ^{2}{\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)*cot(b*x+a)**2*csc(b*x+a),x)

[Out]

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

[Out]

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