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

Optimal. Leaf size=95 \[ \frac{3 i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{4 d \sin (a+b x)}{b^2}+\frac{4 (c+d x) \cos (a+b x)}{b}-\frac{6 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.108855, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4431, 4408, 3296, 2637, 4183, 2279, 2391} \[ \frac{3 i d \text{PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac{3 i d \text{PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac{4 d \sin (a+b x)}{b^2}+\frac{4 (c+d x) \cos (a+b x)}{b}-\frac{6 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A]  time = 0.341125, size = 171, normalized size = 1.8 \[ \frac{3 d \left (i \left (\text{PolyLog}\left (2,-e^{i (a+b x)}\right )-\text{PolyLog}\left (2,e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )\right )}{b^2}-\frac{4 d \sin (a+b x)}{b^2}-\frac{3 a d \log \left (\tan \left (\frac{1}{2} (a+b x)\right )\right )}{b^2}+\frac{4 c \cos (a+b x)}{b}+\frac{3 c \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{3 c \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b}+\frac{4 d x \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*Cos[a + b*x])/b + (4*d*x*Cos[a + b*x])/b - (3*c*Log[Cos[(a + b*x)/2]])/b + (3*c*Log[Sin[(a + b*x)/2]])/b
- (3*a*d*Log[Tan[(a + b*x)/2]])/b^2 + (3*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))])))/b^2 - (4*d*Sin[a + b*x])/b^2

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Maple [B]  time = 0.249, size = 205, normalized size = 2.2 \begin{align*} 2\,{\frac{ \left ( dxb+bc+id \right ){{\rm e}^{i \left ( bx+a \right ) }}}{{b}^{2}}}+2\,{\frac{ \left ( dxb+bc-id \right ){{\rm e}^{-i \left ( bx+a \right ) }}}{{b}^{2}}}-6\,{\frac{c{\it Artanh} \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{b}}+3\,{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{b}}+3\,{\frac{d\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{2}}}-{\frac{3\,id{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-3\,{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{b}}-3\,{\frac{d\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) a}{{b}^{2}}}+{\frac{3\,id{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+6\,{\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)*csc(b*x+a)^2*sin(3*b*x+3*a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c{\left (8 \, \cos \left (b x + a\right ) - 3 \, \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right ) + 3 \, \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right )\right )}}{2 \, b} + \frac{-\frac{1}{2} \,{\left (6 i \, b x \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) + 6 i \, b x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 8 \, b x \cos \left (b x + a\right ) + 3 \, b x \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 \, b x \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 ) - 6 i \,{\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \,{\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + 8 \, \sin \left (b x + a\right )\right )} d}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c*(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) + sin
(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 + (4*
b*x*cos(b*x + a) + 3*b^2*integrate(x*sin(b*x + a)/(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1), x) +
3*b^2*integrate(x*sin(b*x + a)/(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1), x) - 4*sin(b*x + a))*d
/b^2

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Fricas [B]  time = 0.58448, size = 814, normalized size = 8.57 \begin{align*} \frac{8 \,{\left (b d x + b c\right )} \cos \left (b x + a\right ) - 3 i \, d{\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 3 i \, d{\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 3 i \, d{\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 3 i \, d{\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 3 \,{\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 3 \,{\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + 3 \,{\left (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 ) + 3 \,{\left (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 ) + 3 \,{\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + 3 \,{\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 8 \, d \sin \left (b x + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*csc(b*x+a)**2*sin(3*b*x+3*a),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 )} \csc \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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