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

Optimal. Leaf size=115 \[ -\frac{3 i d^3 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 d^2 (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \cot (a+b x)}{2 b^2}-\frac{(c+d x)^3 \csc ^2(a+b x)}{2 b}-\frac{3 i d (c+d x)^2}{2 b^2} \]

[Out]

(((-3*I)/2)*d*(c + d*x)^2)/b^2 - (3*d*(c + d*x)^2*Cot[a + b*x])/(2*b^2) - ((c + d*x)^3*Csc[a + b*x]^2)/(2*b) +
(3*d^2*(c + d*x)*Log[1 - E^((2*I)*(a + b*x))])/b^3 - (((3*I)/2)*d^3*PolyLog[2, E^((2*I)*(a + b*x))])/b^4

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Rubi [A]  time = 0.173609, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4410, 4184, 3717, 2190, 2279, 2391} \[ -\frac{3 i d^3 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 d^2 (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^3}-\frac{3 d (c+d x)^2 \cot (a+b x)}{2 b^2}-\frac{(c+d x)^3 \csc ^2(a+b x)}{2 b}-\frac{3 i d (c+d x)^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/2)*d*(c + d*x)^2)/b^2 - (3*d*(c + d*x)^2*Cot[a + b*x])/(2*b^2) - ((c + d*x)^3*Csc[a + b*x]^2)/(2*b) +
(3*d^2*(c + d*x)*Log[1 - E^((2*I)*(a + b*x))])/b^3 - (((3*I)/2)*d^3*PolyLog[2, E^((2*I)*(a + b*x))])/b^4

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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

Mathematica [B]  time = 6.41234, size = 277, normalized size = 2.41 \[ -\frac{3 d^3 \csc (a) \sec (a) \left (\frac{\tan (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\tan ^2(a)+1}}+b^2 x^2 e^{i \tan ^{-1}(\tan (a))}\right )}{2 b^4 \sqrt{\sec ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac{3 \csc (a) \csc (a+b x) \left (c^2 d \sin (b x)+2 c d^2 x \sin (b x)+d^3 x^2 \sin (b x)\right )}{2 b^2}+\frac{3 c d^2 \csc (a) (\sin (a) \log (\sin (a) \cos (b x)+\cos (a) \sin (b x))-b x \cos (a))}{b^3 \left (\sin ^2(a)+\cos ^2(a)\right )}-\frac{(c+d x)^3 \csc ^2(a+b x)}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.164, size = 409, normalized size = 3.6 \begin{align*}{\frac{2\,b{d}^{3}{x}^{3}{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{d}^{3}{x}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}+6\,bc{d}^{2}{x}^{2}{{\rm e}^{2\,i \left ( bx+a \right ) }}-6\,ic{d}^{2}x{{\rm e}^{2\,i \left ( bx+a \right ) }}+6\,b{c}^{2}dx{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{c}^{2}d{{\rm e}^{2\,i \left ( bx+a \right ) }}+3\,i{d}^{3}{x}^{2}+2\,b{c}^{3}{{\rm e}^{2\,i \left ( bx+a \right ) }}+6\,ic{d}^{2}x+3\,i{c}^{2}d}{{b}^{2} \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) ^{2}}}+3\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}}-6\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+3\,{\frac{{d}^{2}c\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{3}}}-{\frac{3\,i{d}^{3}{x}^{2}}{{b}^{2}}}-{\frac{6\,i{d}^{3}ax}{{b}^{3}}}-{\frac{3\,i{d}^{3}{a}^{2}}{{b}^{4}}}+3\,{\frac{{d}^{3}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) x}{{b}^{3}}}+3\,{\frac{{d}^{3}\ln \left ( 1-{{\rm e}^{i \left ( bx+a \right ) }} \right ) a}{{b}^{4}}}-{\frac{3\,i{d}^{3}{\it polylog} \left ( 2,{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+3\,{\frac{{d}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) x}{{b}^{3}}}-{\frac{3\,i{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-3\,{\frac{{d}^{3}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{4}}}+6\,{\frac{{d}^{3}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*cos(b*x+a)*csc(b*x+a)^3,x)

[Out]

(2*b*d^3*x^3*exp(2*I*(b*x+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6*b*c*d^2*x^2*exp(2*I*(b*x+a))-6*I*c*d^2*x*exp(2*I*
(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))-3*I*c^2*d*exp(2*I*(b*x+a))+3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))+6*I*c*d^
2*x+3*I*c^2*d)/b^2/(exp(2*I*(b*x+a))-1)^2+3*d^2/b^3*c*ln(exp(I*(b*x+a))-1)-6*d^2/b^3*c*ln(exp(I*(b*x+a)))+3*d^
2/b^3*c*ln(exp(I*(b*x+a))+1)-3*I*d^3/b^2*x^2-6*I*d^3/b^3*a*x-3*I*d^3/b^4*a^2+3*d^3/b^3*ln(1-exp(I*(b*x+a)))*x+
3*d^3/b^4*ln(1-exp(I*(b*x+a)))*a-3*I*d^3*polylog(2,exp(I*(b*x+a)))/b^4+3*d^3/b^3*ln(exp(I*(b*x+a))+1)*x-3*I*d^
3/b^4*polylog(2,-exp(I*(b*x+a)))-3*d^3/b^4*a*ln(exp(I*(b*x+a))-1)+6*d^3/b^4*a*ln(exp(I*(b*x+a)))

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Maxima [B]  time = 2.27893, size = 1411, normalized size = 12.27 \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)^3*cos(b*x+a)*csc(b*x+a)^3,x, algorithm="maxima")

[Out]

(6*b^2*c^2*d + (6*b*d^3*x + 6*b*c*d^2 + 6*(b*d^3*x + b*c*d^2)*cos(4*b*x + 4*a) - 12*(b*d^3*x + b*c*d^2)*cos(2*
b*x + 2*a) + (6*I*b*d^3*x + 6*I*b*c*d^2)*sin(4*b*x + 4*a) + (-12*I*b*d^3*x - 12*I*b*c*d^2)*sin(2*b*x + 2*a))*a
rctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*b*c*d^2*cos(4*b*x + 4*a) - 12*b*c*d^2*cos(2*b*x + 2*a) + 6*I*b*c*d
^2*sin(4*b*x + 4*a) - 12*I*b*c*d^2*sin(2*b*x + 2*a) + 6*b*c*d^2)*arctan2(sin(b*x + a), cos(b*x + a) - 1) - (6*
b*d^3*x*cos(4*b*x + 4*a) - 12*b*d^3*x*cos(2*b*x + 2*a) + 6*I*b*d^3*x*sin(4*b*x + 4*a) - 12*I*b*d^3*x*sin(2*b*x
+ 2*a) + 6*b*d^3*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 6*(b^2*d^3*x^2 + 2*b^2*c*d^2*x)*cos(4*b*x + 4*
a) + (-4*I*b^3*d^3*x^3 - 4*I*b^3*c^3 - 6*b^2*c^2*d + (-12*I*b^3*c*d^2 + 6*b^2*d^3)*x^2 - 12*(I*b^3*c^2*d - b^2
*c*d^2)*x)*cos(2*b*x + 2*a) - (6*d^3*cos(4*b*x + 4*a) - 12*d^3*cos(2*b*x + 2*a) + 6*I*d^3*sin(4*b*x + 4*a) - 1
2*I*d^3*sin(2*b*x + 2*a) + 6*d^3)*dilog(-e^(I*b*x + I*a)) - (6*d^3*cos(4*b*x + 4*a) - 12*d^3*cos(2*b*x + 2*a)
+ 6*I*d^3*sin(4*b*x + 4*a) - 12*I*d^3*sin(2*b*x + 2*a) + 6*d^3)*dilog(e^(I*b*x + I*a)) + (-3*I*b*d^3*x - 3*I*b
*c*d^2 + (-3*I*b*d^3*x - 3*I*b*c*d^2)*cos(4*b*x + 4*a) + (6*I*b*d^3*x + 6*I*b*c*d^2)*cos(2*b*x + 2*a) + 3*(b*d
^3*x + b*c*d^2)*sin(4*b*x + 4*a) - 6*(b*d^3*x + b*c*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2
+ 2*cos(b*x + a) + 1) + (-3*I*b*d^3*x - 3*I*b*c*d^2 + (-3*I*b*d^3*x - 3*I*b*c*d^2)*cos(4*b*x + 4*a) + (6*I*b*
d^3*x + 6*I*b*c*d^2)*cos(2*b*x + 2*a) + 3*(b*d^3*x + b*c*d^2)*sin(4*b*x + 4*a) - 6*(b*d^3*x + b*c*d^2)*sin(2*b
*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + (-6*I*b^2*d^3*x^2 - 12*I*b^2*c*d^2*x)*s
in(4*b*x + 4*a) + (4*b^3*d^3*x^3 + 4*b^3*c^3 - 6*I*b^2*c^2*d + 6*(2*b^3*c*d^2 + I*b^2*d^3)*x^2 + (12*b^3*c^2*d
+ 12*I*b^2*c*d^2)*x)*sin(2*b*x + 2*a))/(-2*I*b^4*cos(4*b*x + 4*a) + 4*I*b^4*cos(2*b*x + 2*a) + 2*b^4*sin(4*b*
x + 4*a) - 4*b^4*sin(2*b*x + 2*a) - 2*I*b^4)

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Fricas [B]  time = 0.609607, size = 1445, normalized size = 12.57 \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)^3*cos(b*x+a)*csc(b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3 + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos
(b*x + a)*sin(b*x + a) + (-3*I*d^3*cos(b*x + a)^2 + 3*I*d^3)*dilog(cos(b*x + a) + I*sin(b*x + a)) + (3*I*d^3*c
os(b*x + a)^2 - 3*I*d^3)*dilog(cos(b*x + a) - I*sin(b*x + a)) + (3*I*d^3*cos(b*x + a)^2 - 3*I*d^3)*dilog(-cos(
b*x + a) + I*sin(b*x + a)) + (-3*I*d^3*cos(b*x + a)^2 + 3*I*d^3)*dilog(-cos(b*x + a) - I*sin(b*x + a)) - 3*(b*
d^3*x + b*c*d^2 - (b*d^3*x + b*c*d^2)*cos(b*x + a)^2)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - 3*(b*d^3*x + b*
c*d^2 - (b*d^3*x + b*c*d^2)*cos(b*x + a)^2)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - 3*(b*c*d^2 - a*d^3 - (b*c
*d^2 - a*d^3)*cos(b*x + a)^2)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 3*(b*c*d^2 - a*d^3 - (b*c*d^
2 - a*d^3)*cos(b*x + a)^2)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2) - 3*(b*d^3*x + a*d^3 - (b*d^3*x +
a*d^3)*cos(b*x + a)^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3 - (b*d^3*x + a*d^3)*cos(b
*x + a)^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1))/(b^4*cos(b*x + a)^2 - b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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