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

Optimal. Leaf size=83 \[ -\frac{i d^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \cot (a+b x)}{b}-\frac{i (c+d x)^2}{b} \]

[Out]

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

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Rubi [A]  time = 0.135961, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4184, 3717, 2190, 2279, 2391} \[ -\frac{i d^2 \text{PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac{2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac{(c+d x)^2 \cot (a+b x)}{b}-\frac{i (c+d x)^2}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csc[a + b*x]^2,x]

[Out]

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

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

Mathematica [B]  time = 4.82682, size = 181, normalized size = 2.18 \[ \frac{\csc (a) \left (d^2 \left (-\sin (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )-i b x \left (\pi -2 \tan ^{-1}(\tan (a))\right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )-b^2 x^2 \cos (a) e^{i \tan ^{-1}(\tan (a))} \sqrt{\sec ^2(a)}\right )+b^2 \sin (b x) (c+d x)^2 \csc (a+b x)-2 b c d (b x \cos (a)-\sin (a) \log (\sin (a+b x)))\right )}{b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Csc[a]*(-2*b*c*d*(b*x*Cos[a] - Log[Sin[a + b*x]]*Sin[a]) + d^2*(-(b^2*E^(I*ArcTan[Tan[a]])*x^2*Cos[a]*Sqrt[Se
c[a]^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*Poly
Log[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))])*Sin[a]) + b^2*(c + d*x)^2*Csc[a + b*x]*Sin[b*x]))/b^3

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Maple [B]  time = 0.043, size = 276, normalized size = 3.3 \begin{align*}{\frac{-2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a \right ) }}-1 \right ) }}-4\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{2}}}+2\,{\frac{cd\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+1 \right ) }{{b}^{2}}}-{\frac{2\,i{d}^{2}{x}^{2}}{b}}-{\frac{4\,i{d}^{2}ax}{{b}^{2}}}-{\frac{2\,i{d}^{2}{a}^{2}}{{b}^{3}}}+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}}}-{\frac{2\,i{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+4\,{\frac{{d}^{2}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}-2\,{\frac{{d}^{2}a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-1 \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.46171, size = 749, normalized size = 9.02 \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)^2*csc(b*x+a)^2,x, algorithm="maxima")

[Out]

-(2*b^2*c^2 + (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))*arctan2(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)*arctan2(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)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x)*cos(2*b*x + 2*a) + (2*d^2
*cos(2*b*x + 2*a) + 2*I*d^2*sin(2*b*x + 2*a) - 2*d^2)*dilog(-e^(I*b*x + I*a)) + (2*d^2*cos(2*b*x + 2*a) + 2*I*
d^2*sin(2*b*x + 2*a) - 2*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)*sin(2*b*x + 2*a))/(-I*b^3*co
s(2*b*x + 2*a) + b^3*sin(2*b*x + 2*a) + I*b^3)

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Fricas [B]  time = 1.79532, size = 1026, normalized size = 12.36 \begin{align*} \frac{-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 ) -{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \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*csc(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

Integral((c + d*x)**2*csc(a + b*x)**2, 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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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