3.38 \(\int (c+d x)^2 (a+b \cot (e+f x)) \, dx\)

Optimal. Leaf size=112 \[ -\frac{i b d (c+d x) \text{PolyLog}\left (2,e^{2 i (e+f x)}\right )}{f^2}+\frac{b d^2 \text{PolyLog}\left (3,e^{2 i (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b (c+d x)^3}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.209533, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3722, 3717, 2190, 2531, 2282, 6589} \[ -\frac{i b d (c+d x) \text{PolyLog}\left (2,e^{2 i (e+f x)}\right )}{f^2}+\frac{b d^2 \text{PolyLog}\left (3,e^{2 i (e+f x)}\right )}{2 f^3}+\frac{a (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (c+d x)^2 (a+b \cot (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \cot (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^3}{3 d}+b \int (c+d x)^2 \cot (e+f x) \, dx\\ &=\frac{a (c+d x)^3}{3 d}-\frac{i b (c+d x)^3}{3 d}-(2 i b) \int \frac{e^{2 i (e+f x)} (c+d x)^2}{1-e^{2 i (e+f x)}} \, dx\\ &=\frac{a (c+d x)^3}{3 d}-\frac{i b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{(2 b d) \int (c+d x) \log \left (1-e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{i b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b d (c+d x) \text{Li}_2\left (e^{2 i (e+f x)}\right )}{f^2}+\frac{\left (i b d^2\right ) \int \text{Li}_2\left (e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{i b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b d (c+d x) \text{Li}_2\left (e^{2 i (e+f x)}\right )}{f^2}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^3}\\ &=\frac{a (c+d x)^3}{3 d}-\frac{i b (c+d x)^3}{3 d}+\frac{b (c+d x)^2 \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac{i b d (c+d x) \text{Li}_2\left (e^{2 i (e+f x)}\right )}{f^2}+\frac{b d^2 \text{Li}_3\left (e^{2 i (e+f x)}\right )}{2 f^3}\\ \end{align*}

Mathematica [B]  time = 2.48925, size = 406, normalized size = 3.62 \[ \frac{-3 i b c d f \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )+6 i b d^2 f x \text{PolyLog}\left (2,-e^{-i (e+f x)}\right )+6 i b d^2 f x \text{PolyLog}\left (2,e^{-i (e+f x)}\right )+6 b d^2 \text{PolyLog}\left (3,-e^{-i (e+f x)}\right )+6 b d^2 \text{PolyLog}\left (3,e^{-i (e+f x)}\right )+3 a c^2 f^3 x+3 a c d f^3 x^2+a d^2 f^3 x^3+3 b c^2 f^2 \log (\sin (e+f x))+3 b c d f^3 x^2 \cot (e)-3 b c d f^3 x^2 e^{i \tan ^{-1}(\tan (e))} \cot (e) \sqrt{\sec ^2(e)}-6 i b c d f^2 x \tan ^{-1}(\tan (e))+6 b c d f^2 x \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )+6 b c d f \tan ^{-1}(\tan (e)) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )-6 b c d f \tan ^{-1}(\tan (e)) \log \left (\sin \left (\tan ^{-1}(\tan (e))+f x\right )\right )+3 i \pi b c d f^2 x+3 \pi b c d f \log \left (1+e^{-2 i f x}\right )-3 \pi b c d f \log (\cos (f x))+3 b d^2 f^2 x^2 \log \left (1-e^{-i (e+f x)}\right )+3 b d^2 f^2 x^2 \log \left (1+e^{-i (e+f x)}\right )+i b d^2 f^3 x^3}{3 f^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*a*c^2*f^3*x + (3*I)*b*c*d*f^2*Pi*x + 3*a*c*d*f^3*x^2 + a*d^2*f^3*x^3 + I*b*d^2*f^3*x^3 - (6*I)*b*c*d*f^2*x*
ArcTan[Tan[e]] + 3*b*c*d*f^3*x^2*Cot[e] + 3*b*c*d*f*Pi*Log[1 + E^((-2*I)*f*x)] + 3*b*d^2*f^2*x^2*Log[1 - E^((-
I)*(e + f*x))] + 3*b*d^2*f^2*x^2*Log[1 + E^((-I)*(e + f*x))] + 6*b*c*d*f^2*x*Log[1 - E^((2*I)*(f*x + ArcTan[Ta
n[e]]))] + 6*b*c*d*f*ArcTan[Tan[e]]*Log[1 - E^((2*I)*(f*x + ArcTan[Tan[e]]))] - 3*b*c*d*f*Pi*Log[Cos[f*x]] + 3
*b*c^2*f^2*Log[Sin[e + f*x]] - 6*b*c*d*f*ArcTan[Tan[e]]*Log[Sin[f*x + ArcTan[Tan[e]]]] + (6*I)*b*d^2*f*x*PolyL
og[2, -E^((-I)*(e + f*x))] + (6*I)*b*d^2*f*x*PolyLog[2, E^((-I)*(e + f*x))] - (3*I)*b*c*d*f*PolyLog[2, E^((2*I
)*(f*x + ArcTan[Tan[e]]))] + 6*b*d^2*PolyLog[3, -E^((-I)*(e + f*x))] + 6*b*d^2*PolyLog[3, E^((-I)*(e + f*x))]
- 3*b*c*d*E^(I*ArcTan[Tan[e]])*f^3*x^2*Cot[e]*Sqrt[Sec[e]^2])/(3*f^3)

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Maple [B]  time = 0.339, size = 516, normalized size = 4.6 \begin{align*} acd{x}^{2}-2\,{\frac{b{c}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{f}}+{\frac{b{c}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{f}}+2\,{\frac{bcd\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) e}{{f}^{2}}}+{\frac{a{d}^{2}{x}^{3}}{3}}+a{c}^{2}x-ibcd{x}^{2}-2\,{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+{\frac{b{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ){x}^{2}}{f}}-{\frac{b{d}^{2}\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ){e}^{2}}{{f}^{3}}}+{\frac{b{d}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ){x}^{2}}{f}}+{\frac{{\frac{4\,i}{3}}b{d}^{2}{e}^{3}}{{f}^{3}}}+{\frac{b{c}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) }{f}}+2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+2\,{\frac{b{d}^{2}{\it polylog} \left ( 3,{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{3}}}+2\,{\frac{bcd\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}+1 \right ) x}{f}}+2\,{\frac{bcd\ln \left ( 1-{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{f}}-2\,{\frac{bcde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{{f}^{2}}}+4\,{\frac{bcde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{i}{3}}b{d}^{2}{x}^{3}+ib{c}^{2}x+{\frac{b{d}^{2}{e}^{2}\ln \left ({{\rm e}^{i \left ( fx+e \right ) }}-1 \right ) }{{f}^{3}}}-{\frac{2\,ib{d}^{2}{\it polylog} \left ( 2,{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}-{\frac{2\,ibcd{e}^{2}}{{f}^{2}}}+{\frac{2\,ib{d}^{2}{e}^{2}x}{{f}^{2}}}-{\frac{2\,ib{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) x}{{f}^{2}}}-{\frac{2\,ibcd{\it polylog} \left ( 2,{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{2\,ibcd{\it polylog} \left ( 2,-{{\rm e}^{i \left ( fx+e \right ) }} \right ) }{{f}^{2}}}-{\frac{4\,ibcdex}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.59511, size = 702, normalized size = 6.27 \begin{align*} \frac{6 \,{\left (f x + e\right )} a c^{2} + \frac{2 \,{\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac{6 \,{\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac{6 \,{\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac{6 \,{\left (f x + e\right )}^{2} a c d}{f} - \frac{12 \,{\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sin \left (f x + e\right )\right ) + \frac{6 \, b d^{2} e^{2} \log \left (\sin \left (f x + e\right )\right )}{f^{2}} - \frac{12 \, b c d e \log \left (\sin \left (f x + e\right )\right )}{f} + \frac{-2 i \,{\left (f x + e\right )}^{3} b d^{2} + 12 \, b d^{2}{\rm Li}_{3}(-e^{\left (i \, f x + i \, e\right )}) + 12 \, b d^{2}{\rm Li}_{3}(e^{\left (i \, f x + i \, e\right )}) +{\left (6 i \, b d^{2} e - 6 i \, b c d f\right )}{\left (f x + e\right )}^{2} +{\left (6 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (-12 i \, b d^{2} e + 12 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) +{\left (-6 i \,{\left (f x + e\right )}^{2} b d^{2} +{\left (12 i \, b d^{2} e - 12 i \, b c d f\right )}{\left (f x + e\right )}\right )} \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right ) +{\left (-12 i \,{\left (f x + e\right )} b d^{2} + 12 i \, b d^{2} e - 12 i \, b c d f\right )}{\rm Li}_2\left (-e^{\left (i \, f x + i \, e\right )}\right ) +{\left (-12 i \,{\left (f x + e\right )} b d^{2} + 12 i \, b d^{2} e - 12 i \, b c d f\right )}{\rm Li}_2\left (e^{\left (i \, f x + i \, e\right )}\right ) + 3 \,{\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \,{\left (b d^{2} e - b c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + 3 \,{\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \,{\left (b d^{2} e - b c d f\right )}{\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )}{f^{2}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*cot(f*x+e)),x, algorithm="maxima")

[Out]

1/6*(6*(f*x + e)*a*c^2 + 2*(f*x + e)^3*a*d^2/f^2 - 6*(f*x + e)^2*a*d^2*e/f^2 + 6*(f*x + e)*a*d^2*e^2/f^2 + 6*(
f*x + e)^2*a*c*d/f - 12*(f*x + e)*a*c*d*e/f + 6*b*c^2*log(sin(f*x + e)) + 6*b*d^2*e^2*log(sin(f*x + e))/f^2 -
12*b*c*d*e*log(sin(f*x + e))/f + (-2*I*(f*x + e)^3*b*d^2 + 12*b*d^2*polylog(3, -e^(I*f*x + I*e)) + 12*b*d^2*po
lylog(3, e^(I*f*x + I*e)) + (6*I*b*d^2*e - 6*I*b*c*d*f)*(f*x + e)^2 + (6*I*(f*x + e)^2*b*d^2 + (-12*I*b*d^2*e
+ 12*I*b*c*d*f)*(f*x + e))*arctan2(sin(f*x + e), cos(f*x + e) + 1) + (-6*I*(f*x + e)^2*b*d^2 + (12*I*b*d^2*e -
12*I*b*c*d*f)*(f*x + e))*arctan2(sin(f*x + e), -cos(f*x + e) + 1) + (-12*I*(f*x + e)*b*d^2 + 12*I*b*d^2*e - 1
2*I*b*c*d*f)*dilog(-e^(I*f*x + I*e)) + (-12*I*(f*x + e)*b*d^2 + 12*I*b*d^2*e - 12*I*b*c*d*f)*dilog(e^(I*f*x +
I*e)) + 3*((f*x + e)^2*b*d^2 - 2*(b*d^2*e - b*c*d*f)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*
x + e) + 1) + 3*((f*x + e)^2*b*d^2 - 2*(b*d^2*e - b*c*d*f)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*
cos(f*x + e) + 1))/f^2)/f

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Fricas [C]  time = 1.81055, size = 1025, normalized size = 9.15 \begin{align*} \frac{4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x + 3 \, b d^{2}{\rm polylog}\left (3, \cos \left (2 \, f x + 2 \, e\right ) + i \, \sin \left (2 \, f x + 2 \, e\right )\right ) + 3 \, b d^{2}{\rm polylog}\left (3, \cos \left (2 \, f x + 2 \, e\right ) - i \, \sin \left (2 \, f x + 2 \, e\right )\right ) +{\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )}{\rm Li}_2\left (\cos \left (2 \, f x + 2 \, e\right ) + i \, \sin \left (2 \, f x + 2 \, e\right )\right ) +{\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )}{\rm Li}_2\left (\cos \left (2 \, f x + 2 \, e\right ) - i \, \sin \left (2 \, f x + 2 \, e\right )\right ) + 6 \,{\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (2 \, f x + 2 \, e\right ) + \frac{1}{2} i \, \sin \left (2 \, f x + 2 \, e\right ) + \frac{1}{2}\right ) + 6 \,{\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (2 \, f x + 2 \, e\right ) - \frac{1}{2} i \, \sin \left (2 \, f x + 2 \, e\right ) + \frac{1}{2}\right ) + 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\cos \left (2 \, f x + 2 \, e\right ) + i \, \sin \left (2 \, f x + 2 \, e\right ) + 1\right ) + 6 \,{\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-\cos \left (2 \, f x + 2 \, e\right ) - i \, \sin \left (2 \, f x + 2 \, e\right ) + 1\right )}{12 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*(a+b*cot(f*x+e)),x)

[Out]

Integral((a + b*cot(e + f*x))*(c + d*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}{\left (b \cot \left (f x + e\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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