3.248 \(\int (e+f x)^3 \coth ^{-1}(\cot (a+b x)) \, dx\)

Optimal. Leaf size=302 \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f^3 \text{PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac{3 f^3 \text{PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f} \]

[Out]

((e + f*x)^4*ArcCoth[Cot[a + b*x]])/(4*f) + ((I/4)*(e + f*x)^4*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)^3*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (3*f*
(e + f*x)^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (3*f*(e + f*x)^2*PolyLog[3, I*E^((2*I)*(a + b*x))]
)/(8*b^2) + (((3*I)/8)*f^2*(e + f*x)*PolyLog[4, (-I)*E^((2*I)*(a + b*x))])/b^3 - (((3*I)/8)*f^2*(e + f*x)*Poly
Log[4, I*E^((2*I)*(a + b*x))])/b^3 - (3*f^3*PolyLog[5, (-I)*E^((2*I)*(a + b*x))])/(16*b^4) + (3*f^3*PolyLog[5,
I*E^((2*I)*(a + b*x))])/(16*b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.235077, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6254, 4181, 2531, 6609, 2282, 6589} \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f^3 \text{PolyLog}\left (5,-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac{3 f^3 \text{PolyLog}\left (5,i e^{2 i (a+b x)}\right )}{16 b^4}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*ArcCoth[Cot[a + b*x]],x]

[Out]

((e + f*x)^4*ArcCoth[Cot[a + b*x]])/(4*f) + ((I/4)*(e + f*x)^4*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)^3*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (3*f*
(e + f*x)^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (3*f*(e + f*x)^2*PolyLog[3, I*E^((2*I)*(a + b*x))]
)/(8*b^2) + (((3*I)/8)*f^2*(e + f*x)*PolyLog[4, (-I)*E^((2*I)*(a + b*x))])/b^3 - (((3*I)/8)*f^2*(e + f*x)*Poly
Log[4, I*E^((2*I)*(a + b*x))])/b^3 - (3*f^3*PolyLog[5, (-I)*E^((2*I)*(a + b*x))])/(16*b^4) + (3*f^3*PolyLog[5,
I*E^((2*I)*(a + b*x))])/(16*b^4)

Rule 6254

Int[ArcCoth[Cot[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m + 1)*ArcCoth[
Cot[a + b*x]])/(f*(m + 1)), x] - Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sec[2*a + 2*b*x], x], x] /; FreeQ[{
a, b, e, f}, x] && IGtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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 (e+f x)^3 \coth ^{-1}(\cot (a+b x)) \, dx &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}-\frac{b \int (e+f x)^4 \sec (2 a+2 b x) \, dx}{4 f}\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}+\frac{1}{2} \int (e+f x)^3 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac{1}{2} \int (e+f x)^3 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{3 f (e+f x)^2 \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{\left (3 f^2\right ) \int (e+f x) \text{Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac{\left (3 f^2\right ) \int (e+f x) \text{Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{3 f (e+f x)^2 \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}+\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (i e^{i (2 a+2 b x)}\right ) \, dx}{8 b^3}\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{3 f (e+f x)^2 \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{\left (3 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}+\frac{\left (3 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{16 b^4}\\ &=\frac{(e+f x)^4 \coth ^{-1}(\cot (a+b x))}{4 f}+\frac{i (e+f x)^4 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{4 f}-\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{3 f (e+f x)^2 \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{3 f (e+f x)^2 \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{3 f^3 \text{Li}_5\left (-i e^{2 i (a+b x)}\right )}{16 b^4}+\frac{3 f^3 \text{Li}_5\left (i e^{2 i (a+b x)}\right )}{16 b^4}\\ \end{align*}

Mathematica [B]  time = 0.286991, size = 654, normalized size = 2.17 \[ \frac{1}{4} x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right ) \coth ^{-1}(\cot (a+b x))+\frac{6 b^2 e^2 f \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b^2 e^2 f \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+12 b^2 e f^2 x \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-12 b^2 e f^2 x \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )-4 i b^3 (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+4 i b^3 (e+f x)^3 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b^2 f^3 x^2 \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b^2 f^3 x^2 \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+6 i b e f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-6 i b e f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )+6 i b f^3 x \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-6 i b f^3 x \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )-3 f^3 \text{PolyLog}\left (5,-i e^{2 i (a+b x)}\right )+3 f^3 \text{PolyLog}\left (5,i e^{2 i (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1+i e^{2 i (a+b x)}\right )-8 b^4 e^3 x \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e^3 x \log \left (1+i e^{2 i (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1-i e^{2 i (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1+i e^{2 i (a+b x)}\right )}{16 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^3*ArcCoth[Cot[a + b*x]],x]

[Out]

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*ArcCoth[Cot[a + b*x]])/4 + (-8*b^4*e^3*x*Log[1 - I*E^((2*I)*(a
+ b*x))] - 12*b^4*e^2*f*x^2*Log[1 - I*E^((2*I)*(a + b*x))] - 8*b^4*e*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] -
2*b^4*f^3*x^4*Log[1 - I*E^((2*I)*(a + b*x))] + 8*b^4*e^3*x*Log[1 + I*E^((2*I)*(a + b*x))] + 12*b^4*e^2*f*x^2*L
og[1 + I*E^((2*I)*(a + b*x))] + 8*b^4*e*f^2*x^3*Log[1 + I*E^((2*I)*(a + b*x))] + 2*b^4*f^3*x^4*Log[1 + I*E^((2
*I)*(a + b*x))] - (4*I)*b^3*(e + f*x)^3*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + (4*I)*b^3*(e + f*x)^3*PolyLog[2
, I*E^((2*I)*(a + b*x))] + 6*b^2*e^2*f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] + 12*b^2*e*f^2*x*PolyLog[3, (-I)*E
^((2*I)*(a + b*x))] + 6*b^2*f^3*x^2*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] - 6*b^2*e^2*f*PolyLog[3, I*E^((2*I)*(
a + b*x))] - 12*b^2*e*f^2*x*PolyLog[3, I*E^((2*I)*(a + b*x))] - 6*b^2*f^3*x^2*PolyLog[3, I*E^((2*I)*(a + b*x))
] + (6*I)*b*e*f^2*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b*f^3*x*PolyLog[4, (-I)*E^((2*I)*(a + b*x))] -
(6*I)*b*e*f^2*PolyLog[4, I*E^((2*I)*(a + b*x))] - (6*I)*b*f^3*x*PolyLog[4, I*E^((2*I)*(a + b*x))] - 3*f^3*Poly
Log[5, (-I)*E^((2*I)*(a + b*x))] + 3*f^3*PolyLog[5, I*E^((2*I)*(a + b*x))])/(16*b^4)

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Maple [C]  time = 6.052, size = 7429, normalized size = 24.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*arccoth(cot(b*x+a)),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{16} \,{\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \frac{1}{16} \,{\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \log \left (2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \sin \left (2 \, b x + 2 \, a\right )^{2} - 4 \, \sin \left (2 \, b x + 2 \, a\right ) + 2\right ) - \int \frac{{\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) +{\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) +{\left (b f^{3} x^{4} + 4 \, b e f^{2} x^{3} + 6 \, b e^{2} f x^{2} + 4 \, b e^{3} x\right )} \cos \left (2 \, b x + 2 \, a\right )}{2 \,{\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arccoth(cot(b*x+a)),x, algorithm="maxima")

[Out]

1/16*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 + 4*sin(2
*b*x + 2*a) + 2) - 1/16*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x
+ 2*a)^2 - 4*sin(2*b*x + 2*a) + 2) - integrate(1/2*((b*f^3*x^4 + 4*b*e*f^2*x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*c
os(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f^3*x^4 + 4*b*e*f^2*x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*sin(4*b*x + 4*a)*si
n(2*b*x + 2*a) + (b*f^3*x^4 + 4*b*e*f^2*x^3 + 6*b*e^2*f*x^2 + 4*b*e^3*x)*cos(2*b*x + 2*a))/(cos(4*b*x + 4*a)^2
+ sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1), x)

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Fricas [C]  time = 2.68441, size = 3677, normalized size = 12.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arccoth(cot(b*x+a)),x, algorithm="fricas")

[Out]

-1/32*(3*f^3*polylog(5, I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 3*f^3*polylog(5, I*cos(2*b*x + 2*a) - sin(2*b
*x + 2*a)) + 3*f^3*polylog(5, -I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - 3*f^3*polylog(5, -I*cos(2*b*x + 2*a) -
sin(2*b*x + 2*a)) - (4*I*b^3*f^3*x^3 + 12*I*b^3*e*f^2*x^2 + 12*I*b^3*e^2*f*x + 4*I*b^3*e^3)*dilog(I*cos(2*b*x
+ 2*a) + sin(2*b*x + 2*a)) - (4*I*b^3*f^3*x^3 + 12*I*b^3*e*f^2*x^2 + 12*I*b^3*e^2*f*x + 4*I*b^3*e^3)*dilog(I*
cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - (-4*I*b^3*f^3*x^3 - 12*I*b^3*e*f^2*x^2 - 12*I*b^3*e^2*f*x - 4*I*b^3*e^3
)*dilog(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - (-4*I*b^3*f^3*x^3 - 12*I*b^3*e*f^2*x^2 - 12*I*b^3*e^2*f*x -
4*I*b^3*e^3)*dilog(-I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - 4*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^
2 + 4*b^4*e^3*x)*log((cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1)/(cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1)) - 2*
(4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) + 2*(
4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) + 2*(b
^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^2 + 4*b^4*e^3*x + 4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 -
a^4*f^3)*log(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) - 2*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^2
+ 4*b^4*e^3*x + 4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(I*cos(2*b*x + 2*a) - sin(2*b*x +
2*a) + 1) + 2*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^2 + 4*b^4*e^3*x + 4*a*b^3*e^3 - 6*a^2*b^2*e^2*f +
4*a^3*b*e*f^2 - a^4*f^3)*log(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) - 2*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 +
6*b^4*e^2*f*x^2 + 4*b^4*e^3*x + 4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(-I*cos(2*b*x + 2
*a) - sin(2*b*x + 2*a) + 1) - 2*(4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(-cos(2*b*x + 2*a
) + I*sin(2*b*x + 2*a) + I) + 2*(4*a*b^3*e^3 - 6*a^2*b^2*e^2*f + 4*a^3*b*e*f^2 - a^4*f^3)*log(-cos(2*b*x + 2*a
) - I*sin(2*b*x + 2*a) + I) - (-6*I*b*f^3*x - 6*I*b*e*f^2)*polylog(4, I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) -
(-6*I*b*f^3*x - 6*I*b*e*f^2)*polylog(4, I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - (6*I*b*f^3*x + 6*I*b*e*f^2)*
polylog(4, -I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) - (6*I*b*f^3*x + 6*I*b*e*f^2)*polylog(4, -I*cos(2*b*x + 2*a
) - sin(2*b*x + 2*a)) - 6*(b^2*f^3*x^2 + 2*b^2*e*f^2*x + b^2*e^2*f)*polylog(3, I*cos(2*b*x + 2*a) + sin(2*b*x
+ 2*a)) + 6*(b^2*f^3*x^2 + 2*b^2*e*f^2*x + b^2*e^2*f)*polylog(3, I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) - 6*(b
^2*f^3*x^2 + 2*b^2*e*f^2*x + b^2*e^2*f)*polylog(3, -I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + 6*(b^2*f^3*x^2 +
2*b^2*e*f^2*x + b^2*e^2*f)*polylog(3, -I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)))/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right )^{3} \operatorname{acoth}{\left (\cot{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*acoth(cot(b*x+a)),x)

[Out]

Integral((e + f*x)**3*acoth(cot(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{3} \operatorname{arcoth}\left (\cot \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arccoth(cot(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*arccoth(cot(b*x + a)), x)