3.250 \(\int (e+f x) \coth ^{-1}(\cot (a+b x)) \, dx\)
Optimal. Leaf size=162 \[ \frac{f \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{f \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{i (e+f x) \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x) \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f} \]
[Out]
((e + f*x)^2*ArcCoth[Cot[a + b*x]])/(2*f) + ((I/2)*(e + f*x)^2*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (f*PolyLo
g[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (f*PolyLog[3, I*E^((2*I)*(a + b*x))])/(8*b^2)
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Rubi [A] time = 0.110149, antiderivative size = 162, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used =
{6254, 4181, 2531, 2282, 6589} \[ \frac{f \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{f \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{i (e+f x) \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x) \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[(e + f*x)*ArcCoth[Cot[a + b*x]],x]
[Out]
((e + f*x)^2*ArcCoth[Cot[a + b*x]])/(2*f) + ((I/2)*(e + f*x)^2*ArcTan[E^((2*I)*(a + b*x))])/f - ((I/4)*(e + f*
x)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (f*PolyLo
g[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (f*PolyLog[3, I*E^((2*I)*(a + b*x))])/(8*b^2)
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 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) \coth ^{-1}(\cot (a+b x)) \, dx &=\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f}-\frac{b \int (e+f x)^2 \sec (2 a+2 b x) \, dx}{2 f}\\ &=\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac{1}{2} \int (e+f x) \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac{1}{2} \int (e+f x) \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}-\frac{i (e+f x) \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x) \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{(i f) \int \text{Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac{(i f) \int \text{Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}\\ &=\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}-\frac{i (e+f x) \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x) \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2}-\frac{f \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2}\\ &=\frac{(e+f x)^2 \coth ^{-1}(\cot (a+b x))}{2 f}+\frac{i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}-\frac{i (e+f x) \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x) \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac{f \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.120911, size = 263, normalized size = 1.62 \[ -b e \left (\frac{i \text{PolyLog}\left (2,-i e^{i (2 a+2 b x)}\right )}{4 b^2}-\frac{i \text{PolyLog}\left (2,i e^{i (2 a+2 b x)}\right )}{4 b^2}-\frac{i x \tan ^{-1}\left (e^{2 i a+2 i b x}\right )}{b}\right )+\frac{f \left (2 i b x \text{PolyLog}(2,-\sin (2 (a+b x))+i \cos (2 (a+b x)))-2 i b x \text{PolyLog}(2,\sin (2 (a+b x))-i \cos (2 (a+b x)))-\text{PolyLog}(3,-\sin (2 (a+b x))+i \cos (2 (a+b x)))+\text{PolyLog}(3,\sin (2 (a+b x))-i \cos (2 (a+b x)))+4 i b^2 x^2 \tan ^{-1}(\cos (2 (a+b x))+i \sin (2 (a+b x)))\right )}{8 b^2}+e x \coth ^{-1}(\cot (a+b x))+\frac{1}{2} f x^2 \coth ^{-1}(\cot (a+b x)) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(e + f*x)*ArcCoth[Cot[a + b*x]],x]
[Out]
e*x*ArcCoth[Cot[a + b*x]] + (f*x^2*ArcCoth[Cot[a + b*x]])/2 - b*e*(((-I)*x*ArcTan[E^((2*I)*a + (2*I)*b*x)])/b
+ ((I/4)*PolyLog[2, (-I)*E^(I*(2*a + 2*b*x))])/b^2 - ((I/4)*PolyLog[2, I*E^(I*(2*a + 2*b*x))])/b^2) + (f*((4*I
)*b^2*x^2*ArcTan[Cos[2*(a + b*x)] + I*Sin[2*(a + b*x)]] + (2*I)*b*x*PolyLog[2, I*Cos[2*(a + b*x)] - Sin[2*(a +
b*x)]] - (2*I)*b*x*PolyLog[2, (-I)*Cos[2*(a + b*x)] + Sin[2*(a + b*x)]] - PolyLog[3, I*Cos[2*(a + b*x)] - Sin
[2*(a + b*x)]] + PolyLog[3, (-I)*Cos[2*(a + b*x)] + Sin[2*(a + b*x)]]))/(8*b^2)
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Maple [C] time = 10.737, size = 2543, normalized size = 15.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((f*x+e)*arccoth(cot(b*x+a)),x)
[Out]
-1/4*ln(exp(2*I*(b*x+a))-I)*x^2*f-1/2*ln(exp(2*I*(b*x+a))-I)*x*e-1/2*I/b^2*f*a*dilog(((-I)^(1/2)-exp(I*(b*x+a)
))/(-I)^(1/2))-1/2*I/b^2*f*a*dilog(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))+1/2*I/b*e*(I*b*x+I*a)*ln(((-I)^(1/2
)-exp(I*(b*x+a)))/(-I)^(1/2))-1/2*I/b*e*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2*I/b*e*dilog(1-exp(I*(b*x+a))*(-
1)^(3/4))+1/2*I/b*e*dilog(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))+1/8*f*polylog(3,-I*exp(2*I*(b*x+a)))/b^2-1/8
*f*polylog(3,I*exp(2*I*(b*x+a)))/b^2+1/4/b^2*f*(I*b*x+I*a)^2*ln(1-I*exp(2*I*(b*x+a)))+1/4/b^2*f*(I*b*x+I*a)*po
lylog(2,I*exp(2*I*(b*x+a)))-1/4/b^2*f*(I*b*x+I*a)^2*ln(1+I*exp(2*I*(b*x+a)))-1/4/b^2*f*(I*b*x+I*a)*polylog(2,-
I*exp(2*I*(b*x+a)))+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^3*x^2+1/4*I*Pi*x*e*csgn((1-I)
*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^3+1/8*I*Pi*f*csgn((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))
^3*x^2-1/4*I*Pi*x*e*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2-1/4*I*Pi*
x*e-1/8*I*Pi*f*x^2-1/8*I*Pi*f*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2
*x^2+1/2*I/b*e*dilog(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))+1/4/b^2*f*a^2*ln(-exp(2*I*(b*x+a))+I)-1/2/b*e*a*l
n(-exp(2*I*(b*x+a))+I)+1/2/b*a*e*ln(exp(2*I*(b*x+a))+I)-1/4/b^2*f*a^2*ln(exp(2*I*(b*x+a))+I)+1/2*I/b*e*(I*b*x+
I*a)*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))-1/2*I/b*e*(I*b*x+I*a)*ln(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2*I/b*
e*(I*b*x+I*a)*ln(1-exp(I*(b*x+a))*(-1)^(3/4))+1/2*I/b^2*f*a*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))+(1/4*f*x^2+1/2*
e*x)*ln(exp(2*I*(b*x+a))+I)+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*csgn((1-I)*(exp(2*I*(
b*x+a))+I)/(exp(2*I*(b*x+a))-1))^2*x^2+1/4*I*Pi*x*e*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))+I)/(
exp(2*I*(b*x+a))-1))^2+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^3+1/8*I*Pi*f*csgn(I*(exp
(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^2*x^2+1/2*I/b^2*f*a*dilog(1-exp(I*(b*x+a))
*(-1)^(3/4))+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))-I
)/(exp(2*I*(b*x+a))-1))+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(
b*x+a))-I)/(exp(2*I*(b*x+a))-1))*x^2-1/2*I/b^2*f*a*(I*b*x+I*a)*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))-1/2*
I/b^2*f*a*(I*b*x+I*a)*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I))*csg
n(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2-1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I*(exp(2*I*(b*x+
a))-I)/(exp(2*I*(b*x+a))-1))^2*x^2+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))*csgn((1+I)*(ex
p(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))*x^2+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*csg
n((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^2+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I
*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^2+1/8*I*Pi*f*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(
2*I*(b*x+a))-1))^2*x^2+1/2*I/b^2*f*a*(I*b*x+I*a)*ln(1+exp(I*(b*x+a))*(-1)^(3/4))+1/2*I/b^2*f*a*(I*b*x+I*a)*ln(
1-exp(I*(b*x+a))*(-1)^(3/4))-1/4*I*Pi*x*e*csgn((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^2-1/8*I*Pi*f*c
sgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))^3*x^2-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+
a))-1))^3+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))*csgn((1+I)*(exp(2*I*(b*x+a))-I)/(exp(
2*I*(b*x+a))-1))+1/4*I*Pi*x*e*csgn((1+I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2+1/8*I*Pi*f*csgn((1+I)*(e
xp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2*x^2-1/8*I*Pi*f*csgn((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1)
)^2*x^2-1/8*I*Pi*f*csgn((1+I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^3*x^2-1/4*I*Pi*x*e*csgn((1+I)*(exp(2*
I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^3-1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I/(exp(2*I*(b*x+a))-1))*csg
n(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*x^2-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-
1))*csgn((1+I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*
(b*x+a))-1))*csgn((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I))*c
sgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))-1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+
a))-I)/(exp(2*I*(b*x+a))-1))*csgn((1+I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))-1))^2*x^2-1/8*I*Pi*f*csgn(I*(ex
p(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*csgn((1-I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))-1))*x^2
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{8} \,{\left (f x^{2} + 2 \, e 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}{8} \,{\left (f x^{2} + 2 \, e 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 x^{2} + 2 \, b e x\right )} \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) +{\left (b f x^{2} + 2 \, b e x\right )} \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) +{\left (b f x^{2} + 2 \, b e x\right )} \cos \left (2 \, b x + 2 \, a\right )}{\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}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)*arccoth(cot(b*x+a)),x, algorithm="maxima")
[Out]
1/8*(f*x^2 + 2*e*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) - 1/8*(f*x^2 + 2
*e*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(((b*f*x^2 + 2*b*e*
x)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f*x^2 + 2*b*e*x)*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + (b*f*x^2 + 2*b*
e*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 = 2.13758, size = 1724, normalized size = 10.64 \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)*arccoth(cot(b*x+a)),x, algorithm="fricas")
[Out]
1/16*((2*I*b*f*x + 2*I*b*e)*dilog(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + (2*I*b*f*x + 2*I*b*e)*dilog(I*cos(2
*b*x + 2*a) - sin(2*b*x + 2*a)) + (-2*I*b*f*x - 2*I*b*e)*dilog(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a)) + (-2*I
*b*f*x - 2*I*b*e)*dilog(-I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) + 4*(b^2*f*x^2 + 2*b^2*e*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*(2*a*b*e - a^2*f)*log(cos(2*b*x + 2
*a) + I*sin(2*b*x + 2*a) + I) - 2*(2*a*b*e - a^2*f)*log(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) - 2*(b^2*f*
x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log(I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) + 2*(b^2*f*x^2 + 2*b^2*e*x +
2*a*b*e - a^2*f)*log(I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a) + 1) - 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)
*log(-I*cos(2*b*x + 2*a) + sin(2*b*x + 2*a) + 1) + 2*(b^2*f*x^2 + 2*b^2*e*x + 2*a*b*e - a^2*f)*log(-I*cos(2*b*
x + 2*a) - sin(2*b*x + 2*a) + 1) + 2*(2*a*b*e - a^2*f)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + I) - 2*(2*
a*b*e - a^2*f)*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + I) + f*polylog(3, I*cos(2*b*x + 2*a) + sin(2*b*x +
2*a)) - f*polylog(3, I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)) + f*polylog(3, -I*cos(2*b*x + 2*a) + sin(2*b*x +
2*a)) - f*polylog(3, -I*cos(2*b*x + 2*a) - sin(2*b*x + 2*a)))/b^2
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right ) \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)*acoth(cot(b*x+a)),x)
[Out]
Integral((e + f*x)*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 )} \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)*arccoth(cot(b*x+a)),x, algorithm="giac")
[Out]
integrate((f*x + e)*arccoth(cot(b*x + a)), x)