3.232 \(\int (e+f x)^2 \coth ^{-1}(\tan (a+b x)) \, dx\)

Optimal. Leaf size=234 \[ \frac{f (e+f x) \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17167, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6252, 4181, 2531, 6609, 2282, 6589} \[ \frac{f (e+f x) \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*ArcCoth[Tan[a + b*x]],x]

[Out]

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

Rule 6252

Int[ArcCoth[Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m + 1)*ArcCoth[
Tan[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)^2 \coth ^{-1}(\tan (a+b x)) \, dx &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}-\frac{b \int (e+f x)^3 \sec (2 a+2 b x) \, dx}{3 f}\\ &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}+\frac{1}{2} \int (e+f x)^2 \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac{1}{2} \int (e+f x)^2 \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{(i f) \int (e+f x) \text{Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}-\frac{(i f) \int (e+f x) \text{Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{2 b}\\ &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f^2 \int \text{Li}_3\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}+\frac{f^2 \int \text{Li}_3\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b^2}\\ &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{\left (i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^3}-\frac{\left (i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^3}\\ &=\frac{(e+f x)^3 \coth ^{-1}(\tan (a+b x))}{3 f}+\frac{i (e+f x)^3 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac{f (e+f x) \text{Li}_3\left (-i e^{2 i (a+b x)}\right )}{4 b^2}-\frac{f (e+f x) \text{Li}_3\left (i e^{2 i (a+b x)}\right )}{4 b^2}+\frac{i f^2 \text{Li}_4\left (-i e^{2 i (a+b x)}\right )}{8 b^3}-\frac{i f^2 \text{Li}_4\left (i e^{2 i (a+b x)}\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.188441, size = 409, normalized size = 1.75 \[ \frac{1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \coth ^{-1}(\tan (a+b x))+\frac{-6 i b^2 (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 i (a+b x)}\right )+6 i b^2 (e+f x)^2 \text{PolyLog}\left (2,i e^{2 i (a+b x)}\right )+6 b e f \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b e f \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+6 b f^2 x \text{PolyLog}\left (3,-i e^{2 i (a+b x)}\right )-6 b f^2 x \text{PolyLog}\left (3,i e^{2 i (a+b x)}\right )+3 i f^2 \text{PolyLog}\left (4,-i e^{2 i (a+b x)}\right )-3 i f^2 \text{PolyLog}\left (4,i e^{2 i (a+b x)}\right )-12 b^3 e^2 x \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e^2 x \log \left (1+i e^{2 i (a+b x)}\right )-12 b^3 e f x^2 \log \left (1-i e^{2 i (a+b x)}\right )+12 b^3 e f x^2 \log \left (1+i e^{2 i (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1-i e^{2 i (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1+i e^{2 i (a+b x)}\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*ArcCoth[Tan[a + b*x]],x]

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcCoth[Tan[a + b*x]])/3 + (-12*b^3*e^2*x*Log[1 - I*E^((2*I)*(a + b*x))] - 12*b
^3*e*f*x^2*Log[1 - I*E^((2*I)*(a + b*x))] - 4*b^3*f^2*x^3*Log[1 - I*E^((2*I)*(a + b*x))] + 12*b^3*e^2*x*Log[1
+ I*E^((2*I)*(a + b*x))] + 12*b^3*e*f*x^2*Log[1 + I*E^((2*I)*(a + b*x))] + 4*b^3*f^2*x^3*Log[1 + I*E^((2*I)*(a
+ b*x))] - (6*I)*b^2*(e + f*x)^2*PolyLog[2, (-I)*E^((2*I)*(a + b*x))] + (6*I)*b^2*(e + f*x)^2*PolyLog[2, I*E^
((2*I)*(a + b*x))] + 6*b*e*f*PolyLog[3, (-I)*E^((2*I)*(a + b*x))] + 6*b*f^2*x*PolyLog[3, (-I)*E^((2*I)*(a + b*
x))] - 6*b*e*f*PolyLog[3, I*E^((2*I)*(a + b*x))] - 6*b*f^2*x*PolyLog[3, I*E^((2*I)*(a + b*x))] + (3*I)*f^2*Pol
yLog[4, (-I)*E^((2*I)*(a + b*x))] - (3*I)*f^2*PolyLog[4, I*E^((2*I)*(a + b*x))])/(24*b^3)

________________________________________________________________________________________

Maple [C]  time = 13.508, size = 5543, normalized size = 23.7 \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)^2*arccoth(tan(b*x+a)),x)

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{12} \,{\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} 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}{12} \,{\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} 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{2 \,{\left ({\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \cos \left (4 \, b x + 4 \, a\right ) \cos \left (2 \, b x + 2 \, a\right ) +{\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) +{\left (b f^{2} x^{3} + 3 \, b e f x^{2} + 3 \, b e^{2} x\right )} \cos \left (2 \, b x + 2 \, a\right )\right )}}{3 \,{\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)^2*arccoth(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/12*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*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)
- 1/12*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*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(2/3*((b*f^2*x^3 + 3*b*e*f*x^2 + 3*b*e^2*x)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f^2*x^3 + 3*
b*e*f*x^2 + 3*b*e^2*x)*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + (b*f^2*x^3 + 3*b*e*f*x^2 + 3*b*e^2*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)

________________________________________________________________________________________

Fricas [C]  time = 2.20852, size = 3357, normalized size = 14.35 \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)^2*arccoth(tan(b*x+a)),x, algorithm="fricas")

[Out]

1/48*(3*I*f^2*polylog(4, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 3*I*f^2*polylog(4, (I
*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) - 3*I*f^2*polylog(4, (-I*tan(b*x + a)^2 + 2*tan(b*
x + a) + I)/(tan(b*x + a)^2 + 1)) - 3*I*f^2*polylog(4, (-I*tan(b*x + a)^2 - 2*tan(b*x + a) + I)/(tan(b*x + a)^
2 + 1)) + (6*I*b^2*f^2*x^2 + 12*I*b^2*e*f*x + 6*I*b^2*e^2)*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I
+ 1)/(tan(b*x + a)^2 + 1) + 1) + (6*I*b^2*f^2*x^2 + 12*I*b^2*e*f*x + 6*I*b^2*e^2)*dilog(-((I + 1)*tan(b*x + a
)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + (-6*I*b^2*f^2*x^2 - 12*I*b^2*e*f*x - 6*I*b^2*e^2)*di
log(-(-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + (-6*I*b^2*f^2*x^2 - 12*I*b
^2*e*f*x - 6*I*b^2*e^2)*dilog(-(-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) -
4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I + 1)*tan(b*x + a)^
2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I + 1)*tan(b
*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) - 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I
+ 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I - 1)/(tan(b*x + a)^2 + 1)) + 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3
*e^2*x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x +
a)^2 + 1)) - 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log((-(I - 1
)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 4*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*
x + 3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log((-(I - 1)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^
2 + 1)) + 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I - 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I + 1)/(tan
(b*x + a)^2 + 1)) - 4*(3*a*b^2*e^2 - 3*a^2*b*e*f + a^3*f^2)*log(((I - 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I
+ 1)/(tan(b*x + a)^2 + 1)) + 8*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x)*log((tan(b*x + a) + 1)/(tan(b*x +
a) - 1)) + 6*(b*f^2*x + b*e*f)*polylog(3, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) - 6*(b
*f^2*x + b*e*f)*polylog(3, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) + 6*(b*f^2*x + b*e*f)
*polylog(3, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) - 6*(b*f^2*x + b*e*f)*polylog(3, (-
I*tan(b*x + a)^2 - 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)))/b^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*acoth(tan(b*x+a)),x)

[Out]

Integral((e + f*x)**2*acoth(tan(a + b*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{2} \operatorname{arcoth}\left (\tan \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)^2*arccoth(tan(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*arccoth(tan(b*x + a)), x)