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

Optimal. Leaf size=229 \[ \frac{i f (e+f x) \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f (e+f x) \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}+\frac{i f^2 \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}+\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f} \]

[Out]

((e + f*x)^3*ArcTan[E^(2*a + 2*b*x)])/(3*f) + ((e + f*x)^3*ArcTan[Coth[a + b*x]])/(3*f) - ((I/4)*(e + f*x)^2*P
olyLog[2, (-I)*E^(2*a + 2*b*x)])/b + ((I/4)*(e + f*x)^2*PolyLog[2, I*E^(2*a + 2*b*x)])/b + ((I/4)*f*(e + f*x)*
PolyLog[3, (-I)*E^(2*a + 2*b*x)])/b^2 - ((I/4)*f*(e + f*x)*PolyLog[3, I*E^(2*a + 2*b*x)])/b^2 - ((I/8)*f^2*Pol
yLog[4, (-I)*E^(2*a + 2*b*x)])/b^3 + ((I/8)*f^2*PolyLog[4, I*E^(2*a + 2*b*x)])/b^3

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Rubi [A]  time = 0.15292, antiderivative size = 229, 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 = {5185, 4180, 2531, 6609, 2282, 6589} \[ \frac{i f (e+f x) \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f (e+f x) \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f^2 \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}+\frac{i f^2 \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{i (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}+\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*ArcTan[Coth[a + b*x]],x]

[Out]

((e + f*x)^3*ArcTan[E^(2*a + 2*b*x)])/(3*f) + ((e + f*x)^3*ArcTan[Coth[a + b*x]])/(3*f) - ((I/4)*(e + f*x)^2*P
olyLog[2, (-I)*E^(2*a + 2*b*x)])/b + ((I/4)*(e + f*x)^2*PolyLog[2, I*E^(2*a + 2*b*x)])/b + ((I/4)*f*(e + f*x)*
PolyLog[3, (-I)*E^(2*a + 2*b*x)])/b^2 - ((I/4)*f*(e + f*x)*PolyLog[3, I*E^(2*a + 2*b*x)])/b^2 - ((I/8)*f^2*Pol
yLog[4, (-I)*E^(2*a + 2*b*x)])/b^3 + ((I/8)*f^2*PolyLog[4, I*E^(2*a + 2*b*x)])/b^3

Rule 5185

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, 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 \tan ^{-1}(\coth (a+b x)) \, dx &=\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}+\frac{b \int (e+f x)^3 \text{sech}(2 a+2 b x) \, dx}{3 f}\\ &=\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}-\frac{1}{2} i \int (e+f x)^2 \log \left (1-i e^{2 a+2 b x}\right ) \, dx+\frac{1}{2} i \int (e+f x)^2 \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac{(i f) \int (e+f x) \text{Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{2 b}-\frac{(i f) \int (e+f x) \text{Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac{i f (e+f x) \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f (e+f x) \text{Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac{\left (i f^2\right ) \int \text{Li}_3\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b^2}+\frac{\left (i f^2\right ) \int \text{Li}_3\left (i e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac{i f (e+f x) \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f (e+f x) \text{Li}_3\left (i e^{2 a+2 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^{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^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac{(e+f x)^3 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{3 f}+\frac{(e+f x)^3 \tan ^{-1}(\coth (a+b x))}{3 f}-\frac{i (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}+\frac{i (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}+\frac{i f (e+f x) \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f (e+f x) \text{Li}_3\left (i e^{2 a+2 b x}\right )}{4 b^2}-\frac{i f^2 \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}+\frac{i f^2 \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 2.6565, size = 375, normalized size = 1.64 \[ \frac{1}{3} x \left (3 e^2+3 e f x+f^2 x^2\right ) \tan ^{-1}(\coth (a+b x))+\frac{i \left (-6 b^2 (e+f x)^2 \text{PolyLog}\left (2,-i e^{2 (a+b x)}\right )+6 b^2 (e+f x)^2 \text{PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b e f \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b e f \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )+6 b f^2 x \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b f^2 x \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )-3 f^2 \text{PolyLog}\left (4,-i e^{2 (a+b x)}\right )+3 f^2 \text{PolyLog}\left (4,i e^{2 (a+b x)}\right )+12 b^3 e^2 x \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e^2 x \log \left (1+i e^{2 (a+b x)}\right )+12 b^3 e f x^2 \log \left (1-i e^{2 (a+b x)}\right )-12 b^3 e f x^2 \log \left (1+i e^{2 (a+b x)}\right )+4 b^3 f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-4 b^3 f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*ArcTan[Coth[a + b*x]],x]

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2)*ArcTan[Coth[a + b*x]])/3 + ((I/24)*(12*b^3*e^2*x*Log[1 - I*E^(2*(a + b*x))] + 1
2*b^3*e*f*x^2*Log[1 - I*E^(2*(a + b*x))] + 4*b^3*f^2*x^3*Log[1 - I*E^(2*(a + b*x))] - 12*b^3*e^2*x*Log[1 + I*E
^(2*(a + b*x))] - 12*b^3*e*f*x^2*Log[1 + I*E^(2*(a + b*x))] - 4*b^3*f^2*x^3*Log[1 + I*E^(2*(a + b*x))] - 6*b^2
*(e + f*x)^2*PolyLog[2, (-I)*E^(2*(a + b*x))] + 6*b^2*(e + f*x)^2*PolyLog[2, I*E^(2*(a + b*x))] + 6*b*e*f*Poly
Log[3, (-I)*E^(2*(a + b*x))] + 6*b*f^2*x*PolyLog[3, (-I)*E^(2*(a + b*x))] - 6*b*e*f*PolyLog[3, I*E^(2*(a + b*x
))] - 6*b*f^2*x*PolyLog[3, I*E^(2*(a + b*x))] - 3*f^2*PolyLog[4, (-I)*E^(2*(a + b*x))] + 3*f^2*PolyLog[4, I*E^
(2*(a + b*x))]))/b^3

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Maple [C]  time = 7.741, size = 5425, 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*arctan(coth(b*x+a)),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \,{\left (f^{2} x^{3} + 3 \, e f x^{2} + 3 \, e^{2} x\right )} \arctan \left (e^{\left (2 \, b x + 2 \, a\right )} + 1, e^{\left (2 \, b x + 2 \, a\right )} - 1\right ) + \int \frac{2 \,{\left (b f^{2} x^{3} e^{\left (2 \, a\right )} + 3 \, b e f x^{2} e^{\left (2 \, a\right )} + 3 \, b e^{2} x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{3 \,{\left (e^{\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*arctan(coth(b*x+a)),x, algorithm="maxima")

[Out]

1/3*(f^2*x^3 + 3*e*f*x^2 + 3*e^2*x)*arctan2(e^(2*b*x + 2*a) + 1, e^(2*b*x + 2*a) - 1) + integrate(2/3*(b*f^2*x
^3*e^(2*a) + 3*b*e*f*x^2*e^(2*a) + 3*b*e^2*x*e^(2*a))*e^(2*b*x)/(e^(4*b*x + 4*a) + 1), x)

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Fricas [C]  time = 2.62113, size = 2901, normalized size = 12.67 \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*arctan(coth(b*x+a)),x, algorithm="fricas")

[Out]

1/6*(6*I*f^2*polylog(4, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 6*I*f^2*polylog(4, -1/2*sqrt(4*I)*(co
sh(b*x + a) + sinh(b*x + a))) - 6*I*f^2*polylog(4, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 6*I*f^2*p
olylog(4, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*(b^3*f^2*x^3 + 3*b^3*e*f*x^2 + 3*b^3*e^2*x)*arc
tan(cosh(b*x + a)/sinh(b*x + a)) + (3*I*b^2*f^2*x^2 + 6*I*b^2*e*f*x + 3*I*b^2*e^2)*dilog(1/2*sqrt(4*I)*(cosh(b
*x + a) + sinh(b*x + a))) + (3*I*b^2*f^2*x^2 + 6*I*b^2*e*f*x + 3*I*b^2*e^2)*dilog(-1/2*sqrt(4*I)*(cosh(b*x + a
) + sinh(b*x + a))) + (-3*I*b^2*f^2*x^2 - 6*I*b^2*e*f*x - 3*I*b^2*e^2)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + s
inh(b*x + a))) + (-3*I*b^2*f^2*x^2 - 6*I*b^2*e*f*x - 3*I*b^2*e^2)*dilog(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(
b*x + a))) + (I*b^3*f^2*x^3 + 3*I*b^3*e*f*x^2 + 3*I*b^3*e^2*x + 3*I*a*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log
(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^3*f^2*x^3 + 3*I*b^3*e*f*x^2 + 3*I*b^3*e^2*x + 3*I*a
*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^3*f^2*x^
3 - 3*I*b^3*e*f*x^2 - 3*I*b^3*e^2*x - 3*I*a*b^2*e^2 + 3*I*a^2*b*e*f - I*a^3*f^2)*log(1/2*sqrt(-4*I)*(cosh(b*x
+ a) + sinh(b*x + a)) + 1) + (-I*b^3*f^2*x^3 - 3*I*b^3*e*f*x^2 - 3*I*b^3*e^2*x - 3*I*a*b^2*e^2 + 3*I*a^2*b*e*f
- I*a^3*f^2)*log(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-3*I*a*b^2*e^2 + 3*I*a^2*b*e*f - I*a
^3*f^2)*log(I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-3*I*a*b^2*e^2 + 3*I*a^2*b*e*f - I*a^3*f^2)*lo
g(-I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (3*I*a*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log(I*sqrt(-
4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (3*I*a*b^2*e^2 - 3*I*a^2*b*e*f + I*a^3*f^2)*log(-I*sqrt(-4*I) + 2*
cosh(b*x + a) + 2*sinh(b*x + a)) + (-6*I*b*f^2*x - 6*I*b*e*f)*polylog(3, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b
*x + a))) + (-6*I*b*f^2*x - 6*I*b*e*f)*polylog(3, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (6*I*b*f^2
*x + 6*I*b*e*f)*polylog(3, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (6*I*b*f^2*x + 6*I*b*e*f)*polylog
(3, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))))/b^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*atan(coth(b*x+a)),x)

[Out]

Timed out

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

[Out]

integrate((f*x + e)^2*arctan(coth(b*x + a)), x)