3.193 \(\int x \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx\)

Optimal. Leaf size=113 \[ \frac{i \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2}-\frac{i x \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))+\frac{1}{6} i b x^3 \]

[Out]

(I/6)*b*x^3 + (x^2*ArcCot[c + (I + c)*Tanh[a + b*x]])/2 - (I/4)*x^2*Log[1 + I*c*E^(2*a + 2*b*x)] - ((I/4)*x*Po
lyLog[2, (-I)*c*E^(2*a + 2*b*x)])/b + ((I/8)*PolyLog[3, (-I)*c*E^(2*a + 2*b*x)])/b^2

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Rubi [A]  time = 0.197445, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {5196, 2184, 2190, 2531, 2282, 6589} \[ \frac{i \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2}-\frac{i x \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))+\frac{1}{6} i b x^3 \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCot[c + (I + c)*Tanh[a + b*x]],x]

[Out]

(I/6)*b*x^3 + (x^2*ArcCot[c + (I + c)*Tanh[a + b*x]])/2 - (I/4)*x^2*Log[1 + I*c*E^(2*a + 2*b*x)] - ((I/4)*x*Po
lyLog[2, (-I)*c*E^(2*a + 2*b*x)])/b + ((I/8)*PolyLog[3, (-I)*c*E^(2*a + 2*b*x)])/b^2

Rule 5196

Int[ArcCot[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
 + 1)*ArcCot[c + d*Tanh[a + b*x]])/(f*(m + 1)), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d + c*E^(2
*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - d)^2, -1]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && 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 x \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx &=\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{2} b \int \frac{x^2}{-i+c e^{2 a+2 b x}} \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{2} (i b c) \int \frac{e^{2 a+2 b x} x^2}{-i+c e^{2 a+2 b x}} \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{2} i \int x \log \left (1+i c e^{2 a+2 b x}\right ) \, dx\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac{i \int \text{Li}_2\left (-i c e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac{1}{6} i b x^3+\frac{1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac{i x \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac{i \text{Li}_3\left (-i c e^{2 a+2 b x}\right )}{8 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0864655, size = 102, normalized size = 0.9 \[ \frac{1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))-\frac{i \left (-2 b x \text{PolyLog}\left (2,\frac{i e^{-2 (a+b x)}}{c}\right )-\text{PolyLog}\left (3,\frac{i e^{-2 (a+b x)}}{c}\right )+2 b^2 x^2 \log \left (1-\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCot[c + (I + c)*Tanh[a + b*x]],x]

[Out]

(x^2*ArcCot[c + (I + c)*Tanh[a + b*x]])/2 - ((I/8)*(2*b^2*x^2*Log[1 - I/(c*E^(2*(a + b*x)))] - 2*b*x*PolyLog[2
, I/(c*E^(2*(a + b*x)))] - PolyLog[3, I/(c*E^(2*(a + b*x)))]))/b^2

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Maple [C]  time = 8.242, size = 1513, normalized size = 13.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(c+(I+c)*tanh(b*x+a)),x)

[Out]

1/8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^3-1/2/b^2*a^2/(I+c)*ln(exp(b*x+a))+1/2/b/(I+c)*
x*a^2-1/4*I/b^2*ln(1+I*c*exp(2*b*x+2*a))*a^2-1/4*I/b^2*polylog(2,-I*c*exp(2*b*x+2*a))*a+1/2*I/b^2*a^2*ln(1+I*e
xp(b*x+a)*(I*c)^(1/2))+1/2*I/b^2*a^2*ln(1-I*exp(b*x+a)*(I*c)^(1/2))+1/2*I/b^2*a*dilog(1+I*exp(b*x+a)*(I*c)^(1/
2))+1/2*I/b^2*a*dilog(1-I*exp(b*x+a)*(I*c)^(1/2))+1/8*I*polylog(3,-I*c*exp(2*b*x+2*a))/b^2-1/3*I/b^2*c/(I+c)*a
^3+1/6*I*b*c/(I+c)*x^3-1/8*Pi*x^2*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^3+1/2*I/b
*a*ln(1+I*exp(b*x+a)*(I*c)^(1/2))*x-1/8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))*csgn((2*exp
(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^2-1/4*I*x^2*ln(1+I*c*exp(2*b*x+2*a))+1/4*I*x^2*ln(2*exp(2*b*x+2*a)*c-2*
I)-1/8*Pi*x^2*csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^3-1/8*Pi*x^2*csgn((2*exp(2*b*x+
2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^3+1/3/b^2/(I+c)*a^3-1/6*b/(I+c)*x^3+1/2*I/b*a*ln(1-I*exp(b*x+a)*(I*c)^(1/2))*x
-1/4*I*x^2*ln(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)-1/4*I*x*polylog(2,-I*c*exp(2*b*x+2*a))/b-1/4*I/b^2*a^2*ln
(-exp(2*b*x+2*a)*c+I)-1/8*Pi*x^2*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))*csgn((2*I*
exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))+1/8*Pi*x^2*csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a
)+1))^2+1/8*Pi*x^2*csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2-1/8*Pi*x^2*csgn(I/(exp(2
*b*x+2*a)+1))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^2+1/8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)
/(exp(2*b*x+2*a)+1))*csgn((2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))+1/8*Pi*x^2*csgn(I*(2*I*exp(2*b*x+2*a)+2
*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))*csgn((2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2+1/8*
Pi*x^2*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c))*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b
*x+2*a)+1))^2+1/8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+
2*a)+1))^2-1/8*Pi*x^2*csgn(I*(2*exp(2*b*x+2*a)*c-2*I))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))^2-1
/2*I/b*ln(1+I*c*exp(2*b*x+2*a))*x*a+1/2*I/b^2*c*a^2/(I+c)*ln(exp(b*x+a))-1/2*I/b*c/(I+c)*x*a^2+1/8*Pi*x^2*csgn
(I/(exp(2*b*x+2*a)+1))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I))*csgn(I*(2*exp(2*b*x+2*a)*c-2*I)/(exp(2*b*x+2*a)+1))-1/
8*Pi*x^2*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c))*csgn(I*(2*I*exp(2*b*x+2*a)
+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))

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Maxima [A]  time = 5.78706, size = 144, normalized size = 1.27 \begin{align*} -{\left (\frac{2 \, x^{3}}{3 i \, c - 3} - \frac{2 \, b^{2} x^{2} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) -{\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{3}{\left (-i \, c + 1\right )}}\right )} b{\left (c + i\right )} + \frac{1}{2} \, x^{2} \operatorname{arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(I+c)*tanh(b*x+a)),x, algorithm="maxima")

[Out]

-(2*x^3/(3*I*c - 3) - (2*b^2*x^2*log(I*c*e^(2*b*x + 2*a) + 1) + 2*b*x*dilog(-I*c*e^(2*b*x + 2*a)) - polylog(3,
 -I*c*e^(2*b*x + 2*a)))/(b^3*(2*I*c - 2)))*b*(c + I) + 1/2*x^2*arccot((c + I)*tanh(b*x + a) + c)

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Fricas [C]  time = 2.17546, size = 720, normalized size = 6.37 \begin{align*} \frac{2 i \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (\frac{{\left (c e^{\left (2 \, b x + 2 \, a\right )} - i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c + i}\right ) + 2 i \, a^{3} - 6 i \, b x{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) - 6 i \, b x{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) - 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (b x + a\right )} + i \, \sqrt{-4 i \, c}}{2 \, c}\right ) - 3 i \, a^{2} \log \left (\frac{2 \, c e^{\left (b x + a\right )} - i \, \sqrt{-4 i \, c}}{2 \, c}\right ) +{\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) +{\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 6 i \,{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \,{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(I+c)*tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(2*I*b^3*x^3 + 3*I*b^2*x^2*log((c*e^(2*b*x + 2*a) - I)*e^(-2*b*x - 2*a)/(c + I)) + 2*I*a^3 - 6*I*b*x*dilo
g(1/2*sqrt(-4*I*c)*e^(b*x + a)) - 6*I*b*x*dilog(-1/2*sqrt(-4*I*c)*e^(b*x + a)) - 3*I*a^2*log(1/2*(2*c*e^(b*x +
 a) + I*sqrt(-4*I*c))/c) - 3*I*a^2*log(1/2*(2*c*e^(b*x + a) - I*sqrt(-4*I*c))/c) + (-3*I*b^2*x^2 + 3*I*a^2)*lo
g(1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) + (-3*I*b^2*x^2 + 3*I*a^2)*log(-1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) + 6*I*po
lylog(3, 1/2*sqrt(-4*I*c)*e^(b*x + a)) + 6*I*polylog(3, -1/2*sqrt(-4*I*c)*e^(b*x + a)))/b^2

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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(x*acot(c+(I+c)*tanh(b*x+a)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(I+c)*tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arccot((c + I)*tanh(b*x + a) + c), x)