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

Optimal. Leaf size=116 \[ \frac{i \text{PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{8 b^2}-\frac{i x \text{PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} i x^2 \log \left (1-i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac{1}{6} i b x^3 \]

[Out]

(I/6)*b*x^3 + (x^2*ArcTan[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 = 0.196986, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5195, 2184, 2190, 2531, 2282, 6589} \[ \frac{i \text{PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{8 b^2}-\frac{i x \text{PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac{1}{4} i x^2 \log \left (1-i c e^{2 a+2 b x}\right )+\frac{1}{2} x^2 \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac{1}{6} i b x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/6)*b*x^3 + (x^2*ArcTan[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 5195

Int[ArcTan[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTan[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 \tan ^{-1}(c-(i-c) \tanh (a+b x)) \, dx &=\frac{1}{2} x^2 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 = 4.98064, size = 102, normalized size = 0.88 \[ \frac{1}{2} x^2 \tan ^{-1}(c+(c-i) \tanh (a+b x))-\frac{i \left (-2 b x \text{PolyLog}\left (2,-\frac{i e^{-2 (a+b x)}}{c}\right )-\text{PolyLog}\left (3,-\frac{i e^{-2 (a+b x)}}{c}\right )+2 b^2 x^2 \log \left (1+\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*ArcTan[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 = 3.744, size = 1534, normalized size = 13.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*Pi*x^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))^3+1/8*Pi*x^2*csgn((2*
exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))^3-1/8*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/(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+1/3/b^2/(I-c)*a^3-1/6*b*x^3/(I-c)+1/2/b/(I-c)*x*a^2+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+1/8*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))-1/4*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*exp(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*Pi*x^2*cs
gn(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))^2-1/8*Pi*x^2*csgn(I*(-2*I*e
xp(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*P
i*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-
1/8*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))^2-1/4*I/b^2*a^2*ln(exp(2*b*x+2*a)*c+I)+1/8*I*polylog(3,I*c*exp(2*b*x+2*a))/
b^2+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(I*(2*exp(2*b*x+2*a)*c+2*I)/
(exp(2*b*x+2*a)+1))^3-1/8*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))*c
sgn(I*(-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*I*exp(2*b*x+2*a)+2*exp(
2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2+1/4*I*x^2*ln(-2*exp(2*b*x+2*a)*c-2*I)-1/4*I*x^2*ln(2*I*exp(2*b*x+2*a)-2*ex
p(2*b*x+2*a)*c)-1/4*I*x^2*ln(1-I*c*exp(2*b*x+2*a))+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*(-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))+1/2*I/b*c/(
I-c)*x*a^2-1/2*I/b^2*c*a^2/(I-c)*ln(exp(b*x+a))-1/2/b^2*a^2/(I-c)*ln(exp(b*x+a))-1/2*I/b*ln(1-I*c*exp(2*b*x+2*
a))*x*a+1/2*I/b*a*ln(1-I*exp(b*x+a)*(-I*c)^(1/2))*x+1/2*I/b*a*ln(1+I*exp(b*x+a)*(-I*c)^(1/2))*x+1/3*I/b^2*c/(I
-c)*a^3-1/6*I*c*b*x^3/(I-c)-1/4*I*x*polylog(2,I*c*exp(2*b*x+2*a))/b

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Maxima [A]  time = 5.92242, size = 144, normalized size = 1.24 \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} \arctan \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*arctan(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*arctan((c - I)*tanh(b*x + a) + c)

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Fricas [C]  time = 1.95827, size = 710, normalized size = 6.12 \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*arctan(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*dil
og(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)*log(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*polylog
(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]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{i b \int \frac{x^{2}}{i c e^{2 a} e^{2 b x} - 1}\, dx}{2} + \frac{i x^{2} \log{\left (- i c + \frac{i c}{e^{2 a} e^{2 b x} + 1} - \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 + \frac{1}{e^{2 a} e^{2 b x} + 1} - \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} \right )}}{4} - \frac{i x^{2} \log{\left (i c - \frac{i c}{e^{2 a} e^{2 b x} + 1} + \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 - \frac{1}{e^{2 a} e^{2 b x} + 1} + \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*b*Integral(x**2/(I*c*exp(2*a)*exp(2*b*x) - 1), x)/2 + I*x**2*log(-I*c + I*c/(exp(2*a)*exp(2*b*x) + 1) - I*c
*exp(a)*exp(b*x)/(exp(a)*exp(b*x) + exp(-a)*exp(-b*x)) + 1 + 1/(exp(2*a)*exp(2*b*x) + 1) - exp(a)*exp(b*x)/(ex
p(a)*exp(b*x) + exp(-a)*exp(-b*x)))/4 - I*x**2*log(I*c - I*c/(exp(2*a)*exp(2*b*x) + 1) + I*c*exp(a)*exp(b*x)/(
exp(a)*exp(b*x) + exp(-a)*exp(-b*x)) + 1 - 1/(exp(2*a)*exp(2*b*x) + 1) + exp(a)*exp(b*x)/(exp(a)*exp(b*x) + ex
p(-a)*exp(-b*x)))/4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \arctan \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*arctan(c-(I-c)*tanh(b*x+a)),x, algorithm="giac")

[Out]

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