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

Optimal. Leaf size=82 \[ -\frac {i \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+x \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac {1}{2} i b x^2 \]

[Out]

1/2*I*b*x^2+x*arctan(c-(I-c)*tanh(b*x+a))-1/2*I*x*ln(1-I*c*exp(2*b*x+2*a))-1/4*I*polylog(2,I*c*exp(2*b*x+2*a))
/b

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Rubi [A]  time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5187, 2184, 2190, 2279, 2391} \[ -\frac {i \text {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+x \tan ^{-1}(c-(-c+i) \tanh (a+b x))+\frac {1}{2} i b x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/2)*b*x^2 + x*ArcTan[c - (I - c)*Tanh[a + b*x]] - (I/2)*x*Log[1 - I*c*E^(2*a + 2*b*x)] - ((I/4)*PolyLog[2, I
*c*E^(2*a + 2*b*x)])/b

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5187

Int[ArcTan[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcTan[c + d*Tanh[a + b*x]], x] - Dist
[b, Int[x/(c - d + c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[(c - d)^2, -1]

Rubi steps

\begin {align*} \int \tan ^{-1}(c-(i-c) \tanh (a+b x)) \, dx &=x \tan ^{-1}(c-(i-c) \tanh (a+b x))-b \int \frac {x}{i+c e^{2 a+2 b x}} \, dx\\ &=\frac {1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \tanh (a+b x))-(i b c) \int \frac {e^{2 a+2 b x} x}{i+c e^{2 a+2 b x}} \, dx\\ &=\frac {1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \tanh (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+\frac {1}{2} i \int \log \left (1-i c e^{2 a+2 b x}\right ) \, dx\\ &=\frac {1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \tanh (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )+\frac {i \operatorname {Subst}\left (\int \frac {\log (1-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b}\\ &=\frac {1}{2} i b x^2+x \tan ^{-1}(c-(i-c) \tanh (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 a+2 b x}\right )-\frac {i \text {Li}_2\left (i c e^{2 a+2 b x}\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 1.85, size = 71, normalized size = 0.87 \[ x \tan ^{-1}(c+(c-i) \tanh (a+b x))-\frac {i \left (2 b x \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-\text {Li}_2\left (-\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcTan[c + (-I + c)*Tanh[a + b*x]] - ((I/4)*(2*b*x*Log[1 + I/(c*E^(2*(a + b*x)))] - PolyLog[2, (-I)/(c*E^(2*
(a + b*x)))]))/b

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fricas [B]  time = 0.55, size = 187, normalized size = 2.28 \[ \frac {i \, b^{2} x^{2} + i \, b x \log \left (-\frac {{\left (c e^{\left (2 \, b x + 2 \, a\right )} + i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c - i}\right ) - i \, a^{2} + {\left (-i \, b x - i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + {\left (-i \, b x - i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) + i \, a \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - i \, {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - i \, {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*b^2*x^2 + I*b*x*log(-(c*e^(2*b*x + 2*a) + I)*e^(-2*b*x - 2*a)/(c - I)) - I*a^2 + (-I*b*x - I*a)*log(1/2
*sqrt(4*I*c)*e^(b*x + a) + 1) + (-I*b*x - I*a)*log(-1/2*sqrt(4*I*c)*e^(b*x + a) + 1) + I*a*log(1/2*(2*c*e^(b*x
+ a) + I*sqrt(4*I*c))/c) + I*a*log(1/2*(2*c*e^(b*x + a) - I*sqrt(4*I*c))/c) - I*dilog(1/2*sqrt(4*I*c)*e^(b*x
+ a)) - I*dilog(-1/2*sqrt(4*I*c)*e^(b*x + a)))/b

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \arctan \left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.46, size = 1351, normalized size = 16.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b/(c-I)/(I-c)*dilog(1/2*(I+(c-I)*tanh(b*x+a)+c)/c)*c+1/2/b/(c-I)/(I-c)*dilog(((c-I)*tanh(b*x+a)+c-I)/(-2*
I+2*c))*c+1/b/(c-I)*arctan((c-I)*tanh(b*x+a)+c)/(2*I-2*c)*ln((c-I)*tanh(b*x+a)+c-I)-1/4*I/b/(c-I)/(I-c)*dilog(
((c-I)*tanh(b*x+a)+c-I)/(-2*I+2*c))+1/8*I/b/(c-I)/(I-c)*ln((c-I)*tanh(b*x+a)+c-I)^2+1/4*I/b/(c-I)/(I-c)*dilog(
1/2*(I+(c-I)*tanh(b*x+a)+c)/c)-1/b/(c-I)*arctan((c-I)*tanh(b*x+a)+c)/(2*I-2*c)*ln((c-I)*tanh(b*x+a)-c+I)+1/2/b
/(c-I)/(I-c)*dilog(-1/2*I*(I+(c-I)*tanh(b*x+a)+c))*c-1/4/b/(c-I)/(I-c)*ln((c-I)*tanh(b*x+a)+c-I)^2*c-1/4*I/b/(
c-I)/(I-c)*dilog(-1/2*I*(I+(c-I)*tanh(b*x+a)+c))-1/4*I/b/(c-I)/(I-c)*ln(1/2*(I+(c-I)*tanh(b*x+a)+c)/c)*ln((c-I
)*tanh(b*x+a)-c+I)*c^2+1/4*I/b/(c-I)/(I-c)*ln(((c-I)*tanh(b*x+a)+c-I)/(-2*I+2*c))*ln((c-I)*tanh(b*x+a)-c+I)*c^
2+1/4*I/b/(c-I)/(I-c)*ln((c-I)*tanh(b*x+a)+c-I)*ln(-1/2*I*(I+(c-I)*tanh(b*x+a)+c))*c^2-1/8*I/b/(c-I)/(I-c)*ln(
(c-I)*tanh(b*x+a)+c-I)^2*c^2-1/4*I/b/(c-I)/(I-c)*dilog(1/2*(I+(c-I)*tanh(b*x+a)+c)/c)*c^2+1/4*I/b/(c-I)/(I-c)*
dilog(((c-I)*tanh(b*x+a)+c-I)/(-2*I+2*c))*c^2-1/4*I/b/(c-I)/(I-c)*ln((c-I)*tanh(b*x+a)+c-I)*ln(-1/2*I*(I+(c-I)
*tanh(b*x+a)+c))+1/4*I/b/(c-I)/(I-c)*ln(1/2*(I+(c-I)*tanh(b*x+a)+c)/c)*ln((c-I)*tanh(b*x+a)-c+I)-1/4*I/b/(c-I)
/(I-c)*ln(((c-I)*tanh(b*x+a)+c-I)/(-2*I+2*c))*ln((c-I)*tanh(b*x+a)-c+I)+1/4*I/b/(c-I)/(I-c)*dilog(-1/2*I*(I+(c
-I)*tanh(b*x+a)+c))*c^2+1/b/(c-I)*arctan((c-I)*tanh(b*x+a)+c)/(2*I-2*c)*ln((c-I)*tanh(b*x+a)-c+I)*c^2-1/b/(c-I
)*arctan((c-I)*tanh(b*x+a)+c)/(2*I-2*c)*ln((c-I)*tanh(b*x+a)+c-I)*c^2-1/2/b/(c-I)/(I-c)*ln(1/2*(I+(c-I)*tanh(b
*x+a)+c)/c)*ln((c-I)*tanh(b*x+a)-c+I)*c+1/2/b/(c-I)/(I-c)*ln(((c-I)*tanh(b*x+a)+c-I)/(-2*I+2*c))*ln((c-I)*tanh
(b*x+a)-c+I)*c+1/2/b/(c-I)/(I-c)*ln((c-I)*tanh(b*x+a)+c-I)*ln(-1/2*I*(I+(c-I)*tanh(b*x+a)+c))*c-2*I/b/(c-I)*ar
ctan((c-I)*tanh(b*x+a)+c)/(2*I-2*c)*ln((c-I)*tanh(b*x+a)-c+I)*c+2*I/b/(c-I)*arctan((c-I)*tanh(b*x+a)+c)/(2*I-2
*c)*ln((c-I)*tanh(b*x+a)+c-I)*c

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maxima [A]  time = 1.95, size = 80, normalized size = 0.98 \[ -2 \, b {\left (c - i\right )} {\left (\frac {2 \, x^{2}}{2 i \, c + 2} - \frac {2 \, b x \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right )}{-2 \, b^{2} {\left (-i \, c - 1\right )}}\right )} + x \arctan \left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*(c - I)*(2*x^2/(2*I*c + 2) - (2*b*x*log(-I*c*e^(2*b*x + 2*a) + 1) + dilog(I*c*e^(2*b*x + 2*a)))/(b^2*(2*I
*c + 2))) + x*arctan((c - I)*tanh(b*x + a) + c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {atan}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c-\mathrm {i}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(c + tanh(a + b*x)*(c - 1i)),x)

[Out]

int(atan(c + tanh(a + b*x)*(c - 1i)), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: CoercionFailed

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