3.1.91 \(\int \text {ArcTan}(c-(i-c) \tanh (a+b x)) \, dx\) [91]

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

[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 = 0.08, 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 = {5295, 2215, 2221, 2317, 2438} \begin {gather*} x \text {ArcTan}(c-(-c+i) \tanh (a+b x))-\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 )+\frac {1}{2} i b x^2 \end {gather*}

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 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 5295

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 \text {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 = 1.10, size = 71, normalized size = 0.87 \begin {gather*} x \text {ArcTan}(c+(-i+c) \tanh (a+b x))-\frac {i \left (2 b x \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-\text {PolyLog}\left (2,-\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{4 b} \end {gather*}

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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 552 vs. \(2 (68 ) = 136\).
time = 0.29, size = 553, normalized size = 6.74

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right )}{2 i-2 c}-\frac {2 i \arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) c}{2 i-2 c}+\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) c^{2}}{2 i-2 c}+\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right )}{2 i-2 c}+\frac {2 i \arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) c}{2 i-2 c}-\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) c^{2}}{2 i-2 c}+\left (i-c \right )^{2} \left (-\frac {i \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right )^{2}}{8 \left (i-c \right )}+\frac {i \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-\frac {i \left (\left (c -i\right ) \tanh \left (b x +a \right )+c +i\right )}{2}\right )}{4 i-4 c}+\frac {i \dilog \left (-\frac {i \left (\left (c -i\right ) \tanh \left (b x +a \right )+c +i\right )}{2}\right )}{4 i-4 c}-\frac {i \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) \ln \left (\frac {\left (c -i\right ) \tanh \left (b x +a \right )+c +i}{2 c}\right )}{4 \left (i-c \right )}-\frac {i \dilog \left (\frac {\left (c -i\right ) \tanh \left (b x +a \right )+c +i}{2 c}\right )}{4 \left (i-c \right )}+\frac {i \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) \ln \left (\frac {-i+\left (c -i\right ) \tanh \left (b x +a \right )+c}{-2 i+2 c}\right )}{4 i-4 c}+\frac {i \dilog \left (\frac {-i+\left (c -i\right ) \tanh \left (b x +a \right )+c}{-2 i+2 c}\right )}{4 i-4 c}\right )}{b \left (c -i\right )}\)	\(553\)
default	\(\frac {-\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right )}{2 i-2 c}-\frac {2 i \arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) c}{2 i-2 c}+\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) c^{2}}{2 i-2 c}+\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right )}{2 i-2 c}+\frac {2 i \arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) c}{2 i-2 c}-\frac {\arctan \left (\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) c^{2}}{2 i-2 c}+\left (i-c \right )^{2} \left (-\frac {i \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right )^{2}}{8 \left (i-c \right )}+\frac {i \ln \left (-i+\left (c -i\right ) \tanh \left (b x +a \right )+c \right ) \ln \left (-\frac {i \left (\left (c -i\right ) \tanh \left (b x +a \right )+c +i\right )}{2}\right )}{4 i-4 c}+\frac {i \dilog \left (-\frac {i \left (\left (c -i\right ) \tanh \left (b x +a \right )+c +i\right )}{2}\right )}{4 i-4 c}-\frac {i \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) \ln \left (\frac {\left (c -i\right ) \tanh \left (b x +a \right )+c +i}{2 c}\right )}{4 \left (i-c \right )}-\frac {i \dilog \left (\frac {\left (c -i\right ) \tanh \left (b x +a \right )+c +i}{2 c}\right )}{4 \left (i-c \right )}+\frac {i \ln \left (\left (c -i\right ) \tanh \left (b x +a \right )-c +i\right ) \ln \left (\frac {-i+\left (c -i\right ) \tanh \left (b x +a \right )+c}{-2 i+2 c}\right )}{4 i-4 c}+\frac {i \dilog \left (\frac {-i+\left (c -i\right ) \tanh \left (b x +a \right )+c}{-2 i+2 c}\right )}{4 i-4 c}\right )}{b \left (c -i\right )}\)	\(553\)
risch	\(\text {Expression too large to display}\)	\(1230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.16, size = 80, normalized size = 0.98 \begin {gather*} -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 ) \end {gather*}

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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (58) = 116\).
time = 3.17, size = 187, normalized size = 2.28 \begin {gather*} \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} \end {gather*}

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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: CoercionFailed} \end {gather*}

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 >> Cannot convert _t0**2*exp(2*a) + 1 of type <class 'sympy.core.add.Add'> to
QQ_I[b,_t0,exp(a)]

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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