3.1.54 \(\int \text {ArcTan}(c+(1+i c) \tan (a+b x)) \, dx\) [54]

Optimal. Leaf size=85 \[ -\frac {b x^2}{2}+x \text {ArcTan}(c+(1+i c) \tan (a+b x))-\frac {1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac {\text {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5271, 2215, 2221, 2317, 2438} \begin {gather*} x \text {ArcTan}(c+(1+i c) \tan (a+b x))-\frac {\text {Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac {b x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

-1/2*(b*x^2) + x*ArcTan[c + (1 + I*c)*Tan[a + b*x]] - (I/2)*x*Log[1 - I*c*E^((2*I)*a + (2*I)*b*x)] - PolyLog[2
, I*c*E^((2*I)*a + (2*I)*b*x)]/(4*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 5271

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

Rubi steps

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

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Mathematica [A]
time = 6.30, size = 75, normalized size = 0.88 \begin {gather*} x \text {ArcTan}(c+(1+i c) \tan (a+b x))-\frac {1}{2} i x \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )+\frac {\text {PolyLog}\left (2,-\frac {i e^{-2 i (a+b x)}}{c}\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

x*ArcTan[c + (1 + I*c)*Tan[a + b*x]] - (I/2)*x*Log[1 + I/(c*E^((2*I)*(a + b*x)))] + PolyLog[2, (-I)/(c*E^((2*I
)*(a + b*x)))]/(4*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. 599 vs. \(2 (69 ) = 138\).
time = 0.26, size = 600, normalized size = 7.06

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c^{2}}{2 i-2 c}+\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c^{2}}{2 i-2 c}-\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2 i-2 c}-\left (i c +1\right )^{2} \left (-\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )^{2}}{8 \left (i-c \right )}+\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )}{4 i-4 c}+\frac {i \dilog \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )}{4 i-4 c}-\frac {i \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )}{4 \left (i-c \right )}-\frac {i \dilog \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )}{4 \left (i-c \right )}+\frac {i \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )}{4 i-4 c}+\frac {i \dilog \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )}{4 i-4 c}\right )}{b \left (i c +1\right )}\)	\(600\)
default	\(\frac {-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c^{2}}{2 i-2 c}+\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) c}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )}{2 i-2 c}+\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c^{2}}{2 i-2 c}-\frac {2 i \arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) c}{2 i-2 c}-\frac {\arctan \left (c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2 i-2 c}-\left (i c +1\right )^{2} \left (-\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right )^{2}}{8 \left (i-c \right )}+\frac {i \ln \left (-i+c +\left (i c +1\right ) \tan \left (b x +a \right )\right ) \ln \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )}{4 i-4 c}+\frac {i \dilog \left (-\frac {i \left (c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right )}{2}\right )}{4 i-4 c}-\frac {i \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )}{4 \left (i-c \right )}-\frac {i \dilog \left (\frac {c +\left (i c +1\right ) \tan \left (b x +a \right )+i}{2 c}\right )}{4 \left (i-c \right )}+\frac {i \ln \left (-c +\left (i c +1\right ) \tan \left (b x +a \right )+i\right ) \ln \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )}{4 i-4 c}+\frac {i \dilog \left (\frac {-i+c +\left (i c +1\right ) \tan \left (b x +a \right )}{-2 i+2 c}\right )}{4 i-4 c}\right )}{b \left (i c +1\right )}\)	\(600\)
risch	\(\text {Expression too large to display}\)	\(1248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(c+(1+I*c)*tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (60) = 120\).
time = 0.52, size = 448, normalized size = 5.27 \begin {gather*} -\frac {{\left (i \, c + 1\right )} {\left (\frac {4 i \, {\left (b x + a\right )} \log \left (-\frac {2 \, {\left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right )}}{2 i \, c^{2} - 2 \, {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 i}\right )}{i \, c + 1} - \frac {i \, {\left (4 \, {\left (b x + a\right )} {\left (\log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right ) - \log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - i\right )\right )} + i \, \log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right )^{2} - 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - i\right ) \log \left (-\frac {1}{2} \, {\left (c - i\right )} \tan \left (b x + a\right ) + \frac {1}{2} i \, c + \frac {1}{2}\right ) + 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - i\right ) \log \left (-\frac {{\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c + i}{2 \, c} + 1\right ) - 2 i \, \log \left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right ) \log \left (-\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) - 2 i \, {\rm Li}_2\left (\frac {1}{2} \, {\left (c - i\right )} \tan \left (b x + a\right ) - \frac {1}{2} i \, c + \frac {1}{2}\right ) + 2 i \, {\rm Li}_2\left (\frac {{\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c + i}{2 \, c}\right ) - 2 i \, {\rm Li}_2\left (\frac {1}{2} i \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{i \, c + 1}\right )} - 8 \, {\left (b x + a\right )} \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right ) + 4 \, {\left (-i \, b x - i \, a\right )} \log \left (-\frac {2 \, {\left (-i \, c^{2} + {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) - 2 \, c + i\right )}}{2 i \, c^{2} - 2 \, {\left (c^{2} - 2 i \, c - 1\right )} \tan \left (b x + a\right ) + 2 i}\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")

[Out]

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

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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. 202 vs. \(2 (60) = 120\).
time = 1.21, size = 202, normalized size = 2.38 \begin {gather*} -\frac {b^{2} x^{2} - i \, b x \log \left (-\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) - a^{2} - {\left (-i \, b x - i \, a\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - {\left (-i \, b x - i \, a\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - i \, a \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: CoercionFailed >> Cannot convert _t0**2 + exp(2*I*a) of type <class 'sympy.core.add.Add'> to
QQ_I[b,_t0,exp(I*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+(1+I*c)*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(arctan((I*c + 1)*tan(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 {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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