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

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

[Out]

(b*x^2)/2 + x*ArcTan[c + (1 - I*c)*Cot[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)

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Rubi [A]  time = 0.131453, antiderivative size = 85, normalized size of antiderivative = 1., 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 = {5165, 2184, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )+x \tan ^{-1}(c+(1-i c) \cot (a+b x))+\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^2)/2 + x*ArcTan[c + (1 - I*c)*Cot[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 5165

Int[ArcTan[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)], x_Symbol] :> Simp[x*ArcTan[c + d*Cot[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]

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]

Rubi steps

\begin{align*} \int \tan ^{-1}(c+(1-i c) \cot (a+b x)) \, dx &=x \tan ^{-1}(c+(1-i c) \cot (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) \cot (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) \cot (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) \cot (a+b x))+\frac{1}{2} i x \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{\operatorname{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) \cot (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*}

Mathematica [B]  time = 15.8397, size = 929, normalized size = 10.93 \[ x \tan ^{-1}(c+(1-i c) \cot (a+b x))-\frac{i x \csc ^2(a+b x) \left (2 b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))+i \log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))-i \log \left (\frac{\sec (b x) ((1-i c) \cos (a+b x)+(c-i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \log (i \tan (b x)+1)+i \text{PolyLog}(2,i \sin (2 b x)-\cos (2 b x))+i \text{PolyLog}\left (2,\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-i \text{PolyLog}\left (2,\frac{1}{2} \sec (b x) ((i c+1) \cos (a)-(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))\right )\right ) (\cos (b x)-i \sin (b x)) (\cos (b x)+i \sin (b x))}{(\cot (a+b x)+i) (i c+(c+i) \cot (a+b x)+1) \left (i \log (i \tan (b x)+1) \tan (b x) \cos ^2(a)+2 i b x+\log \left (1-\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )+\log \left (\frac{1}{2} \sec (b x) ((-i c-1) \cos (a)+(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))+1\right )+i \log (i \tan (b x)+1) \sin ^2(a) \tan (b x)+2 b x \tan (b x)+i \log \left (1-\frac{\sec (b x) ((c-i) \cos (a)+i (c+i) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)-i \log \left (\frac{1}{2} \sec (b x) ((-i c-1) \cos (a)+(c+i) \sin (a)) (\cos (a+b x)+i \sin (a+b x))+1\right ) \tan (b x)-i \log (1-i \tan (b x)) \tan (b x)+\frac{(c-i) \cos (a+b x) (\log (1-i \tan (b x))-\log (i \tan (b x)+1))}{(c+i) \cos (a+b x)+(i c+1) \sin (a+b x)}+\frac{(c+i) (\log (1-i \tan (b x))-\log (i \tan (b x)+1)) \sin (a+b x)}{(1-i c) \cos (a+b x)+(c-i) \sin (a+b x)}+\frac{i \log \left (\frac{\sec (b x) ((1-i c) \cos (a+b x)+(c-i) \sin (a+b x))}{2 \cos (a)-2 i \sin (a)}\right ) \sec ^2(b x)}{\tan (b x)-i}-\frac{i \log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c+i) \cos (a+b x)+(i c+1) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{\tan (b x)+i}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

x*ArcTan[c + (1 - I*c)*Cot[a + b*x]] - (I*x*Csc[a + b*x]^2*(2*b*x*Log[2*Cos[b*x]*(Cos[b*x] - I*Sin[b*x])] + I*
Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a + b*x]))/(2*c)]*Log[1 - I*Tan[b*x]]
- I*Log[(Sec[b*x]*((1 - I*c)*Cos[a + b*x] + (-I + c)*Sin[a + b*x]))/(2*Cos[a] - (2*I)*Sin[a])]*Log[1 + I*Tan[b
*x]] + I*PolyLog[2, -Cos[2*b*x] + I*Sin[2*b*x]] + I*PolyLog[2, (Sec[b*x]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*
(Cos[a + b*x] - I*Sin[a + b*x]))/(2*c)] - I*PolyLog[2, (Sec[b*x]*((1 + I*c)*Cos[a] - (I + c)*Sin[a])*(Cos[a +
b*x] + I*Sin[a + b*x]))/2])*(Cos[b*x] - I*Sin[b*x])*(Cos[b*x] + I*Sin[b*x]))/((I + Cot[a + b*x])*(1 + I*c + (I
+ c)*Cot[a + b*x])*((2*I)*b*x + Log[1 - (Sec[b*x]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*(Cos[a + b*x] - I*Sin[
a + b*x]))/(2*c)] + Log[1 + (Sec[b*x]*((-1 - I*c)*Cos[a] + (I + c)*Sin[a])*(Cos[a + b*x] + I*Sin[a + b*x]))/2]
+ ((-I + c)*Cos[a + b*x]*(Log[1 - I*Tan[b*x]] - Log[1 + I*Tan[b*x]]))/((I + c)*Cos[a + b*x] + (1 + I*c)*Sin[a
+ b*x]) + ((I + c)*(Log[1 - I*Tan[b*x]] - Log[1 + I*Tan[b*x]])*Sin[a + b*x])/((1 - I*c)*Cos[a + b*x] + (-I +
c)*Sin[a + b*x]) + 2*b*x*Tan[b*x] + I*Log[1 - (Sec[b*x]*((-I + c)*Cos[a] + I*(I + c)*Sin[a])*(Cos[a + b*x] - I
*Sin[a + b*x]))/(2*c)]*Tan[b*x] - I*Log[1 + (Sec[b*x]*((-1 - I*c)*Cos[a] + (I + c)*Sin[a])*(Cos[a + b*x] + I*S
in[a + b*x]))/2]*Tan[b*x] - I*Log[1 - I*Tan[b*x]]*Tan[b*x] + I*Cos[a]^2*Log[1 + I*Tan[b*x]]*Tan[b*x] + I*Log[1
+ I*Tan[b*x]]*Sin[a]^2*Tan[b*x] + (I*Log[(Sec[b*x]*((1 - I*c)*Cos[a + b*x] + (-I + c)*Sin[a + b*x]))/(2*Cos[a
] - (2*I)*Sin[a])]*Sec[b*x]^2)/(-I + Tan[b*x]) - (I*Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((I + c)*Cos[a + b*x] +
(1 + I*c)*Sin[a + b*x]))/(2*c)]*Sec[b*x]^2)/(I + Tan[b*x])))

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Maple [B]  time = 0.137, size = 1495, normalized size = 17.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-arctan(-c-(1-I*c)*cot(b*x+a)),x)

[Out]

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

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Maxima [B]  time = 1.57339, size = 616, normalized size = 7.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.96799, size = 309, normalized size = 3.64 \begin{align*} \frac{2 \, b^{2} x^{2} + 2 i \, b x \log \left (-\frac{{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} + i}\right ) - 2 \, a^{2} +{\left (2 i \, b x + 2 i \, a\right )} \log \left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 2 i \, a \log \left (\frac{c e^{\left (2 i \, b x + 2 i \, a\right )} + i}{c}\right ) +{\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} i b \int \frac{x}{c e^{2 i a} e^{2 i b x} + i}\, dx + \frac{i x \log{\left (- i c - \frac{i c}{e^{2 i a} e^{2 i b x} - 1} - \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} - e^{- i a} e^{- i b x}} + 1 + \frac{1}{e^{2 i a} e^{2 i b x} - 1} + \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} - e^{- i a} e^{- i b x}} \right )}}{2} - \frac{i x \log{\left (i c + \frac{i c}{e^{2 i a} e^{2 i b x} - 1} + \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} - e^{- i a} e^{- i b x}} + 1 - \frac{1}{e^{2 i a} e^{2 i b x} - 1} - \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} - e^{- i a} e^{- i b x}} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-atan(-c-(1-I*c)*cot(b*x+a)),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\arctan \left (-{\left (-i \, c + 1\right )} \cot \left (b x + a\right ) - c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-arctan(-c-(1-I*c)*cot(b*x+a)),x, algorithm="giac")

[Out]

integrate(-arctan(-(-I*c + 1)*cot(b*x + a) - c), x)