3.168 \(\int \cot ^{-1}(c-(1-i c) \tan (a+b x)) \, dx\)
Optimal. Leaf size=86 \[ -\frac{\text{PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1+i c e^{2 i a+2 i b x}\right )+x \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac{b x^2}{2} \]
[Out]
-(b*x^2)/2 + x*ArcCot[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)
________________________________________________________________________________________
Rubi [A] time = 0.141577, antiderivative size = 86, normalized size of antiderivative = 1.,
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 =
{5164, 2184, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{2} i x \log \left (1+i c e^{2 i a+2 i b x}\right )+x \cot ^{-1}(c-(1-i c) \tan (a+b x))-\frac{b x^2}{2} \]
Antiderivative was successfully verified.
[In]
Int[ArcCot[c - (1 - I*c)*Tan[a + b*x]],x]
[Out]
-(b*x^2)/2 + x*ArcCot[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 5164
Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]], x_Symbol] :> Simp[x*ArcCot[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]
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 \cot ^{-1}(c-(1-i c) \tan (a+b x)) \, dx &=x \cot ^{-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 \cot ^{-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 \cot ^{-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 \cot ^{-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{\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 \cot ^{-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*}
Mathematica [B] time = 3.02039, size = 847, normalized size = 9.85 \[ x \cot ^{-1}(c+i (c+i) \tan (a+b x))-\frac{i x \left (-2 i b x \log (2 \cos (b x) (\cos (b x)-i \sin (b x)))+\log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \log (1-i \tan (b x))-\log \left (\frac{1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \log (i \tan (b x)+1)+\text{PolyLog}(2,i \sin (2 b x)-\cos (2 b x))+\text{PolyLog}\left (2,\frac{\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )-\text{PolyLog}\left (2,\frac{1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right )\right ) \sec (a+b x) (\cos (b x)+i \sin (b x)) (i \cos (b x)+\sin (b x))}{((c-i) \cos (a+b x)+i (c+i) \sin (a+b x)) \left (-\frac{\log \left (\frac{1}{2} \sec (b x) (\cos (a)+i \sin (a)) ((i c+1) \cos (a+b x)-(c+i) \sin (a+b x))\right ) \sec ^2(b x)}{\tan (b x)-i}+\frac{\log \left (1-\frac{1}{2} (\cos (a)+i \sin (a)) ((c+i) \cos (a)+(i c+1) \sin (a)) (\tan (b x)-i)\right ) \sec ^2(b x)}{\tan (b x)-i}+\frac{\log \left (\frac{\sec (b x) (\cos (a)-i \sin (a)) ((c-i) \cos (a+b x)+i (c+i) \sin (a+b x))}{2 c}\right ) \sec ^2(b x)}{\tan (b x)+i}-2 b x+i \log \left (1-\frac{\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right )+2 i b x \tan (b x)-\log \left (1-\frac{\sec (b x) ((c+i) \cos (a)+(i c+1) \sin (a)) (\cos (a+b x)-i \sin (a+b x))}{2 c}\right ) \tan (b x)+\log (1-i \tan (b x)) \tan (b x)-\log (i \tan (b x)+1) \tan (b x)+\frac{i (c+i) \cos (a+b x) (\log (1-i \tan (b x))-\log (i \tan (b x)+1))}{(c-i) \cos (a+b x)+i (c+i) \sin (a+b x)}+\frac{(i c+1) (\log (1-i \tan (b x))-\log (i \tan (b x)+1)) \sin (a+b x)}{(-i c-1) \cos (a+b x)+(c+i) \sin (a+b x)}\right ) (\tan (a+b x)-i)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcCot[c - (1 - I*c)*Tan[a + b*x]],x]
[Out]
x*ArcCot[c + I*(I + c)*Tan[a + b*x]] - (I*x*((-2*I)*b*x*Log[2*Cos[b*x]*(Cos[b*x] - I*Sin[b*x])] + Log[(Sec[b*x
]*(Cos[a] - I*Sin[a])*((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x]))/(2*c)]*Log[1 - I*Tan[b*x]] - Log[(Sec[
b*x]*(Cos[a] + I*Sin[a])*((1 + I*c)*Cos[a + b*x] - (I + c)*Sin[a + b*x]))/2]*Log[1 + I*Tan[b*x]] + PolyLog[2,
-Cos[2*b*x] + I*Sin[2*b*x]] + PolyLog[2, (Sec[b*x]*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(Cos[a + b*x] - I*Sin[a
+ b*x]))/(2*c)] - PolyLog[2, ((Cos[a] + I*Sin[a])*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(-I + Tan[b*x]))/2])*Se
c[a + b*x]*(Cos[b*x] + I*Sin[b*x])*(I*Cos[b*x] + Sin[b*x]))/(((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x])*
(-2*b*x + I*Log[1 - (Sec[b*x]*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(Cos[a + b*x] - I*Sin[a + b*x]))/(2*c)] + (I
*(I + c)*Cos[a + b*x]*(Log[1 - I*Tan[b*x]] - Log[1 + I*Tan[b*x]]))/((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a +
b*x]) + ((1 + 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*I)*b*x*Tan[b*x] - Log[1 - (Sec[b*x]*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(Cos[a + b*x] - I
*Sin[a + b*x]))/(2*c)]*Tan[b*x] + Log[1 - I*Tan[b*x]]*Tan[b*x] - Log[1 + I*Tan[b*x]]*Tan[b*x] - (Log[(Sec[b*x]
*(Cos[a] + I*Sin[a])*((1 + I*c)*Cos[a + b*x] - (I + c)*Sin[a + b*x]))/2]*Sec[b*x]^2)/(-I + Tan[b*x]) + (Log[1
- ((Cos[a] + I*Sin[a])*((I + c)*Cos[a] + (1 + I*c)*Sin[a])*(-I + Tan[b*x]))/2]*Sec[b*x]^2)/(-I + Tan[b*x]) + (
Log[(Sec[b*x]*(Cos[a] - I*Sin[a])*((-I + c)*Cos[a + b*x] + I*(I + c)*Sin[a + b*x]))/(2*c)]*Sec[b*x]^2)/(I + Ta
n[b*x]))*(-I + Tan[a + b*x]))
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Maple [B] time = 0.112, size = 1681, normalized size = 19.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arccot(c-(1-I*c)*tan(b*x+a)),x)
[Out]
-1/4/b/(-1+I*c)/(I+c)*ln((-1+I*c)*tan(b*x+a)+c+I)^2*c-1/2/b/(-1+I*c)/(I+c)*dilog(-1/2*(-(-1+I*c)*tan(b*x+a)-c+
I)/c)*c+1/2/b/(-1+I*c)/(I+c)*dilog((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*c+1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan
(b*x+a))/(2*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)-1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln((-1+I*
c)*tan(b*x+a)+c+I)-1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)+1/4*I/b/(-1+I*c)/(I+c)*dilo
g((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))-1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*I*((-1+I*c)*tan(b*x+a)+c+I))-1/8*I/
b/(-1+I*c)/(I+c)*ln((-1+I*c)*tan(b*x+a)+c+I)^2+1/2/b/(-1+I*c)/(I+c)*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*ln((
-1+I*c)*tan(b*x+a)+c+I)*c-1/2/b/(-1+I*c)/(I+c)*dilog(-1/2*I*((-1+I*c)*tan(b*x+a)+c+I))*c+1/4*I/b/(-1+I*c)/(I+c
)*dilog(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*c^2-1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*ln
(-(-1+I*c)*tan(b*x+a)+c+I)-1/4*I/b/(-1+I*c)/(I+c)*dilog((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*c^2-1/4*I/b/(-1
+I*c)/(I+c)*ln(-1/2*I*((-1+I*c)*tan(b*x+a)+c+I))*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))+1/4*I/b/(-1+I*c)/(I+c)*
ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*ln((-1+I*c)*tan(b*x+a)+c+I)-1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(
2*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c^2+1/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln((-1+I*c)*tan
(b*x+a)+c+I)*c^2-1/2/b/(-1+I*c)/(I+c)*ln(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c+1/2
/b/(-1+I*c)/(I+c)*ln((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c-1/2/b/(-1+I*c)/(I+c
)*ln(-1/2*I*((-1+I*c)*tan(b*x+a)+c+I))*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*c+1/4*I/b/(-1+I*c)/(I+c)*ln((-(-1
+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*ln(-(-1+I*c)*tan(b*x+a)+c+I)+1/4*I/b/(-1+I*c)/(I+c)*dilog(-1/2*I*((-1+I*c)*t
an(b*x+a)+c+I))*c^2+1/8*I/b/(-1+I*c)/(I+c)*ln((-1+I*c)*tan(b*x+a)+c+I)^2*c^2+1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*I*
((-1+I*c)*tan(b*x+a)+c+I))*ln(-1/2*I*(-(-1+I*c)*tan(b*x+a)-c+I))*c^2-1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*I*(-(-1+I*
c)*tan(b*x+a)-c+I))*ln((-1+I*c)*tan(b*x+a)+c+I)*c^2+1/4*I/b/(-1+I*c)/(I+c)*ln(-1/2*(-(-1+I*c)*tan(b*x+a)-c+I)/
c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c^2-1/4*I/b/(-1+I*c)/(I+c)*ln((-(-1+I*c)*tan(b*x+a)-c-I)/(-2*I-2*c))*ln(-(-1+I
*c)*tan(b*x+a)+c+I)*c^2-2*I/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln(-(-1+I*c)*tan(b*x+a)+c+I)*c+
2*I/b/(-1+I*c)*arccot(c+(-1+I*c)*tan(b*x+a))/(2*I+2*c)*ln((-1+I*c)*tan(b*x+a)+c+I)*c
________________________________________________________________________________________
Maxima [B] time = 1.60214, size = 608, normalized size = 7.07 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccot(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 - 2*(c^2 + 2*I*c - 1)*tan(b*x + a) + 2*I)/(2*I*c^2 - 2*(c^2 + 2*I*c
- 1)*tan(b*x + a) - 4*c - 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)*arccot((-I*c + 1)*tan(b*x + a) - c) + 4*(-I*b*x - I*a)*log((2*I*c^2 - 2*(
c^2 + 2*I*c - 1)*tan(b*x + a) + 2*I)/(2*I*c^2 - 2*(c^2 + 2*I*c - 1)*tan(b*x + a) - 4*c - 2*I)))/b
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Fricas [B] time = 2.59641, size = 552, normalized size = 6.42 \begin{align*} -\frac{b^{2} x^{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 ) - 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{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccot(c-(1-I*c)*tan(b*x+a)),x, algorithm="fricas")
[Out]
-1/2*(b^2*x^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)) - a^2 - (-I*b*x - I*a)*log(
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
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(acot(c-(1-I*c)*tan(b*x+a)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arccot}\left (-{\left (-i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arccot(c-(1-I*c)*tan(b*x+a)),x, algorithm="giac")
[Out]
integrate(arccot(-(-I*c + 1)*tan(b*x + a) + c), x)