3.62 \(\int x \tan ^{-1}(c+d \cot (a+b x)) \, dx\)
Optimal. Leaf size=303 \[ \frac{i \text{PolyLog}\left (3,\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{8 b^2}-\frac{i \text{PolyLog}\left (3,\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b}-\frac{x \text{PolyLog}\left (2,\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1-\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{1}{2} x^2 \tan ^{-1}(d \cot (a+b x)+c) \]
[Out]
(x^2*ArcTan[c + d*Cot[a + b*x]])/2 + (I/4)*x^2*Log[1 - ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)]
- (I/4)*x^2*Log[1 - ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))] + (x*PolyLog[2, ((1 + I*c - d)*
E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/(4*b) - (x*PolyLog[2, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c +
I*(1 - d))])/(4*b) + ((I/8)*PolyLog[3, ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/b^2 - ((I/8)*Po
lyLog[3, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))])/b^2
________________________________________________________________________________________
Rubi [A] time = 0.40045, antiderivative size = 303, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used =
{5177, 2190, 2531, 2282, 6589} \[ \frac{i \text{PolyLog}\left (3,\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{8 b^2}-\frac{i \text{PolyLog}\left (3,\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b}-\frac{x \text{PolyLog}\left (2,\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac{1}{4} i x^2 \log \left (1-\frac{(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{1}{2} x^2 \tan ^{-1}(d \cot (a+b x)+c) \]
Antiderivative was successfully verified.
[In]
Int[x*ArcTan[c + d*Cot[a + b*x]],x]
[Out]
(x^2*ArcTan[c + d*Cot[a + b*x]])/2 + (I/4)*x^2*Log[1 - ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)]
- (I/4)*x^2*Log[1 - ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))] + (x*PolyLog[2, ((1 + I*c - d)*
E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/(4*b) - (x*PolyLog[2, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c +
I*(1 - d))])/(4*b) + ((I/8)*PolyLog[3, ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/b^2 - ((I/8)*Po
lyLog[3, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))])/b^2
Rule 5177
Int[ArcTan[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTan[c + d*Cot[a + b*x]])/(f*(m + 1)), x] + (Dist[(b*(1 + I*c - d))/(f*(m + 1)), Int[((e + f*x)^(m + 1)
*E^(2*I*a + 2*I*b*x))/(1 + I*c + d - (1 + I*c - d)*E^(2*I*a + 2*I*b*x)), x], x] - Dist[(b*(1 - I*c + d))/(f*(m
+ 1)), Int[((e + f*x)^(m + 1)*E^(2*I*a + 2*I*b*x))/(1 - I*c - d - (1 - I*c + d)*E^(2*I*a + 2*I*b*x)), x], x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c - I*d)^2, -1]
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 2531
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]
Rule 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 6589
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]
Rubi steps
\begin{align*} \int x \tan ^{-1}(c+d \cot (a+b x)) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(c+d \cot (a+b x))+\frac{1}{2} (b (1+i c-d)) \int \frac{e^{2 i a+2 i b x} x^2}{1+i c+d+(-1-i c+d) e^{2 i a+2 i b x}} \, dx-\frac{1}{2} (b (1-i c+d)) \int \frac{e^{2 i a+2 i b x} x^2}{1-i c-d+(-1+i c-d) e^{2 i a+2 i b x}} \, dx\\ &=\frac{1}{2} x^2 \tan ^{-1}(c+d \cot (a+b x))+\frac{1}{4} i x^2 \log \left (1-\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{1}{2} i \int x \log \left (1+\frac{(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx-\frac{1}{2} i \int x \log \left (1+\frac{(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx\\ &=\frac{1}{2} x^2 \tan ^{-1}(c+d \cot (a+b x))+\frac{1}{4} i x^2 \log \left (1-\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{x \text{Li}_2\left (\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac{\int \text{Li}_2\left (-\frac{(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx}{4 b}-\frac{\int \text{Li}_2\left (-\frac{(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx}{4 b}\\ &=\frac{1}{2} x^2 \tan ^{-1}(c+d \cot (a+b x))+\frac{1}{4} i x^2 \log \left (1-\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{x \text{Li}_2\left (\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{(-1-i c+d) x}{1+i c+d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{(c+i (1+d)) x}{c-i (-1+d)}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=\frac{1}{2} x^2 \tan ^{-1}(c+d \cot (a+b x))+\frac{1}{4} i x^2 \log \left (1-\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )-\frac{1}{4} i x^2 \log \left (1-\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac{x \text{Li}_2\left (\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}+\frac{i \text{Li}_3\left (\frac{(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{8 b^2}-\frac{i \text{Li}_3\left (\frac{(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.538076, size = 270, normalized size = 0.89 \[ \frac{1}{2} x^2 \tan ^{-1}(d \cot (a+b x)+c)+\frac{i \left (-2 i b x \text{PolyLog}\left (2,\frac{(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )+2 i b x \text{PolyLog}\left (2,\frac{(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )+\text{PolyLog}\left (3,\frac{(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )-\text{PolyLog}\left (3,\frac{(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )+2 b^2 x^2 \log \left (1-\frac{(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )-2 b^2 x^2 \log \left (1-\frac{(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*ArcTan[c + d*Cot[a + b*x]],x]
[Out]
(x^2*ArcTan[c + d*Cot[a + b*x]])/2 + ((I/8)*(2*b^2*x^2*Log[1 - ((c + I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(
1 + d))] - 2*b^2*x^2*Log[1 - ((c + I*(1 + d))*E^((2*I)*(a + b*x)))/(I + c - I*d)] - (2*I)*b*x*PolyLog[2, ((c +
I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(1 + d))] + (2*I)*b*x*PolyLog[2, ((c + I*(1 + d))*E^((2*I)*(a + b*x))
)/(I + c - I*d)] + PolyLog[3, ((c + I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(1 + d))] - PolyLog[3, ((c + I*(1
+ d))*E^((2*I)*(a + b*x)))/(I + c - I*d)]))/b^2
________________________________________________________________________________________
Maple [C] time = 24.825, size = 7556, normalized size = 24.9 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*arctan(c+d*cot(b*x+a)),x)
[Out]
result too large to display
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arctan(c+d*cot(b*x+a)),x, algorithm="maxima")
[Out]
-1/4*x^2*arctan2(-c*cos(2*b*x + 2*a) + (d + 1)*sin(2*b*x + 2*a) + c, (d + 1)*cos(2*b*x + 2*a) + c*sin(2*b*x +
2*a) + d - 1) - 1/4*x^2*arctan2(-c*cos(2*b*x + 2*a) + (d - 1)*sin(2*b*x + 2*a) + c, -(d - 1)*cos(2*b*x + 2*a)
- c*sin(2*b*x + 2*a) - d - 1) + 2*b*d*integrate((2*(c^2 + d^2 + 1)*x^2*cos(2*b*x + 2*a)^2 + 2*c*d*x^2*sin(2*b*
x + 2*a) + 2*(c^2 + d^2 + 1)*x^2*sin(2*b*x + 2*a)^2 - (c^2 - d^2 + 1)*x^2*cos(2*b*x + 2*a) - (2*c*d*x^2*sin(2*
b*x + 2*a) + (c^2 - d^2 + 1)*x^2*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^2*cos(2*b*x + 2*a) - (c^2 - d^2
+ 1)*x^2*sin(2*b*x + 2*a))*sin(4*b*x + 4*a))/(c^4 + d^4 + 2*(c^2 - 1)*d^2 + (c^4 + d^4 + 2*(c^2 - 1)*d^2 + 2*
c^2 + 1)*cos(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4 + 2*
(c^2 - 1)*d^2 + 2*c^2 + 1)*sin(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*sin(2*b*x + 2*a)^2
+ 2*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 + 1)*d^2 + 2*c^2 - 2*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(c*d^3 +
(c^3 + c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) - 4*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) + 4*(2*c*d^3
- 2*(c^3 + c)*d + 2*(c*d^3 + (c^3 + c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 + 2*c^2 + 1)*sin(2*b*x + 2*a))*sin(4*b
*x + 4*a) + 8*(c*d^3 + (c^3 + c)*d)*sin(2*b*x + 2*a) + 1), x)
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Fricas [C] time = 2.88313, size = 3298, normalized size = 10.88 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arctan(c+d*cot(b*x+a)),x, algorithm="fricas")
[Out]
1/16*(8*b^2*x^2*arctan(d*cot(b*x + a) + c) + 2*b*x*dilog(-(c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2
*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) + 2*b*x*dilog(-(c^2
+ d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a) + 2*d + 1)/(
c^2 + d^2 + 2*d + 1) + 1) - 2*b*x*dilog(-(c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2
*c*d + I*d^2 - I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1) + 1) - 2*b*x*dilog(-(c^2 + d^2 - (c^2 - 2*
I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d +
1) + 1) + 2*I*a^2*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2 + d^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I
*d^2 + 2*I*d + I)*sin(2*b*x + 2*a) + 1/2) - 2*I*a^2*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2 + d^2 - 2*d + 1)*
cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 2*a) + 1/2) - 2*I*a^2*log(-1/2*c^2 + I*c*d + 1/
2*d^2 + 1/2*(c^2 + d^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2*b*x + 2*a) - 1/2) +
2*I*a^2*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 - 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*
I*d + I)*sin(2*b*x + 2*a) - 1/2) + (2*I*b^2*x^2 - 2*I*a^2)*log((c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*
x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1)) + (-2*I*b^2*x^2 + 2
*I*a^2)*log((c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x +
2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1)) + (-2*I*b^2*x^2 + 2*I*a^2)*log((c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*c
os(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1)) + (2*I*b^2*x
^2 - 2*I*a^2)*log((c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*
b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1)) + I*polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c
^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*d + 1)) - I*polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*cos
(2*b*x + 2*a) + (I*c^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*d + 1)) - I*polylog(3, ((c^2 - 2*
I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*d + 1)) + I*
polylog(3, ((c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a))/(c^2 +
d^2 - 2*d + 1)))/b^2
________________________________________________________________________________________
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(x*atan(c+d*cot(b*x+a)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \arctan \left (d \cot \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*arctan(c+d*cot(b*x+a)),x, algorithm="giac")
[Out]
integrate(x*arctan(d*cot(b*x + a) + c), x)