3.180 \(\int x \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx\)
Optimal. Leaf size=124 \[ \frac {i \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{8 b^2}+\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {b x^3}{6} \]
[Out]
1/6*b*x^3+1/2*x^2*(Pi-arccot(-c+(1+I*c)*cot(b*x+a)))+1/4*I*x^2*ln(1+I*c*exp(2*I*a+2*I*b*x))+1/4*x*polylog(2,-I
*c*exp(2*I*a+2*I*b*x))/b+1/8*I*polylog(3,-I*c*exp(2*I*a+2*I*b*x))/b^2
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Rubi [A] time = 0.22, antiderivative size = 124, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used =
{5174, 2184, 2190, 2531, 2282, 6589} \[ \frac {i \text {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{8 b^2}+\frac {x \text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {b x^3}{6} \]
Antiderivative was successfully verified.
[In]
Int[x*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
[Out]
(b*x^3)/6 + (x^2*ArcCot[c - (1 + I*c)*Cot[a + b*x]])/2 + (I/4)*x^2*Log[1 + I*c*E^((2*I)*a + (2*I)*b*x)] + (x*P
olyLog[2, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/(4*b) + ((I/8)*PolyLog[3, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/b^2
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 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 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 5174
Int[ArcCot[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCot[c + d*Cot[a + b*x]])/(f*(m + 1)), x] + Dist[(I*b)/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - I*d - c*
E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - I*d)^2, -1]
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 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx &=\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{2} (i b) \int \frac {x^2}{-i (-1-i c)+c-c e^{2 i a+2 i b x}} \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{2} (b c) \int \frac {e^{2 i a+2 i b x} x^2}{-i (-1-i c)+c-c e^{2 i a+2 i b x}} \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {1}{2} i \int x \log \left (1-\frac {c e^{2 i a+2 i b x}}{-i (-1-i c)+c}\right ) \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {\int \text {Li}_2\left (\frac {c e^{2 i a+2 i b x}}{-i (-1-i c)+c}\right ) \, dx}{4 b}\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 110, normalized size = 0.89 \[ \frac {i \left (2 b^2 x^2 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \text {Li}_2\left (\frac {i e^{-2 i (a+b x)}}{c}\right )+\text {Li}_3\left (\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2}+\frac {1}{2} x^2 \cot ^{-1}(c+(-1-i c) \cot (a+b x)) \]
Antiderivative was successfully verified.
[In]
Integrate[x*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
[Out]
(x^2*ArcCot[c + (-1 - I*c)*Cot[a + b*x]])/2 + ((I/8)*(2*b^2*x^2*Log[1 - I/(c*E^((2*I)*(a + b*x)))] + (2*I)*b*x
*PolyLog[2, I/(c*E^((2*I)*(a + b*x)))] + PolyLog[3, I/(c*E^((2*I)*(a + b*x)))]))/b^2
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fricas [C] time = 0.69, size = 151, normalized size = 1.22 \[ \frac {4 \, b^{3} x^{3} + 12 \, \pi b^{2} x^{2} + 6 i \, b^{2} x^{2} \log \left (\frac {{\left (c - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 4 \, a^{3} + 6 \, b x {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 6 i \, a^{2} \log \left (\frac {c e^{\left (2 i \, b x + 2 i \, a\right )} - i}{c}\right ) + {\left (6 i \, b^{2} x^{2} - 6 i \, a^{2}\right )} \log \left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) + 3 i \, {\rm polylog}\left (3, -i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="fricas")
[Out]
1/24*(4*b^3*x^3 + 12*pi*b^2*x^2 + 6*I*b^2*x^2*log((c - I)*e^(2*I*b*x + 2*I*a)/(c*e^(2*I*b*x + 2*I*a) - I)) + 4
*a^3 + 6*b*x*dilog(-I*c*e^(2*I*b*x + 2*I*a)) + 6*I*a^2*log((c*e^(2*I*b*x + 2*I*a) - I)/c) + (6*I*b^2*x^2 - 6*I
*a^2)*log(I*c*e^(2*I*b*x + 2*I*a) + 1) + 3*I*polylog(3, -I*c*e^(2*I*b*x + 2*I*a)))/b^2
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\pi - \operatorname {arccot}\left ({\left (i \, c + 1\right )} \cot \left (b x + a\right ) - c\right )\right )} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="giac")
[Out]
integrate((pi - arccot((I*c + 1)*cot(b*x + a) - c))*x, x)
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maple [C] time = 6.07, size = 1492, normalized size = 12.03 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(Pi-arccot(-c+(1+I*c)*cot(b*x+a))),x)
[Out]
1/2*I*x^2*ln(exp(I*(b*x+a)))+1/4*I*x^2*ln(1+I*c*exp(2*I*(b*x+a)))+1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(ex
p(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))+1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/
(exp(2*I*(b*x+a))-1))^3-1/8*x^2*Pi*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^3+1/8*x^2*Pi*csgn(I*(exp(
2*I*(b*x+a))*c-I))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2+1/8*x^2*Pi*csgn(I/(exp(2*I*(b*x+a))-1
))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a
))-1))^2-1/8*x^2*Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn(I*exp(2*
I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/4*I*x^2*ln(exp(2*I*(b*x+a))*c-I)+1/8*I*pol
ylog(3,-I*exp(2*I*(b*x+a))*c)/b^2-1/2*I/b*a*ln(1+I*exp(I*(b*x+a))*(I*c)^(1/2))*x-1/2*I/b*a*ln(1-I*exp(I*(b*x+a
))*(I*c)^(1/2))*x-1/8*x^2*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))*csgn((exp(2*I*(b*x+a))*c-I)/(
exp(2*I*(b*x+a))-1))-1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(c-I
)/(exp(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I
*(b*x+a))-1))^2+1/8*x^2*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))*csgn((exp(2*I*(b*x+a))*c-I)/(ex
p(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^3+1/8*x^2*Pi*csgn(I*(c-I)/
(exp(2*I*(b*x+a))-1))^3+1/8*x^2*Pi*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn((exp(2*
I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^3+1/8*x^2*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*exp(2*I*(b*x+a)))-1/4*x^2*P
i*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^2+1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))^3-1/8*x^2*Pi*csgn(I*(e
xp(2*I*(b*x+a))*c-I))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))+1/4*I/b
^2*ln(1+I*c*exp(2*I*(b*x+a)))*a^2+1/4*I/b^2*a^2*ln(-exp(2*I*(b*x+a))*c+I)-1/2*I/b^2*a^2*ln(1+I*exp(I*(b*x+a))*
(I*c)^(1/2))-1/2*I/b^2*a^2*ln(1-I*exp(I*(b*x+a))*(I*c)^(1/2))+1/2*I/b*ln(1+I*c*exp(2*I*(b*x+a)))*x*a+1/6*b*x^3
+1/4*x*polylog(2,-I*exp(2*I*(b*x+a))*c)/b+1/4*I*x^2*ln(c-I)-1/2/b^2*a*dilog(1+I*exp(I*(b*x+a))*(I*c)^(1/2))-1/
2/b^2*a*dilog(1-I*exp(I*(b*x+a))*(I*c)^(1/2))+1/4/b^2*polylog(2,-I*exp(2*I*(b*x+a))*c)*a+1/8*x^2*Pi*csgn(I*(c-
I))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))+1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I
*(c-I)/(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))+1/8*x^2*Pi*csgn((exp(2*I*(b*x
+a))*c-I)/(exp(2*I*(b*x+a))-1))^2
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is c-1 zero or nonzero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\left (\Pi +\mathrm {acot}\left (c-\mathrm {cot}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))),x)
[Out]
int(x*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))), x)
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(pi-acot(-c+(1+I*c)*cot(b*x+a))),x)
[Out]
Exception raised: CoercionFailed
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