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

Optimal. Leaf size=123 \[ -\frac {i \text {Li}_3\left (i c e^{2 i a+2 i b x}\right )}{8 b^2}-\frac {x \text {Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c+(1-i c) \cot (a+b x))-\frac {b x^3}{6} \]

[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 = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, 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*
PolyLog[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 = 0.11, size = 110, normalized size = 0.89 \[ \frac {1}{2} x^2 \cot ^{-1}(c+(1-i c) \cot (a+b x))-\frac {i \left (2 b^2 x^2 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \text {Li}_2\left (-\frac {i e^{-2 i (a+b x)}}{c}\right )+\text {Li}_3\left (-\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2} \]

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 = 0.43, size = 152, normalized size = 1.24 \[ -\frac {4 \, b^{3} x^{3} - 12 \, \pi b^{2} x^{2} - 6 i \, b^{2} x^{2} \log \left (\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c + i}\right ) + 4 \, a^{3} + 6 \, b x {\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 6 i \, a^{2} \log \left (\frac {c e^{\left (2 i \, b x + 2 i \, a\right )} + i}{c}\right ) - {\left (-6 i \, b^{2} x^{2} + 6 i \, a^{2}\right )} \log \left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) + 3 i \, {\rm polylog}\left (3, i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{2}} \]

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*e^(2*I*b*x + 2*I*a) + I)*e^(-2*I*b*x - 2*I*a)/(c + 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 = 6.12, size = 1491, normalized size = 12.12 \[ \text {result too large to display} \]

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]

-1/4*x*polylog(2,I*c*exp(2*I*(b*x+a)))/b-1/8*I*polylog(3,I*c*exp(2*I*(b*x+a)))/b^2-1/4/b^2*polylog(2,I*c*exp(2
*I*(b*x+a)))*a+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
))-1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))^3-1/4*I*x^2*ln(I+c)+1/8*x^2*Pi*csgn(I*exp(2*
I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))^2-1/8*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(exp(2*I*(b*x+a))*(I+c)/(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+1/4*I*x^2*ln(exp(2*I*(b*x+a))*c+I
)-1/4*I/b^2*ln(1-I*exp(2*I*(b*x+a))*c)*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*a*ln(1+I*exp(I*(b*x+a))*(-I*c)
^(1/2))*x-1/2*I*x^2*ln(exp(I*(b*x+a)))+1/8*x^2*Pi*csgn(I*(exp(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))-1/8*x^2*Pi*csgn(I*(I+c))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn
(I*(I+c)/(exp(2*I*(b*x+a))-1))-1/4*I*x^2*ln(1-I*exp(2*I*(b*x+a))*c)-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*exp(2*I*(b*x+a))*(I
+c)/(exp(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))-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*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^2+1/2*I/b*a*ln(1-I*exp(I
*(b*x+a))*(-I*c)^(1/2))*x-1/2*I/b*ln(1-I*exp(2*I*(b*x+a))*c)*x*a+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*exp(2*I*(b*x+a)))^3+1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(
b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))^2+1/8*x^2*Pi*csgn(I*(I+c)/(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(I
+c)/(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))^3-1/6*b*x^3+1/8*x^2*P
i*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*(I+c)/(exp(2*I*(b*x+a))-1))^3+1/8*x^2*Pi*csgn(I*(I+c))*csgn(I*(I+c)/(exp(2*I*(b*x+a))-1))^2+1/8*x^
2*Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(I+c)/(exp(2*I*(b*x+a))-1))^2+1/8*x^2*Pi*csgn(exp(2*I*(b*x+a))*(I+c)/
(exp(2*I*(b*x+a))-1))^2-1/8*x^2*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*(I+c)/(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*
(b*x+a))*(I+c)/(exp(2*I*(b*x+a))-1))

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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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