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

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

[Out]

1/6*b*x^3+1/2*x^2*arccot(c+(1+I*c)*tan(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 = {5172, 2184, 2190, 2531, 2282, 6589} \[ \frac {i \text {PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{8 b^2}+\frac {x \text {PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{4} i x^2 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x))+\frac {b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

(b*x^3)/6 + (x^2*ArcCot[c + (1 + I*c)*Tan[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 5172

Int[ArcCot[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCot[c + d*Tan[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) \tan (a+b x)) \, dx &=\frac {1}{2} x^2 \cot ^{-1}(c+(1+i c) \tan (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) \tan (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) \tan (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) \tan (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) \tan (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) \tan (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 \[ \frac {i \left (2 b^2 x^2 \log \left (1+\frac {i e^{-2 i (a+b x)}}{c}\right )+2 i b x \text {Li}_2\left (-\frac {i e^{-2 i (a+b x)}}{c}\right )+\text {Li}_3\left (-\frac {i e^{-2 i (a+b x)}}{c}\right )\right )}{8 b^2}+\frac {1}{2} x^2 \cot ^{-1}(c+(1+i c) \tan (a+b x)) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCot[c + (1 + I*c)*Tan[a + b*x]],x]

[Out]

(x^2*ArcCot[c + (1 + I*c)*Tan[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 = 0.72, size = 268, normalized size = 2.18 \[ \frac {2 \, b^{3} x^{3} - 3 i \, b^{2} x^{2} \log \left (\frac {{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 2 \, a^{3} + 6 \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) + {\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + {\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(2*b^3*x^3 - 3*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)) + 2*a^3 + 6*b*x*di
log(1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 6*b*x*dilog(-1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 3*I*a^2*log(1/2*(2*c*e^
(I*b*x + I*a) + I*sqrt(4*I*c))/c) + 3*I*a^2*log(1/2*(2*c*e^(I*b*x + I*a) - I*sqrt(4*I*c))/c) + (3*I*b^2*x^2 -
3*I*a^2)*log(1/2*sqrt(4*I*c)*e^(I*b*x + I*a) + 1) + (3*I*b^2*x^2 - 3*I*a^2)*log(-1/2*sqrt(4*I*c)*e^(I*b*x + I*
a) + 1) + 6*I*polylog(3, 1/2*sqrt(4*I*c)*e^(I*b*x + I*a)) + 6*I*polylog(3, -1/2*sqrt(4*I*c)*e^(I*b*x + I*a)))/
b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arccot((I*c + 1)*tan(b*x + a) + c), x)

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maple [C]  time = 5.68, 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*arccot(c+(1+I*c)*tan(b*x+a)),x)

[Out]

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

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maxima [B]  time = 0.34, size = 218, normalized size = 1.77 \[ \frac {\frac {{\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {arccot}\left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b} - \frac {2 \, {\left (-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (-6 i \, {\left (b x + a\right )}^{2} + 12 i \, {\left (b x + a\right )} a\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )} {\left (-i \, c - 1\right )}}{b {\left (12 \, c - 12 i\right )}}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(c+(1+I*c)*tan(b*x+a)),x, algorithm="maxima")

[Out]

1/2*(((b*x + a)^2 - 2*(b*x + a)*a)*arccot((I*c + 1)*tan(b*x + a) + c)/b - 2*(-4*I*(b*x + a)^3 + 12*I*(b*x + a)
^2*a - 6*I*b*x*dilog(I*c*e^(2*I*b*x + 2*I*a)) + (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a)*arctan2(c*cos(2*b*x + 2*
a), c*sin(2*b*x + 2*a) + 1) + 3*((b*x + a)^2 - 2*(b*x + a)*a)*log(c^2*cos(2*b*x + 2*a)^2 + c^2*sin(2*b*x + 2*a
)^2 + 2*c*sin(2*b*x + 2*a) + 1) + 3*polylog(3, I*c*e^(2*I*b*x + 2*I*a)))*(-I*c - 1)/(b*(12*c - 12*I)))/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {acot}\left (c+\mathrm {tan}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acot(c + tan(a + b*x)*(c*1i + 1)),x)

[Out]

int(x*acot(c + tan(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*acot(c+(1+I*c)*tan(b*x+a)),x)

[Out]

Exception raised: CoercionFailed

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