3.3.45 \(\int x \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx\) [245]

Optimal. Leaf size=134 \[ \frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x \text {PolyLog}\left (2,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {\text {PolyLog}\left (3,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{8 b^2} \]

[Out]

1/6*I*b*x^3+1/2*x^2*arccoth(1+I*d-d*tan(b*x+a))-1/4*x^2*ln(1+(1+I*d)*exp(2*I*a+2*I*b*x))+1/4*I*x*polylog(2,-(1
+I*d)*exp(2*I*a+2*I*b*x))/b-1/8*polylog(3,-(1+I*d)*exp(2*I*a+2*I*b*x))/b^2

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 134, 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 = {6399, 2215, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {\text {Li}_3\left (-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{8 b^2}+\frac {i x \text {Li}_2\left (-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\tan (a+b x))+i d+1)+\frac {1}{6} i b x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*ArcCoth[1 + I*d - d*Tan[a + b*x]],x]

[Out]

(I/6)*b*x^3 + (x^2*ArcCoth[1 + I*d - d*Tan[a + b*x]])/2 - (x^2*Log[1 + (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/4 +
((I/4)*x*PolyLog[2, -((1 + I*d)*E^((2*I)*a + (2*I)*b*x))])/b - PolyLog[3, -((1 + I*d)*E^((2*I)*a + (2*I)*b*x)
)]/(8*b^2)

Rule 2215

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 6399

Int[ArcCoth[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[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 6724

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 \coth ^{-1}(1+i d-d \tan (a+b x)) \, dx &=\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))+\frac {1}{2} (i b) \int \frac {x^2}{1+(1+i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{2} (b (i-d)) \int \frac {e^{2 i a+2 i b x} x^2}{1+(1+i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {1}{2} \int x \log \left (1+(1+i d) e^{2 i a+2 i b x}\right ) \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x \text {Li}_2\left (-(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {i \int \text {Li}_2\left (-(1+i d) e^{2 i a+2 i b x}\right ) \, dx}{4 b}\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x \text {Li}_2\left (-(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {\text {Subst}\left (\int \frac {\text {Li}_2((-1-i d) x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^2}\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x \text {Li}_2\left (-(1+i d) e^{2 i a+2 i b x}\right )}{4 b}-\frac {\text {Li}_3\left (-(1+i d) e^{2 i a+2 i b x}\right )}{8 b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 120, normalized size = 0.90 \begin {gather*} \frac {1}{2} x^2 \coth ^{-1}(1+i d-d \tan (a+b x))-\frac {2 b^2 x^2 \log \left (1-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+2 i b x \text {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+\text {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCoth[1 + I*d - d*Tan[a + b*x]],x]

[Out]

(x^2*ArcCoth[1 + I*d - d*Tan[a + b*x]])/2 - (2*b^2*x^2*Log[1 - I/((-I + d)*E^((2*I)*(a + b*x)))] + (2*I)*b*x*P
olyLog[2, I/((-I + d)*E^((2*I)*(a + b*x)))] + PolyLog[3, I/((-I + d)*E^((2*I)*(a + b*x)))])/(8*b^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.43, size = 2351, normalized size = 17.54

method	result	size
risch	\(\text {Expression too large to display}\)	\(2351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccoth(1+I*d-d*tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (94) = 188\).
time = 0.27, size = 247, normalized size = 1.84 \begin {gather*} -\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {arcoth}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{b} + \frac {-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 ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), -d \sin \left (2 \, b x + 2 \, a\right ) + \cos \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 ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="maxima")

[Out]

-1/24*(12*((b*x + a)^2 - 2*(b*x + a)*a)*arccoth(d*tan(b*x + a) - I*d - 1)/b + (-4*I*(b*x + a)^3 + 12*I*(b*x +
a)^2*a - 6*I*b*x*dilog((-I*d - 1)*e^(2*I*b*x + 2*I*a)) - 6*(-I*(b*x + a)^2 + 2*I*(b*x + a)*a)*arctan2(d*cos(2*
b*x + 2*a) + sin(2*b*x + 2*a), -d*sin(2*b*x + 2*a) + cos(2*b*x + 2*a) + 1) + 3*((b*x + a)^2 - 2*(b*x + a)*a)*l
og((d^2 + 1)*cos(2*b*x + 2*a)^2 + (d^2 + 1)*sin(2*b*x + 2*a)^2 - 2*d*sin(2*b*x + 2*a) + 2*cos(2*b*x + 2*a) + 1
) + 3*polylog(3, (-I*d - 1)*e^(2*I*b*x + 2*I*a)))/b)/b

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (94) = 188\).
time = 0.37, size = 292, normalized size = 2.18 \begin {gather*} \frac {2 i \, b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 2 i \, a^{3} + 6 i \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 6 \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(2*I*b^3*x^3 - 3*b^2*x^2*log(d*e^(2*I*b*x + 2*I*a)/((d - I)*e^(2*I*b*x + 2*I*a) - I)) + 2*I*a^3 + 6*I*b*x
*dilog(1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) + 6*I*b*x*dilog(-1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) - 3*a^2*lo
g(1/2*(2*(d - I)*e^(I*b*x + I*a) + I*sqrt(-4*I*d - 4))/(d - I)) - 3*a^2*log(1/2*(2*(d - I)*e^(I*b*x + I*a) - I
*sqrt(-4*I*d - 4))/(d - I)) - 3*(b^2*x^2 - a^2)*log(1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a) + 1) - 3*(b^2*x^2 - a
^2)*log(-1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a) + 1) - 6*polylog(3, 1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)) - 6*po
lylog(3, -1/2*sqrt(-4*I*d - 4)*e^(I*b*x + I*a)))/b^2

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {acoth}{\left (- d \tan {\left (a + b x \right )} + i d + 1 \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acoth(1+I*d-d*tan(b*x+a)),x)

[Out]

Integral(x*acoth(-d*tan(a + b*x) + I*d + 1), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccoth(1+I*d-d*tan(b*x+a)),x, algorithm="giac")

[Out]

integrate(x*arccoth(-d*tan(b*x + a) + I*d + 1), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {acoth}\left (1-d\,\mathrm {tan}\left (a+b\,x\right )+d\,1{}\mathrm {i}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*acoth(d*1i - d*tan(a + b*x) + 1),x)

[Out]

int(x*acoth(d*1i - d*tan(a + b*x) + 1), x)

________________________________________________________________________________________