\(\int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx\) [326]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \text {arctanh}(1+i d-d \tan (a+b x))-\frac {1}{4} x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )+\frac {i x \operatorname {PolyLog}\left (2,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{4 b}-\frac {\operatorname {PolyLog}\left (3,-\left ((1+i d) e^{2 i a+2 i b x}\right )\right )}{8 b^2} \] Output:

1/6*I*b*x^3-1/2*x^2*arctanh(-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*polyl 
og(3,-(1+I*d)*exp(2*I*a+2*I*b*x))/b^2
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.90 \[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {1}{2} x^2 \text {arctanh}(1+i d-d \tan (a+b x))-\frac {2 b^2 x^2 \log \left (1-\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+2 i b x \operatorname {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )+\operatorname {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{-i+d}\right )}{8 b^2} \] Input:

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

Output:

(x^2*ArcTanh[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*PolyLog[2, I/((-I + d)*E^((2*I)*(a + b* 
x)))] + PolyLog[3, I/((-I + d)*E^((2*I)*(a + b*x)))])/(8*b^2)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6817, 2615, 2620, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x \text {arctanh}(d (-\tan (a+b x))+i d+1) \, dx\)

\(\Big \downarrow \) 6817

\(\displaystyle \frac {1}{2} i b \int \frac {x^2}{e^{2 i a+2 i b x} (i d+1)+1}dx+\frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1+i d) \int \frac {e^{2 i a+2 i b x} x^2}{e^{2 i a+2 i b x} (i d+1)+1}dx\right )+\frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1+i d) \left (\frac {x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}-\frac {\int x \log \left (e^{2 i a+2 i b x} (i d+1)+1\right )dx}{b (-d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1+i d) \left (\frac {x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {i \int \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )dx}{2 b}}{b (-d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{2} i b \left (\frac {x^3}{3}-(1+i d) \left (\frac {x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )de^{2 i a+2 i b x}}{4 b^2}}{b (-d+i)}\right )\right )+\frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{2} x^2 \text {arctanh}(d (-\tan (a+b x))+i d+1)+\frac {1}{2} i b \left (\frac {x^3}{3}-(1+i d) \left (\frac {x^2 \log \left (1+(1+i d) e^{2 i a+2 i b x}\right )}{2 b (-d+i)}-\frac {\frac {i x \operatorname {PolyLog}\left (2,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{2 b}-\frac {\operatorname {PolyLog}\left (3,-\left ((i d+1) e^{2 i a+2 i b x}\right )\right )}{4 b^2}}{b (-d+i)}\right )\right )\)

Input:

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

Output:

(x^2*ArcTanh[1 + I*d - d*Tan[a + b*x]])/2 + (I/2)*b*(x^3/3 - (1 + I*d)*((x 
^2*Log[1 + (1 + I*d)*E^((2*I)*a + (2*I)*b*x)])/(2*b*(I - d)) - (((I/2)*x*P 
olyLog[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))]/(4*b^2))/(b*(I - d))))
 

Defintions of rubi rules used

rule 2615

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] - Simp[ 
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 2620

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] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 6817

Int[ArcTanh[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ 
.), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTanh[c + d*Tan[a + b*x]]/(f*(m 
+ 1))), x] + Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 3.81 (sec) , antiderivative size = 2289, normalized size of antiderivative = 17.08

method	result	size
risch	\(\text {Expression too large to display}\)	\(2289\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (93) = 186\).

Time = 0.11 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.19 \[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\frac {2 i \, b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (-\frac {d e^{\left (2 i \, b x + 2 i \, a\right )}}{{\left (d - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 2 i \, a^{3} + 6 i \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) + 6 i \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} + i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) - 3 \, a^{2} \log \left (\frac {2 \, {\left (d - i\right )} e^{\left (i \, b x + i \, a\right )} - i \, \sqrt {-4 i \, d - 4}}{2 \, {\left (d - i\right )}}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 6 \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right ) - 6 \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, d - 4} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, b^{2}} \] Input:

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

Output:

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*log 
(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*polylog(3, -1/2*sqrt(-4*I*d - 4)*e^(I*b*x + 
 I*a)))/b^2
 

Sympy [F]

\[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\int x \operatorname {atanh}{\left (- d \tan {\left (a + b x \right )} + i d + 1 \right )}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (93) = 186\).

Time = 0.05 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.84 \[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=-\frac {\frac {12 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d - 1\right )}{b} + \frac {-4 i \, {\left (b x + a\right )}^{3} + 12 i \, {\left (b x + a\right )}^{2} a - 6 i \, b x {\rm Li}_2\left ({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 6 \, {\left (-i \, {\left (b x + a\right )}^{2} + 2 i \, {\left (b x + a\right )} a\right )} \arctan \left (d \cos \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, a\right ), -d \sin \left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a\right )} \log \left ({\left (d^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (d^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, d \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \, {\rm Li}_{3}({\left (-i \, d - 1\right )} e^{\left (2 i \, b x + 2 i \, a\right )})}{b}}{24 \, b} \] Input:

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

Output:

-1/24*(12*((b*x + a)^2 - 2*(b*x + a)*a)*arctanh(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)*log((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
 

Giac [F]

\[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\int { -x \operatorname {artanh}\left (d \tan \left (b x + a\right ) - i \, d - 1\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=\int x\,\mathrm {atanh}\left (1-d\,\mathrm {tan}\left (a+b\,x\right )+d\,1{}\mathrm {i}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x \text {arctanh}(1+i d-d \tan (a+b x)) \, dx=-\left (\int \mathit {atanh} \left (\tan \left (b x +a \right ) d -d i -1\right ) x d x \right ) \] Input:

int(-x*atanh(-1-I*d+d*tan(b*x+a)),x)
 

Output:

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