3.3.37 \(\int x \coth ^{-1}(c+d \tan (a+b x)) \, dx\) [237]

3.3.37.1 Optimal result

Integrand size = 13, antiderivative size = 295 \[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\frac {1}{2} x^2 \coth ^{-1}(c+d \tan (a+b x))+\frac {1}{4} x^2 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{4} x^2 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x \operatorname {PolyLog}\left (2,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (2,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (3,-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{8 b^2}-\frac {\operatorname {PolyLog}\left (3,-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{8 b^2} \]

output

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

3.3.37.2 Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.88 \[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\frac {4 b^2 x^2 \coth ^{-1}(c+d \tan (a+b x))+2 b^2 x^2 \log \left (1+\frac {(-1+c+i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-2 b^2 x^2 \log \left (1+\frac {(1+c+i d) e^{-2 i (a+b x)}}{1+c-i d}\right )+2 i b x \operatorname {PolyLog}\left (2,\frac {(1-c-i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-2 i b x \operatorname {PolyLog}\left (2,\frac {(-1-c-i d) e^{-2 i (a+b x)}}{1+c-i d}\right )+\operatorname {PolyLog}\left (3,\frac {(1-c-i d) e^{-2 i (a+b x)}}{-1+c-i d}\right )-\operatorname {PolyLog}\left (3,\frac {(-1-c-i d) e^{-2 i (a+b x)}}{1+c-i d}\right )}{8 b^2} \]

input

Integrate[x*ArcCoth[c + d*Tan[a + b*x]],x]
 

output

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

3.3.37.3 Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.36, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6822, 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.

\(\displaystyle \int x \coth ^{-1}(d \tan (a+b x)+c) \, dx\)

\(\Big \downarrow \) 6822

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

input

Int[x*ArcCoth[c + d*Tan[a + b*x]],x]
 

output

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

3.3.37.3.1 Defintions of rubi rules used

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 6822

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] + (-Simp[I*b*((1 + c - I*d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1) 
*(E^(2*I*a + 2*I*b*x)/(1 + c + I*d + (1 + c - I*d)*E^(2*I*a + 2*I*b*x))), x 
], x] + Simp[I*b*((1 - c + I*d)/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*(E^(2* 
I*a + 2*I*b*x)/(1 - c - I*d + (1 - c + I*d)*E^(2*I*a + 2*I*b*x))), x], x]) 
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(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]
 

3.3.37.4 Maple [C] (warning: unable to verify)

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

Time = 2.92 (sec) , antiderivative size = 6481, normalized size of antiderivative = 21.97

method	result	size
risch	\(\text {Expression too large to display}\)	\(6481\)

input

int(x*arccoth(c+d*tan(b*x+a)),x,method=_RETURNVERBOSE)
 

output

result too large to display
 

3.3.37.5 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. 1688 vs. \(2 (209) = 418\).

Time = 0.33 (sec) , antiderivative size = 1688, normalized size of antiderivative = 5.72 \[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\text {Too large to display} \]

input

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

output

1/16*(4*b^2*x^2*log((d*tan(b*x + a) + c + 1)/(d*tan(b*x + a) + c - 1)) - 2 
*I*b*x*dilog(2*((I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c + 1)*d + ( 
I*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/((c^2 + d 
^2 + 2*c + 1)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) + 2*I*b*x*dilog(2 
*((-I*(c + 1)*d - d^2)*tan(b*x + a)^2 - c^2 + I*(c + 1)*d + (-I*c^2 - 2*(c 
 + 1)*d + I*d^2 - 2*I*c - I)*tan(b*x + a) - 2*c - 1)/((c^2 + d^2 + 2*c + 1 
)*tan(b*x + a)^2 + c^2 + d^2 + 2*c + 1) + 1) + 2*I*b*x*dilog(2*((I*(c - 1) 
*d - d^2)*tan(b*x + a)^2 - c^2 - I*(c - 1)*d + (I*c^2 - 2*(c - 1)*d - I*d^ 
2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a) 
^2 + c^2 + d^2 - 2*c + 1) + 1) - 2*I*b*x*dilog(2*((-I*(c - 1)*d - d^2)*tan 
(b*x + a)^2 - c^2 + I*(c - 1)*d + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - 
I)*tan(b*x + a) + 2*c - 1)/((c^2 + d^2 - 2*c + 1)*tan(b*x + a)^2 + c^2 + d 
^2 - 2*c + 1) + 1) - 2*a^2*log(((I*(c + 1)*d + d^2)*tan(b*x + a)^2 - c^2 + 
 I*(c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) - 2*c - 1)/(tan(b* 
x + a)^2 + 1)) - 2*a^2*log(((I*(c + 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*( 
c + 1)*d + (I*c^2 + I*d^2 + 2*I*c + I)*tan(b*x + a) + 2*c + 1)/(tan(b*x + 
a)^2 + 1)) + 2*a^2*log(((I*(c - 1)*d + d^2)*tan(b*x + a)^2 - c^2 + I*(c - 
1)*d + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) + 2*c - 1)/(tan(b*x + a)^2 
 + 1)) + 2*a^2*log(((I*(c - 1)*d - d^2)*tan(b*x + a)^2 + c^2 + I*(c - 1)*d 
 + (I*c^2 + I*d^2 - 2*I*c + I)*tan(b*x + a) - 2*c + 1)/(tan(b*x + a)^2 ...
 

3.3.37.6 Sympy [F]

\[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int x \operatorname {acoth}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]

input

integrate(x*acoth(c+d*tan(b*x+a)),x)
 

output

Integral(x*acoth(c + d*tan(a + b*x)), x)
 

3.3.37.7 Maxima [F]

\[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int { x \operatorname {arcoth}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]

input

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

output

-2*b*d*integrate(-(2*(c^2 + d^2 - 1)*x^2*cos(2*b*x + 2*a)^2 + 2*c*d*x^2*si 
n(2*b*x + 2*a) + 2*(c^2 + d^2 - 1)*x^2*sin(2*b*x + 2*a)^2 + (c^2 - d^2 - 1 
)*x^2*cos(2*b*x + 2*a) - (2*c*d*x^2*sin(2*b*x + 2*a) - (c^2 - d^2 - 1)*x^2 
*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^2*cos(2*b*x + 2*a) + (c^2 - 
 d^2 - 1)*x^2*sin(2*b*x + 2*a))*sin(4*b*x + 4*a))/(c^4 + d^4 + 2*(c^2 + 1) 
*d^2 + (c^4 + d^4 + 2*(c^2 + 1)*d^2 - 2*c^2 + 1)*cos(4*b*x + 4*a)^2 + 4*(c 
^4 + d^4 + 2*(c^2 - 1)*d^2 - 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4 + 
2*(c^2 + 1)*d^2 - 2*c^2 + 1)*sin(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 - 
1)*d^2 - 2*c^2 + 1)*sin(2*b*x + 2*a)^2 - 2*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 - 
 1)*d^2 - 2*c^2 + 2*(c^4 - d^4 - 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(c*d^3 + 
(c^3 - c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + 4*(c^4 - d^4 - 2*c^2 
 + 1)*cos(2*b*x + 2*a) - 4*(2*c*d^3 - 2*(c^3 - c)*d - 2*(c*d^3 + (c^3 - c) 
*d)*cos(2*b*x + 2*a) - (c^4 - d^4 - 2*c^2 + 1)*sin(2*b*x + 2*a))*sin(4*b*x 
 + 4*a) + 8*(c*d^3 + (c^3 - c)*d)*sin(2*b*x + 2*a) + 1), x) + 1/8*x^2*log( 
(c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c + 1)*d*sin(2*b*x + 2*a) + 
(c^2 + d^2 + 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 + 2*c 
+ 1)*cos(2*b*x + 2*a) + 2*c + 1) - 1/8*x^2*log((c^2 + d^2 - 2*c + 1)*cos(2 
*b*x + 2*a)^2 + 4*(c - 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 - 2*c + 1)*sin(2 
*b*x + 2*a)^2 + c^2 + d^2 + 2*(c^2 - d^2 - 2*c + 1)*cos(2*b*x + 2*a) - 2*c 
 + 1)
 

3.3.37.8 Giac [F]

\[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int { x \operatorname {arcoth}\left (d \tan \left (b x + a\right ) + c\right ) \,d x } \]

input

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

output

integrate(x*arccoth(d*tan(b*x + a) + c), x)
 

3.3.37.9 Mupad [F(-1)]

Timed out. \[ \int x \coth ^{-1}(c+d \tan (a+b x)) \, dx=\int x\,\mathrm {acoth}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]

input

int(x*acoth(c + d*tan(a + b*x)),x)
 

output

int(x*acoth(c + d*tan(a + b*x)), x)