\(\int (a+b \coth ^{-1}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [274]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-2 a e x-2 b e x \coth ^{-1}(c x)+\frac {e \left (a+b \coth ^{-1}(c x)\right )^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e} \]

[Out]

-2*a*e*x-2*b*e*x*arccoth(c*x)+e*(a+b*arccoth(c*x))^2/b/c-b*e*ln(-c^2*x^2+1)/c+x*(a+b*arccoth(c*x))*(d+e*ln(-c^
2*x^2+1))+1/4*b*(d+e*ln(-c^2*x^2+1))^2/c/e

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6221, 2525, 2437, 2338, 6128, 6022, 266, 6096} \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=x \left (a+b \coth ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {e \left (a+b \coth ^{-1}(c x)\right )^2}{b c}-2 a e x+\frac {b \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{4 c e}-\frac {b e \log \left (1-c^2 x^2\right )}{c}-2 b e x \coth ^{-1}(c x) \]

[In]

Int[(a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]),x]

[Out]

-2*a*e*x - 2*b*e*x*ArcCoth[c*x] + (e*(a + b*ArcCoth[c*x])^2)/(b*c) - (b*e*Log[1 - c^2*x^2])/c + x*(a + b*ArcCo
th[c*x])*(d + e*Log[1 - c^2*x^2]) + (b*(d + e*Log[1 - c^2*x^2])^2)/(4*c*e)

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 6022

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCoth[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6096

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6128

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcCoth[c*x])^p, x], x] - Dist[d*(f^2/e), Int[(f*x)^(m - 2)*((a + b*ArcCoth[c*x])
^p/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 6221

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.)), x_Symbol] :> Simp[x*(d + e*
Log[f + g*x^2])*(a + b*ArcCoth[c*x]), x] + (-Dist[b*c, Int[x*((d + e*Log[f + g*x^2])/(1 - c^2*x^2)), x], x] -
Dist[2*e*g, Int[x^2*((a + b*ArcCoth[c*x])/(f + g*x^2)), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x]

Rubi steps \begin{align*} \text {integral}& = x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-(b c) \int \frac {x \left (d+e \log \left (1-c^2 x^2\right )\right )}{1-c^2 x^2} \, dx+\left (2 c^2 e\right ) \int \frac {x^2 \left (a+b \coth ^{-1}(c x)\right )}{1-c^2 x^2} \, dx \\ & = x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {1}{2} (b c) \text {Subst}\left (\int \frac {d+e \log \left (1-c^2 x\right )}{1-c^2 x} \, dx,x,x^2\right )-(2 e) \int \left (a+b \coth ^{-1}(c x)\right ) \, dx+(2 e) \int \frac {a+b \coth ^{-1}(c x)}{1-c^2 x^2} \, dx \\ & = -2 a e x+\frac {e \left (a+b \coth ^{-1}(c x)\right )^2}{b c}+x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \text {Subst}\left (\int \frac {d+e \log (x)}{x} \, dx,x,1-c^2 x^2\right )}{2 c}-(2 b e) \int \coth ^{-1}(c x) \, dx \\ & = -2 a e x-2 b e x \coth ^{-1}(c x)+\frac {e \left (a+b \coth ^{-1}(c x)\right )^2}{b c}+x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e}+(2 b c e) \int \frac {x}{1-c^2 x^2} \, dx \\ & = -2 a e x-2 b e x \coth ^{-1}(c x)+\frac {e \left (a+b \coth ^{-1}(c x)\right )^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.38 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=a d x-2 a e x+b d x \coth ^{-1}(c x)-2 b e x \coth ^{-1}(c x)+\frac {b e \coth ^{-1}(c x)^2}{c}+\frac {2 a e \text {arctanh}(c x)}{c}+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+a e x \log \left (1-c^2 x^2\right )+b e x \coth ^{-1}(c x) \log \left (1-c^2 x^2\right )+\frac {b e \log ^2\left (1-c^2 x^2\right )}{4 c} \]

[In]

Integrate[(a + b*ArcCoth[c*x])*(d + e*Log[1 - c^2*x^2]),x]

[Out]

a*d*x - 2*a*e*x + b*d*x*ArcCoth[c*x] - 2*b*e*x*ArcCoth[c*x] + (b*e*ArcCoth[c*x]^2)/c + (2*a*e*ArcTanh[c*x])/c
+ (b*d*Log[1 - c^2*x^2])/(2*c) - (b*e*Log[1 - c^2*x^2])/c + a*e*x*Log[1 - c^2*x^2] + b*e*x*ArcCoth[c*x]*Log[1
- c^2*x^2] + (b*e*Log[1 - c^2*x^2]^2)/(4*c)

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {4 b e \ln \left (-c^{2} x^{2}+1\right ) x \,\operatorname {arccoth}\left (c x \right ) c +4 b \,\operatorname {arccoth}\left (c x \right ) x c d -8 e b x \,\operatorname {arccoth}\left (c x \right ) c +4 a e x \ln \left (-c^{2} x^{2}+1\right ) c +4 a c d x -8 a e c x +4 e b \operatorname {arccoth}\left (c x \right )^{2}+e b \ln \left (-c^{2} x^{2}+1\right )^{2}+8 \,\operatorname {arccoth}\left (c x \right ) a e +2 \ln \left (-c^{2} x^{2}+1\right ) b d -4 \ln \left (-c^{2} x^{2}+1\right ) b e}{4 c}\)	\(141\)
risch	\(\text {Expression too large to display}\)	\(1104\)
default	\(\text {Expression too large to display}\)	\(2096\)
parts	\(\text {Expression too large to display}\)	\(2096\)

[In]

int((a+b*arccoth(c*x))*(d+e*ln(-c^2*x^2+1)),x,method=_RETURNVERBOSE)

[Out]

1/4*(4*b*e*ln(-c^2*x^2+1)*x*arccoth(c*x)*c+4*b*arccoth(c*x)*x*c*d-8*e*b*x*arccoth(c*x)*c+4*a*e*x*ln(-c^2*x^2+1
)*c+4*a*c*d*x-8*a*e*c*x+4*e*b*arccoth(c*x)^2+e*b*ln(-c^2*x^2+1)^2+8*arccoth(c*x)*a*e+2*ln(-c^2*x^2+1)*b*d-4*ln
(-c^2*x^2+1)*b*e)/c

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.25 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\frac {b e \log \left (-c^{2} x^{2} + 1\right )^{2} + b e \log \left (\frac {c x + 1}{c x - 1}\right )^{2} + 4 \, {\left (a c d - 2 \, a c e\right )} x + 2 \, {\left (2 \, a c e x + b d - 2 \, b e\right )} \log \left (-c^{2} x^{2} + 1\right ) + 2 \, {\left (b c e x \log \left (-c^{2} x^{2} + 1\right ) + 2 \, a e + {\left (b c d - 2 \, b c e\right )} x\right )} \log \left (\frac {c x + 1}{c x - 1}\right )}{4 \, c} \]

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/4*(b*e*log(-c^2*x^2 + 1)^2 + b*e*log((c*x + 1)/(c*x - 1))^2 + 4*(a*c*d - 2*a*c*e)*x + 2*(2*a*c*e*x + b*d - 2
*b*e)*log(-c^2*x^2 + 1) + 2*(b*c*e*x*log(-c^2*x^2 + 1) + 2*a*e + (b*c*d - 2*b*c*e)*x)*log((c*x + 1)/(c*x - 1))
)/c

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\begin {cases} a d x + a e x \log {\left (- c^{2} x^{2} + 1 \right )} - 2 a e x + \frac {2 a e \operatorname {acoth}{\left (c x \right )}}{c} + b d x \operatorname {acoth}{\left (c x \right )} + b e x \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {acoth}{\left (c x \right )} - 2 b e x \operatorname {acoth}{\left (c x \right )} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{4 c} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}}{c} + \frac {b e \operatorname {acoth}^{2}{\left (c x \right )}}{c} & \text {for}\: c \neq 0 \\d x \left (a + \frac {i \pi b}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*acoth(c*x))*(d+e*ln(-c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x + a*e*x*log(-c**2*x**2 + 1) - 2*a*e*x + 2*a*e*acoth(c*x)/c + b*d*x*acoth(c*x) + b*e*x*log(-c*
*2*x**2 + 1)*acoth(c*x) - 2*b*e*x*acoth(c*x) + b*d*log(-c**2*x**2 + 1)/(2*c) + b*e*log(-c**2*x**2 + 1)**2/(4*c
) - b*e*log(-c**2*x**2 + 1)/c + b*e*acoth(c*x)**2/c, Ne(c, 0)), (d*x*(a + I*pi*b/2), True))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.71 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-{\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} b e \operatorname {arcoth}\left (c x\right ) - {\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} a e + a d x + \frac {{\left (2 \, c x \operatorname {arcoth}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} + \frac {{\left ({\left (i \, \pi + 2 \, \log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) + {\left (i \, \pi - 2\right )} \log \left (c x - 1\right )\right )} b e}{2 \, c} \]

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="maxima")

[Out]

-(c^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3) - x*log(-c^2*x^2 + 1))*b*e*arccoth(c*x) - (c^2*(2*x/c^2
- log(c*x + 1)/c^3 + log(c*x - 1)/c^3) - x*log(-c^2*x^2 + 1))*a*e + a*d*x + 1/2*(2*c*x*arccoth(c*x) + log(-c^2
*x^2 + 1))*b*d/c + 1/2*((I*pi + 2*log(c*x - 1) - 2)*log(c*x + 1) + (I*pi - 2)*log(c*x - 1))*b*e/c

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.90 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=-\frac {1}{2} \, b e x \log \left (-c x + 1\right )^{2} - \frac {1}{2} \, {\left (-i \, \pi b e - b d - 2 \, a e + 2 \, b e\right )} x \log \left (c x + 1\right ) + \frac {1}{2} \, {\left (b e x + \frac {b e}{c}\right )} \log \left (c x + 1\right )^{2} - \frac {b e \log \left (c x - 1\right )^{2}}{2 \, c} - \frac {1}{2} \, {\left (-i \, \pi b d + 2 i \, \pi b e - 2 \, a d + 4 \, a e\right )} x - \frac {1}{2} \, {\left ({\left (-i \, \pi b e + b d - 2 \, a e - 2 \, b e\right )} x - \frac {2 \, b e \log \left (c x - 1\right )}{c}\right )} \log \left (-c x + 1\right ) + \frac {{\left (i \, \pi b e + b d + 2 \, a e - 2 \, b e\right )} \log \left (c x + 1\right )}{2 \, c} + \frac {{\left (-i \, \pi b e + b d - 2 \, a e - 2 \, b e\right )} \log \left (c x - 1\right )}{2 \, c} \]

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="giac")

[Out]

-1/2*b*e*x*log(-c*x + 1)^2 - 1/2*(-I*pi*b*e - b*d - 2*a*e + 2*b*e)*x*log(c*x + 1) + 1/2*(b*e*x + b*e/c)*log(c*
x + 1)^2 - 1/2*b*e*log(c*x - 1)^2/c - 1/2*(-I*pi*b*d + 2*I*pi*b*e - 2*a*d + 4*a*e)*x - 1/2*((-I*pi*b*e + b*d -
2*a*e - 2*b*e)*x - 2*b*e*log(c*x - 1)/c)*log(-c*x + 1) + 1/2*(I*pi*b*e + b*d + 2*a*e - 2*b*e)*log(c*x + 1)/c
+ 1/2*(-I*pi*b*e + b*d - 2*a*e - 2*b*e)*log(c*x - 1)/c

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.03 \[ \int \left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx=\ln \left (\frac {1}{c\,x}+1\right )\,\left (\frac {b\,d\,x}{2}-b\,e\,x+\frac {b\,e\,x\,\ln \left (1-c^2\,x^2\right )}{2}\right )+\ln \left (1-\frac {1}{c\,x}\right )\,\left (\frac {b\,d\,x^2-b\,c^2\,d\,x^4}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}-\frac {2\,b\,e\,x^2-2\,b\,c^2\,e\,x^4}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}+\frac {\ln \left (1-c^2\,x^2\right )\,\left (b\,e\,x^2-b\,c^2\,e\,x^4\right )}{2\,\left (c\,x^2+x\right )\,\left (c\,x-1\right )}-\frac {b\,e\,\ln \left (\frac {1}{c\,x}+1\right )}{2\,c}\right )+a\,x\,\left (d-2\,e\right )+\frac {\ln \left (c\,x+1\right )\,\left (2\,a\,e+b\,d-2\,b\,e\right )}{2\,c}-\frac {\ln \left (c\,x-1\right )\,\left (2\,a\,e-b\,d+2\,b\,e\right )}{2\,c}+\frac {b\,e\,{\ln \left (\frac {1}{c\,x}+1\right )}^2}{4\,c}+\frac {b\,e\,{\ln \left (1-\frac {1}{c\,x}\right )}^2}{4\,c}+\frac {b\,e\,{\ln \left (1-c^2\,x^2\right )}^2}{4\,c}+a\,e\,x\,\ln \left (1-c^2\,x^2\right ) \]

[In]

int((a + b*acoth(c*x))*(d + e*log(1 - c^2*x^2)),x)

[Out]

log(1/(c*x) + 1)*((b*d*x)/2 - b*e*x + (b*e*x*log(1 - c^2*x^2))/2) + log(1 - 1/(c*x))*((b*d*x^2 - b*c^2*d*x^4)/
(2*(x + c*x^2)*(c*x - 1)) - (2*b*e*x^2 - 2*b*c^2*e*x^4)/(2*(x + c*x^2)*(c*x - 1)) + (log(1 - c^2*x^2)*(b*e*x^2
- b*c^2*e*x^4))/(2*(x + c*x^2)*(c*x - 1)) - (b*e*log(1/(c*x) + 1))/(2*c)) + a*x*(d - 2*e) + (log(c*x + 1)*(2*
a*e + b*d - 2*b*e))/(2*c) - (log(c*x - 1)*(2*a*e - b*d + 2*b*e))/(2*c) + (b*e*log(1/(c*x) + 1)^2)/(4*c) + (b*e
*log(1 - 1/(c*x))^2)/(4*c) + (b*e*log(1 - c^2*x^2)^2)/(4*c) + a*e*x*log(1 - c^2*x^2)