\(\int \frac {(a+b \text {arctanh}(c x)) (d+e \log (1-c^2 x^2))}{x^3} \, dx\) [528]

Optimal result

Integrand size = 27, antiderivative size = 157 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=-a c^2 e \log (x)+\frac {1}{2} (a+b) c^2 e \log (1-c x)+\frac {1}{2} (a-b) c^2 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x}+\frac {1}{2} b c^2 \text {arctanh}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 e \operatorname {PolyLog}(2,-c x)-\frac {1}{2} b c^2 e \operatorname {PolyLog}(2,c x) \] Output:

-a*c^2*e*ln(x)+1/2*(a+b)*c^2*e*ln(-c*x+1)+1/2*(a-b)*c^2*e*ln(c*x+1)-1/2*b* 
c*(d+e*ln(-c^2*x^2+1))/x+1/2*b*c^2*arctanh(c*x)*(d+e*ln(-c^2*x^2+1))-1/2*( 
a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^2+1/2*b*c^2*e*polylog(2,-c*x)-1/2 
*b*c^2*e*polylog(2,c*x)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\frac {1}{2} \left (-\frac {a d}{x^2}-2 a c^2 e \log (x)+(a+b) c^2 e \log (1-c x)+(a-b) c^2 e \log (1+c x)-\frac {b d (2 \text {arctanh}(c x)+c x (2+c x \log (1-c x)-c x \log (1+c x)))}{2 x^2}-\frac {e \left (a+b c x+\left (b-b c^2 x^2\right ) \text {arctanh}(c x)\right ) \log \left (1-c^2 x^2\right )}{x^2}+b c^2 e (\operatorname {PolyLog}(2,-c x)-\operatorname {PolyLog}(2,c x))\right ) \] Input:

Integrate[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^3,x]
 

Output:

(-((a*d)/x^2) - 2*a*c^2*e*Log[x] + (a + b)*c^2*e*Log[1 - c*x] + (a - b)*c^ 
2*e*Log[1 + c*x] - (b*d*(2*ArcTanh[c*x] + c*x*(2 + c*x*Log[1 - c*x] - c*x* 
Log[1 + c*x])))/(2*x^2) - (e*(a + b*c*x + (b - b*c^2*x^2)*ArcTanh[c*x])*Lo 
g[1 - c^2*x^2])/x^2 + b*c^2*e*(PolyLog[2, -(c*x)] - PolyLog[2, c*x]))/2
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {6647, 2009}

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.

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

\(\Big \downarrow \) 6647

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

\(\Big \downarrow \) 2009

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

Input:

Int[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^3,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6647

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]* 
(e_.))*(x_)^(m_.), x_Symbol] :> With[{u = IntHide[x^m*(a + b*ArcTanh[c*x]), 
 x]}, Simp[(d + e*Log[f + g*x^2])   u, x] - Simp[2*e*g   Int[ExpandIntegran 
d[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && Inte 
gerQ[m] && NeQ[m, -1]
 

Maple [F]

\[\int \frac {\left (a +b \,\operatorname {arctanh}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{3}}d x\]

Input:

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^3,x)
 

Output:

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^3,x)
 

Fricas [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^3,x, algorithm="frica 
s")
 

Output:

integral((b*d*arctanh(c*x) + a*d + (b*e*arctanh(c*x) + a*e)*log(-c^2*x^2 + 
 1))/x^3, x)
 

Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{3}}\, dx \] Input:

integrate((a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1))/x**3,x)
 

Output:

Integral((a + b*atanh(c*x))*(d + e*log(-c**2*x**2 + 1))/x**3, x)
 

Maxima [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^3,x, algorithm="maxim 
a")
 

Output:

1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d + 
 1/2*(c^2*(log(c^2*x^2 - 1) - log(x^2)) - log(-c^2*x^2 + 1)/x^2)*a*e + 1/4 
*b*e*(log(-c*x + 1)^2/x^2 - 2*integrate(-((c*x - 1)*log(c*x + 1)^2 - c*x*l 
og(-c*x + 1))/(c*x^4 - x^3), x)) - 1/2*a*d/x^2
 

Giac [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^3,x, algorithm="giac" 
)
 

Output:

integrate((b*arctanh(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^3} \,d x \] Input:

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^3,x)
 

Output:

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^3, x)
 

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^3} \, dx=\frac {\mathit {atanh} \left (c x \right ) b \,c^{2} d \,x^{2}-\mathit {atanh} \left (c x \right ) b d +2 \left (\int \frac {\mathit {atanh} \left (c x \right ) \mathrm {log}\left (-c^{2} x^{2}+1\right )}{x^{3}}d x \right ) b e \,x^{2}+\mathrm {log}\left (-c^{2} x^{2}+1\right ) a \,c^{2} e \,x^{2}-\mathrm {log}\left (-c^{2} x^{2}+1\right ) a e -2 \,\mathrm {log}\left (x \right ) a \,c^{2} e \,x^{2}-a d -b c d x}{2 x^{2}} \] Input:

int((a+b*atanh(c*x))*(d+e*log(-c^2*x^2+1))/x^3,x)
 

Output:

(atanh(c*x)*b*c**2*d*x**2 - atanh(c*x)*b*d + 2*int((atanh(c*x)*log( - c**2 
*x**2 + 1))/x**3,x)*b*e*x**2 + log( - c**2*x**2 + 1)*a*c**2*e*x**2 - log( 
- c**2*x**2 + 1)*a*e - 2*log(x)*a*c**2*e*x**2 - a*d - b*c*d*x)/(2*x**2)