\(\int \frac {(a+b \text {arctanh}(c x)) (d+e \log (1-c^2 x^2))}{x^5} \, dx\) [530]

Optimal result

Integrand size = 27, antiderivative size = 244 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}-\frac {1}{4} b c^4 e \text {arctanh}(c x)+\frac {b c^2 e \text {arctanh}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \text {arctanh}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \operatorname {PolyLog}(2,-c x)-\frac {1}{4} b c^4 e \operatorname {PolyLog}(2,c x) \] Output:

1/4*a*c^2*e/x^2+5/12*b*c^3*e/x-1/4*b*c^4*e*arctanh(c*x)+1/4*b*c^2*e*arctan 
h(c*x)/x^2-1/2*a*c^4*e*ln(x)+1/12*(3*a+4*b)*c^4*e*ln(-c*x+1)+1/12*(3*a-4*b 
)*c^4*e*ln(c*x+1)-1/12*b*c*(d+e*ln(-c^2*x^2+1))/x^3-1/4*b*c^3*(d+e*ln(-c^2 
*x^2+1))/x+1/4*b*c^4*arctanh(c*x)*(d+e*ln(-c^2*x^2+1))-1/4*(a+b*arctanh(c* 
x))*(d+e*ln(-c^2*x^2+1))/x^4+1/4*b*c^4*e*polylog(2,-c*x)-1/4*b*c^4*e*polyl 
og(2,c*x)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=-\frac {a d}{4 x^4}+\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} \left (3 a c^4 e+4 b c^4 e\right ) \log (1-c x)-\frac {1}{2} b c^4 e \left (-\frac {\text {arctanh}(c x)}{2 c^2 x^2}+\frac {1}{2} \left (-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+b c^4 d \left (-\frac {\text {arctanh}(c x)}{4 c^4 x^4}+\frac {1}{4} \left (-\frac {1}{3 c^3 x^3}-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+\frac {1}{12} \left (3 a c^4 e-4 b c^4 e\right ) \log (1+c x)+\frac {e \left (-3 a-b c x-3 b c^3 x^3-3 b \text {arctanh}(c x)+3 b c^4 x^4 \text {arctanh}(c x)\right ) \log \left (1-c^2 x^2\right )}{12 x^4}-\frac {1}{4} b c^4 e (-\operatorname {PolyLog}(2,-c x)+\operatorname {PolyLog}(2,c x)) \] Input:

Integrate[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]
 

Output:

-1/4*(a*d)/x^4 + (a*c^2*e)/(4*x^2) + (b*c^3*e)/(6*x) - (a*c^4*e*Log[x])/2 
+ ((3*a*c^4*e + 4*b*c^4*e)*Log[1 - c*x])/12 - (b*c^4*e*(-1/2*ArcTanh[c*x]/ 
(c^2*x^2) + (-(1/(c*x)) - Log[1 - c*x]/2 + Log[1 + c*x]/2)/2))/2 + b*c^4*d 
*(-1/4*ArcTanh[c*x]/(c^4*x^4) + (-1/3*1/(c^3*x^3) - 1/(c*x) - Log[1 - c*x] 
/2 + Log[1 + c*x]/2)/4) + ((3*a*c^4*e - 4*b*c^4*e)*Log[1 + c*x])/12 + (e*( 
-3*a - b*c*x - 3*b*c^3*x^3 - 3*b*ArcTanh[c*x] + 3*b*c^4*x^4*ArcTanh[c*x])* 
Log[1 - c^2*x^2])/(12*x^4) - (b*c^4*e*(-PolyLog[2, -(c*x)] + PolyLog[2, c* 
x]))/4
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.96, 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.

\(\displaystyle \int \frac {(a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )}{x^5} \, dx\)

\(\Big \downarrow \) 6647

\(\displaystyle 2 c^2 e \int \left (-\frac {3 b c^3 x^3+b c x+3 a}{12 x^3 \left (1-c^2 x^2\right )}-\frac {b \left (c^2 x^2+1\right ) \text {arctanh}(c x)}{4 x^3}\right )dx-\frac {(a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}+\frac {1}{4} b c^4 \text {arctanh}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {(a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}+2 c^2 e \left (\frac {1}{24} c^2 (3 a+4 b) \log (1-c x)+\frac {1}{24} c^2 (3 a-4 b) \log (c x+1)-\frac {1}{4} a c^2 \log (x)+\frac {a}{8 x^2}-\frac {1}{8} b c^2 \text {arctanh}(c x)+\frac {b \text {arctanh}(c x)}{8 x^2}+\frac {1}{8} b c^2 \operatorname {PolyLog}(2,-c x)-\frac {1}{8} b c^2 \operatorname {PolyLog}(2,c x)+\frac {5 b c}{24 x}\right )+\frac {1}{4} b c^4 \text {arctanh}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x}\)

Input:

Int[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]
 

Output:

-1/12*(b*c*(d + e*Log[1 - c^2*x^2]))/x^3 - (b*c^3*(d + e*Log[1 - c^2*x^2]) 
)/(4*x) + (b*c^4*ArcTanh[c*x]*(d + e*Log[1 - c^2*x^2]))/4 - ((a + b*ArcTan 
h[c*x])*(d + e*Log[1 - c^2*x^2]))/(4*x^4) + 2*c^2*e*(a/(8*x^2) + (5*b*c)/( 
24*x) - (b*c^2*ArcTanh[c*x])/8 + (b*ArcTanh[c*x])/(8*x^2) - (a*c^2*Log[x]) 
/4 + ((3*a + 4*b)*c^2*Log[1 - c*x])/24 + ((3*a - 4*b)*c^2*Log[1 + c*x])/24 
 + (b*c^2*PolyLog[2, -(c*x)])/8 - (b*c^2*PolyLog[2, c*x])/8)
 

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]

\[\int \frac {\left (a +b \,\operatorname {arctanh}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{5}}d x\]

Input:

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)
 

Output:

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)
 

Fricas [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^5,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^5, x)
 

Sympy [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \] Input:

integrate((a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1))/x**5,x)
 

Output:

Integral((a + b*atanh(c*x))*(d + e*log(-c**2*x**2 + 1))/x**5, x)
 

Maxima [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="maxim 
a")
 

Output:

1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c 
- 6*arctanh(c*x)/x^4)*b*d + 1/4*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/ 
x^2)*c^2 - log(-c^2*x^2 + 1)/x^4)*a*e + 1/8*b*e*(log(-c*x + 1)^2/x^4 - 4*i 
ntegrate(-1/2*(2*(c*x - 1)*log(c*x + 1)^2 - c*x*log(-c*x + 1))/(c*x^6 - x^ 
5), x)) - 1/4*a*d/x^4
 

Giac [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \] Input:

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="giac" 
)
 

Output:

integrate((b*arctanh(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^5, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^5} \,d x \] Input:

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^5,x)
 

Output:

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^5, x)
 

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\frac {3 \mathit {atanh} \left (c x \right ) b \,c^{4} d \,x^{4}-3 \mathit {atanh} \left (c x \right ) b d +12 \left (\int \frac {\mathit {atanh} \left (c x \right ) \mathrm {log}\left (-c^{2} x^{2}+1\right )}{x^{5}}d x \right ) b e \,x^{4}+3 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a \,c^{4} e \,x^{4}-3 \,\mathrm {log}\left (-c^{2} x^{2}+1\right ) a e -6 \,\mathrm {log}\left (x \right ) a \,c^{4} e \,x^{4}+3 a \,c^{2} e \,x^{2}-3 a d -3 b \,c^{3} d \,x^{3}-b c d x}{12 x^{4}} \] Input:

int((a+b*atanh(c*x))*(d+e*log(-c^2*x^2+1))/x^5,x)
 

Output:

(3*atanh(c*x)*b*c**4*d*x**4 - 3*atanh(c*x)*b*d + 12*int((atanh(c*x)*log( - 
 c**2*x**2 + 1))/x**5,x)*b*e*x**4 + 3*log( - c**2*x**2 + 1)*a*c**4*e*x**4 
- 3*log( - c**2*x**2 + 1)*a*e - 6*log(x)*a*c**4*e*x**4 + 3*a*c**2*e*x**2 - 
 3*a*d - 3*b*c**3*d*x**3 - b*c*d*x)/(12*x**4)