Integrand size = 27, antiderivative size = 244 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}-\frac {1}{4} b c^4 e \text {arctanh}(c x)+\frac {b c^2 e \text {arctanh}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \text {arctanh}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \operatorname {PolyLog}(2,-c x)-\frac {1}{4} b c^4 e \operatorname {PolyLog}(2,c x) \]
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)
Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=-\frac {a d}{4 x^4}+\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} \left (3 a c^4 e+4 b c^4 e\right ) \log (1-c x)-\frac {1}{2} b c^4 e \left (-\frac {\text {arctanh}(c x)}{2 c^2 x^2}+\frac {1}{2} \left (-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+b c^4 d \left (-\frac {\text {arctanh}(c x)}{4 c^4 x^4}+\frac {1}{4} \left (-\frac {1}{3 c^3 x^3}-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+\frac {1}{12} \left (3 a c^4 e-4 b c^4 e\right ) \log (1+c x)+\frac {e \left (-3 a-b c x-3 b c^3 x^3-3 b \text {arctanh}(c x)+3 b c^4 x^4 \text {arctanh}(c x)\right ) \log \left (1-c^2 x^2\right )}{12 x^4}-\frac {1}{4} b c^4 e (-\operatorname {PolyLog}(2,-c x)+\operatorname {PolyLog}(2,c x)) \]
-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
Time = 0.54 (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 |
\(\displaystyle -\frac {(a+b \text {arctanh}(c x)) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}+2 c^2 e \left (\frac {1}{24} c^2 (3 a+4 b) \log (1-c x)+\frac {1}{24} c^2 (3 a-4 b) \log (c x+1)-\frac {1}{4} a c^2 \log (x)+\frac {a}{8 x^2}-\frac {1}{8} b c^2 \text {arctanh}(c x)+\frac {b \text {arctanh}(c x)}{8 x^2}+\frac {1}{8} b c^2 \operatorname {PolyLog}(2,-c x)-\frac {1}{8} b c^2 \operatorname {PolyLog}(2,c x)+\frac {5 b c}{24 x}\right )+\frac {1}{4} b c^4 \text {arctanh}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x}\) |
-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)
3.6.31.3.1 Defintions of rubi rules used
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]
\[\int \frac {\left (a +b \,\operatorname {arctanh}\left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{5}}d x\]
\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]
\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \]
\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]
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
\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^5} \,d x \]