\(\int \frac {(a+b \text {arctanh}(c x)) (d+e \log (f+g x^2))}{x^3} \, dx\) [536]

Optimal result

Integrand size = 24, antiderivative size = 511 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx=\frac {b c e \sqrt {g} \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{\sqrt {f}}+\frac {a e g \log (x)}{f}+\frac {b e \left (c^2 f+g\right ) \text {arctanh}(c x) \log \left (\frac {2}{1+c x}\right )}{f}-\frac {b e \left (c^2 f+g\right ) \text {arctanh}(c x) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-\sqrt {g}\right ) (1+c x)}\right )}{2 f}-\frac {b e \left (c^2 f+g\right ) \text {arctanh}(c x) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+\sqrt {g}\right ) (1+c x)}\right )}{2 f}-\frac {a e g \log \left (f+g x^2\right )}{2 f}-\frac {b c \left (d+e \log \left (f+g x^2\right )\right )}{2 x}+\frac {1}{2} b c^2 \text {arctanh}(c x) \left (d+e \log \left (f+g x^2\right )\right )-\frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (f+g x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 e \operatorname {PolyLog}(2,-c x)-\frac {b e \left (c^2 f+g\right ) \operatorname {PolyLog}(2,-c x)}{2 f}-\frac {1}{2} b c^2 e \operatorname {PolyLog}(2,c x)+\frac {b e \left (c^2 f+g\right ) \operatorname {PolyLog}(2,c x)}{2 f}-\frac {b e \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 f}+\frac {b e \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-\sqrt {g}\right ) (1+c x)}\right )}{4 f}+\frac {b e \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{\left (c \sqrt {-f}+\sqrt {g}\right ) (1+c x)}\right )}{4 f} \] Output:

b*c*e*g^(1/2)*arctan(g^(1/2)*x/f^(1/2))/f^(1/2)+a*e*g*ln(x)/f+b*e*(c^2*f+g 
)*arctanh(c*x)*ln(2/(c*x+1))/f-1/2*b*e*(c^2*f+g)*arctanh(c*x)*ln(2*c*((-f) 
^(1/2)-g^(1/2)*x)/(c*(-f)^(1/2)-g^(1/2))/(c*x+1))/f-1/2*b*e*(c^2*f+g)*arct 
anh(c*x)*ln(2*c*((-f)^(1/2)+g^(1/2)*x)/(c*(-f)^(1/2)+g^(1/2))/(c*x+1))/f-1 
/2*a*e*g*ln(g*x^2+f)/f-1/2*b*c*(d+e*ln(g*x^2+f))/x+1/2*b*c^2*arctanh(c*x)* 
(d+e*ln(g*x^2+f))-1/2*(a+b*arctanh(c*x))*(d+e*ln(g*x^2+f))/x^2+1/2*b*c^2*e 
*polylog(2,-c*x)-1/2*b*e*(c^2*f+g)*polylog(2,-c*x)/f-1/2*b*c^2*e*polylog(2 
,c*x)+1/2*b*e*(c^2*f+g)*polylog(2,c*x)/f-1/2*b*e*(c^2*f+g)*polylog(2,1-2/( 
c*x+1))/f+1/4*b*e*(c^2*f+g)*polylog(2,1-2*c*((-f)^(1/2)-g^(1/2)*x)/(c*(-f) 
^(1/2)-g^(1/2))/(c*x+1))/f+1/4*b*e*(c^2*f+g)*polylog(2,1-2*c*((-f)^(1/2)+g 
^(1/2)*x)/(c*(-f)^(1/2)+g^(1/2))/(c*x+1))/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.06 (sec) , antiderivative size = 1211, normalized size of antiderivative = 2.37 \[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*a*d*f - 2*b*c*d*f*x + 4*b*c*e*Sqrt[f]*Sqrt[g]*x^2*ArcTan[(Sqrt[g]*x)/S 
qrt[f]] - 2*b*d*f*ArcTanh[c*x] + 2*b*c^2*d*f*x^2*ArcTanh[c*x] + (4*I)*b*c^ 
2*e*f*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f + g)]]*ArcTanh[(c*g*x)/Sqrt[-(c^2*f*g 
)]] + (4*I)*b*e*g*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f + g)]]*ArcTanh[(c*g*x)/Sq 
rt[-(c^2*f*g)]] + 4*b*e*g*x^2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + 
4*b*c^2*e*f*x^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + (2*I)*b*c^2*e* 
f*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f + g)]]*Log[(c^2*(1 + E^(2*ArcTanh[c*x]))* 
f + (-1 + E^(2*ArcTanh[c*x]))*g - 2*Sqrt[-(c^2*f*g)])/(E^(2*ArcTanh[c*x])* 
(c^2*f + g))] + (2*I)*b*e*g*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f + g)]]*Log[(c^2 
*(1 + E^(2*ArcTanh[c*x]))*f + (-1 + E^(2*ArcTanh[c*x]))*g - 2*Sqrt[-(c^2*f 
*g)])/(E^(2*ArcTanh[c*x])*(c^2*f + g))] - 2*b*c^2*e*f*x^2*ArcTanh[c*x]*Log 
[(c^2*(1 + E^(2*ArcTanh[c*x]))*f + (-1 + E^(2*ArcTanh[c*x]))*g - 2*Sqrt[-( 
c^2*f*g)])/(E^(2*ArcTanh[c*x])*(c^2*f + g))] - 2*b*e*g*x^2*ArcTanh[c*x]*Lo 
g[(c^2*(1 + E^(2*ArcTanh[c*x]))*f + (-1 + E^(2*ArcTanh[c*x]))*g - 2*Sqrt[- 
(c^2*f*g)])/(E^(2*ArcTanh[c*x])*(c^2*f + g))] - (2*I)*b*c^2*e*f*x^2*ArcSin 
[Sqrt[(c^2*f)/(c^2*f + g)]]*Log[(c^2*(1 + E^(2*ArcTanh[c*x]))*f + (-1 + E^ 
(2*ArcTanh[c*x]))*g + 2*Sqrt[-(c^2*f*g)])/(E^(2*ArcTanh[c*x])*(c^2*f + g)) 
] - (2*I)*b*e*g*x^2*ArcSin[Sqrt[(c^2*f)/(c^2*f + g)]]*Log[(c^2*(1 + E^(2*A 
rcTanh[c*x]))*f + (-1 + E^(2*ArcTanh[c*x]))*g + 2*Sqrt[-(c^2*f*g)])/(E^(2* 
ArcTanh[c*x])*(c^2*f + g))] - 2*b*c^2*e*f*x^2*ArcTanh[c*x]*Log[(c^2*(1 ...
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 487, normalized size of antiderivative = 0.95, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 6647

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

\(\Big \downarrow \) 2009

\(\displaystyle -2 e g \left (\frac {a \log \left (f+g x^2\right )}{4 f}-\frac {a \log (x)}{2 f}-\frac {b c \arctan \left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g}}-\frac {b \text {arctanh}(c x) \left (c^2 f+g\right ) \log \left (\frac {2}{c x+1}\right )}{2 f g}+\frac {b \text {arctanh}(c x) \left (c^2 f+g\right ) \log \left (\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{(c x+1) \left (c \sqrt {-f}-\sqrt {g}\right )}\right )}{4 f g}+\frac {b \text {arctanh}(c x) \left (c^2 f+g\right ) \log \left (\frac {2 c \left (\sqrt {-f}+\sqrt {g} x\right )}{(c x+1) \left (c \sqrt {-f}+\sqrt {g}\right )}\right )}{4 f g}+\frac {b \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{4 f g}-\frac {b \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-f}-\sqrt {g} x\right )}{\left (c \sqrt {-f}-\sqrt {g}\right ) (c x+1)}\right )}{8 f g}-\frac {b \left (c^2 f+g\right ) \operatorname {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {g} x+\sqrt {-f}\right )}{\left (\sqrt {-f} c+\sqrt {g}\right ) (c x+1)}\right )}{8 f g}+\frac {b \operatorname {PolyLog}(2,-c x)}{4 f}-\frac {b \operatorname {PolyLog}(2,c x)}{4 f}\right )-\frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (f+g x^2\right )\right )}{2 x^2}+\frac {1}{2} b c^2 \text {arctanh}(c x) \left (d+e \log \left (f+g x^2\right )\right )-\frac {b c \left (d+e \log \left (f+g x^2\right )\right )}{2 x}\)

Input:

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

Output:

-1/2*(b*c*(d + e*Log[f + g*x^2]))/x + (b*c^2*ArcTanh[c*x]*(d + e*Log[f + g 
*x^2]))/2 - ((a + b*ArcTanh[c*x])*(d + e*Log[f + g*x^2]))/(2*x^2) - 2*e*g* 
(-1/2*(b*c*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(Sqrt[f]*Sqrt[g]) - (a*Log[x])/(2* 
f) - (b*(c^2*f + g)*ArcTanh[c*x]*Log[2/(1 + c*x)])/(2*f*g) + (b*(c^2*f + g 
)*ArcTanh[c*x]*Log[(2*c*(Sqrt[-f] - Sqrt[g]*x))/((c*Sqrt[-f] - Sqrt[g])*(1 
 + c*x))])/(4*f*g) + (b*(c^2*f + g)*ArcTanh[c*x]*Log[(2*c*(Sqrt[-f] + Sqrt 
[g]*x))/((c*Sqrt[-f] + Sqrt[g])*(1 + c*x))])/(4*f*g) + (a*Log[f + g*x^2])/ 
(4*f) + (b*PolyLog[2, -(c*x)])/(4*f) - (b*PolyLog[2, c*x])/(4*f) + (b*(c^2 
*f + g)*PolyLog[2, 1 - 2/(1 + c*x)])/(4*f*g) - (b*(c^2*f + g)*PolyLog[2, 1 
 - (2*c*(Sqrt[-f] - Sqrt[g]*x))/((c*Sqrt[-f] - Sqrt[g])*(1 + c*x))])/(8*f* 
g) - (b*(c^2*f + g)*PolyLog[2, 1 - (2*c*(Sqrt[-f] + Sqrt[g]*x))/((c*Sqrt[- 
f] + Sqrt[g])*(1 + c*x))])/(8*f*g))
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(960\) vs. \(2(445)=890\).

Time = 12.71 (sec) , antiderivative size = 961, normalized size of antiderivative = 1.88

method	result	size
risch	\(-\frac {b c d}{2 x}-\frac {a e g \ln \left (g \,x^{2}+f \right )}{2 f}+\frac {a e g \ln \left (x \right )}{f}-\frac {b \,c^{2} d \ln \left (-c x +1\right )}{4}-\frac {a d}{2 x^{2}}-\frac {\ln \left (c x +1\right ) b d}{4 x^{2}}+\frac {b d \ln \left (-c x +1\right )}{4 x^{2}}-\frac {g b e \ln \left (c x +1\right ) \ln \left (\frac {c \sqrt {-f g}-\left (c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4 f}+\frac {g e b c \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{\sqrt {f g}}+\frac {g b e \ln \left (-c x +1\right ) \ln \left (\frac {c \sqrt {-f g}-\left (-c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4 f}+\frac {g b e \ln \left (-c x +1\right ) \ln \left (\frac {c \sqrt {-f g}+\left (-c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4 f}-\frac {g b e \operatorname {dilog}\left (c x +1\right )}{2 f}-\frac {b e \ln \left (c x +1\right ) \ln \left (\frac {c \sqrt {-f g}+\left (c x +1\right ) g -g}{c \sqrt {-f g}-g}\right ) c^{2}}{4}-\frac {b e \ln \left (c x +1\right ) \ln \left (\frac {c \sqrt {-f g}-\left (c x +1\right ) g +g}{c \sqrt {-f g}+g}\right ) c^{2}}{4}-\frac {g b e \operatorname {dilog}\left (\frac {c \sqrt {-f g}-\left (c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4 f}-\frac {g b e \operatorname {dilog}\left (\frac {c \sqrt {-f g}+\left (c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4 f}+\frac {g b e \operatorname {dilog}\left (-c x +1\right )}{2 f}+\frac {b e \ln \left (-c x +1\right ) \ln \left (\frac {c \sqrt {-f g}-\left (-c x +1\right ) g +g}{c \sqrt {-f g}+g}\right ) c^{2}}{4}+\frac {b e \ln \left (-c x +1\right ) \ln \left (\frac {c \sqrt {-f g}+\left (-c x +1\right ) g -g}{c \sqrt {-f g}-g}\right ) c^{2}}{4}+\frac {g b e \operatorname {dilog}\left (\frac {c \sqrt {-f g}-\left (-c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4 f}+\frac {g b e \operatorname {dilog}\left (\frac {c \sqrt {-f g}+\left (-c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4 f}+\left (-\frac {b e \ln \left (c x +1\right )}{4 x^{2}}-\frac {e \left (b \,c^{2} x^{2} \ln \left (-c x +1\right )-b \,c^{2} \ln \left (c x +1\right ) x^{2}+2 b c x -b \ln \left (-c x +1\right )+2 a \right )}{4 x^{2}}\right ) \ln \left (g \,x^{2}+f \right )-\frac {b e \,c^{2} \operatorname {dilog}\left (\frac {c \sqrt {-f g}-\left (c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4}-\frac {b e \,c^{2} \operatorname {dilog}\left (\frac {c \sqrt {-f g}+\left (c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4}+\frac {b e \,c^{2} \operatorname {dilog}\left (\frac {c \sqrt {-f g}-\left (-c x +1\right ) g +g}{c \sqrt {-f g}+g}\right )}{4}+\frac {b e \,c^{2} \operatorname {dilog}\left (\frac {c \sqrt {-f g}+\left (-c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4}-\frac {d b \,c^{2} \ln \left (c x \right )}{4}+\frac {d b \,c^{2} \ln \left (c x +1\right )}{4}+\frac {d b \,c^{2} \ln \left (-c x \right )}{4}-\frac {g b e \ln \left (c x +1\right ) \ln \left (\frac {c \sqrt {-f g}+\left (c x +1\right ) g -g}{c \sqrt {-f g}-g}\right )}{4 f}\)	\(961\)

Input:

int((a+b*arctanh(c*x))*(d+e*ln(g*x^2+f))/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*b*c*d/x-1/2*a*e*g*ln(g*x^2+f)/f+a*e*g*ln(x)/f-1/4*b*c^2*d*ln(-c*x+1)- 
1/2*a*d/x^2-1/4/x^2*ln(c*x+1)*b*d+1/4/x^2*b*d*ln(-c*x+1)-1/4*g*b*e/f*ln(c* 
x+1)*ln((c*(-f*g)^(1/2)-(c*x+1)*g+g)/(c*(-f*g)^(1/2)+g))+g*e*b*c/(f*g)^(1/ 
2)*arctan(x*g/(f*g)^(1/2))+1/4*g*b*e/f*ln(-c*x+1)*ln((c*(-f*g)^(1/2)-(-c*x 
+1)*g+g)/(c*(-f*g)^(1/2)+g))+1/4*g*b*e/f*ln(-c*x+1)*ln((c*(-f*g)^(1/2)+(-c 
*x+1)*g-g)/(c*(-f*g)^(1/2)-g))-1/2*g*b*e/f*dilog(c*x+1)-1/4*b*e*ln(c*x+1)* 
ln((c*(-f*g)^(1/2)+(c*x+1)*g-g)/(c*(-f*g)^(1/2)-g))*c^2-1/4*b*e*ln(c*x+1)* 
ln((c*(-f*g)^(1/2)-(c*x+1)*g+g)/(c*(-f*g)^(1/2)+g))*c^2-1/4*g*b*e/f*dilog( 
(c*(-f*g)^(1/2)-(c*x+1)*g+g)/(c*(-f*g)^(1/2)+g))-1/4*g*b*e/f*dilog((c*(-f* 
g)^(1/2)+(c*x+1)*g-g)/(c*(-f*g)^(1/2)-g))+1/2*g*b*e/f*dilog(-c*x+1)+1/4*b* 
e*ln(-c*x+1)*ln((c*(-f*g)^(1/2)-(-c*x+1)*g+g)/(c*(-f*g)^(1/2)+g))*c^2+1/4* 
b*e*ln(-c*x+1)*ln((c*(-f*g)^(1/2)+(-c*x+1)*g-g)/(c*(-f*g)^(1/2)-g))*c^2+1/ 
4*g*b*e/f*dilog((c*(-f*g)^(1/2)-(-c*x+1)*g+g)/(c*(-f*g)^(1/2)+g))+1/4*g*b* 
e/f*dilog((c*(-f*g)^(1/2)+(-c*x+1)*g-g)/(c*(-f*g)^(1/2)-g))+(-1/4*b*e/x^2* 
ln(c*x+1)-1/4*e*(b*c^2*x^2*ln(-c*x+1)-b*c^2*ln(c*x+1)*x^2+2*b*c*x-b*ln(-c* 
x+1)+2*a)/x^2)*ln(g*x^2+f)-1/4*b*e*c^2*dilog((c*(-f*g)^(1/2)-(c*x+1)*g+g)/ 
(c*(-f*g)^(1/2)+g))-1/4*b*e*c^2*dilog((c*(-f*g)^(1/2)+(c*x+1)*g-g)/(c*(-f* 
g)^(1/2)-g))+1/4*b*e*c^2*dilog((c*(-f*g)^(1/2)-(-c*x+1)*g+g)/(c*(-f*g)^(1/ 
2)+g))+1/4*b*e*c^2*dilog((c*(-f*g)^(1/2)+(-c*x+1)*g-g)/(c*(-f*g)^(1/2)-g)) 
-1/4*d*b*c^2*ln(c*x)+1/4*d*b*c^2*ln(c*x+1)+1/4*d*b*c^2*ln(-c*x)-1/4*g*b...
 

Fricas [F]

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

integrate((a+b*arctanh(c*x))*(d+e*log(g*x^2+f))/x^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((a+b*arctanh(c*x))*(d+e*log(g*x^2+f))/x^3,x, algorithm="maxima")
 

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*(g*(log(g*x^2 + f)/f - log(x^2)/f) + log(g*x^2 + f)/x^2)*a*e - 1/4*(2 
*c^2*g*integrate(x^2*log(c*x + 1)/(g*x^3 + f*x), x) - 2*c^2*g*integrate(x^ 
2*log(-c*x + 1)/(g*x^3 + f*x), x) + 2*I*c*g*(log(I*g*x/sqrt(f*g) + 1) - lo 
g(-I*g*x/sqrt(f*g) + 1))/sqrt(f*g) - 2*g*integrate(log(c*x + 1)/(g*x^3 + f 
*x), x) + 2*g*integrate(log(-c*x + 1)/(g*x^3 + f*x), x) + (2*c*x - (c^2*x^ 
2 - 1)*log(c*x + 1) + (c^2*x^2 - 1)*log(-c*x + 1))*log(g*x^2 + f)/x^2)*b*e 
 - 1/2*a*d/x^2
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \text {arctanh}(c x)) \left (d+e \log \left (f+g x^2\right )\right )}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*b*c*e*g*x**2 - atanh(c 
*x)*log(f + g*x**2)*b*c**2*e*f**2 + atanh(c*x)*log(f + g*x**2)*b*e*f*g + a 
tanh(c*x)*b*c**4*d*f**2*x**2 + atanh(c*x)*b*c**4*e*f**2*x**2 - atanh(c*x)* 
b*c**2*d*f**2 - atanh(c*x)*b*c**2*d*f*g*x**2 - atanh(c*x)*b*c**2*e*f**2 - 
atanh(c*x)*b*c**2*e*f*g*x**2 + atanh(c*x)*b*d*f*g + atanh(c*x)*b*e*f*g + 2 
*int(atanh(c*x)/(c**4*f**2*x**5 + c**4*f*g*x**7 - c**2*f**2*x**3 - 2*c**2* 
f*g*x**5 - c**2*g**2*x**7 + f*g*x**3 + g**2*x**5),x)*b*c**4*e*f**4*x**2 - 
4*int(atanh(c*x)/(c**4*f**2*x**5 + c**4*f*g*x**7 - c**2*f**2*x**3 - 2*c**2 
*f*g*x**5 - c**2*g**2*x**7 + f*g*x**3 + g**2*x**5),x)*b*c**2*e*f**3*g*x**2 
 + 2*int(atanh(c*x)/(c**4*f**2*x**5 + c**4*f*g*x**7 - c**2*f**2*x**3 - 2*c 
**2*f*g*x**5 - c**2*g**2*x**7 + f*g*x**3 + g**2*x**5),x)*b*e*f**2*g**2*x** 
2 - 2*int(atanh(c*x)/(c**4*f**2*x**3 + c**4*f*g*x**5 - c**2*f**2*x - 2*c** 
2*f*g*x**3 - c**2*g**2*x**5 + f*g*x + g**2*x**3),x)*b*c**6*e*f**4*x**2 + 4 
*int(atanh(c*x)/(c**4*f**2*x**3 + c**4*f*g*x**5 - c**2*f**2*x - 2*c**2*f*g 
*x**3 - c**2*g**2*x**5 + f*g*x + g**2*x**3),x)*b*c**4*e*f**3*g*x**2 - 2*in 
t(atanh(c*x)/(c**4*f**2*x**3 + c**4*f*g*x**5 - c**2*f**2*x - 2*c**2*f*g*x* 
*3 - c**2*g**2*x**5 + f*g*x + g**2*x**3),x)*b*c**2*e*f**2*g**2*x**2 - int( 
log(f + g*x**2)/(c**4*f**2*x**4 + c**4*f*g*x**6 - c**2*f**2*x**2 - 2*c**2* 
f*g*x**4 - c**2*g**2*x**6 + f*g*x**2 + g**2*x**4),x)*b*c**5*e*f**4*x**2 + 
int(log(f + g*x**2)/(c**4*f**2*x**4 + c**4*f*g*x**6 - c**2*f**2*x**2 - ...