3.1.26 \(\int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx\) [26]

3.1.26.1 Optimal result
3.1.26.2 Mathematica [A] (verified)
3.1.26.3 Rubi [A] (verified)
3.1.26.4 Maple [A] (verified)
3.1.26.5 Fricas [F]
3.1.26.6 Sympy [F]
3.1.26.7 Maxima [F]
3.1.26.8 Giac [F]
3.1.26.9 Mupad [F(-1)]

3.1.26.1 Optimal result

Integrand size = 20, antiderivative size = 160 \[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=-\frac {b c d^3}{2 x}+a c^3 d^3 x+\frac {1}{2} b c^2 d^3 \text {arctanh}(c x)+b c^3 d^3 x \text {arctanh}(c x)-\frac {d^3 (a+b \text {arctanh}(c x))}{2 x^2}-\frac {3 c d^3 (a+b \text {arctanh}(c x))}{x}+3 a c^2 d^3 \log (x)+3 b c^2 d^3 \log (x)-b c^2 d^3 \log \left (1-c^2 x^2\right )-\frac {3}{2} b c^2 d^3 \operatorname {PolyLog}(2,-c x)+\frac {3}{2} b c^2 d^3 \operatorname {PolyLog}(2,c x) \]

output
-1/2*b*c*d^3/x+a*c^3*d^3*x+1/2*b*c^2*d^3*arctanh(c*x)+b*c^3*d^3*x*arctanh( 
c*x)-1/2*d^3*(a+b*arctanh(c*x))/x^2-3*c*d^3*(a+b*arctanh(c*x))/x+3*a*c^2*d 
^3*ln(x)+3*b*c^2*d^3*ln(x)-b*c^2*d^3*ln(-c^2*x^2+1)-3/2*b*c^2*d^3*polylog( 
2,-c*x)+3/2*b*c^2*d^3*polylog(2,c*x)
 
3.1.26.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.03 \[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=\frac {d^3 \left (-2 a-12 a c x-2 b c x+4 a c^3 x^3-2 b \text {arctanh}(c x)-12 b c x \text {arctanh}(c x)+4 b c^3 x^3 \text {arctanh}(c x)+12 a c^2 x^2 \log (x)+12 b c^2 x^2 \log (c x)-b c^2 x^2 \log (1-c x)+b c^2 x^2 \log (1+c x)-4 b c^2 x^2 \log \left (1-c^2 x^2\right )-6 b c^2 x^2 \operatorname {PolyLog}(2,-c x)+6 b c^2 x^2 \operatorname {PolyLog}(2,c x)\right )}{4 x^2} \]

input
Integrate[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^3,x]
 
output
(d^3*(-2*a - 12*a*c*x - 2*b*c*x + 4*a*c^3*x^3 - 2*b*ArcTanh[c*x] - 12*b*c* 
x*ArcTanh[c*x] + 4*b*c^3*x^3*ArcTanh[c*x] + 12*a*c^2*x^2*Log[x] + 12*b*c^2 
*x^2*Log[c*x] - b*c^2*x^2*Log[1 - c*x] + b*c^2*x^2*Log[1 + c*x] - 4*b*c^2* 
x^2*Log[1 - c^2*x^2] - 6*b*c^2*x^2*PolyLog[2, -(c*x)] + 6*b*c^2*x^2*PolyLo 
g[2, c*x]))/(4*x^2)
 
3.1.26.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6502, 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.

\(\displaystyle \int \frac {(c d x+d)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx\)

\(\Big \downarrow \) 6502

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

\(\Big \downarrow \) 2009

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

input
Int[((d + c*d*x)^3*(a + b*ArcTanh[c*x]))/x^3,x]
 
output
-1/2*(b*c*d^3)/x + a*c^3*d^3*x + (b*c^2*d^3*ArcTanh[c*x])/2 + b*c^3*d^3*x* 
ArcTanh[c*x] - (d^3*(a + b*ArcTanh[c*x]))/(2*x^2) - (3*c*d^3*(a + b*ArcTan 
h[c*x]))/x + 3*a*c^2*d^3*Log[x] + 3*b*c^2*d^3*Log[x] - b*c^2*d^3*Log[1 - c 
^2*x^2] - (3*b*c^2*d^3*PolyLog[2, -(c*x)])/2 + (3*b*c^2*d^3*PolyLog[2, c*x 
])/2
 

3.1.26.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6502
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e 
_.)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^p, ( 
f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] 
 && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
 
3.1.26.4 Maple [A] (verified)

Time = 1.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86

method result size
parts \(d^{3} a \left (c^{3} x -\frac {3 c}{x}-\frac {1}{2 x^{2}}+3 c^{2} \ln \left (x \right )\right )+d^{3} b \,c^{2} \left (c x \,\operatorname {arctanh}\left (c x \right )+3 \ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {3 \operatorname {dilog}\left (c x +1\right )}{2}-\frac {3 \ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (c x \right )}{2}-\frac {3 \ln \left (c x +1\right )}{4}-\frac {5 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+3 \ln \left (c x \right )\right )\) \(137\)
derivativedivides \(c^{2} \left (d^{3} a \left (c x +3 \ln \left (c x \right )-\frac {3}{c x}-\frac {1}{2 c^{2} x^{2}}\right )+d^{3} b \left (c x \,\operatorname {arctanh}\left (c x \right )+3 \ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {3 \operatorname {dilog}\left (c x +1\right )}{2}-\frac {3 \ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (c x \right )}{2}-\frac {3 \ln \left (c x +1\right )}{4}-\frac {5 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+3 \ln \left (c x \right )\right )\right )\) \(140\)
default \(c^{2} \left (d^{3} a \left (c x +3 \ln \left (c x \right )-\frac {3}{c x}-\frac {1}{2 c^{2} x^{2}}\right )+d^{3} b \left (c x \,\operatorname {arctanh}\left (c x \right )+3 \ln \left (c x \right ) \operatorname {arctanh}\left (c x \right )-\frac {3 \,\operatorname {arctanh}\left (c x \right )}{c x}-\frac {\operatorname {arctanh}\left (c x \right )}{2 c^{2} x^{2}}-\frac {3 \operatorname {dilog}\left (c x +1\right )}{2}-\frac {3 \ln \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (c x \right )}{2}-\frac {3 \ln \left (c x +1\right )}{4}-\frac {5 \ln \left (c x -1\right )}{4}-\frac {1}{2 c x}+3 \ln \left (c x \right )\right )\right )\) \(140\)
risch \(-\frac {c^{3} d^{3} b x \ln \left (-c x +1\right )}{2}-\frac {5 c^{2} d^{3} b \ln \left (-c x +1\right )}{4}-b \,c^{2} d^{3}-\frac {b c \,d^{3}}{2 x}+\frac {7 c^{2} d^{3} b \ln \left (-c x \right )}{4}+\frac {d^{3} b \ln \left (-c x +1\right )}{4 x^{2}}+\frac {3 c \,d^{3} b \ln \left (-c x +1\right )}{2 x}+\frac {3 c^{2} d^{3} \operatorname {dilog}\left (-c x +1\right ) b}{2}+a \,c^{3} d^{3} x -a \,c^{2} d^{3}-\frac {d^{3} a}{2 x^{2}}-\frac {3 c \,d^{3} a}{x}+3 c^{2} d^{3} \ln \left (-c x \right ) a +\frac {b \,c^{3} d^{3} \ln \left (c x +1\right ) x}{2}-\frac {3 b \,c^{2} d^{3} \ln \left (c x +1\right )}{4}+\frac {5 b \,c^{2} d^{3} \ln \left (c x \right )}{4}-\frac {b \,d^{3} \ln \left (c x +1\right )}{4 x^{2}}-\frac {3 b c \,d^{3} \ln \left (c x +1\right )}{2 x}-\frac {3 b \,c^{2} d^{3} \operatorname {dilog}\left (c x +1\right )}{2}\) \(258\)

input
int((c*d*x+d)^3*(a+b*arctanh(c*x))/x^3,x,method=_RETURNVERBOSE)
 
output
d^3*a*(c^3*x-3*c/x-1/2/x^2+3*c^2*ln(x))+d^3*b*c^2*(c*x*arctanh(c*x)+3*ln(c 
*x)*arctanh(c*x)-3/c/x*arctanh(c*x)-1/2/c^2/x^2*arctanh(c*x)-3/2*dilog(c*x 
+1)-3/2*ln(c*x)*ln(c*x+1)-3/2*dilog(c*x)-3/4*ln(c*x+1)-5/4*ln(c*x-1)-1/2/c 
/x+3*ln(c*x))
 
3.1.26.5 Fricas [F]

\[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=\int { \frac {{\left (c d x + d\right )}^{3} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^3,x, algorithm="fricas")
 
output
integral((a*c^3*d^3*x^3 + 3*a*c^2*d^3*x^2 + 3*a*c*d^3*x + a*d^3 + (b*c^3*d 
^3*x^3 + 3*b*c^2*d^3*x^2 + 3*b*c*d^3*x + b*d^3)*arctanh(c*x))/x^3, x)
 
3.1.26.6 Sympy [F]

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

input
integrate((c*d*x+d)**3*(a+b*atanh(c*x))/x**3,x)
 
output
d**3*(Integral(a*c**3, x) + Integral(a/x**3, x) + Integral(3*a*c/x**2, x) 
+ Integral(3*a*c**2/x, x) + Integral(b*c**3*atanh(c*x), x) + Integral(b*at 
anh(c*x)/x**3, x) + Integral(3*b*c*atanh(c*x)/x**2, x) + Integral(3*b*c**2 
*atanh(c*x)/x, x))
 
3.1.26.7 Maxima [F]

\[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=\int { \frac {{\left (c d x + d\right )}^{3} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^3,x, algorithm="maxima")
 
output
a*c^3*d^3*x + 1/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*b*c^2*d^3 + 3/2 
*b*c^2*d^3*integrate((log(c*x + 1) - log(-c*x + 1))/x, x) + 3*a*c^2*d^3*lo 
g(x) - 3/2*(c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*b*c*d^3 + 
1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d^3 
 - 3*a*c*d^3/x - 1/2*a*d^3/x^2
 
3.1.26.8 Giac [F]

\[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=\int { \frac {{\left (c d x + d\right )}^{3} {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input
integrate((c*d*x+d)^3*(a+b*arctanh(c*x))/x^3,x, algorithm="giac")
 
output
integrate((c*d*x + d)^3*(b*arctanh(c*x) + a)/x^3, x)
 
3.1.26.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d+c d x)^3 (a+b \text {arctanh}(c x))}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^3}{x^3} \,d x \]

input
int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x^3,x)
 
output
int(((a + b*atanh(c*x))*(d + c*d*x)^3)/x^3, x)