\(\int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx\) [9]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {b c d}{12 x^3}-\frac {b c^2 d}{6 x^2}-\frac {b c^3 d}{4 x}-\frac {d (a+b \text {arctanh}(c x))}{4 x^4}-\frac {c d (a+b \text {arctanh}(c x))}{3 x^3}+\frac {1}{3} b c^4 d \log (x)-\frac {7}{24} b c^4 d \log (1-c x)-\frac {1}{24} b c^4 d \log (1+c x) \] Output:

-1/12*b*c*d/x^3-1/6*b*c^2*d/x^2-1/4*b*c^3*d/x-1/4*d*(a+b*arctanh(c*x))/x^4 
-1/3*c*d*(a+b*arctanh(c*x))/x^3+1/3*b*c^4*d*ln(x)-7/24*b*c^4*d*ln(-c*x+1)- 
1/24*b*c^4*d*ln(c*x+1)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {d \left (6 a+8 a c x+2 b c x+4 b c^2 x^2+6 b c^3 x^3+2 b (3+4 c x) \text {arctanh}(c x)-8 b c^4 x^4 \log (x)+7 b c^4 x^4 \log (1-c x)+b c^4 x^4 \log (1+c x)\right )}{24 x^4} \] Input:

Integrate[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^5,x]
 

Output:

-1/24*(d*(6*a + 8*a*c*x + 2*b*c*x + 4*b*c^2*x^2 + 6*b*c^3*x^3 + 2*b*(3 + 4 
*c*x)*ArcTanh[c*x] - 8*b*c^4*x^4*Log[x] + 7*b*c^4*x^4*Log[1 - c*x] + b*c^4 
*x^4*Log[1 + c*x]))/x^4
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6498, 27, 523, 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.

\(\displaystyle \int \frac {(c d x+d) (a+b \text {arctanh}(c x))}{x^5} \, dx\)

\(\Big \downarrow \) 6498

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 523

\(\displaystyle \frac {1}{12} b c d \int \left (-\frac {7 c^4}{2 (c x-1)}-\frac {c^4}{2 (c x+1)}+\frac {4 c^3}{x}+\frac {3 c^2}{x^2}+\frac {4 c}{x^3}+\frac {3}{x^4}\right )dx-\frac {d (a+b \text {arctanh}(c x))}{4 x^4}-\frac {c d (a+b \text {arctanh}(c x))}{3 x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d (a+b \text {arctanh}(c x))}{4 x^4}-\frac {c d (a+b \text {arctanh}(c x))}{3 x^3}+\frac {1}{12} b c d \left (4 c^3 \log (x)-\frac {7}{2} c^3 \log (1-c x)-\frac {1}{2} c^3 \log (c x+1)-\frac {3 c^2}{x}-\frac {2 c}{x^2}-\frac {1}{x^3}\right )\)

Input:

Int[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^5,x]
 

Output:

-1/4*(d*(a + b*ArcTanh[c*x]))/x^4 - (c*d*(a + b*ArcTanh[c*x]))/(3*x^3) + ( 
b*c*d*(-x^(-3) - (2*c)/x^2 - (3*c^2)/x + 4*c^3*Log[x] - (7*c^3*Log[1 - c*x 
])/2 - (c^3*Log[1 + c*x])/2))/12
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 523

Int[((x_)^(m_.)*((c_) + (d_.)*(x_)))/((a_) + (b_.)*(x_)^2), x_Symbol] :> In 
t[ExpandIntegrand[x^m*((c + d*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d} 
, x] && IntegerQ[m]
 

rule 2009

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

rule 6498

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*( 
x_))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x)^q, x]}, Simp[( 
a + b*ArcTanh[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/(1 - c^2*x 
^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && Intege 
rQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0 
]))
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85

method	result	size
parts	\(a d \left (-\frac {1}{4 x^{4}}-\frac {c}{3 x^{3}}\right )+d b \,c^{4} \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {7 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{4 c x}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x +1\right )}{24}\right )\)	\(94\)
derivativedivides	\(c^{4} \left (a d \left (-\frac {1}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+d b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {7 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{4 c x}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x +1\right )}{24}\right )\right )\)	\(100\)
default	\(c^{4} \left (a d \left (-\frac {1}{3 c^{3} x^{3}}-\frac {1}{4 c^{4} x^{4}}\right )+d b \left (-\frac {\operatorname {arctanh}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\operatorname {arctanh}\left (c x \right )}{4 c^{4} x^{4}}-\frac {7 \ln \left (c x -1\right )}{24}-\frac {1}{12 c^{3} x^{3}}-\frac {1}{6 c^{2} x^{2}}-\frac {1}{4 c x}+\frac {\ln \left (c x \right )}{3}-\frac {\ln \left (c x +1\right )}{24}\right )\right )\)	\(100\)
parallelrisch	\(\frac {4 b \,c^{4} d \ln \left (x \right ) x^{4}-4 \ln \left (c x -1\right ) x^{4} b \,c^{4} d -d b \,\operatorname {arctanh}\left (c x \right ) x^{4} c^{4}-2 d \,x^{4} c^{4} b -3 b \,c^{3} d \,x^{3}-2 b \,c^{2} d \,x^{2}-4 b c d x \,\operatorname {arctanh}\left (c x \right )-4 a d x c -b c d x -3 \,\operatorname {arctanh}\left (c x \right ) b d -3 a d}{12 x^{4}}\)	\(113\)
risch	\(-\frac {d b \left (4 c x +3\right ) \ln \left (c x +1\right )}{24 x^{4}}+\frac {d \left (8 b \,c^{4} \ln \left (-x \right ) x^{4}-7 b \,x^{4} \ln \left (-c x +1\right ) c^{4}-b \,c^{4} \ln \left (c x +1\right ) x^{4}-6 b \,c^{3} x^{3}-4 b \,c^{2} x^{2}+4 b c x \ln \left (-c x +1\right )-8 a c x -2 b c x +3 b \ln \left (-c x +1\right )-6 a \right )}{24 x^{4}}\)	\(125\)

Input:

int((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x,method=_RETURNVERBOSE)
 

Output:

a*d*(-1/4/x^4-1/3*c/x^3)+d*b*c^4*(-1/3*arctanh(c*x)/c^3/x^3-1/4*arctanh(c* 
x)/c^4/x^4-7/24*ln(c*x-1)-1/12/c^3/x^3-1/6/c^2/x^2-1/4/c/x+1/3*ln(c*x)-1/2 
4*ln(c*x+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {b c^{4} d x^{4} \log \left (c x + 1\right ) + 7 \, b c^{4} d x^{4} \log \left (c x - 1\right ) - 8 \, b c^{4} d x^{4} \log \left (x\right ) + 6 \, b c^{3} d x^{3} + 4 \, b c^{2} d x^{2} + 2 \, {\left (4 \, a + b\right )} c d x + 6 \, a d + {\left (4 \, b c d x + 3 \, b d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, x^{4}} \] Input:

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x, algorithm="fricas")
 

Output:

-1/24*(b*c^4*d*x^4*log(c*x + 1) + 7*b*c^4*d*x^4*log(c*x - 1) - 8*b*c^4*d*x 
^4*log(x) + 6*b*c^3*d*x^3 + 4*b*c^2*d*x^2 + 2*(4*a + b)*c*d*x + 6*a*d + (4 
*b*c*d*x + 3*b*d)*log(-(c*x + 1)/(c*x - 1)))/x^4
 

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.17 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=\begin {cases} - \frac {a c d}{3 x^{3}} - \frac {a d}{4 x^{4}} + \frac {b c^{4} d \log {\left (x \right )}}{3} - \frac {b c^{4} d \log {\left (x - \frac {1}{c} \right )}}{3} - \frac {b c^{4} d \operatorname {atanh}{\left (c x \right )}}{12} - \frac {b c^{3} d}{4 x} - \frac {b c^{2} d}{6 x^{2}} - \frac {b c d \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} - \frac {b c d}{12 x^{3}} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} & \text {for}\: c \neq 0 \\- \frac {a d}{4 x^{4}} & \text {otherwise} \end {cases} \] Input:

integrate((c*d*x+d)*(a+b*atanh(c*x))/x**5,x)
 

Output:

Piecewise((-a*c*d/(3*x**3) - a*d/(4*x**4) + b*c**4*d*log(x)/3 - b*c**4*d*l 
og(x - 1/c)/3 - b*c**4*d*atanh(c*x)/12 - b*c**3*d/(4*x) - b*c**2*d/(6*x**2 
) - b*c*d*atanh(c*x)/(3*x**3) - b*c*d/(12*x**3) - b*d*atanh(c*x)/(4*x**4), 
 Ne(c, 0)), (-a*d/(4*x**4), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b c d + \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b d - \frac {a c d}{3 \, x^{3}} - \frac {a d}{4 \, x^{4}} \] Input:

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x, algorithm="maxima")
 

Output:

-1/6*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3 
)*b*c*d + 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 - 1/3*a*c*d/x^3 - 1/4*a*d/x^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (94) = 188\).

Time = 0.12 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.65 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {1}{3} \, {\left (b c^{3} d \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{3} d \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {6 \, {\left (c x + 1\right )}^{3} b c^{3} d}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2} b c^{3} d}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )} b c^{3} d}{c x - 1} + b c^{3} d\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {12 \, {\left (c x + 1\right )}^{3} a c^{3} d}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} a c^{3} d}{{\left (c x - 1\right )}^{2}} + \frac {8 \, {\left (c x + 1\right )} a c^{3} d}{c x - 1} + 2 \, a c^{3} d + \frac {5 \, {\left (c x + 1\right )}^{3} b c^{3} d}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2} b c^{3} d}{{\left (c x - 1\right )}^{2}} + \frac {7 \, {\left (c x + 1\right )} b c^{3} d}{c x - 1} + 2 \, b c^{3} d}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \] Input:

integrate((c*d*x+d)*(a+b*arctanh(c*x))/x^5,x, algorithm="giac")
 

Output:

1/3*(b*c^3*d*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^3*d*log(-(c*x + 1)/(c*x - 
 1)) + (6*(c*x + 1)^3*b*c^3*d/(c*x - 1)^3 + 3*(c*x + 1)^2*b*c^3*d/(c*x - 1 
)^2 + 4*(c*x + 1)*b*c^3*d/(c*x - 1) + b*c^3*d)*log(-(c*x + 1)/(c*x - 1))/( 
(c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 
 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1) + (12*(c*x + 1)^3*a*c^3*d/(c*x - 1)^3 + 
 6*(c*x + 1)^2*a*c^3*d/(c*x - 1)^2 + 8*(c*x + 1)*a*c^3*d/(c*x - 1) + 2*a*c 
^3*d + 5*(c*x + 1)^3*b*c^3*d/(c*x - 1)^3 + 10*(c*x + 1)^2*b*c^3*d/(c*x - 1 
)^2 + 7*(c*x + 1)*b*c^3*d/(c*x - 1) + 2*b*c^3*d)/((c*x + 1)^4/(c*x - 1)^4 
+ 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x 
 - 1) + 1))*c
 

Mupad [B] (verification not implemented)

Time = 3.46 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {b\,c^4\,d\,\ln \left (x\right )}{3}-\frac {a\,c\,d}{3\,x^3}-\frac {b\,c\,d}{12\,x^3}-\frac {b\,d\,\mathrm {atanh}\left (c\,x\right )}{4\,x^4}-\frac {b\,c^4\,d\,\ln \left (c^2\,x^2-1\right )}{6}-\frac {b\,c^2\,d}{6\,x^2}-\frac {b\,c^3\,d}{4\,x}-\frac {a\,d}{4\,x^4}-\frac {b\,c^5\,d\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )}{4\,\sqrt {-c^2}}-\frac {b\,c\,d\,\mathrm {atanh}\left (c\,x\right )}{3\,x^3} \] Input:

int(((a + b*atanh(c*x))*(d + c*d*x))/x^5,x)
 

Output:

(b*c^4*d*log(x))/3 - (a*c*d)/(3*x^3) - (b*c*d)/(12*x^3) - (b*d*atanh(c*x)) 
/(4*x^4) - (b*c^4*d*log(c^2*x^2 - 1))/6 - (b*c^2*d)/(6*x^2) - (b*c^3*d)/(4 
*x) - (a*d)/(4*x^4) - (b*c^5*d*atan((c^2*x)/(-c^2)^(1/2)))/(4*(-c^2)^(1/2) 
) - (b*c*d*atanh(c*x))/(3*x^3)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int \frac {(d+c d x) (a+b \text {arctanh}(c x))}{x^5} \, dx=\frac {d \left (-\mathit {atanh} \left (c x \right ) b \,c^{4} x^{4}-4 \mathit {atanh} \left (c x \right ) b c x -3 \mathit {atanh} \left (c x \right ) b -4 \,\mathrm {log}\left (c^{2} x -c \right ) b \,c^{4} x^{4}+4 \,\mathrm {log}\left (x \right ) b \,c^{4} x^{4}-4 a c x -3 a -3 b \,c^{3} x^{3}-2 b \,c^{2} x^{2}-b c x \right )}{12 x^{4}} \] Input:

int((c*d*x+d)*(a+b*atanh(c*x))/x^5,x)
 

Output:

(d*( - atanh(c*x)*b*c**4*x**4 - 4*atanh(c*x)*b*c*x - 3*atanh(c*x)*b - 4*lo 
g(c**2*x - c)*b*c**4*x**4 + 4*log(x)*b*c**4*x**4 - 4*a*c*x - 3*a - 3*b*c** 
3*x**3 - 2*b*c**2*x**2 - b*c*x))/(12*x**4)