\(\int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx\) [37]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 18, antiderivative size = 167 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\frac {b d}{2 (d e+f-c f) (d e-(1+c) f) (e+f x)}-\frac {a+b \text {arctanh}(c+d x)}{2 f (e+f x)^2}-\frac {b d^2 \log (1-c-d x)}{4 f (d e+f-c f)^2}+\frac {b d^2 \log (1+c+d x)}{4 f (d e-f-c f)^2}-\frac {b d^2 (d e-c f) \log (e+f x)}{(d e+f-c f)^2 (d e-(1+c) f)^2} \] Output:

1/2*b*d/(-c*f+d*e+f)/(d*e-(1+c)*f)/(f*x+e)-1/2*(a+b*arctanh(d*x+c))/f/(f*x 
+e)^2-1/4*b*d^2*ln(-d*x-c+1)/f/(-c*f+d*e+f)^2+1/4*b*d^2*ln(d*x+c+1)/f/(-c* 
f+d*e-f)^2-b*d^2*(-c*f+d*e)*ln(f*x+e)/(-c*f+d*e+f)^2/(d*e-(1+c)*f)^2
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\frac {1}{4} \left (-\frac {2 a}{f (e+f x)^2}+\frac {2 b d}{\left (d^2 e^2-2 c d e f+\left (-1+c^2\right ) f^2\right ) (e+f x)}-\frac {2 b \text {arctanh}(c+d x)}{f (e+f x)^2}-\frac {b d^2 \log (1-c-d x)}{f (d e+f-c f)^2}+\frac {b d^2 \log (1+c+d x)}{f (-d e+f+c f)^2}-\frac {4 b d^2 (d e-c f) \log (e+f x)}{\left (d^2 e^2-2 c d e f+\left (-1+c^2\right ) f^2\right )^2}\right ) \] Input:

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

Output:

((-2*a)/(f*(e + f*x)^2) + (2*b*d)/((d^2*e^2 - 2*c*d*e*f + (-1 + c^2)*f^2)* 
(e + f*x)) - (2*b*ArcTanh[c + d*x])/(f*(e + f*x)^2) - (b*d^2*Log[1 - c - d 
*x])/(f*(d*e + f - c*f)^2) + (b*d^2*Log[1 + c + d*x])/(f*(-(d*e) + f + c*f 
)^2) - (4*b*d^2*(d*e - c*f)*Log[e + f*x])/(d^2*e^2 - 2*c*d*e*f + (-1 + c^2 
)*f^2)^2)/4
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6659, 2081, 1141, 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.

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

\(\Big \downarrow \) 6659

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

\(\Big \downarrow \) 2081

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

\(\Big \downarrow \) 1141

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(a + b*ArcTanh[c + d*x])/(f*(e + f*x)^2) - (b*d^3*(-(f/(d^2*(d*e + f 
- c*f)*(d*e - (1 + c)*f)*(e + f*x))) + Log[1 - c - d*x]/(2*d*(d*e + f - c* 
f)^2) - Log[1 + c + d*x]/(2*d*(d*e - (1 + c)*f)^2) + (2*f*(d*e - c*f)*Log[ 
e + f*x])/(d*(d*e + f - c*f)^2*(d*e - (1 + c)*f)^2)))/(2*f)
 

Defintions of rubi rules used

rule 1141
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ 
Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p   Int[ExpandIntegrand[ 
(d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 
1] ||  !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 
0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
 

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

rule 2081
Int[(u_)^(m_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum 
[v, x]^p, x] /; FreeQ[{m, p}, x] && LinearQ[u, x] && QuadraticQ[v, x] &&  ! 
(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])
 

rule 6659
Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^( 
m_), x_Symbol] :> Simp[(e + f*x)^(m + 1)*((a + b*ArcTanh[c + d*x])^p/(f*(m 
+ 1))), x] - Simp[b*d*(p/(f*(m + 1)))   Int[(e + f*x)^(m + 1)*((a + b*ArcTa 
nh[c + d*x])^(p - 1)/(1 - (c + d*x)^2)), x], x] /; FreeQ[{a, b, c, d, e, f} 
, x] && IGtQ[p, 0] && ILtQ[m, -1]
 
Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.18

method result size
parts \(-\frac {a}{2 \left (f x +e \right )^{2} f}+\frac {b \left (-\frac {d^{3} \operatorname {arctanh}\left (d x +c \right )}{2 \left (f \left (d x +c \right )-c f +d e \right )^{2} f}+\frac {d^{3} \left (\frac {f}{\left (c f -d e +f \right ) \left (c f -d e -f \right ) \left (f \left (d x +c \right )-c f +d e \right )}+\frac {2 f \left (c f -d e \right ) \ln \left (f \left (d x +c \right )-c f +d e \right )}{\left (c f -d e +f \right )^{2} \left (c f -d e -f \right )^{2}}-\frac {\ln \left (d x +c -1\right )}{2 \left (c f -d e -f \right )^{2}}+\frac {\ln \left (d x +c +1\right )}{2 \left (c f -d e +f \right )^{2}}\right )}{2 f}\right )}{d}\) \(197\)
derivativedivides \(\frac {-\frac {a \,d^{3}}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-b \,d^{3} \left (\frac {\operatorname {arctanh}\left (d x +c \right )}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-\frac {-\frac {\ln \left (d x +c -1\right )}{2 \left (c f -d e -f \right )^{2}}+\frac {\ln \left (d x +c +1\right )}{2 \left (c f -d e +f \right )^{2}}-\frac {f}{\left (c f -d e -f \right ) \left (c f -d e +f \right ) \left (c f -d e -f \left (d x +c \right )\right )}+\frac {2 f \left (c f -d e \right ) \ln \left (c f -d e -f \left (d x +c \right )\right )}{\left (c f -d e -f \right )^{2} \left (c f -d e +f \right )^{2}}}{2 f}\right )}{d}\) \(214\)
default \(\frac {-\frac {a \,d^{3}}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-b \,d^{3} \left (\frac {\operatorname {arctanh}\left (d x +c \right )}{2 \left (c f -d e -f \left (d x +c \right )\right )^{2} f}-\frac {-\frac {\ln \left (d x +c -1\right )}{2 \left (c f -d e -f \right )^{2}}+\frac {\ln \left (d x +c +1\right )}{2 \left (c f -d e +f \right )^{2}}-\frac {f}{\left (c f -d e -f \right ) \left (c f -d e +f \right ) \left (c f -d e -f \left (d x +c \right )\right )}+\frac {2 f \left (c f -d e \right ) \ln \left (c f -d e -f \left (d x +c \right )\right )}{\left (c f -d e -f \right )^{2} \left (c f -d e +f \right )^{2}}}{2 f}\right )}{d}\) \(214\)
parallelrisch \(-\frac {-b \,d^{5} e^{3} f^{2}+b \,d^{3} e \,f^{4}+a \,d^{6} e^{4} f -2 a \,d^{4} e^{2} f^{3}-4 a \,c^{3} d^{3} e \,f^{4}+6 a \,c^{2} d^{4} e^{2} f^{3}-4 a c \,d^{5} e^{3} f^{2}+4 a c \,d^{3} e \,f^{4}-b \,c^{2} d^{3} e \,f^{4}+2 b c \,d^{4} e^{2} f^{3}-x^{2} \operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d^{4} f^{5}-x^{2} \operatorname {arctanh}\left (d x +c \right ) b \,d^{6} e^{2} f^{3}+2 x^{2} \operatorname {arctanh}\left (d x +c \right ) b c \,d^{4} f^{5}-2 x^{2} \operatorname {arctanh}\left (d x +c \right ) b \,d^{5} e \,f^{4}-2 \ln \left (f x +e \right ) b c \,d^{4} e^{2} f^{3}-4 x \,\operatorname {arctanh}\left (d x +c \right ) b \,d^{5} e^{2} f^{3}+4 \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{3} e \,f^{4}-4 \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{3} d^{3} e \,f^{4}+5 \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d^{4} e^{2} f^{3}-2 \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{5} e^{3} f^{2}+2 \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{4} e^{2} f^{3}+2 x b c \,d^{4} e \,f^{4}-2 x \,\operatorname {arctanh}\left (d x +c \right ) b \,d^{6} e^{3} f^{2}-2 x \,\operatorname {arctanh}\left (d x +c \right ) b \,d^{4} e \,f^{4}+2 \ln \left (d x +c -1\right ) x^{2} b c \,d^{4} f^{5}-2 \ln \left (d x +c -1\right ) x^{2} b \,d^{5} e \,f^{4}-2 \ln \left (f x +e \right ) x^{2} b c \,d^{4} f^{5}+2 \ln \left (f x +e \right ) x^{2} b \,d^{5} e \,f^{4}-4 \ln \left (d x +c -1\right ) x b \,d^{5} e^{2} f^{3}+4 \ln \left (f x +e \right ) x b \,d^{5} e^{2} f^{3}+2 \ln \left (d x +c -1\right ) b c \,d^{4} e^{2} f^{3}+4 \ln \left (d x +c -1\right ) x b c \,d^{4} e \,f^{4}-4 \ln \left (f x +e \right ) x b c \,d^{4} e \,f^{4}-2 x \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d^{4} e \,f^{4}+4 x \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{5} e^{2} f^{3}+4 x \,\operatorname {arctanh}\left (d x +c \right ) b c \,d^{4} e \,f^{4}+2 x^{2} \operatorname {arctanh}\left (d x +c \right ) b c \,d^{5} e \,f^{4}-2 \ln \left (d x +c -1\right ) b \,d^{5} e^{3} f^{2}+2 \ln \left (f x +e \right ) b \,d^{5} e^{3} f^{2}-x b \,c^{2} d^{3} f^{5}-x^{2} \operatorname {arctanh}\left (d x +c \right ) b \,d^{4} f^{5}+\operatorname {arctanh}\left (d x +c \right ) b \,c^{4} d^{2} f^{5}-2 \,\operatorname {arctanh}\left (d x +c \right ) b \,d^{5} e^{3} f^{2}-2 \,\operatorname {arctanh}\left (d x +c \right ) b \,c^{2} d^{2} f^{5}-3 \,\operatorname {arctanh}\left (d x +c \right ) b \,d^{4} e^{2} f^{3}-x b \,d^{5} e^{2} f^{3}+\operatorname {arctanh}\left (d x +c \right ) b \,d^{2} f^{5}+x b \,d^{3} f^{5}+a \,d^{2} f^{5}+a \,c^{4} d^{2} f^{5}-2 a \,c^{2} d^{2} f^{5}}{2 \left (c^{4} f^{4}-4 c^{3} d e \,f^{3}+6 c^{2} d^{2} e^{2} f^{2}-4 c \,d^{3} e^{3} f +d^{4} e^{4}-2 c^{2} f^{4}+4 c d e \,f^{3}-2 d^{2} e^{2} f^{2}+f^{4}\right ) \left (f x +e \right )^{2} f^{2} d^{2}}\) \(937\)
risch \(\text {Expression too large to display}\) \(1289\)

Input:

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

Output:

-1/2*a/(f*x+e)^2/f+b/d*(-1/2*d^3/(f*(d*x+c)-c*f+d*e)^2/f*arctanh(d*x+c)+1/ 
2*d^3/f*(f/(c*f-d*e+f)/(c*f-d*e-f)/(f*(d*x+c)-c*f+d*e)+2*f*(c*f-d*e)/(c*f- 
d*e+f)^2/(c*f-d*e-f)^2*ln(f*(d*x+c)-c*f+d*e)-1/2/(c*f-d*e-f)^2*ln(d*x+c-1) 
+1/2/(c*f-d*e+f)^2*ln(d*x+c+1)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (159) = 318\).

Time = 0.69 (sec) , antiderivative size = 834, normalized size of antiderivative = 4.99 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19859 vs. \(2 (141) = 282\).

Time = 11.08 (sec) , antiderivative size = 19859, normalized size of antiderivative = 118.92 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((-(a + b*atanh(c))/(2*e**2*f + 4*e*f**2*x + 2*f**3*x**2), Eq(d, 
0)), ((a*x + b*c*atanh(c + d*x)/d + b*x*atanh(c + d*x) + b*log(c/d + x + 1 
/d)/d - b*atanh(c + d*x)/d)/e**3, Eq(f, 0)), (-4*a*f**2/(8*e**2*f**3 + 16* 
e*f**4*x + 8*f**5*x**2) + b*d**2*e**2*atanh(d*e/f + d*x - 1)/(8*e**2*f**3 
+ 16*e*f**4*x + 8*f**5*x**2) + 2*b*d**2*e*f*x*atanh(d*e/f + d*x - 1)/(8*e* 
*2*f**3 + 16*e*f**4*x + 8*f**5*x**2) + b*d**2*f**2*x**2*atanh(d*e/f + d*x 
- 1)/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) - b*d*e*f/(8*e**2*f**3 + 16 
*e*f**4*x + 8*f**5*x**2) - b*d*f**2*x/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5* 
x**2) - 4*b*f**2*atanh(d*e/f + d*x - 1)/(8*e**2*f**3 + 16*e*f**4*x + 8*f** 
5*x**2) - b*f**2/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2), Eq(c, (d*e - f 
)/f)), (-4*a*f**2/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) + b*d**2*e**2* 
atanh(d*e/f + d*x + 1)/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) + 2*b*d** 
2*e*f*x*atanh(d*e/f + d*x + 1)/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) + 
 b*d**2*f**2*x**2*atanh(d*e/f + d*x + 1)/(8*e**2*f**3 + 16*e*f**4*x + 8*f* 
*5*x**2) - b*d*e*f/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) - b*d*f**2*x/ 
(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) - 4*b*f**2*atanh(d*e/f + d*x + 1 
)/(8*e**2*f**3 + 16*e*f**4*x + 8*f**5*x**2) + b*f**2/(8*e**2*f**3 + 16*e*f 
**4*x + 8*f**5*x**2), Eq(c, (d*e + f)/f)), (-a*c**4*f**4/(2*c**4*e**2*f**5 
 + 4*c**4*e*f**6*x + 2*c**4*f**7*x**2 - 8*c**3*d*e**3*f**4 - 16*c**3*d*e** 
2*f**5*x - 8*c**3*d*e*f**6*x**2 + 12*c**2*d**2*e**4*f**3 + 24*c**2*d**2...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.74 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\frac {1}{4} \, {\left (d {\left (\frac {d \log \left (d x + c + 1\right )}{d^{2} e^{2} f - 2 \, {\left (c + 1\right )} d e f^{2} + {\left (c^{2} + 2 \, c + 1\right )} f^{3}} - \frac {d \log \left (d x + c - 1\right )}{d^{2} e^{2} f - 2 \, {\left (c - 1\right )} d e f^{2} + {\left (c^{2} - 2 \, c + 1\right )} f^{3}} - \frac {4 \, {\left (d^{2} e - c d f\right )} \log \left (f x + e\right )}{d^{4} e^{4} - 4 \, c d^{3} e^{3} f + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} e^{2} f^{2} - 4 \, {\left (c^{3} - c\right )} d e f^{3} + {\left (c^{4} - 2 \, c^{2} + 1\right )} f^{4}} + \frac {2}{d^{2} e^{3} - 2 \, c d e^{2} f + {\left (c^{2} - 1\right )} e f^{2} + {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + {\left (c^{2} - 1\right )} f^{3}\right )} x}\right )} - \frac {2 \, \operatorname {artanh}\left (d x + c\right )}{f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f}\right )} b - \frac {a}{2 \, {\left (f^{3} x^{2} + 2 \, e f^{2} x + e^{2} f\right )}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2567 vs. \(2 (159) = 318\).

Time = 0.19 (sec) , antiderivative size = 2567, normalized size of antiderivative = 15.37 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((c + 1)*d - (c - 1)*d)*((b*d^2*e - b*c*d*f)*log(-(d*x + c + 1)*d*e/( 
d*x + c - 1) + d*e + (d*x + c + 1)*c*f/(d*x + c - 1) - c*f - (d*x + c + 1) 
*f/(d*x + c - 1) - f)/(d^4*e^4 - 4*c*d^3*e^3*f + 6*c^2*d^2*e^2*f^2 - 4*c^3 
*d*e*f^3 + c^4*f^4 - 2*d^2*e^2*f^2 + 4*c*d*e*f^3 - 2*c^2*f^4 + f^4) - ((d* 
x + c + 1)*b*d^2*e/(d*x + c - 1) - b*d^2*e - (d*x + c + 1)*b*c*d*f/(d*x + 
c - 1) + b*c*d*f + (d*x + c + 1)*b*d*f/(d*x + c - 1))*log(-(d*x + c + 1)/( 
d*x + c - 1))/((d*x + c + 1)^2*d^4*e^4/(d*x + c - 1)^2 - 2*(d*x + c + 1)*d 
^4*e^4/(d*x + c - 1) + d^4*e^4 - 4*(d*x + c + 1)^2*c*d^3*e^3*f/(d*x + c - 
1)^2 + 8*(d*x + c + 1)*c*d^3*e^3*f/(d*x + c - 1) - 4*c*d^3*e^3*f + 6*(d*x 
+ c + 1)^2*c^2*d^2*e^2*f^2/(d*x + c - 1)^2 - 12*(d*x + c + 1)*c^2*d^2*e^2* 
f^2/(d*x + c - 1) + 6*c^2*d^2*e^2*f^2 - 4*(d*x + c + 1)^2*c^3*d*e*f^3/(d*x 
 + c - 1)^2 + 8*(d*x + c + 1)*c^3*d*e*f^3/(d*x + c - 1) - 4*c^3*d*e*f^3 + 
(d*x + c + 1)^2*c^4*f^4/(d*x + c - 1)^2 - 2*(d*x + c + 1)*c^4*f^4/(d*x + c 
 - 1) + c^4*f^4 + 4*(d*x + c + 1)^2*d^3*e^3*f/(d*x + c - 1)^2 - 4*(d*x + c 
 + 1)*d^3*e^3*f/(d*x + c - 1) - 12*(d*x + c + 1)^2*c*d^2*e^2*f^2/(d*x + c 
- 1)^2 + 12*(d*x + c + 1)*c*d^2*e^2*f^2/(d*x + c - 1) + 12*(d*x + c + 1)^2 
*c^2*d*e*f^3/(d*x + c - 1)^2 - 12*(d*x + c + 1)*c^2*d*e*f^3/(d*x + c - 1) 
- 4*(d*x + c + 1)^2*c^3*f^4/(d*x + c - 1)^2 + 4*(d*x + c + 1)*c^3*f^4/(d*x 
 + c - 1) + 6*(d*x + c + 1)^2*d^2*e^2*f^2/(d*x + c - 1)^2 - 2*d^2*e^2*f^2 
- 12*(d*x + c + 1)^2*c*d*e*f^3/(d*x + c - 1)^2 + 4*c*d*e*f^3 + 6*(d*x +...
 

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.50 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx=\frac {b\,d^2\,\ln \left (c+d\,x+1\right )}{4\,c^2\,f^3-8\,c\,d\,e\,f^2+8\,c\,f^3+4\,d^2\,e^2\,f-8\,d\,e\,f^2+4\,f^3}-\frac {\ln \left (e+f\,x\right )\,\left (b\,d^3\,e-b\,c\,d^2\,f\right )}{c^4\,f^4-4\,c^3\,d\,e\,f^3+6\,c^2\,d^2\,e^2\,f^2-2\,c^2\,f^4-4\,c\,d^3\,e^3\,f+4\,c\,d\,e\,f^3+d^4\,e^4-2\,d^2\,e^2\,f^2+f^4}-\frac {b\,\ln \left (c+d\,x+1\right )}{4\,f\,\left (e^2+2\,e\,f\,x+f^2\,x^2\right )}-\frac {b\,d^2\,\ln \left (c+d\,x-1\right )}{4\,c^2\,f^3-8\,c\,d\,e\,f^2-8\,c\,f^3+4\,d^2\,e^2\,f+8\,d\,e\,f^2+4\,f^3}-\frac {\frac {-a\,c^2\,f^2+2\,a\,c\,d\,e\,f-a\,d^2\,e^2+b\,d\,e\,f+a\,f^2}{-c^2\,f^2+2\,c\,d\,e\,f-d^2\,e^2+f^2}+\frac {b\,d\,f^2\,x}{-c^2\,f^2+2\,c\,d\,e\,f-d^2\,e^2+f^2}}{2\,e^2\,f+4\,e\,f^2\,x+2\,f^3\,x^2}+\frac {b\,\ln \left (1-d\,x-c\right )}{2\,f\,\left (2\,e^2+4\,e\,f\,x+2\,f^2\,x^2\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 2159, normalized size of antiderivative = 12.93 \[ \int \frac {a+b \text {arctanh}(c+d x)}{(e+f x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*atanh(c + d*x)*b*c**4*e*f**5*x + 2*atanh(c + d*x)*b*c**4*f**6*x**2 - 16 
*atanh(c + d*x)*b*c**3*d*e**2*f**4*x - 8*atanh(c + d*x)*b*c**3*d*e*f**5*x* 
*2 + 24*atanh(c + d*x)*b*c**2*d**2*e**3*f**3*x + 12*atanh(c + d*x)*b*c**2* 
d**2*e**2*f**4*x**2 - 8*atanh(c + d*x)*b*c**2*e*f**5*x - 4*atanh(c + d*x)* 
b*c**2*f**6*x**2 - 16*atanh(c + d*x)*b*c*d**3*e**4*f**2*x - 8*atanh(c + d* 
x)*b*c*d**3*e**3*f**3*x**2 + 16*atanh(c + d*x)*b*c*d*e**2*f**4*x + 8*atanh 
(c + d*x)*b*c*d*e*f**5*x**2 + 4*atanh(c + d*x)*b*d**4*e**5*f*x + 2*atanh(c 
 + d*x)*b*d**4*e**4*f**2*x**2 - 8*atanh(c + d*x)*b*d**2*e**3*f**3*x - 4*at 
anh(c + d*x)*b*d**2*e**2*f**4*x**2 + 4*atanh(c + d*x)*b*e*f**5*x + 2*atanh 
(c + d*x)*b*f**6*x**2 + log(c + d*x - 1)*b*c**4*e**2*f**4 + 2*log(c + d*x 
- 1)*b*c**4*e*f**5*x + log(c + d*x - 1)*b*c**4*f**6*x**2 - 4*log(c + d*x - 
 1)*b*c**3*d*e**3*f**3 - 8*log(c + d*x - 1)*b*c**3*d*e**2*f**4*x - 4*log(c 
 + d*x - 1)*b*c**3*d*e*f**5*x**2 + 5*log(c + d*x - 1)*b*c**2*d**2*e**4*f** 
2 + 10*log(c + d*x - 1)*b*c**2*d**2*e**3*f**3*x + 5*log(c + d*x - 1)*b*c** 
2*d**2*e**2*f**4*x**2 - 2*log(c + d*x - 1)*b*c**2*e**2*f**4 - 4*log(c + d* 
x - 1)*b*c**2*e*f**5*x - 2*log(c + d*x - 1)*b*c**2*f**6*x**2 - 2*log(c + d 
*x - 1)*b*c*d**3*e**5*f - 4*log(c + d*x - 1)*b*c*d**3*e**4*f**2*x - 2*log( 
c + d*x - 1)*b*c*d**3*e**3*f**3*x**2 - 2*log(c + d*x - 1)*b*c*d**2*e**4*f* 
*2 - 4*log(c + d*x - 1)*b*c*d**2*e**3*f**3*x - 2*log(c + d*x - 1)*b*c*d**2 
*e**2*f**4*x**2 + 4*log(c + d*x - 1)*b*c*d*e**3*f**3 + 8*log(c + d*x - ...