\(\int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx\) [356]

Optimal result

Integrand size = 19, antiderivative size = 128 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{c (1-a x)}-\frac {3 a^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c} \] Output:

-1/3*(-a^2*x^2+1)^(1/2)/c/x^3-a*(-a^2*x^2+1)^(1/2)/c/x^2-8/3*a^2*(-a^2*x^2 
+1)^(1/2)/c/x+2*a^3*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)-3*a^3*arctanh((-a^2*x^2+ 
1)^(1/2))/c
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=-\frac {1+3 a x+7 a^2 x^2-9 a^3 x^3-14 a^4 x^4+9 a^3 x^3 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{3 c x^3 \sqrt {1-a^2 x^2}} \] Input:

Integrate[E^ArcTanh[a*x]/(x^4*(c - a*c*x)),x]
 

Output:

-1/3*(1 + 3*a*x + 7*a^2*x^2 - 9*a^3*x^3 - 14*a^4*x^4 + 9*a^3*x^3*Sqrt[1 - 
a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(c*x^3*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {6678, 27, 564, 2338, 27, 2338, 25, 27, 534, 243, 73, 221}

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.

\(\displaystyle \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx\)

\(\Big \downarrow \) 6678

\(\displaystyle c \int \frac {\sqrt {1-a^2 x^2}}{c^2 x^4 (1-a x)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {1-a^2 x^2}}{x^4 (1-a x)^2}dx}{c}\)

\(\Big \downarrow \) 564

\(\displaystyle \frac {\int \frac {2 a^3 x^3+2 a^2 x^2+2 a x+1}{x^4 \sqrt {1-a^2 x^2}}dx+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {-\frac {1}{3} \int -\frac {2 \left (3 x^2 a^3+4 x a^2+3 a\right )}{x^3 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2}{3} \int \frac {3 x^2 a^3+4 x a^2+3 a}{x^3 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\frac {2}{3} \left (-\frac {1}{2} \int -\frac {a^2 (9 a x+8)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} \int \frac {a^2 (9 a x+8)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} a^2 \int \frac {9 a x+8}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} a^2 \left (9 a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {8 \sqrt {1-a^2 x^2}}{x}\right )-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} a^2 \left (\frac {9}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {8 \sqrt {1-a^2 x^2}}{x}\right )-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} a^2 \left (-\frac {9 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {8 \sqrt {1-a^2 x^2}}{x}\right )-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {2}{3} \left (\frac {1}{2} a^2 \left (-9 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {8 \sqrt {1-a^2 x^2}}{x}\right )-\frac {3 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{1-a x}}{c}\)

Input:

Int[E^ArcTanh[a*x]/(x^4*(c - a*c*x)),x]
 

Output:

(-1/3*Sqrt[1 - a^2*x^2]/x^3 + (2*a^3*Sqrt[1 - a^2*x^2])/(1 - a*x) + (2*((- 
3*a*Sqrt[1 - a^2*x^2])/(2*x^2) + (a^2*((-8*Sqrt[1 - a^2*x^2])/x - 9*a*ArcT 
anh[Sqrt[1 - a^2*x^2]]))/2))/3)/c
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 564

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b 
^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(x^m/Sqrt[a + b 
*x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- 
n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 
2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

rule 6678

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)* 
(x_))^(m_.), x_Symbol] :> Simp[c^n   Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - 
 a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c + d, 
0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1 
, 0]) && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {8 a^{4} x^{4}+3 a^{3} x^{3}-7 a^{2} x^{2}-3 a x -1}{3 x^{3} \sqrt {-a^{2} x^{2}+1}\, c}+\frac {a^{3} \left (-3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}\right )}{c}\)	\(116\)
default	\(-\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{3 x^{3}}+\frac {8 a^{2} \sqrt {-a^{2} x^{2}+1}}{3 x}-2 a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )+2 a^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {2 a^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}}{c}\)	\(142\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x,method=_RETURNVERBOSE)
 

Output:

1/3*(8*a^4*x^4+3*a^3*x^3-7*a^2*x^2-3*a*x-1)/x^3/(-a^2*x^2+1)^(1/2)/c+a^3*( 
-3*arctanh(1/(-a^2*x^2+1)^(1/2))-2/a/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^ 
(1/2))/c
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=\frac {6 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 9 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a c x^{4} - c x^{3}\right )}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="fricas")
 

Output:

1/3*(6*a^4*x^4 - 6*a^3*x^3 + 9*(a^4*x^4 - a^3*x^3)*log((sqrt(-a^2*x^2 + 1) 
 - 1)/x) - (14*a^3*x^3 - 5*a^2*x^2 - 2*a*x - 1)*sqrt(-a^2*x^2 + 1))/(a*c*x 
^4 - c*x^3)
 

Sympy [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=- \frac {\int \frac {a x}{a x^{5} \sqrt {- a^{2} x^{2} + 1} - x^{4} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{a x^{5} \sqrt {- a^{2} x^{2} + 1} - x^{4} \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \] Input:

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**4/(-a*c*x+c),x)
 

Output:

-(Integral(a*x/(a*x**5*sqrt(-a**2*x**2 + 1) - x**4*sqrt(-a**2*x**2 + 1)), 
x) + Integral(1/(a*x**5*sqrt(-a**2*x**2 + 1) - x**4*sqrt(-a**2*x**2 + 1)), 
 x))/c
 

Maxima [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=\int { -\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1} {\left (a c x - c\right )} x^{4}} \,d x } \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="maxima")
 

Output:

-integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)*x^4), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (113) = 226\).

Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.21 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=-\frac {{\left (a^{4} + \frac {5 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2}}{x} + \frac {27 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}} - \frac {129 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{2} x^{3}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} - \frac {3 \, a^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} - \frac {\frac {33 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c^{2}}{x} + \frac {6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c^{2}}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{2}}{x^{3}}}{24 \, a^{2} c^{3} {\left | a \right |}} \] Input:

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="giac")
 

Output:

-1/24*(a^4 + 5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2/x + 27*(sqrt(-a^2*x^2 + 
 1)*abs(a) + a)^2/x^2 - 129*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^2*x^3))*a 
^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c*((sqrt(-a^2*x^2 + 1)*abs(a) + 
a)/(a^2*x) - 1)*abs(a)) - 3*a^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 
 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1/24*(33*(sqrt(-a^2*x^2 + 1)*abs(a) + a)* 
a^4*c^2/x + 6*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^2/x^2 + (sqrt(-a^2*x 
^2 + 1)*abs(a) + a)^3*c^2/x^3)/(a^2*c^3*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=-\frac {\sqrt {1-a^2\,x^2}}{3\,c\,x^3}-\frac {a\,\sqrt {1-a^2\,x^2}}{c\,x^2}-\frac {8\,a^2\,\sqrt {1-a^2\,x^2}}{3\,c\,x}-\frac {2\,a^4\,\sqrt {1-a^2\,x^2}}{\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {a^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{c} \] Input:

int((a*x + 1)/(x^4*(1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)
 

Output:

(a^3*atan((1 - a^2*x^2)^(1/2)*1i)*3i)/c - (1 - a^2*x^2)^(1/2)/(3*c*x^3) - 
(a*(1 - a^2*x^2)^(1/2))/(c*x^2) - (8*a^2*(1 - a^2*x^2)^(1/2))/(3*c*x) - (2 
*a^4*(1 - a^2*x^2)^(1/2))/(((c*(-a^2)^(1/2))/a - c*x*(-a^2)^(1/2))*(-a^2)^ 
(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \frac {e^{\text {arctanh}(a x)}}{x^4 (c-a c x)} \, dx=\frac {a^{3} \left (72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}+\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{7}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{6}+27 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}-162 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}+27 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+1\right )}{24 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3} c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x)
 

Output:

(a**3*(72*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**4 - 72*log(tan(asin(a*x) 
/2))*tan(asin(a*x)/2)**3 + tan(asin(a*x)/2)**7 + 5*tan(asin(a*x)/2)**6 + 2 
7*tan(asin(a*x)/2)**5 - 162*tan(asin(a*x)/2)**4 + 27*tan(asin(a*x)/2)**2 + 
 5*tan(asin(a*x)/2) + 1))/(24*tan(asin(a*x)/2)**3*c*(tan(asin(a*x)/2) - 1) 
)