\(\int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx\) [1152]

Optimal result

Integrand size = 27, antiderivative size = 135 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx=\frac {2 a^3 (1+a x)}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{3 c x^3}-\frac {a \sqrt {c-a^2 c x^2}}{c x^2}-\frac {8 a^2 \sqrt {c-a^2 c x^2}}{3 c x}-\frac {3 a^3 \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.75 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (1+2 a x+5 a^2 x^2-14 a^3 x^3\right )}{3 c x^3 (-1+a x)}+\frac {3 a^3 \log (x)}{\sqrt {c}}-\frac {3 a^3 \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )}{\sqrt {c}} \] Input:

Integrate[E^(2*ArcTanh[a*x])/(x^4*Sqrt[c - a^2*c*x^2]),x]
 

Output:

(Sqrt[c - a^2*c*x^2]*(1 + 2*a*x + 5*a^2*x^2 - 14*a^3*x^3))/(3*c*x^3*(-1 + 
a*x)) + (3*a^3*Log[x])/Sqrt[c] - (3*a^3*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2 
]])/Sqrt[c]
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {6701, 528, 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.

\(\displaystyle \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 528

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[E^(2*ArcTanh[a*x])/(x^4*Sqrt[c - a^2*c*x^2]),x]
 

Output:

c*((2*a^3*(1 + a*x))/(c*Sqrt[c - a^2*c*x^2]) + (-1/3*Sqrt[c - a^2*c*x^2]/( 
c*x^3) + (2*((-3*a*Sqrt[c - a^2*c*x^2])/(2*x^2) + (a^2*c*((-8*Sqrt[c - a^2 
*c*x^2])/(c*x) - (9*a*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/Sqrt[c]))/2))/ 
(3*c))/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 528

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

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 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 6701

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {8 a^{4} x^{4}+3 a^{3} x^{3}-7 a^{2} x^{2}-3 a x -1}{3 x^{3} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {3 a^{3} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}-\frac {2 a^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{c \left (x -\frac {1}{a}\right )}\)	\(132\)
default	\(-\frac {\sqrt {-a^{2} c \,x^{2}+c}}{3 c \,x^{3}}-\frac {8 a^{2} \sqrt {-a^{2} c \,x^{2}+c}}{3 c x}+2 a \left (-\frac {\sqrt {-a^{2} c \,x^{2}+c}}{2 c \,x^{2}}-\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2 \sqrt {c}}\right )-\frac {2 a^{3} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{\sqrt {c}}-\frac {2 a^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}{c \left (x -\frac {1}{a}\right )}\)	\(188\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.59 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx=\left [\frac {9 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, {\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c}}{6 \, {\left (a c x^{4} - c x^{3}\right )}}, \frac {9 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) - {\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c}}{3 \, {\left (a c x^{4} - c x^{3}\right )}}\right ] \] Input:

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

Output:

[1/6*(9*(a^4*x^4 - a^3*x^3)*sqrt(c)*log(-(a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + 
c)*sqrt(c) - 2*c)/x^2) - 2*(14*a^3*x^3 - 5*a^2*x^2 - 2*a*x - 1)*sqrt(-a^2* 
c*x^2 + c))/(a*c*x^4 - c*x^3), 1/3*(9*(a^4*x^4 - a^3*x^3)*sqrt(-c)*arctan( 
sqrt(-a^2*c*x^2 + c)*sqrt(-c)/c) - (14*a^3*x^3 - 5*a^2*x^2 - 2*a*x - 1)*sq 
rt(-a^2*c*x^2 + c))/(a*c*x^4 - c*x^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

undef
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int \frac {e^{2 \text {arctanh}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx=\frac {\sqrt {c}\, a^{3} \left (72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}+\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{7}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{6}+27 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}-162 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}+27 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )+1\right )}{24 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3} c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:

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

Output:

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