\(\int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{(c-a^2 c x^2)^{5/4}} \, dx\) [1321]

Optimal result

Integrand size = 27, antiderivative size = 105 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2}}{a^2 c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac {\sqrt [4]{1-a^2 x^2} \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2 c \sqrt [4]{c-a^2 c x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2} \left (\frac {1}{a^2 \sqrt {1-a x}}+\frac {\text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2}\right )}{c \sqrt [4]{c-a^2 c x^2}} \] Input:

Integrate[(E^(ArcTanh[a*x]/2)*x)/(c - a^2*c*x^2)^(5/4),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6703, 6700, 87, 73, 219}

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.

\(\displaystyle \int \frac {x e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (1-a^2 x^2\right )^{5/4}}dx}{c \sqrt [4]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt [4]{1-a^2 x^2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a^2}+\frac {1}{a^2 \sqrt {1-a x}}\right )}{c \sqrt [4]{c-a^2 c x^2}}\)

Input:

Int[(E^(ArcTanh[a*x]/2)*x)/(c - a^2*c*x^2)^(5/4),x]
 

Output:

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

Defintions of rubi rules used

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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6700

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

rule 6703

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

Maple [F]

\[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{4}}}d x\]

Input:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x)
 

Output:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x, alg 
orithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x, alg 
orithm="maxima")
 

Output:

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

Giac [F]

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

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x, alg 
orithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((x*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(5/4),x)
 

Output:

int((x*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(5/4), x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.58 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {-\sqrt {-a^{2} x^{2}+1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {a x +1}}{\sqrt {2}}\right )}{2}\right )\right )+2 \sqrt {a x +1}}{2 c^{\frac {5}{4}} \sqrt {-a^{2} x^{2}+1}\, a^{2}} \] Input:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x/(-a^2*c*x^2+c)^(5/4),x)
 

Output:

( - sqrt( - a**2*x**2 + 1)*sqrt(2)*log(tan(asin(sqrt(a*x + 1)/sqrt(2))/2)) 
 + 2*sqrt(a*x + 1))/(2*c**(1/4)*sqrt( - a**2*x**2 + 1)*a**2*c)