\(\int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{(c-a^2 c x^2)^{5/4}} \, dx\) [1322]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2}}{a c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac {\sqrt [4]{1-a^2 x^2} \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{\sqrt {2} a c \sqrt [4]{c-a^2 c x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt [4]{1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1-a x)\right )}{a c \sqrt {1-a x} \sqrt [4]{c-a^2 c x^2}} \] Input:

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

Output:

((1 - a^2*x^2)^(1/4)*Hypergeometric2F1[-1/2, 1, 1/2, (1 - a*x)/2])/(a*c*Sq 
rt[1 - a*x]*(c - a^2*c*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6693, 6690, 61, 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.

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

\(\Big \downarrow \) 6693

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

\(\Big \downarrow \) 6690

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 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 6690

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

rule 6693

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

int(((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 = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.53 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{5/4}} \, dx=\frac {\sqrt {-a x +1}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {a x +1}}{\sqrt {2}}\right )}{2}\right )\right )-\sqrt {-a x +1}\, \sqrt {2}+2}{2 c^{\frac {5}{4}} \sqrt {-a x +1}\, a} \] Input:

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

Output:

(c**(3/4)*(sqrt( - a*x + 1)*sqrt(2)*log(tan(asin(sqrt(a*x + 1)/sqrt(2))/2) 
) - sqrt( - a*x + 1)*sqrt(2) + 2))/(2*sqrt( - a*x + 1)*a*c**2)