\(\int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{(c-a^2 c x^2)^{9/8}} \, dx\) [1325]

Optimal result

Integrand size = 29, antiderivative size = 252 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {40 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{21 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}-\frac {4 x^2 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac {8 (1-a x)^{5/8} \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{21 a^4 c \sqrt [8]{c-a^2 c x^2}}+\frac {64 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )}{105 a^4 c \sqrt [8]{c-a^2 c x^2}} \] Output:

40/21*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)/a^4/c/(-a*x+1)^(3/8)/(-a^2*c*x^2+c) 
^(1/8)-4/7*x^2*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)/a^2/c/(-a*x+1)^(3/8)/(-a^2 
*c*x^2+c)^(1/8)+8/21*(-a*x+1)^(5/8)*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)/a^4/c 
/(-a^2*c*x^2+c)^(1/8)+64/105*2^(1/8)*(-a*x+1)^(5/8)*(-a^2*x^2+1)^(1/8)*hyp 
ergeom([5/8, 7/8],[13/8],-1/2*a*x+1/2)/a^4/c/(-a^2*c*x^2+c)^(1/8)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=-\frac {4 \sqrt [8]{1-a^2 x^2} \left (5 \sqrt [8]{1+a x} \left (-12+2 a x+3 a^2 x^2\right )+16 \sqrt [8]{2} (-1+a x) \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )\right )}{105 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \] Input:

Integrate[(E^(ArcTanh[a*x]/2)*x^3)/(c - a^2*c*x^2)^(9/8),x]
 

Output:

(-4*(1 - a^2*x^2)^(1/8)*(5*(1 + a*x)^(1/8)*(-12 + 2*a*x + 3*a^2*x^2) + 16* 
2^(1/8)*(-1 + a*x)*Hypergeometric2F1[5/8, 7/8, 13/8, (1 - a*x)/2]))/(105*a 
^4*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.58, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {6703, 6700, 111, 27, 163, 79}

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.

\(\displaystyle \int \frac {x^3 e^{\frac {1}{2} \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{9/8}} \, dx\)

\(\Big \downarrow \) 6703

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (1-a^2 x^2\right )^{9/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 6700

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \int \frac {x^3}{(1-a x)^{11/8} (a x+1)^{7/8}}dx}{c \sqrt [8]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 111

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (-\frac {4 \int -\frac {x (a x+4)}{2 (1-a x)^{11/8} (a x+1)^{7/8}}dx}{7 a^2}-\frac {4 x^2 \sqrt [8]{a x+1}}{7 a^2 (1-a x)^{3/8}}\right )}{c \sqrt [8]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (\frac {2 \int \frac {x (a x+4)}{(1-a x)^{11/8} (a x+1)^{7/8}}dx}{7 a^2}-\frac {4 x^2 \sqrt [8]{a x+1}}{7 a^2 (1-a x)^{3/8}}\right )}{c \sqrt [8]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 163

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (\frac {2 \left (\frac {4 (6-a x) \sqrt [8]{a x+1}}{3 a^2 (1-a x)^{3/8}}-\frac {8 \int \frac {1}{(1-a x)^{3/8} (a x+1)^{7/8}}dx}{3 a}\right )}{7 a^2}-\frac {4 x^2 \sqrt [8]{a x+1}}{7 a^2 (1-a x)^{3/8}}\right )}{c \sqrt [8]{c-a^2 c x^2}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\sqrt [8]{1-a^2 x^2} \left (\frac {2 \left (\frac {32 \sqrt [8]{2} (1-a x)^{5/8} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},\frac {7}{8},\frac {13}{8},\frac {1}{2} (1-a x)\right )}{15 a^2}+\frac {4 \sqrt [8]{a x+1} (6-a x)}{3 a^2 (1-a x)^{3/8}}\right )}{7 a^2}-\frac {4 x^2 \sqrt [8]{a x+1}}{7 a^2 (1-a x)^{3/8}}\right )}{c \sqrt [8]{c-a^2 c x^2}}\)

Input:

Int[(E^(ArcTanh[a*x]/2)*x^3)/(c - a^2*c*x^2)^(9/8),x]
 

Output:

((1 - a^2*x^2)^(1/8)*((-4*x^2*(1 + a*x)^(1/8))/(7*a^2*(1 - a*x)^(3/8)) + ( 
2*((4*(6 - a*x)*(1 + a*x)^(1/8))/(3*a^2*(1 - a*x)^(3/8)) + (32*2^(1/8)*(1 
- a*x)^(5/8)*Hypergeometric2F1[5/8, 7/8, 13/8, (1 - a*x)/2])/(15*a^2)))/(7 
*a^2)))/(c*(c - a^2*c*x^2)^(1/8))
 

Defintions of rubi rules used

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 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 111

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

rule 163

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n 
+ 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* 
(m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + 
d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f 
*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* 
d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* 
d*(b*c - a*d)*(m + 1)*(m + n + 3))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 
1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 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]

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

Input:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x)
 

Output:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \] Input:

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x, a 
lgorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Timed out} \] Input:

integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(1/2)*x**3/(-a**2*c*x**2+c)**(9 
/8),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int { \frac {x^{3} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}} \,d x } \] Input:

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x, a 
lgorithm="maxima")
 

Output:

integrate(x^3*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), x 
)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x, a 
lgorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\int \frac {x^3\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{9/8}} \,d x \] Input:

int((x^3*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(9/8),x)
 

Output:

int((x^3*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2))/(c - a^2*c*x^2)^(9/8), x)
 

Reduce [F]

\[ \int \frac {e^{\frac {1}{2} \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{9/8}} \, dx=\frac {\int \frac {\sqrt {a x +1}\, \left (-a^{2} x^{2}+1\right )^{\frac {5}{8}} x^{3}}{a^{4} x^{4}-2 a^{2} x^{2}+1}d x}{c^{\frac {5}{8}} \sqrt {c}} \] Input:

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x)
                                                                                    
                                                                                    
 

Output:

(c**(3/8)*int((sqrt(a*x + 1)*( - a**2*x**2 + 1)**(5/8)*x**3)/(a**4*x**4 - 
2*a**2*x**2 + 1),x))/(sqrt(c)*c)