Integrand size = 27, antiderivative size = 270 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=-\frac {x^2 (1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2}}{a^2 c \sqrt {c-a^2 c x^2}}+\frac {(1-a x)^{\frac {1}{2} (-1-n)} (1+a x)^{\frac {1}{2} (-1+n)} \left (2+2 n+n^2-a n (3+2 n) x\right ) \sqrt {1-a^2 x^2}}{a^4 c \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {2^{\frac {1}{2} (-1+n)} n (1-a x)^{\frac {3-n}{2}} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3-n}{2},\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{a^4 c (3-n) \sqrt {c-a^2 c x^2}} \]
-x^2*(-a*x+1)^(-1/2-1/2*n)*(a*x+1)^(-1/2+1/2*n)*(-a^2*x^2+1)^(1/2)/a^2/c/( -a^2*c*x^2+c)^(1/2)+(-a*x+1)^(-1/2-1/2*n)*(a*x+1)^(-1/2+1/2*n)*(2+2*n+n^2- a*n*(3+2*n)*x)*(-a^2*x^2+1)^(1/2)/a^4/c/(-n^2+1)/(-a^2*c*x^2+c)^(1/2)-2^(- 1/2+1/2*n)*n*(-a*x+1)^(3/2-1/2*n)*hypergeom([3/2-1/2*n, 3/2-1/2*n],[5/2-1/ 2*n],-1/2*a*x+1/2)*(-a^2*x^2+1)^(1/2)/a^4/c/(3-n)/(-a^2*c*x^2+c)^(1/2)
Time = 0.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.69 \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {(1-a x)^{-\frac {1}{2}-\frac {n}{2}} \sqrt {1-a^2 x^2} \left (-4 a^4 x^2 (1+a x)^{\frac {1}{2} (-1+n)}+\frac {4 a^2 (1+a x)^{\frac {1}{2} (-1+n)} \left (-2+n^2 (-1+2 a x)+n (-2+3 a x)\right )}{-1+n^2}+\frac {2^{\frac {3+n}{2}} a^2 n (-1+a x)^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},\frac {5}{2}-\frac {n}{2},\frac {1}{2}-\frac {a x}{2}\right )}{-3+n}\right )}{4 a^6 c \sqrt {c-a^2 c x^2}} \]
((1 - a*x)^(-1/2 - n/2)*Sqrt[1 - a^2*x^2]*(-4*a^4*x^2*(1 + a*x)^((-1 + n)/ 2) + (4*a^2*(1 + a*x)^((-1 + n)/2)*(-2 + n^2*(-1 + 2*a*x) + n*(-2 + 3*a*x) ))/(-1 + n^2) + (2^((3 + n)/2)*a^2*n*(-1 + a*x)^2*Hypergeometric2F1[3/2 - n/2, 3/2 - n/2, 5/2 - n/2, 1/2 - (a*x)/2])/(-3 + n)))/(4*a^6*c*Sqrt[c - a^ 2*c*x^2])
Time = 0.63 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6703, 6700, 111, 25, 162, 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.
| \(\displaystyle \int \frac {x^3 e^{n \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 6700 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \int x^3 (1-a x)^{\frac {1}{2} (-n-3)} (a x+1)^{\frac {n-3}{2}}dx}{c \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (-\frac {\int -x (1-a x)^{\frac {1}{2} (-n-3)} (a x+1)^{\frac {n-3}{2}} (a n x+2)dx}{a^2}-\frac {x^2 (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1}{2} (-n-1)}}{a^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\int x (1-a x)^{\frac {1}{2} (-n-3)} (a x+1)^{\frac {n-3}{2}} (a n x+2)dx}{a^2}-\frac {x^2 (1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{a^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 162 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\frac {n \int (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-3}{2}}dx}{a}+\frac {(a x+1)^{\frac {n-1}{2}} \left (-a (2 n+3) n x+n^2+2 n+2\right ) (1-a x)^{\frac {1}{2} (-n-1)}}{a^2 \left (1-n^2\right )}}{a^2}-\frac {x^2 (1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{a^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
| \(\Big \downarrow \) 79 |
| \(\displaystyle \frac {\sqrt {1-a^2 x^2} \left (\frac {\frac {(1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}} \left (-a (2 n+3) n x+n^2+2 n+2\right )}{a^2 \left (1-n^2\right )}-\frac {2^{\frac {n-1}{2}} n (1-a x)^{\frac {3-n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {3-n}{2},\frac {3-n}{2},\frac {5-n}{2},\frac {1}{2} (1-a x)\right )}{a^2 (3-n)}}{a^2}-\frac {x^2 (1-a x)^{\frac {1}{2} (-n-1)} (a x+1)^{\frac {n-1}{2}}}{a^2}\right )}{c \sqrt {c-a^2 c x^2}}\) |
(Sqrt[1 - a^2*x^2]*(-((x^2*(1 - a*x)^((-1 - n)/2)*(1 + a*x)^((-1 + n)/2))/ a^2) + (((1 - a*x)^((-1 - n)/2)*(1 + a*x)^((-1 + n)/2)*(2 + 2*n + n^2 - a* n*(3 + 2*n)*x))/(a^2*(1 - n^2)) - (2^((-1 + n)/2)*n*(1 - a*x)^((3 - n)/2)* Hypergeometric2F1[(3 - n)/2, (3 - n)/2, (5 - n)/2, (1 - a*x)/2])/(a^2*(3 - n)))/a^2))/(c*Sqrt[c - a^2*c*x^2])
3.14.41.3.1 Defintions of rubi rules used
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]))
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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e *g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b *c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d *(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( b^2*(b*c - a*d)^2*(m + 1)*(m + 2))) Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] , x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] && !LtQ[n, -2]))
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])
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]
\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{3}}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{3} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(-a^2*c*x^2 + c)*x^3*(-(a*x + 1)/(a*x - 1))^(1/2*n)/(a^4*c^2* x^4 - 2*a^2*c^2*x^2 + c^2), x)
\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^{3} e^{n \operatorname {atanh}{\left (a x \right )}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { \frac {x^{3} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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
Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int \frac {x^3\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{{\left (c-a^2\,c\,x^2\right )}^{3/2}} \,d x \]