Integrand size = 25, antiderivative size = 209 \[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=-\frac {n x^2 (1-a x)^{-n/2} (1+a x)^{n/2}}{6 a^3 c}-\frac {x^3 (1-a x)^{-n/2} (1+a x)^{n/2}}{3 a^2 c}+\frac {(1-a x)^{-n/2} (1+a x)^{n/2} \left (6+8 n+n^2+n^3-a n \left (6+n^2\right ) x\right )}{6 a^5 c n}+\frac {2^{-1+\frac {n}{2}} n \left (8+n^2\right ) (1-a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},1-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a x)\right )}{3 a^5 c (2-n)} \]
-1/6*n*x^2*(a*x+1)^(1/2*n)/a^3/c/((-a*x+1)^(1/2*n))-1/3*x^3*(a*x+1)^(1/2*n )/a^2/c/((-a*x+1)^(1/2*n))+1/6*(a*x+1)^(1/2*n)*(6+8*n+n^2+n^3-a*n*(n^2+6)* x)/a^5/c/n/((-a*x+1)^(1/2*n))+1/3*2^(-1+1/2*n)*n*(n^2+8)*(-a*x+1)^(1-1/2*n )*hypergeom([1-1/2*n, 1-1/2*n],[2-1/2*n],-1/2*a*x+1/2)/a^5/c/(2-n)
Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.82 \[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=\frac {(1-a x)^{-n/2} \left (2^{1+\frac {n}{2}} n \left (2-2 n+n^2\right ) (-1+a x) \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a x)\right )-(-2+n) \left ((1+a x)^{n/2} \left (-6+(n+a n x)^2+2 n \left (3+2 a x+a^3 x^3\right )\right )-2^{2+\frac {n}{2}} n (3+n) \operatorname {Hypergeometric2F1}\left (-\frac {n}{2},-\frac {n}{2},1-\frac {n}{2},\frac {1}{2} (1-a x)\right )\right )\right )}{6 a^5 c (-2+n) n} \]
(2^(1 + n/2)*n*(2 - 2*n + n^2)*(-1 + a*x)*Hypergeometric2F1[1 - n/2, -1/2* n, 2 - n/2, (1 - a*x)/2] - (-2 + n)*((1 + a*x)^(n/2)*(-6 + (n + a*n*x)^2 + 2*n*(3 + 2*a*x + a^3*x^3)) - 2^(2 + n/2)*n*(3 + n)*Hypergeometric2F1[-1/2 *n, -1/2*n, 1 - n/2, (1 - a*x)/2]))/(6*a^5*c*(-2 + n)*n*(1 - a*x)^(n/2))
Time = 0.45 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6700, 111, 25, 170, 25, 27, 160, 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.
\(\displaystyle \int \frac {x^4 e^{n \text {arctanh}(a x)}}{c-a^2 c x^2} \, dx\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {\int x^4 (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}}dx}{c}\) |
\(\Big \downarrow \) 111 |
\(\displaystyle \frac {-\frac {\int -x^2 (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}} (a n x+3)dx}{3 a^2}-\frac {x^3 (a x+1)^{n/2} (1-a x)^{-n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int x^2 (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}} (a n x+3)dx}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 170 |
\(\displaystyle \frac {\frac {-\frac {\int -a x (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}} \left (2 n+a \left (n^2+6\right ) x\right )dx}{2 a^2}-\frac {n x^2 (a x+1)^{n/2} (1-a x)^{-n/2}}{2 a}}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\frac {\int a x (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}} \left (2 n+a \left (n^2+6\right ) x\right )dx}{2 a^2}-\frac {n x^2 (1-a x)^{-n/2} (a x+1)^{n/2}}{2 a}}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int x (1-a x)^{-\frac {n}{2}-1} (a x+1)^{\frac {n-2}{2}} \left (2 n+a \left (n^2+6\right ) x\right )dx}{2 a}-\frac {n x^2 (1-a x)^{-n/2} (a x+1)^{n/2}}{2 a}}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 160 |
\(\displaystyle \frac {\frac {\frac {\frac {(1-a x)^{-n/2} (a x+1)^{n/2} \left (-a \left (n^2+6\right ) n x+n^3+n^2+8 n+6\right )}{a^2 n}-\frac {n \left (n^2+8\right ) \int (1-a x)^{-n/2} (a x+1)^{\frac {n-2}{2}}dx}{a}}{2 a}-\frac {n x^2 (1-a x)^{-n/2} (a x+1)^{n/2}}{2 a}}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {\frac {\frac {\frac {2^{n/2} n \left (n^2+8\right ) (1-a x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {2-n}{2},1-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a x)\right )}{a^2 (2-n)}+\frac {(a x+1)^{n/2} \left (-a \left (n^2+6\right ) n x+n^3+n^2+8 n+6\right ) (1-a x)^{-n/2}}{a^2 n}}{2 a}-\frac {n x^2 (1-a x)^{-n/2} (a x+1)^{n/2}}{2 a}}{3 a^2}-\frac {x^3 (1-a x)^{-n/2} (a x+1)^{n/2}}{3 a^2}}{c}\) |
(-1/3*(x^3*(1 + a*x)^(n/2))/(a^2*(1 - a*x)^(n/2)) + (-1/2*(n*x^2*(1 + a*x) ^(n/2))/(a*(1 - a*x)^(n/2)) + (((1 + a*x)^(n/2)*(6 + 8*n + n^2 + n^3 - a*n *(6 + n^2)*x))/(a^2*n*(1 - a*x)^(n/2)) + (2^(n/2)*n*(8 + n^2)*(1 - a*x)^(1 - n/2)*Hypergeometric2F1[(2 - n)/2, 1 - n/2, 2 - n/2, (1 - a*x)/2])/(a^2* (2 - n)))/(2*a))/(3*a^2))/c
3.14.11.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d*(b*c - a*d)*(m + 1))), x] + Simp[(a*d*f*h*m + b*(d* (f*g + e*h) - c*f*h*(m + 2)))/(b^2*d) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]
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])
\[\int \frac {{\mathrm e}^{n \,\operatorname {arctanh}\left (a x \right )} x^{4}}{-a^{2} c \,x^{2}+c}d x\]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=\int { -\frac {x^{4} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c} \,d x } \]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=- \frac {\int \frac {x^{4} e^{n \operatorname {atanh}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=\int { -\frac {x^{4} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c} \,d x } \]
\[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=\int { -\frac {x^{4} \left (-\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c} \,d x } \]
Timed out. \[ \int \frac {e^{n \text {arctanh}(a x)} x^4}{c-a^2 c x^2} \, dx=\int \frac {x^4\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{c-a^2\,c\,x^2} \,d x \]