Integrand size = 25, antiderivative size = 224 \[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\frac {4 \left (c-a^2 c x^2\right )^p}{a^4 (1-2 p) \sqrt {1-a^2 x^2}}-\frac {a x^5 \left (c-a^2 c x^2\right )^p}{2 (2+p) \sqrt {1-a^2 x^2}}+\frac {7 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (1+2 p)}-\frac {3 \left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (3+2 p)}+\frac {a (17+6 p) x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3}{2}-p,\frac {7}{2},a^2 x^2\right )}{10 (2+p)} \]
-3*(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^p/a^4/(3+2*p)+1/10*a*(17+6*p)*x^5*(-a ^2*c*x^2+c)^p*hypergeom([5/2, 3/2-p],[7/2],a^2*x^2)/(2+p)/((-a^2*x^2+1)^p) +4*(-a^2*c*x^2+c)^p/a^4/(1-2*p)/(-a^2*x^2+1)^(1/2)-1/2*a*x^5*(-a^2*c*x^2+c )^p/(2+p)/(-a^2*x^2+1)^(1/2)+7*(-a^2*c*x^2+c)^p*(-a^2*x^2+1)^(1/2)/a^4/(1+ 2*p)
Time = 0.18 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.77 \[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {4 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p}}{a^4 (1-2 p)}+\frac {7 \left (1-a^2 x^2\right )^{\frac {1}{2}+p}}{a^4 (1+2 p)}-\frac {3 \left (1-a^2 x^2\right )^{\frac {3}{2}+p}}{a^4 (3+2 p)}+\frac {3}{5} a x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3}{2}-p,\frac {7}{2},a^2 x^2\right )+\frac {1}{7} a^3 x^7 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {3}{2}-p,\frac {9}{2},a^2 x^2\right )\right ) \]
((c - a^2*c*x^2)^p*((4*(1 - a^2*x^2)^(-1/2 + p))/(a^4*(1 - 2*p)) + (7*(1 - a^2*x^2)^(1/2 + p))/(a^4*(1 + 2*p)) - (3*(1 - a^2*x^2)^(3/2 + p))/(a^4*(3 + 2*p)) + (3*a*x^5*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/5 + (a^ 3*x^7*Hypergeometric2F1[7/2, 3/2 - p, 9/2, a^2*x^2])/7))/(1 - a^2*x^2)^p
Time = 0.59 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.83, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6703, 6698, 543, 354, 86, 363, 278, 2009}
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.
| \(\displaystyle \int x^3 e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 6703 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \int e^{3 \text {arctanh}(a x)} x^3 \left (1-a^2 x^2\right )^pdx\) |
| \(\Big \downarrow \) 6698 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \int x^3 (a x+1)^3 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx\) |
| \(\Big \downarrow \) 543 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\int x^3 \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (3 a^2 x^2+1\right )dx+\int x^4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (x^2 a^3+3 a\right )dx\right )\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {1}{2} \int x^2 \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (3 a^2 x^2+1\right )dx^2+\int x^4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (x^2 a^3+3 a\right )dx\right )\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {1}{2} \int \left (\frac {4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}}{a^2}-\frac {7 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}+\frac {3 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2+\int x^4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (x^2 a^3+3 a\right )dx\right )\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {1}{2} \int \left (\frac {4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}}{a^2}-\frac {7 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}+\frac {3 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2+\frac {a (6 p+17) \int x^4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx}{2 (p+2)}-\frac {a x^5 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{2 (p+2)}\right )\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {1}{2} \int \left (\frac {4 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}}{a^2}-\frac {7 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^2}+\frac {3 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^2}\right )dx^2+\frac {a (6 p+17) x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3}{2}-p,\frac {7}{2},a^2 x^2\right )}{10 (p+2)}-\frac {a x^5 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{2 (p+2)}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {a (6 p+17) x^5 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3}{2}-p,\frac {7}{2},a^2 x^2\right )}{10 (p+2)}-\frac {a x^5 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{2 (p+2)}+\frac {1}{2} \left (\frac {8 \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^4 (1-2 p)}+\frac {14 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}-\frac {6 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}\right )\right )\) |
((c - a^2*c*x^2)^p*(-1/2*(a*x^5*(1 - a^2*x^2)^(-1/2 + p))/(2 + p) + ((8*(1 - a^2*x^2)^(-1/2 + p))/(a^4*(1 - 2*p)) + (14*(1 - a^2*x^2)^(1/2 + p))/(a^ 4*(1 + 2*p)) - (6*(1 - a^2*x^2)^(3/2 + p))/(a^4*(3 + 2*p)))/2 + (a*(17 + 6 *p)*x^5*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/(10*(2 + p))))/(1 - a^2*x^2)^p
3.12.75.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ (n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] && !IntegerQ[2*p] && !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0]) && IGtQ[(n + 1)/2, 0] && !IntegerQ[p - n/2]
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]
\[\int \frac {\left (a x +1\right )^{3} x^{3} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}d x\]
\[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\int \frac {x^{3} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
-(a^2*c^p*(2*p - 1)*x^2 + 2*c^p)*(-a^2*x^2 + 1)^p/(sqrt(-a^2*x^2 + 1)*(4*p ^2 - 1)*a^4) - integrate((a^3*c^p*x^6 + 3*a^2*c^p*x^5 + 3*a*c^p*x^4)*e^(p* log(a*x + 1) + p*log(-a*x + 1))/((a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(-a*x + 1 )), x)
\[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int e^{3 \text {arctanh}(a x)} x^3 \left (c-a^2 c x^2\right )^p \, dx=\int \frac {x^3\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]