Integrand size = 27, antiderivative size = 183 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {2 x^{1+m} (1+a x)}{3 \left (c-a^2 c x^2\right )^{3/2}}+\frac {(1-2 m) x^{1+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{3 c (1+m) \sqrt {c-a^2 c x^2}}+\frac {2 a (1-m) x^{2+m} \sqrt {1-a^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{3 c (2+m) \sqrt {c-a^2 c x^2}} \]
2/3*x^(1+m)*(a*x+1)/(-a^2*c*x^2+c)^(3/2)+1/3*(1-2*m)*x^(1+m)*hypergeom([3/ 2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)*(-a^2*x^2+1)^(1/2)/c/(1+m)/(-a^2*c*x^2+ c)^(1/2)+2/3*a*(1-m)*x^(2+m)*hypergeom([3/2, 1+1/2*m],[2+1/2*m],a^2*x^2)*( -a^2*x^2+1)^(1/2)/c/(2+m)/(-a^2*c*x^2+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.25 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\frac {x^{1+m} \left (\sqrt {-1+a x} \sqrt {1+a x} \sqrt {c-a c x} \operatorname {AppellF1}\left (1+m,\frac {1}{2},-\frac {1}{2},2+m,-a x,a x\right )+(-1+a x) \sqrt {-c (1+a x)} \left (\operatorname {AppellF1}\left (1+m,\frac {1}{2},-\frac {1}{2},2+m,a x,-a x\right )+2 \operatorname {AppellF1}\left (1+m,\frac {3}{2},-\frac {1}{2},2+m,a x,-a x\right )+4 \operatorname {AppellF1}\left (1+m,\frac {5}{2},-\frac {1}{2},2+m,a x,-a x\right )\right )\right )}{8 c^2 (1+m) \sqrt {-(-1+a x)^2} \sqrt {1+a x}} \]
(x^(1 + m)*(Sqrt[-1 + a*x]*Sqrt[1 + a*x]*Sqrt[c - a*c*x]*AppellF1[1 + m, 1 /2, -1/2, 2 + m, -(a*x), a*x] + (-1 + a*x)*Sqrt[-(c*(1 + a*x))]*(AppellF1[ 1 + m, 1/2, -1/2, 2 + m, a*x, -(a*x)] + 2*AppellF1[1 + m, 3/2, -1/2, 2 + m , a*x, -(a*x)] + 4*AppellF1[1 + m, 5/2, -1/2, 2 + m, a*x, -(a*x)])))/(8*c^ 2*(1 + m)*Sqrt[-(-1 + a*x)^2]*Sqrt[1 + a*x])
Time = 0.45 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6701, 558, 25, 557, 279, 278}
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.
| \(\displaystyle \int \frac {x^m e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6701 |
| \(\displaystyle c \int \frac {x^m (a x+1)^2}{\left (c-a^2 c x^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 558 |
| \(\displaystyle c \left (\frac {2 (a x+1) x^{m+1}}{3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {\int -\frac {x^m (-2 m+2 a (1-m) x+1)}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle c \left (\frac {\int \frac {x^m (-2 m+2 a (1-m) x+1)}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {2 (a x+1) x^{m+1}}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 557 |
| \(\displaystyle c \left (\frac {2 a (1-m) \int \frac {x^{m+1}}{\left (c-a^2 c x^2\right )^{3/2}}dx+(1-2 m) \int \frac {x^m}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {2 (a x+1) x^{m+1}}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle c \left (\frac {\frac {2 a (1-m) \sqrt {1-a^2 x^2} \int \frac {x^{m+1}}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}+\frac {(1-2 m) \sqrt {1-a^2 x^2} \int \frac {x^m}{\left (1-a^2 x^2\right )^{3/2}}dx}{c \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 (a x+1) x^{m+1}}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle c \left (\frac {\frac {(1-2 m) \sqrt {1-a^2 x^2} x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{c (m+1) \sqrt {c-a^2 c x^2}}+\frac {2 a (1-m) \sqrt {1-a^2 x^2} x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{c (m+2) \sqrt {c-a^2 c x^2}}}{3 c}+\frac {2 (a x+1) x^{m+1}}{3 c \left (c-a^2 c x^2\right )^{3/2}}\right )\) |
c*((2*x^(1 + m)*(1 + a*x))/(3*c*(c - a^2*c*x^2)^(3/2)) + (((1 - 2*m)*x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, a^2*x^ 2])/(c*(1 + m)*Sqrt[c - a^2*c*x^2]) + (2*a*(1 - m)*x^(2 + m)*Sqrt[1 - a^2* x^2]*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(c*(2 + m)*Sqr t[c - a^2*c*x^2]))/(3*c))
3.12.35.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym bol] :> Simp[c Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e Int[(e*x)^( m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* (f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt Q[n, 1] && !IntegerQ[m] && LtQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ Symbol] :> Simp[c^(n/2) Int[x^m*(c + d*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/2, 0]
\[\int \frac {\left (a x +1\right )^{2} x^{m}}{\left (-a^{2} x^{2}+1\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a^{2} x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=- \int \frac {x^{m}}{- a^{3} c x^{3} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt {- a^{2} c x^{2} + c} + a c x \sqrt {- a^{2} c x^{2} + c} - c \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x x^{m}}{- a^{3} c x^{3} \sqrt {- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt {- a^{2} c x^{2} + c} + a c x \sqrt {- a^{2} c x^{2} + c} - c \sqrt {- a^{2} c x^{2} + c}}\, dx \]
-Integral(x**m/(-a**3*c*x**3*sqrt(-a**2*c*x**2 + c) + a**2*c*x**2*sqrt(-a* *2*c*x**2 + c) + a*c*x*sqrt(-a**2*c*x**2 + c) - c*sqrt(-a**2*c*x**2 + c)), x) - Integral(a*x*x**m/(-a**3*c*x**3*sqrt(-a**2*c*x**2 + c) + a**2*c*x**2 *sqrt(-a**2*c*x**2 + c) + a*c*x*sqrt(-a**2*c*x**2 + c) - c*sqrt(-a**2*c*x* *2 + c)), x)
\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a^{2} x^{2} - 1\right )}} \,d x } \]
\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} {\left (a^{2} x^{2} - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^{3/2}} \, dx=\int -\frac {x^m\,{\left (a\,x+1\right )}^2}{{\left (c-a^2\,c\,x^2\right )}^{3/2}\,\left (a^2\,x^2-1\right )} \,d x \]