Integrand size = 25, antiderivative size = 251 \[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=-\frac {3 x^{1+m} \left (c-a^2 c x^2\right )^p}{(m+2 p) \sqrt {1-a^2 x^2}}-\frac {a x^{2+m} \left (c-a^2 c x^2\right )^p}{(1+m+2 p) \sqrt {1-a^2 x^2}}+\frac {(3+4 m+2 p) x^{1+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {3}{2}-p,\frac {3+m}{2},a^2 x^2\right )}{(1+m) (m+2 p)}+\frac {a (5+4 m+6 p) x^{2+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},\frac {3}{2}-p,\frac {4+m}{2},a^2 x^2\right )}{(2+m) (1+m+2 p)} \]
(3+4*m+2*p)*x^(1+m)*(-a^2*c*x^2+c)^p*hypergeom([3/2-p, 1/2+1/2*m],[3/2+1/2 *m],a^2*x^2)/(1+m)/(m+2*p)/((-a^2*x^2+1)^p)+a*(5+4*m+6*p)*x^(2+m)*(-a^2*c* x^2+c)^p*hypergeom([1+1/2*m, 3/2-p],[2+1/2*m],a^2*x^2)/(2+m)/(1+m+2*p)/((- a^2*x^2+1)^p)-3*x^(1+m)*(-a^2*c*x^2+c)^p/(m+2*p)/(-a^2*x^2+1)^(1/2)-a*x^(2 +m)*(-a^2*c*x^2+c)^p/(1+m+2*p)/(-a^2*x^2+1)^(1/2)
Time = 0.14 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.74 \[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=x^{1+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {3}{2}-p,\frac {3+m}{2},a^2 x^2\right )}{1+m}+a x \left (\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},\frac {3}{2}-p,\frac {4+m}{2},a^2 x^2\right )}{2+m}+a x \left (\frac {3 \operatorname {Hypergeometric2F1}\left (\frac {3+m}{2},\frac {3}{2}-p,\frac {5+m}{2},a^2 x^2\right )}{3+m}+\frac {a x \operatorname {Hypergeometric2F1}\left (\frac {4+m}{2},\frac {3}{2}-p,\frac {6+m}{2},a^2 x^2\right )}{4+m}\right )\right )\right ) \]
(x^(1 + m)*(c - a^2*c*x^2)^p*(Hypergeometric2F1[(1 + m)/2, 3/2 - p, (3 + m )/2, a^2*x^2]/(1 + m) + a*x*((3*Hypergeometric2F1[(2 + m)/2, 3/2 - p, (4 + m)/2, a^2*x^2])/(2 + m) + a*x*((3*Hypergeometric2F1[(3 + m)/2, 3/2 - p, ( 5 + m)/2, a^2*x^2])/(3 + m) + (a*x*Hypergeometric2F1[(4 + m)/2, 3/2 - p, ( 6 + m)/2, a^2*x^2])/(4 + m)))))/(1 - a^2*x^2)^p
Time = 0.74 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {6703, 6698, 559, 25, 2340, 25, 27, 557, 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.
\(\displaystyle \int x^m e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 6703 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \int e^{3 \text {arctanh}(a x)} x^m \left (1-a^2 x^2\right )^pdx\) |
\(\Big \downarrow \) 6698 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \int x^m (a x+1)^3 \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx\) |
\(\Big \downarrow \) 559 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (-\frac {\int -x^m \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (3 (m+2 p+1) x^2 a^4+(4 m+6 p+5) x a^3+(m+2 p+1) a^2\right )dx}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\int x^m \left (1-a^2 x^2\right )^{p-\frac {3}{2}} \left (3 (m+2 p+1) x^2 a^4+(4 m+6 p+5) x a^3+(m+2 p+1) a^2\right )dx}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {-\frac {\int -a^4 x^m ((m+2 p+1) (4 m+2 p+3)+a (m+2 p) (4 m+6 p+5) x) \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx}{a^2 (m+2 p)}-\frac {3 a^2 (m+2 p+1) x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p}}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\frac {\int a^4 x^m ((m+2 p+1) (4 m+2 p+3)+a (m+2 p) (4 m+6 p+5) x) \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx}{a^2 (m+2 p)}-\frac {3 a^2 (m+2 p+1) x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p}}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\frac {a^2 \int x^m ((m+2 p+1) (4 m+2 p+3)+a (m+2 p) (4 m+6 p+5) x) \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx}{m+2 p}-\frac {3 a^2 (m+2 p+1) x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p}}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 557 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\frac {a^2 \left (a (m+2 p) (4 m+6 p+5) \int x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx+(m+2 p+1) (4 m+2 p+3) \int x^m \left (1-a^2 x^2\right )^{p-\frac {3}{2}}dx\right )}{m+2 p}-\frac {3 a^2 (m+2 p+1) x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p}}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {\frac {a^2 \left (\frac {(m+2 p+1) (4 m+2 p+3) x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {3}{2}-p,\frac {m+3}{2},a^2 x^2\right )}{m+1}+\frac {a (m+2 p) (4 m+6 p+5) x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {m+2}{2},\frac {3}{2}-p,\frac {m+4}{2},a^2 x^2\right )}{m+2}\right )}{m+2 p}-\frac {3 a^2 (m+2 p+1) x^{m+1} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p}}{a^2 (m+2 p+1)}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{m+2 p+1}\right )\) |
((c - a^2*c*x^2)^p*(-((a*x^(2 + m)*(1 - a^2*x^2)^(-1/2 + p))/(1 + m + 2*p) ) + ((-3*a^2*(1 + m + 2*p)*x^(1 + m)*(1 - a^2*x^2)^(-1/2 + p))/(m + 2*p) + (a^2*(((1 + m + 2*p)*(3 + 4*m + 2*p)*x^(1 + m)*Hypergeometric2F1[(1 + m)/ 2, 3/2 - p, (3 + m)/2, a^2*x^2])/(1 + m) + (a*(m + 2*p)*(5 + 4*m + 6*p)*x^ (2 + m)*Hypergeometric2F1[(2 + m)/2, 3/2 - p, (4 + m)/2, a^2*x^2])/(2 + m) ))/(m + 2*p))/(a^2*(1 + m + 2*p))))/(1 - a^2*x^2)^p
3.12.74.3.1 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[((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[((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] :> Simp[d^n*(e*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*e^(n - 1)*( m + n + 2*p + 1))), x] + Simp[1/(b*(m + n + 2*p + 1)) Int[(e*x)^m*(a + b* x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1 )*x^n - a*d^n*(m + n - 1)*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && IGtQ[n, 1] && !IntegerQ[m] && NeQ[m + n + 2*p + 1, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
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]
\[\int \frac {\left (a x +1\right )^{3} x^{m} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}d x\]
\[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=\text {Timed out} \]
\[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=\text {Exception raised: TypeError} \]
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. \[ \int e^{3 \text {arctanh}(a x)} x^m \left (c-a^2 c x^2\right )^p \, dx=\int \frac {x^m\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]