3.1176 \(\int e^{3 \tanh ^{-1}(a x)} x^2 (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=222 \[ \frac {(2 p+11) x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2},\frac {3}{2}-p;\frac {5}{2};a^2 x^2\right )}{6 (p+1)}-\frac {3 x^3 \left (c-a^2 c x^2\right )^p}{2 (p+1) \sqrt {1-a^2 x^2}}-\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+3)}+\frac {5 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+1)}+\frac {4 \left (c-a^2 c x^2\right )^p}{a^3 (1-2 p) \sqrt {1-a^2 x^2}} \]

[Out]

-(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)^p/a^3/(3+2*p)+1/6*(11+2*p)*x^3*(-a^2*c*x^2+c)^p*hypergeom([3/2, 3/2-p],[5/2
],a^2*x^2)/(1+p)/((-a^2*x^2+1)^p)+4*(-a^2*c*x^2+c)^p/a^3/(1-2*p)/(-a^2*x^2+1)^(1/2)-3/2*x^3*(-a^2*c*x^2+c)^p/(
1+p)/(-a^2*x^2+1)^(1/2)+5*(-a^2*c*x^2+c)^p*(-a^2*x^2+1)^(1/2)/a^3/(1+2*p)

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6153, 6148, 1652, 459, 364, 446, 77} \[ \frac {(2 p+11) x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2},\frac {3}{2}-p;\frac {5}{2};a^2 x^2\right )}{6 (p+1)}-\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+3)}+\frac {5 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (2 p+1)}-\frac {3 x^3 \left (c-a^2 c x^2\right )^p}{2 (p+1) \sqrt {1-a^2 x^2}}+\frac {4 \left (c-a^2 c x^2\right )^p}{a^3 (1-2 p) \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*x^2*(c - a^2*c*x^2)^p,x]

[Out]

(4*(c - a^2*c*x^2)^p)/(a^3*(1 - 2*p)*Sqrt[1 - a^2*x^2]) - (3*x^3*(c - a^2*c*x^2)^p)/(2*(1 + p)*Sqrt[1 - a^2*x^
2]) + (5*Sqrt[1 - a^2*x^2]*(c - a^2*c*x^2)^p)/(a^3*(1 + 2*p)) - ((1 - a^2*x^2)^(3/2)*(c - a^2*c*x^2)^p)/(a^3*(
3 + 2*p)) + ((11 + 2*p)*x^3*(c - a^2*c*x^2)^p*Hypergeometric2F1[3/2, 3/2 - p, 5/2, a^2*x^2])/(6*(1 + p)*(1 - a
^2*x^2)^p)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((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]))))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 1652

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[x^m*Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2] && IGtQ[m, -2] &&  !
IntegerQ[2*p]

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[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]) &&  !IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} x^2 \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int e^{3 \tanh ^{-1}(a x)} x^2 \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 (1+a x)^3 \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \left (1+3 a^2 x^2\right ) \, dx+\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^3 \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \left (3 a+a^3 x^2\right ) \, dx\\ &=-\frac {3 x^3 \left (c-a^2 c x^2\right )^p}{2 (1+p) \sqrt {1-a^2 x^2}}+\frac {1}{2} \left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \operatorname {Subst}\left (\int x \left (1-a^2 x\right )^{-\frac {3}{2}+p} \left (3 a+a^3 x\right ) \, dx,x,x^2\right )+\frac {\left ((11+2 p) \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^2 \left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \, dx}{2 (1+p)}\\ &=-\frac {3 x^3 \left (c-a^2 c x^2\right )^p}{2 (1+p) \sqrt {1-a^2 x^2}}+\frac {(11+2 p) x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2},\frac {3}{2}-p;\frac {5}{2};a^2 x^2\right )}{6 (1+p)}+\frac {1}{2} \left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \operatorname {Subst}\left (\int \left (\frac {4 \left (1-a^2 x\right )^{-\frac {3}{2}+p}}{a}-\frac {5 \left (1-a^2 x\right )^{-\frac {1}{2}+p}}{a}+\frac {\left (1-a^2 x\right )^{\frac {1}{2}+p}}{a}\right ) \, dx,x,x^2\right )\\ &=\frac {4 \left (c-a^2 c x^2\right )^p}{a^3 (1-2 p) \sqrt {1-a^2 x^2}}-\frac {3 x^3 \left (c-a^2 c x^2\right )^p}{2 (1+p) \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^3 (1+2 p)}-\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^3 (3+2 p)}+\frac {(11+2 p) x^3 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2},\frac {3}{2}-p;\frac {5}{2};a^2 x^2\right )}{6 (1+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 179, normalized size = 0.81 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {3}{5} a^2 x^5 \, _2F_1\left (\frac {5}{2},\frac {3}{2}-p;\frac {7}{2};a^2 x^2\right )+\frac {1}{3} x^3 \, _2F_1\left (\frac {3}{2},\frac {3}{2}-p;\frac {5}{2};a^2 x^2\right )+\frac {\left (a^4 x^4-4 a^2 p^2 x^2 \left (a^2 x^2+3\right )-4 p \left (5 a^2 x^2+3\right )+13 a^2 x^2-26\right ) \left (1-a^2 x^2\right )^{p-\frac {1}{2}}}{a^3 (2 p-1) (2 p+1) (2 p+3)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(3*ArcTanh[a*x])*x^2*(c - a^2*c*x^2)^p,x]

[Out]

((c - a^2*c*x^2)^p*(((1 - a^2*x^2)^(-1/2 + p)*(-26 + 13*a^2*x^2 + a^4*x^4 - 4*a^2*p^2*x^2*(3 + a^2*x^2) - 4*p*
(3 + 5*a^2*x^2)))/(a^3*(-1 + 2*p)*(1 + 2*p)*(3 + 2*p)) + (x^3*Hypergeometric2F1[3/2, 3/2 - p, 5/2, a^2*x^2])/3
+ (3*a^2*x^5*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/5))/(1 - a^2*x^2)^p

________________________________________________________________________________________

fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + x^{2}\right )} {\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{2} - 2 \, a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^2*(-a^2*c*x^2+c)^p,x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*x^2 + 1)*(a*x^3 + x^2)*(-a^2*c*x^2 + c)^p/(a^2*x^2 - 2*a*x + 1), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^2*(-a^2*c*x^2+c)^p,x, algorithm="giac")

[Out]

integrate((a*x + 1)^3*(-a^2*c*x^2 + c)^p*x^2/(-a^2*x^2 + 1)^(3/2), x)

________________________________________________________________________________________

maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right )^{3} x^{2} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^2*(-a^2*c*x^2+c)^p,x)

[Out]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^2*(-a^2*c*x^2+c)^p,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{2}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*x^2*(-a^2*c*x^2+c)^p,x, algorithm="maxima")

[Out]

integrate((a*x + 1)^3*(-a^2*c*x^2 + c)^p*x^2/(-a^2*x^2 + 1)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (c-a^2\,c\,x^2\right )}^p\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(c - a^2*c*x^2)^p*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)

[Out]

int((x^2*(c - a^2*c*x^2)^p*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*x**2*(-a**2*c*x**2+c)**p,x)

[Out]

Integral(x**2*(-c*(a*x - 1)*(a*x + 1))**p*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**(3/2), x)

________________________________________________________________________________________