3.1222 \(\int e^{-\tanh ^{-1}(a x)} x^3 (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=134 \[ -\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6153, 6149, 764, 266, 43, 364} \[ -\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 6149

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] || G
tQ[c, 0]) && ILtQ[(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^{-\tanh ^{-1}(a x)} x^3 \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^{-\tanh ^{-1}(a x)} x^3 \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^3 (1-a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, 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 {1}{2}+p} \, dx-\left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^4 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=-\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\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 {1}{2}+p} \, dx,x,x^2\right )\\ &=-\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\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 {\left (1-a^2 x\right )^{-\frac {1}{2}+p}}{a^2}-\frac {\left (1-a^2 x\right )^{\frac {1}{2}+p}}{a^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (3+2 p)}-\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 119, normalized size = 0.89 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {1}{2} \left (\frac {2 \left (1-a^2 x^2\right )^{p+\frac {3}{2}}}{a^4 (2 p+3)}-\frac {2 \left (1-a^2 x^2\right )^{p+\frac {1}{2}}}{a^4 (2 p+1)}\right )-\frac {1}{5} a x^5 \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{3}}{a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (-a^{2} c \,x^{2}+c \right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,{\left (c-a^2\,c\,x^2\right )}^p\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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