3.1175 \(\int e^{3 \tanh ^{-1}(a x)} x^3 (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=224 \[ \frac{a (6 p+17) x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{3}{2}-p,\frac{7}{2},a^2 x^2\right )}{10 (p+2)}-\frac{3 \left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}+\frac{7 \sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)}-\frac{a x^5 \left (c-a^2 c x^2\right )^p}{2 (p+2) \sqrt{1-a^2 x^2}}+\frac{4 \left (c-a^2 c x^2\right )^p}{a^4 (1-2 p) \sqrt{1-a^2 x^2}} \]

[Out]

(4*(c - a^2*c*x^2)^p)/(a^4*(1 - 2*p)*Sqrt[1 - a^2*x^2]) - (a*x^5*(c - a^2*c*x^2)^p)/(2*(2 + p)*Sqrt[1 - a^2*x^
2]) + (7*Sqrt[1 - a^2*x^2]*(c - a^2*c*x^2)^p)/(a^4*(1 + 2*p)) - (3*(1 - a^2*x^2)^(3/2)*(c - a^2*c*x^2)^p)/(a^4
*(3 + 2*p)) + (a*(17 + 6*p)*x^5*(c - a^2*c*x^2)^p*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/(10*(2 + p)*(
1 - a^2*x^2)^p)

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Rubi [A]  time = 0.355352, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {6153, 6148, 1652, 446, 77, 459, 364} \[ \frac{a (6 p+17) x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{5}{2},\frac{3}{2}-p;\frac{7}{2};a^2 x^2\right )}{10 (p+2)}-\frac{3 \left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}+\frac{7 \sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)}-\frac{a x^5 \left (c-a^2 c x^2\right )^p}{2 (p+2) \sqrt{1-a^2 x^2}}+\frac{4 \left (c-a^2 c x^2\right )^p}{a^4 (1-2 p) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(c - a^2*c*x^2)^p)/(a^4*(1 - 2*p)*Sqrt[1 - a^2*x^2]) - (a*x^5*(c - a^2*c*x^2)^p)/(2*(2 + p)*Sqrt[1 - a^2*x^
2]) + (7*Sqrt[1 - a^2*x^2]*(c - a^2*c*x^2)^p)/(a^4*(1 + 2*p)) - (3*(1 - a^2*x^2)^(3/2)*(c - a^2*c*x^2)^p)/(a^4
*(3 + 2*p)) + (a*(17 + 6*p)*x^5*(c - a^2*c*x^2)^p*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/(10*(2 + p)*(
1 - a^2*x^2)^p)

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]

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 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 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 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 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 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])

Rubi steps

\begin{align*} \int e^{3 \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^{3 \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)^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^3 \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^4 \left (1-a^2 x^2\right )^{-\frac{3}{2}+p} \left (3 a+a^3 x^2\right ) \, dx\\ &=-\frac{a x^5 \left (c-a^2 c x^2\right )^p}{2 (2+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 (1+3 a^2 x\right ) \, dx,x,x^2\right )+\frac{\left (a (17+6 p) \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{3}{2}+p} \, dx}{2 (2+p)}\\ &=-\frac{a x^5 \left (c-a^2 c x^2\right )^p}{2 (2+p) \sqrt{1-a^2 x^2}}+\frac{a (17+6 p) 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{3}{2}-p;\frac{7}{2};a^2 x^2\right )}{10 (2+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^2}-\frac{7 \left (1-a^2 x\right )^{-\frac{1}{2}+p}}{a^2}+\frac{3 \left (1-a^2 x\right )^{\frac{1}{2}+p}}{a^2}\right ) \, dx,x,x^2\right )\\ &=\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 \, _2F_1\left (\frac{5}{2},\frac{3}{2}-p;\frac{7}{2};a^2 x^2\right )}{10 (2+p)}\\ \end{align*}

Mathematica [A]  time = 0.202111, size = 176, normalized size = 0.79 \[ \frac{\left (c-a^2 c x^2\right )^p \left (21 a^5 x^5 \left (1-a^2 x^2\right )^{-p} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{3}{2}-p,\frac{7}{2},a^2 x^2\right )+5 a^7 x^7 \left (1-a^2 x^2\right )^{-p} \text{Hypergeometric2F1}\left (\frac{7}{2},\frac{3}{2}-p,\frac{9}{2},a^2 x^2\right )-\frac{105 \left (1-a^2 x^2\right )^{3/2}}{2 p+3}+\frac{245 \sqrt{1-a^2 x^2}}{2 p+1}+\frac{140}{(1-2 p) \sqrt{1-a^2 x^2}}\right )}{35 a^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((c - a^2*c*x^2)^p*(140/((1 - 2*p)*Sqrt[1 - a^2*x^2]) + (245*Sqrt[1 - a^2*x^2])/(1 + 2*p) - (105*(1 - a^2*x^2)
^(3/2))/(3 + 2*p) + (21*a^5*x^5*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/(1 - a^2*x^2)^p + (5*a^7*x^7*Hy
pergeometric2F1[7/2, 3/2 - p, 9/2, a^2*x^2])/(1 - a^2*x^2)^p))/(35*a^4)

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Maple [F]  time = 0.411, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ) ^{3}{x}^{3} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (a^{2} c^{p}{\left (2 \, p - 1\right )} x^{2} + 2 \, c^{p}\right )}{\left (-a^{2} x^{2} + 1\right )}^{p}}{\sqrt{-a^{2} x^{2} + 1}{\left (4 \, p^{2} - 1\right )} a^{4}} - \int \frac{{\left (a^{3} c^{p} x^{6} + 3 \, a^{2} c^{p} x^{5} + 3 \, a c^{p} x^{4}\right )} e^{\left (p \log \left (a x + 1\right ) + p \log \left (-a x + 1\right )\right )}}{{\left (a^{2} x^{2} - 1\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x^{4} + x^{3}\right )} \sqrt{-a^{2} x^{2} + 1}{\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{2} - 2 \, a x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x^4 + x^3)*sqrt(-a^2*x^2 + 1)*(-a^2*c*x^2 + c)^p/(a^2*x^2 - 2*a*x + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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