3.1177 \(\int e^{3 \tanh ^{-1}(a x)} x (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=138 \[ \frac{3\ 2^{p+\frac{3}{2}} (1-a x)^{p-\frac{1}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (-p-\frac{3}{2},p-\frac{1}{2},p+\frac{1}{2},\frac{1}{2} (1-a x)\right )}{a^2 \left (-2 p^2-p+1\right )}-\frac{(a x+1)^3 \left (c-a^2 c x^2\right )^p}{2 a^2 (p+1) \sqrt{1-a^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.169421, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6153, 6148, 795, 676, 69} \[ \frac{3\ 2^{p+\frac{3}{2}} (1-a x)^{p-\frac{1}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-p-\frac{3}{2},p-\frac{1}{2};p+\frac{1}{2};\frac{1}{2} (1-a x)\right )}{a^2 \left (-2 p^2-p+1\right )}-\frac{(a x+1)^3 \left (c-a^2 c x^2\right )^p}{2 a^2 (p+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + a*x)^3*(c - a^2*c*x^2)^p)/(2*a^2*(1 + p)*Sqrt[1 - a^2*x^2]) + (3*2^(3/2 + p)*(1 - a*x)^(-1/2 + p)*(c -
a^2*c*x^2)^p*Hypergeometric2F1[-3/2 - p, -1/2 + p, 1/2 + p, (1 - a*x)/2])/(a^2*(1 - p - 2*p^2)*(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 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 676

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a^(p + 1)*d^(m - 1)*((d - e*x)/d)^
(p + 1))/(a/d + (c*x)/e)^(p + 1), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a, c, d, e, m}
, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) && GtQ[a, 0] &&  !(IGtQ[m, 0] &&
(IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.362921, size = 134, normalized size = 0.97 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (a x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{3}{2}-p,\frac{5}{2},a^2 x^2\right )+\frac{1}{5} a^3 x^5 \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{3}{2}-p,\frac{7}{2},a^2 x^2\right )+\frac{\left (\frac{3-3 a^2 x^2}{2 p+1}+\frac{4}{1-2 p}\right ) \left (1-a^2 x^2\right )^{p-\frac{1}{2}}}{a^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((c - a^2*c*x^2)^p*(((1 - a^2*x^2)^(-1/2 + p)*(4/(1 - 2*p) + (3 - 3*a^2*x^2)/(1 + 2*p)))/a^2 + a*x^3*Hypergeom
etric2F1[3/2, 3/2 - p, 5/2, a^2*x^2] + (a^3*x^5*Hypergeometric2F1[5/2, 3/2 - p, 7/2, a^2*x^2])/5))/(1 - a^2*x^
2)^p

________________________________________________________________________________________

Maple [F]  time = 0.401, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ) ^{3}x \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*(-a^2*c*x^2+c)^p,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (-a^{2} x^{2} + 1\right )}^{p} c^{p}}{\sqrt{-a^{2} x^{2} + 1} a^{2}{\left (2 \, p - 1\right )}} - \int \frac{{\left (a^{3} c^{p} x^{4} + 3 \, a^{2} c^{p} x^{3} + 3 \, a c^{p} x^{2}\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*(-a^2*c*x^2+c)^p,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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*(-a**2*c*x**2+c)**p,x)

[Out]

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

________________________________________________________________________________________

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}{{\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*(-a^2*c*x^2+c)^p,x, algorithm="giac")

[Out]

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