3.1179 \(\int \frac{e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^p}{x} \, dx\)

Optimal. Leaf size=193 \[ \frac{a (6 p+1) x \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{2}-p,\frac{3}{2},a^2 x^2\right )}{2 p}-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p \text{Hypergeometric2F1}\left (1,p+\frac{1}{2},p+\frac{3}{2},1-a^2 x^2\right )}{2 p+1}-\frac{a x \left (c-a^2 c x^2\right )^p}{2 p \sqrt{1-a^2 x^2}}+\frac{4 \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.291826, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {6153, 6148, 1652, 446, 79, 65, 388, 245} \[ \frac{a (6 p+1) x \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac{1}{2},\frac{3}{2}-p;\frac{3}{2};a^2 x^2\right )}{2 p}-\frac{\sqrt{1-a^2 x^2} \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,p+\frac{1}{2};p+\frac{3}{2};1-a^2 x^2\right )}{2 p+1}-\frac{a x \left (c-a^2 c x^2\right )^p}{2 p \sqrt{1-a^2 x^2}}+\frac{4 \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(c - a^2*c*x^2)^p)/((1 - 2*p)*Sqrt[1 - a^2*x^2]) - (a*x*(c - a^2*c*x^2)^p)/(2*p*Sqrt[1 - a^2*x^2]) + (a*(1
+ 6*p)*x*(c - a^2*c*x^2)^p*Hypergeometric2F1[1/2, 3/2 - p, 3/2, a^2*x^2])/(2*p*(1 - a^2*x^2)^p) - (Sqrt[1 - a^
2*x^2]*(c - a^2*c*x^2)^p*Hypergeometric2F1[1, 1/2 + p, 3/2 + p, 1 - a^2*x^2])/(1 + 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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 65

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

Rule 388

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

Rule 245

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

Rubi steps

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

Mathematica [A]  time = 0.223485, size = 159, normalized size = 0.82 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (-\frac{\left (1-a^2 x^2\right )^{p-\frac{1}{2}} \text{Hypergeometric2F1}\left (1,p-\frac{1}{2},p+\frac{1}{2},1-a^2 x^2\right )}{2 \left (p-\frac{1}{2}\right )}+3 a x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{2}-p,\frac{3}{2},a^2 x^2\right )+\frac{1}{3} a^3 x^3 \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{3}{2}-p,\frac{5}{2},a^2 x^2\right )+\frac{3 \left (1-a^2 x^2\right )^{p-\frac{1}{2}}}{1-2 p}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.396, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{3} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{p}}{x} \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)*(-a^2*c*x^2+c)^p/x,x)

[Out]

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

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Maxima [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}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}\,{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)*(-a^2*c*x^2+c)^p/x,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

[Out]

Integral((-c*(a*x - 1)*(a*x + 1))**p*(a*x + 1)**3/(x*(-(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}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}\,{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)*(-a^2*c*x^2+c)^p/x,x, algorithm="giac")

[Out]

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