3.1181 \(\int \frac {e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=194 \[ \frac {a^2 (9-2 p) \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,p-\frac {1}{2};p+\frac {1}{2};1-a^2 x^2\right )}{2 (1-2 p) \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}}+a^3 (7-6 p) x \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {1}{2},\frac {3}{2}-p;\frac {3}{2};a^2 x^2\right ) \]

[Out]

a^3*(7-6*p)*x*(-a^2*c*x^2+c)^p*hypergeom([1/2, 3/2-p],[3/2],a^2*x^2)/((-a^2*x^2+1)^p)-1/2*(-a^2*c*x^2+c)^p/x^2
/(-a^2*x^2+1)^(1/2)-3*a*(-a^2*c*x^2+c)^p/x/(-a^2*x^2+1)^(1/2)+1/2*a^2*(9-2*p)*(-a^2*c*x^2+c)^p*hypergeom([1, -
1/2+p],[1/2+p],-a^2*x^2+1)/(1-2*p)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 194, 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, 1807, 764, 266, 65, 245} \[ a^3 (7-6 p) x \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {1}{2},\frac {3}{2}-p;\frac {3}{2};a^2 x^2\right )+\frac {a^2 (9-2 p) \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,p-\frac {1}{2};p+\frac {1}{2};1-a^2 x^2\right )}{2 (1-2 p) \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c - a^2*c*x^2)^p/(2*x^2*Sqrt[1 - a^2*x^2]) - (3*a*(c - a^2*c*x^2)^p)/(x*Sqrt[1 - a^2*x^2]) + (a^3*(7 - 6*p)*
x*(c - a^2*c*x^2)^p*Hypergeometric2F1[1/2, 3/2 - p, 3/2, a^2*x^2])/(1 - a^2*x^2)^p + (a^2*(9 - 2*p)*(c - a^2*c
*x^2)^p*Hypergeometric2F1[1, -1/2 + p, 1/2 + p, 1 - a^2*x^2])/(2*(1 - 2*p)*Sqrt[1 - a^2*x^2])

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 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])

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 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 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
 R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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 \frac {e^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^p}{x^3} \, 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^3} \, 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^3} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \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 ) \int \frac {\left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \left (-6 a-a^2 (9-2 p) x-2 a^3 x^2\right )}{x^2} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \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 ) \int \frac {\left (a^2 (9-2 p)+2 a^3 (7-6 p) x\right ) \left (1-a^2 x^2\right )^{-\frac {3}{2}+p}}{x} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+\left (a^3 (7-6 p) \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+\frac {1}{2} \left (a^2 (9-2 p) \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}}{x} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+a^3 (7-6 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 )+\frac {1}{4} \left (a^2 (9-2 p) \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}}{x} \, dx,x,x^2\right )\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{2 x^2 \sqrt {1-a^2 x^2}}-\frac {3 a \left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+a^3 (7-6 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 )+\frac {a^2 (9-2 p) \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,-\frac {1}{2}+p;\frac {1}{2}+p;1-a^2 x^2\right )}{2 (1-2 p) \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 154, normalized size = 0.79 \[ a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (a \left (\frac {\left (1-a^2 x^2\right )^{p-\frac {1}{2}} \left (3 \, _2F_1\left (1,p-\frac {1}{2};p+\frac {1}{2};1-a^2 x^2\right )+\, _2F_1\left (2,p-\frac {1}{2};p+\frac {1}{2};1-a^2 x^2\right )\right )}{1-2 p}+a x \, _2F_1\left (\frac {1}{2},\frac {3}{2}-p;\frac {3}{2};a^2 x^2\right )\right )-\frac {3 \, _2F_1\left (-\frac {1}{2},\frac {3}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(c - a^2*c*x^2)^p*((-3*Hypergeometric2F1[-1/2, 3/2 - p, 1/2, a^2*x^2])/x + a*(a*x*Hypergeometric2F1[1/2, 3/
2 - p, 3/2, a^2*x^2] + ((1 - a^2*x^2)^(-1/2 + p)*(3*Hypergeometric2F1[1, -1/2 + p, 1/2 + p, 1 - a^2*x^2] + Hyp
ergeometric2F1[2, -1/2 + p, 1/2 + p, 1 - a^2*x^2]))/(1 - 2*p))))/(1 - a^2*x^2)^p

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x + 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{5} - 2 \, a x^{4} + x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*(a*x + 1)*(-a^2*c*x^2 + c)^p/(a^2*x^5 - 2*a*x^4 + x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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