3.12.80 \(\int \frac {e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^p}{x^2} \, dx\) [1180]

Optimal. Leaf size=187 \[ \frac {4 a \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+a^2 (5-2 p) x \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {1}{2},\frac {3}{2}-p;\frac {3}{2};a^2 x^2\right )-\frac {3 a \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,\frac {1}{2}+p;\frac {3}{2}+p;1-a^2 x^2\right )}{1+2 p} \]

[Out]

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

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Rubi [A]
time = 0.36, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {6288, 6283, 1821, 1666, 457, 80, 67, 12, 251} \begin {gather*} a^2 (5-2 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 {3 a \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 {4 a \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(c - a^2*c*x^2)^p)/((1 - 2*p)*Sqrt[1 - a^2*x^2]) - (c - a^2*c*x^2)^p/(x*Sqrt[1 - a^2*x^2]) + (a^2*(5 - 2*
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 - (3*a*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 67

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

Rule 80

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 251

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 457

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 1666

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 1821

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 6283

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 6288

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^2} \, 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^2} \, 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^2} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^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 (-3 a-a^2 (5-2 p) x-a^3 x^2\right )}{x} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int a^2 (5-2 p) \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 \frac {\left (1-a^2 x^2\right )^{-\frac {3}{2}+p} \left (-3 a-a^3 x^2\right )}{x} \, dx\\ &=-\frac {\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 ) \text {Subst}\left (\int \frac {\left (1-a^2 x\right )^{-\frac {3}{2}+p} \left (-3 a-a^3 x\right )}{x} \, dx,x,x^2\right )+\left (a^2 (5-2 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 {4 a \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+a^2 (5-2 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}{2} \left (3 a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \text {Subst}\left (\int \frac {\left (1-a^2 x\right )^{-\frac {1}{2}+p}}{x} \, dx,x,x^2\right )\\ &=\frac {4 a \left (c-a^2 c x^2\right )^p}{(1-2 p) \sqrt {1-a^2 x^2}}-\frac {\left (c-a^2 c x^2\right )^p}{x \sqrt {1-a^2 x^2}}+a^2 (5-2 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 {3 a \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*}

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Mathematica [A]
time = 0.19, size = 133, normalized size = 0.71 \begin {gather*} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (-\frac {\, _2F_1\left (-\frac {1}{2},\frac {3}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}+a \left (3 a x \, _2F_1\left (\frac {1}{2},\frac {3}{2}-p;\frac {3}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \left (1+3 \, _2F_1\left (1,-\frac {1}{2}+p;\frac {1}{2}+p;1-a^2 x^2\right )\right )}{1-2 p}\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +1\right )^{3} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} x^{2}}\, dx\]

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^2,x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="fricas")

[Out]

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

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

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**2,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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