3.13.27 \(\int \frac {e^{-\tanh ^{-1}(a x)} (c-a^2 c x^2)^p}{x^2} \, dx\) [1227]

Optimal. Leaf size=112 \[ -\frac {\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}+\frac {a \sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p \, _2F_1\left (1,\frac {1}{2}+p;\frac {3}{2}+p;1-a^2 x^2\right )}{1+2 p} \]

[Out]

-(-a^2*c*x^2+c)^p*hypergeom([-1/2, 1/2-p],[1/2],a^2*x^2)/x/((-a^2*x^2+1)^p)+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 = 0.12, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6288, 6284, 778, 371, 272, 67} \begin {gather*} \frac {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 {\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((c - a^2*c*x^2)^p*Hypergeometric2F1[-1/2, 1/2 - p, 1/2, a^2*x^2])/(x*(1 - a^2*x^2)^p)) + (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 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 272

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 778

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 6284

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] || G
tQ[c, 0]) && ILtQ[(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^{-\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^{-\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) \left (1-a^2 x^2\right )^{-\frac {1}{2}+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 {\left (1-a^2 x^2\right )^{-\frac {1}{2}+p}}{x^2} \, dx-\left (a \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 {1}{2}+p}}{x} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}-\frac {1}{2} \left (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 {\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (-\frac {1}{2},\frac {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}+\frac {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 = 0.03, size = 102, normalized size = 0.91 \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 {1}{2}-p;\frac {1}{2};a^2 x^2\right )}{x}+\frac {a \left (1-a^2 x^2\right )^{\frac {1}{2}+p} \, _2F_1\left (1,\frac {1}{2}+p;\frac {3}{2}+p;1-a^2 x^2\right )}{1+2 p}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c - a^2*c*x^2)^p*(-(Hypergeometric2F1[-1/2, 1/2 - p, 1/2, a^2*x^2]/x) + (a*(1 - a^2*x^2)^(1/2 + p)*Hypergeom
etric2F1[1, 1/2 + p, 3/2 + 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 \[\int \frac {\left (-a^{2} c \,x^{2}+c \right )^{p} \sqrt {-a^{2} x^{2}+1}}{\left (a x +1\right ) x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*(-a^2*c*x^2 + c)^p/(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 {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{x^{2} \left (a x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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\,\sqrt {1-a^2\,x^2}}{x^2\,\left (a\,x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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