3.1012 \(\int e^{\tanh ^{-1}(a x)} x^3 (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=134 \[ \frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6153, 6148, 764, 266, 43, 364} \[ \frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+3)}-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 364

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

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 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 e^{\tanh ^{-1}(a x)} x^3 \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^{\tanh ^{-1}(a x)} x^3 \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^3 (1+a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^3 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx+\left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^4 \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\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 x \left (1-a^2 x\right )^{-\frac {1}{2}+p} \, dx,x,x^2\right )\\ &=\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )+\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 \left (\frac {\left (1-a^2 x\right )^{-\frac {1}{2}+p}}{a^2}-\frac {\left (1-a^2 x\right )^{\frac {1}{2}+p}}{a^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2} \left (c-a^2 c x^2\right )^p}{a^4 (1+2 p)}+\frac {\left (1-a^2 x^2\right )^{3/2} \left (c-a^2 c x^2\right )^p}{a^4 (3+2 p)}+\frac {1}{5} a x^5 \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 105, normalized size = 0.78 \[ \frac {\left (c-a^2 c x^2\right )^p \left (a^5 x^5 \left (1-a^2 x^2\right )^{-p} \, _2F_1\left (\frac {5}{2},\frac {1}{2}-p;\frac {7}{2};a^2 x^2\right )-\frac {5 \sqrt {1-a^2 x^2} \left (a^2 (2 p+1) x^2+2\right )}{4 p^2+8 p+3}\right )}{5 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - a^2*c*x^2)^p*((-5*Sqrt[1 - a^2*x^2]*(2 + a^2*(1 + 2*p)*x^2))/(3 + 8*p + 4*p^2) + (a^5*x^5*Hypergeometric
2F1[5/2, 1/2 - p, 7/2, a^2*x^2])/(1 - a^2*x^2)^p))/(5*a^4)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{3}}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{3} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a c^{p} \int \frac {x^{4} e^{\left (p \log \left (a x + 1\right ) + p \log \left (-a x + 1\right )\right )}}{\sqrt {a x + 1} \sqrt {-a x + 1}}\,{d x} + \frac {{\left (a^{4} c^{p} {\left (2 \, p + 1\right )} x^{4} - a^{2} c^{p} {\left (2 \, p - 1\right )} x^{2} - 2 \, c^{p}\right )} {\left (-a^{2} x^{2} + 1\right )}^{p}}{\sqrt {-a^{2} x^{2} + 1} {\left (4 \, p^{2} + 8 \, p + 3\right )} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c^p*integrate(x^4*e^(p*log(a*x + 1) + p*log(-a*x + 1))/(sqrt(a*x + 1)*sqrt(-a*x + 1)), x) + (a^4*c^p*(2*p +
1)*x^4 - a^2*c^p*(2*p - 1)*x^2 - 2*c^p)*(-a^2*x^2 + 1)^p/(sqrt(-a^2*x^2 + 1)*(4*p^2 + 8*p + 3)*a^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 54.21, size = 272, normalized size = 2.03 \[ - \frac {a a^{2 p} c^{p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, p + \frac {5}{2} \\ p + 1, p + \frac {7}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} c^{p} x^{5} x^{2 p} e^{i \pi p} \Gamma \left (- p - \frac {5}{2}\right ) \Gamma \left (p + \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - p - \frac {5}{2} \\ \frac {1}{2}, - p - \frac {3}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- p - \frac {3}{2}\right ) \Gamma \left (p + 1\right )} - \frac {c^{p} {G_{3, 3}^{2, 2}\left (\begin {matrix} - p - 1, 1 & -1 \\- p - \frac {3}{2}, - p - 1 & 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 \pi a^{4}} - \frac {c^{p} {G_{3, 3}^{1, 3}\left (\begin {matrix} -1, - p - 2, 1 & \\- p - 2 & - p - \frac {3}{2}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{a^{2} x^{2}}} \right )} \Gamma \left (p + \frac {1}{2}\right )}{2 a^{4} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*a**(2*p)*c**p*x**5*x**(2*p)*exp(I*pi*p)*gamma(-p - 5/2)*gamma(p + 1/2)*hyper((1/2, 1, p + 5/2), (p + 1, p +
7/2), a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(-p - 3/2)*gamma(p + 1)) - a*a**(2*p)*c**p*x**5*x**(2*p)*
exp(I*pi*p)*gamma(-p - 5/2)*gamma(p + 1/2)*hyper((1, -p, -p - 5/2), (1/2, -p - 3/2), 1/(a**2*x**2))/(2*sqrt(pi
)*gamma(-p - 3/2)*gamma(p + 1)) - c**p*meijerg(((-p - 1, 1), (-1,)), ((-p - 3/2, -p - 1), (0,)), exp_polar(-I*
pi)/(a**2*x**2))*gamma(p + 1/2)/(2*pi*a**4) - c**p*meijerg(((-1, -p - 2, 1), ()), ((-p - 2,), (-p - 3/2, 0)),
exp_polar(-I*pi)/(a**2*x**2))*gamma(p + 1/2)/(2*a**4*gamma(-p)*gamma(p + 1))

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