3.1131 \(\int e^{2 \tanh ^{-1}(a x)} x^m (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=176 \[ \frac{c^2 (2 m+7) x^{m+1} \sqrt{c-a^2 c x^2} \text{Hypergeometric2F1}\left (-\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{(m+1) (m+6) \sqrt{1-a^2 x^2}}+\frac{2 a c^2 x^{m+2} \sqrt{c-a^2 c x^2} \text{Hypergeometric2F1}\left (-\frac{3}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{(m+2) \sqrt{1-a^2 x^2}}-\frac{x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{m+6} \]

[Out]

-((x^(1 + m)*(c - a^2*c*x^2)^(5/2))/(6 + m)) + (c^2*(7 + 2*m)*x^(1 + m)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[
-3/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*(6 + m)*Sqrt[1 - a^2*x^2]) + (2*a*c^2*x^(2 + m)*Sqrt[c - a^2*c*
x^2]*Hypergeometric2F1[-3/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.314157, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6151, 1809, 808, 365, 364} \[ \frac{c^2 (2 m+7) x^{m+1} \sqrt{c-a^2 c x^2} \, _2F_1\left (-\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{(m+1) (m+6) \sqrt{1-a^2 x^2}}+\frac{2 a c^2 x^{m+2} \sqrt{c-a^2 c x^2} \, _2F_1\left (-\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{(m+2) \sqrt{1-a^2 x^2}}-\frac{x^{m+1} \left (c-a^2 c x^2\right )^{5/2}}{m+6} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^(5/2),x]

[Out]

-((x^(1 + m)*(c - a^2*c*x^2)^(5/2))/(6 + m)) + (c^2*(7 + 2*m)*x^(1 + m)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[
-3/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*(6 + m)*Sqrt[1 - a^2*x^2]) + (2*a*c^2*x^(2 + m)*Sqrt[c - a^2*c*
x^2]*Hypergeometric2F1[-3/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[1 - a^2*x^2])

Rule 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[x^m*(c
+ d*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] ||
GtQ[c, 0]) && IGtQ[n/2, 0]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

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

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])

Rubi steps

\begin{align*} \int e^{2 \tanh ^{-1}(a x)} x^m \left (c-a^2 c x^2\right )^{5/2} \, dx &=c \int x^m (1+a x)^2 \left (c-a^2 c x^2\right )^{3/2} \, dx\\ &=-\frac{x^{1+m} \left (c-a^2 c x^2\right )^{5/2}}{6+m}-\frac{\int x^m \left (-a^2 c (7+2 m)-2 a^3 c (6+m) x\right ) \left (c-a^2 c x^2\right )^{3/2} \, dx}{a^2 (6+m)}\\ &=-\frac{x^{1+m} \left (c-a^2 c x^2\right )^{5/2}}{6+m}+(2 a c) \int x^{1+m} \left (c-a^2 c x^2\right )^{3/2} \, dx+\frac{(c (7+2 m)) \int x^m \left (c-a^2 c x^2\right )^{3/2} \, dx}{6+m}\\ &=-\frac{x^{1+m} \left (c-a^2 c x^2\right )^{5/2}}{6+m}+\frac{\left (2 a c^2 \sqrt{c-a^2 c x^2}\right ) \int x^{1+m} \left (1-a^2 x^2\right )^{3/2} \, dx}{\sqrt{1-a^2 x^2}}+\frac{\left (c^2 (7+2 m) \sqrt{c-a^2 c x^2}\right ) \int x^m \left (1-a^2 x^2\right )^{3/2} \, dx}{(6+m) \sqrt{1-a^2 x^2}}\\ &=-\frac{x^{1+m} \left (c-a^2 c x^2\right )^{5/2}}{6+m}+\frac{c^2 (7+2 m) x^{1+m} \sqrt{c-a^2 c x^2} \, _2F_1\left (-\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{(1+m) (6+m) \sqrt{1-a^2 x^2}}+\frac{2 a c^2 x^{2+m} \sqrt{c-a^2 c x^2} \, _2F_1\left (-\frac{3}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{(2+m) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.220947, size = 180, normalized size = 1.02 \[ \frac{c^2 x^{m+1} \sqrt{c-a^2 c x^2} \left (-\frac{a^4 x^4 \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m+5}{2},\frac{m+7}{2},a^2 x^2\right )}{m+5}-\frac{2 a^3 x^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m}{2}+2,\frac{m}{2}+3,a^2 x^2\right )}{m+4}+\frac{2 a x \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,a^2 x^2\right )}{m+2}+\frac{\text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}\right )}{\sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcTanh[a*x])*x^m*(c - a^2*c*x^2)^(5/2),x]

[Out]

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

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Maple [F]  time = 0.402, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2}{x}^{m}}{-{a}^{2}{x}^{2}+1} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x)

[Out]

int((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}}{\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

-integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^2*x^m/(a^2*x^2 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (a^{4} c^{2} x^{4} + 2 \, a^{3} c^{2} x^{3} - 2 \, a c^{2} x - c^{2}\right )} \sqrt{-a^{2} c x^{2} + c} x^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(-(a^4*c^2*x^4 + 2*a^3*c^2*x^3 - 2*a*c^2*x - c^2)*sqrt(-a^2*c*x^2 + c)*x^m, x)

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Sympy [C]  time = 88.0958, size = 226, normalized size = 1.28 \begin{align*} - \frac{a^{4} c^{\frac{5}{2}} x^{5} x^{m} \Gamma \left (\frac{m}{2} + \frac{5}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{m}{2} + \frac{5}{2} \\ \frac{m}{2} + \frac{7}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{m}{2} + \frac{7}{2}\right )} - \frac{a^{3} c^{\frac{5}{2}} x^{4} x^{m} \Gamma \left (\frac{m}{2} + 2\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{m}{2} + 2 \\ \frac{m}{2} + 3 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{\Gamma \left (\frac{m}{2} + 3\right )} + \frac{a c^{\frac{5}{2}} x^{2} x^{m} \Gamma \left (\frac{m}{2} + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{\Gamma \left (\frac{m}{2} + 2\right )} + \frac{c^{\frac{5}{2}} x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*x**m*(-a**2*c*x**2+c)**(5/2),x)

[Out]

-a**4*c**(5/2)*x**5*x**m*gamma(m/2 + 5/2)*hyper((-1/2, m/2 + 5/2), (m/2 + 7/2,), a**2*x**2*exp_polar(2*I*pi))/
(2*gamma(m/2 + 7/2)) - a**3*c**(5/2)*x**4*x**m*gamma(m/2 + 2)*hyper((-1/2, m/2 + 2), (m/2 + 3,), a**2*x**2*exp
_polar(2*I*pi))/gamma(m/2 + 3) + a*c**(5/2)*x**2*x**m*gamma(m/2 + 1)*hyper((-1/2, m/2 + 1), (m/2 + 2,), a**2*x
**2*exp_polar(2*I*pi))/gamma(m/2 + 2) + c**(5/2)*x*x**m*gamma(m/2 + 1/2)*hyper((-1/2, m/2 + 1/2), (m/2 + 3/2,)
, a**2*x**2*exp_polar(2*I*pi))/(2*gamma(m/2 + 3/2))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}}{\left (a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^m*(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate(-(-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^2*x^m/(a^2*x^2 - 1), x)