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

Optimal. Leaf size=80 \[ \frac {c^2 x^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a c^2 x^{m+2} \, _2F_1\left (-\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6148, 808, 364} \[ \frac {c^2 x^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a c^2 x^{m+2} \, _2F_1\left (-\frac {3}{2},\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x^(1 + m)*Hypergeometric2F1[-3/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(1 + m) + (a*c^2*x^(2 + m)*Hypergeometr
ic2F1[-3/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(2 + m)

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 82, normalized size = 1.02 \[ c^2 \left (\frac {x^{m+1} \, _2F_1\left (-\frac {3}{2},\frac {m+1}{2};\frac {m+1}{2}+1;a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (-\frac {3}{2},\frac {m+2}{2};\frac {m+2}{2}+1;a^2 x^2\right )}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c^2*((x^(1 + m)*Hypergeometric2F1[-3/2, (1 + m)/2, 1 + (1 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeomet
ric2F1[-3/2, (2 + m)/2, 1 + (2 + m)/2, a^2*x^2])/(2 + m))

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a^{3} c^{2} x^{3} + a^{2} c^{2} x^{2} - a c^{2} x - c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.37, size = 227, normalized size = 2.84 \[ \frac {a^{5} c^{2} x^{6+m} \hypergeom \left (\left [\frac {1}{2}, 3+\frac {m}{2}\right ], \left [4+\frac {m}{2}\right ], a^{2} x^{2}\right )}{6+m}-\frac {2 a^{3} c^{2} x^{4+m} \hypergeom \left (\left [\frac {1}{2}, 2+\frac {m}{2}\right ], \left [3+\frac {m}{2}\right ], a^{2} x^{2}\right )}{4+m}+\frac {a \,c^{2} x^{2+m} \hypergeom \left (\left [\frac {1}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], a^{2} x^{2}\right )}{2+m}+\frac {c^{2} a^{4} x^{5+m} \hypergeom \left (\left [\frac {1}{2}, \frac {5}{2}+\frac {m}{2}\right ], \left [\frac {7}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{5+m}-\frac {2 c^{2} a^{2} x^{3+m} \hypergeom \left (\left [\frac {1}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {5}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{3+m}+\frac {c^{2} x^{1+m} \hypergeom \left (\left [\frac {1}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], a^{2} x^{2}\right )}{1+m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 14.39, size = 223, normalized size = 2.79 \[ - \frac {a^{3} c^{2} x^{4} x^{m} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + 3\right )} - \frac {a^{2} c^{2} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {a c^{2} x^{2} x^{m} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {c^{2} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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