3.1004 \(\int e^{\tanh ^{-1}(a x)} x^m (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=136 \[ \frac {x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {6153, 6148, 808, 364} \[ \frac {x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.04, size = 114, normalized size = 0.84 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {x^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+1}{2}+1;a^2 x^2\right )}{m+1}+\frac {a x^{m+2} \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+2}{2}+1;a^2 x^2\right )}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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)^p,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x +1\right ) x^{m} \left (-a^{2} c \,x^{2}+c \right )^{p}}{\sqrt {-a^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)^p,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^m*(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 = 69.57, size = 381, normalized size = 2.80 \[ - \frac {a a^{2 p} c^{p} x^{2} x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p + \frac {1}{2}\right ) \Gamma \left (- \frac {m}{2} - p - 1\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, \frac {m}{2} + p + 1 \\ p + 1, \frac {m}{2} + p + 2 \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (- \frac {m}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a a^{2 p} c^{p} x^{2} x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p + \frac {1}{2}\right ) \Gamma \left (- \frac {m}{2} - p - 1\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - \frac {m}{2} - p - 1 \\ \frac {1}{2}, - \frac {m}{2} - p \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (- \frac {m}{2} - p\right ) \Gamma \left (p + 1\right )} - \frac {a^{2 p} c^{p} x x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p + \frac {1}{2}\right ) \Gamma \left (- \frac {m}{2} - p - \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{2}, 1, \frac {m}{2} + p + \frac {1}{2} \\ p + 1, \frac {m}{2} + p + \frac {3}{2} \end {matrix}\middle | {a^{2} x^{2} e^{2 i \pi }} \right )}}{2 \sqrt {\pi } \Gamma \left (p + 1\right ) \Gamma \left (- \frac {m}{2} - p + \frac {1}{2}\right )} - \frac {a^{2 p} c^{p} x x^{m} x^{2 p} e^{i \pi p} \Gamma \left (p + \frac {1}{2}\right ) \Gamma \left (- \frac {m}{2} - p - \frac {1}{2}\right ) {{}_{3}F_{2}\left (\begin {matrix} 1, - p, - \frac {m}{2} - p - \frac {1}{2} \\ \frac {1}{2}, - \frac {m}{2} - p + \frac {1}{2} \end {matrix}\middle | {\frac {1}{a^{2} x^{2}}} \right )}}{2 \sqrt {\pi } \Gamma \left (p + 1\right ) \Gamma \left (- \frac {m}{2} - p + \frac {1}{2}\right )} \]

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)**p,x)

[Out]

-a*a**(2*p)*c**p*x**2*x**m*x**(2*p)*exp(I*pi*p)*gamma(p + 1/2)*gamma(-m/2 - p - 1)*hyper((1/2, 1, m/2 + p + 1)
, (p + 1, m/2 + p + 2), a**2*x**2*exp_polar(2*I*pi))/(2*sqrt(pi)*gamma(-m/2 - p)*gamma(p + 1)) - a*a**(2*p)*c*
*p*x**2*x**m*x**(2*p)*exp(I*pi*p)*gamma(p + 1/2)*gamma(-m/2 - p - 1)*hyper((1, -p, -m/2 - p - 1), (1/2, -m/2 -
p), 1/(a**2*x**2))/(2*sqrt(pi)*gamma(-m/2 - p)*gamma(p + 1)) - a**(2*p)*c**p*x*x**m*x**(2*p)*exp(I*pi*p)*gamm
a(p + 1/2)*gamma(-m/2 - p - 1/2)*hyper((1/2, 1, m/2 + p + 1/2), (p + 1, m/2 + p + 3/2), a**2*x**2*exp_polar(2*
I*pi))/(2*sqrt(pi)*gamma(p + 1)*gamma(-m/2 - p + 1/2)) - a**(2*p)*c**p*x*x**m*x**(2*p)*exp(I*pi*p)*gamma(p + 1
/2)*gamma(-m/2 - p - 1/2)*hyper((1, -p, -m/2 - p - 1/2), (1/2, -m/2 - p + 1/2), 1/(a**2*x**2))/(2*sqrt(pi)*gam
ma(p + 1)*gamma(-m/2 - p + 1/2))

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