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

Optimal. Leaf size=114 \[ \frac {2 (4 m+5) (a x+1) x^m \sqrt {c-a c x} (-a x)^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};a x+1\right )}{a (2 m+3) \sqrt {1-a^2 x^2}}-\frac {2 c \sqrt {1-a^2 x^2} x^{m+1}}{(2 m+3) \sqrt {c-a c x}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6128, 881, 892, 67, 65} \[ \frac {2 (4 m+5) (a x+1) x^m \sqrt {c-a c x} (-a x)^{-m} \, _2F_1\left (\frac {1}{2},-m;\frac {3}{2};a x+1\right )}{a (2 m+3) \sqrt {1-a^2 x^2}}-\frac {2 c \sqrt {1-a^2 x^2} x^{m+1}}{(2 m+3) \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*x^(1 + m)*Sqrt[1 - a^2*x^2])/((3 + 2*m)*Sqrt[c - a*c*x]) + (2*(5 + 4*m)*x^m*(1 + a*x)*Sqrt[c - a*c*x]*Hy
pergeometric2F1[1/2, -m, 3/2, 1 + a*x])/(a*(3 + 2*m)*(-(a*x))^m*Sqrt[1 - a^2*x^2])

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-((b*c)/d))^IntPart[m]*(b*x)^FracPart[m])/
(-((d*x)/c))^FracPart[m], Int[(-((d*x)/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-(d/(b*c)), 0]

Rule 881

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(d +
e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + p + 2)), x] - Dist[(e*f*(p + 1) - d*g*(2*n + p
+ 3))/(g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m,
n, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1]
&& IntegerQ[2*p]

Rule 892

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + c*x^
2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (
c*x)/e)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !Int
egerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 77, normalized size = 0.68 \[ -\frac {c \sqrt {1-a x} x^{m+1} \left (2 (m+1) \sqrt {a x+1}-(4 m+5) \, _2F_1\left (\frac {1}{2},m+1;m+2;-a x\right )\right )}{(m+1) (2 m+3) \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \sqrt {-a c x +c}\, \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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