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

Optimal. Leaf size=69 \[ \frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt {c-a c x}} \]

[Out]

2/15*c^2*(-a^2*x^2+1)^(3/2)/a^2/(-a*c*x+c)^(3/2)-2/5*c*(-a^2*x^2+1)^(3/2)/a^2/(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6128, 795, 649} \[ \frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x*Sqrt[c - a*c*x],x]

[Out]

(2*c^2*(1 - a^2*x^2)^(3/2))/(15*a^2*(c - a*c*x)^(3/2)) - (2*c*(1 - a^2*x^2)^(3/2))/(5*a^2*Sqrt[c - a*c*x])

Rule 649

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

Rule 795

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

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 \sqrt {c-a c x} \, dx &=c \int \frac {x \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx\\ &=-\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt {c-a c x}}+\frac {c \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx}{5 a}\\ &=\frac {2 c^2 \left (1-a^2 x^2\right )^{3/2}}{15 a^2 (c-a c x)^{3/2}}-\frac {2 c \left (1-a^2 x^2\right )^{3/2}}{5 a^2 \sqrt {c-a c x}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.62 \[ \frac {2 (a x+1)^{3/2} (3 a x-2) \sqrt {c-a c x}}{15 a^2 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a*x]*x*Sqrt[c - a*c*x],x]

[Out]

(2*(1 + a*x)^(3/2)*(-2 + 3*a*x)*Sqrt[c - a*c*x])/(15*a^2*Sqrt[1 - a*x])

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fricas [A]  time = 0.57, size = 49, normalized size = 0.71 \[ -\frac {2 \, {\left (3 \, a^{2} x^{2} + a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{15 \, {\left (a^{3} x - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^2*x^2 + a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^3*x - a^2)

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giac [A]  time = 0.20, size = 54, normalized size = 0.78 \[ -\frac {2 \, c^{2} {\left (\frac {2 \, \sqrt {2}}{a \sqrt {c}} - \frac {3 \, {\left (a c x + c\right )}^{\frac {5}{2}} - 5 \, {\left (a c x + c\right )}^{\frac {3}{2}} c}{a c^{3}}\right )}}{15 \, a {\left | c \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*c^2*(2*sqrt(2)/(a*sqrt(c)) - (3*(a*c*x + c)^(5/2) - 5*(a*c*x + c)^(3/2)*c)/(a*c^3))/(a*abs(c))

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maple [A]  time = 0.03, size = 40, normalized size = 0.58 \[ \frac {2 \left (a x +1\right )^{2} \left (3 a x -2\right ) \sqrt {-a c x +c}}{15 a^{2} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+1)^2*(3*a*x-2)*(-a*c*x+c)^(1/2)/a^2/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.33, size = 83, normalized size = 1.20 \[ \frac {2 \, {\left (3 \, a^{3} \sqrt {c} x^{3} - a^{2} \sqrt {c} x^{2} + 4 \, a \sqrt {c} x + 8 \, \sqrt {c}\right )}}{15 \, \sqrt {a x + 1} a^{2}} + \frac {2 \, {\left (a^{2} \sqrt {c} x^{2} - a \sqrt {c} x - 2 \, \sqrt {c}\right )}}{3 \, \sqrt {a x + 1} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^3*sqrt(c)*x^3 - a^2*sqrt(c)*x^2 + 4*a*sqrt(c)*x + 8*sqrt(c))/(sqrt(a*x + 1)*a^2) + 2/3*(a^2*sqrt(c)*
x^2 - a*sqrt(c)*x - 2*sqrt(c))/(sqrt(a*x + 1)*a^2)

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mupad [B]  time = 0.92, size = 46, normalized size = 0.67 \[ -\frac {\sqrt {c-a\,c\,x}\,\left (\frac {2\,x}{15\,a}-\frac {2\,a\,x^3}{5}+\frac {4}{15\,a^2}-\frac {8\,x^2}{15}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - a*c*x)^(1/2)*((2*x)/(15*a) - (2*a*x^3)/5 + 4/(15*a^2) - (8*x^2)/15))/(1 - a^2*x^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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