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

Optimal. Leaf size=135 \[ \frac {26 c x^2 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}-\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}+\frac {104 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{105 a^3}+\frac {104 c \sqrt {1-a^2 x^2}}{105 a^3 \sqrt {c-a c x}} \]

[Out]

104/105*c*(-a^2*x^2+1)^(1/2)/a^3/(-a*c*x+c)^(1/2)+26/35*c*x^2*(-a^2*x^2+1)^(1/2)/a/(-a*c*x+c)^(1/2)-2/7*c*x^3*
(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(1/2)+104/105*(-a*c*x+c)^(1/2)*(-a^2*x^2+1)^(1/2)/a^3

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Rubi [A]  time = 0.21, antiderivative size = 135, 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, 871, 795, 649} \[ -\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}+\frac {26 c x^2 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}+\frac {104 \sqrt {1-a^2 x^2} \sqrt {c-a c x}}{105 a^3}+\frac {104 c \sqrt {1-a^2 x^2}}{105 a^3 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(104*c*Sqrt[1 - a^2*x^2])/(105*a^3*Sqrt[c - a*c*x]) + (26*c*x^2*Sqrt[1 - a^2*x^2])/(35*a*Sqrt[c - a*c*x]) - (2
*c*x^3*Sqrt[1 - a^2*x^2])/(7*Sqrt[c - a*c*x]) + (104*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2])/(105*a^3)

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 871

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

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 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^2 \sqrt {c-a c x} \, dx &=\frac {\int \frac {x^2 (c-a c x)^{3/2}}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}+\frac {13}{7} \int \frac {x^2 \sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {26 c x^2 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}-\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}-\frac {52 \int \frac {x \sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx}{35 a}\\ &=\frac {26 c x^2 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}-\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}+\frac {104 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{105 a^3}+\frac {52 \int \frac {\sqrt {c-a c x}}{\sqrt {1-a^2 x^2}} \, dx}{105 a^2}\\ &=\frac {104 c \sqrt {1-a^2 x^2}}{105 a^3 \sqrt {c-a c x}}+\frac {26 c x^2 \sqrt {1-a^2 x^2}}{35 a \sqrt {c-a c x}}-\frac {2 c x^3 \sqrt {1-a^2 x^2}}{7 \sqrt {c-a c x}}+\frac {104 \sqrt {c-a c x} \sqrt {1-a^2 x^2}}{105 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.41 \[ -\frac {2 c \sqrt {1-a^2 x^2} \left (15 a^3 x^3-39 a^2 x^2+52 a x-104\right )}{105 a^3 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Sqrt[1 - a^2*x^2]*(-104 + 52*a*x - 39*a^2*x^2 + 15*a^3*x^3))/(105*a^3*Sqrt[c - a*c*x])

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fricas [A]  time = 0.45, size = 58, normalized size = 0.43 \[ \frac {2 \, {\left (15 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 52 \, a x - 104\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{105 \, {\left (a^{4} x - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^3*x^3 - 39*a^2*x^2 + 52*a*x - 104)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^4*x - a^3)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 56, normalized size = 0.41 \[ \frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-a c x +c}\, \left (15 x^{3} a^{3}-39 a^{2} x^{2}+52 a x -104\right )}{105 \left (a x -1\right ) a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(-a^2*x^2+1)^(1/2)*(-a*c*x+c)^(1/2)*(15*a^3*x^3-39*a^2*x^2+52*a*x-104)/(a*x-1)/a^3

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maxima [A]  time = 0.33, size = 62, normalized size = 0.46 \[ -\frac {2 \, {\left (15 \, a^{3} \sqrt {c} x^{3} - 39 \, a^{2} \sqrt {c} x^{2} + 52 \, a \sqrt {c} x - 104 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{105 \, {\left (a^{4} x - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^3*sqrt(c)*x^3 - 39*a^2*sqrt(c)*x^2 + 52*a*sqrt(c)*x - 104*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^4*x
- a^3)

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mupad [B]  time = 0.95, size = 74, normalized size = 0.55 \[ \frac {2\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}\,\left (15\,a^2\,x^2-24\,a\,x+28\right )}{105\,a^3}-\frac {152\,\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{105\,a^3\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(1 - a^2*x^2)^(1/2)*(c - a*c*x)^(1/2)*(15*a^2*x^2 - 24*a*x + 28))/(105*a^3) - (152*(1 - a^2*x^2)^(1/2)*(c -
a*c*x)^(1/2))/(105*a^3*(a*x - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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