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

Optimal. Leaf size=107 \[ \frac {2 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{7 a (c-a c x)^{3/2}}-\frac {8 c^2 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (c-a c x)^{3/2}}+\frac {8 c \left (1-a^2 x^2\right )^{3/2}}{35 a^3 \sqrt {c-a c x}} \]

[Out]

-8/105*c^2*(-a^2*x^2+1)^(3/2)/a^3/(-a*c*x+c)^(3/2)+2/7*c^2*x^2*(-a^2*x^2+1)^(3/2)/a/(-a*c*x+c)^(3/2)+8/35*c*(-
a^2*x^2+1)^(3/2)/a^3/(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6128, 871, 795, 649} \[ \frac {2 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{7 a (c-a c x)^{3/2}}-\frac {8 c^2 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (c-a c x)^{3/2}}+\frac {8 c \left (1-a^2 x^2\right )^{3/2}}{35 a^3 \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*c^2*(1 - a^2*x^2)^(3/2))/(105*a^3*(c - a*c*x)^(3/2)) + (2*c^2*x^2*(1 - a^2*x^2)^(3/2))/(7*a*(c - a*c*x)^(3
/2)) + (8*c*(1 - a^2*x^2)^(3/2))/(35*a^3*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 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 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 &=c \int \frac {x^2 \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx\\ &=\frac {2 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{7 a (c-a c x)^{3/2}}-\frac {(4 c) \int \frac {x \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx}{7 a}\\ &=\frac {2 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{7 a (c-a c x)^{3/2}}+\frac {8 c \left (1-a^2 x^2\right )^{3/2}}{35 a^3 \sqrt {c-a c x}}-\frac {(4 c) \int \frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}} \, dx}{35 a^2}\\ &=-\frac {8 c^2 \left (1-a^2 x^2\right )^{3/2}}{105 a^3 (c-a c x)^{3/2}}+\frac {2 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}{7 a (c-a c x)^{3/2}}+\frac {8 c \left (1-a^2 x^2\right )^{3/2}}{35 a^3 \sqrt {c-a c x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.48 \[ \frac {2 (a x+1)^{3/2} \left (15 a^2 x^2-12 a x+8\right ) \sqrt {c-a c x}}{105 a^3 \sqrt {1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(1 + a*x)^(3/2)*Sqrt[c - a*c*x]*(8 - 12*a*x + 15*a^2*x^2))/(105*a^3*Sqrt[1 - a*x])

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fricas [A]  time = 0.52, size = 58, normalized size = 0.54 \[ -\frac {2 \, {\left (15 \, a^{3} x^{3} + 3 \, a^{2} x^{2} - 4 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{105 \, {\left (a^{4} x - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(15*a^3*x^3 + 3*a^2*x^2 - 4*a*x + 8)*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((a*x+1)/(-a^2*x^2+1)^(1/2)*x^2*(-a*c*x+c)^(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 = 48, normalized size = 0.45 \[ \frac {2 \left (a x +1\right )^{2} \left (15 a^{2} x^{2}-12 a x +8\right ) \sqrt {-a c x +c}}{105 a^{3} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(a*x+1)^2*(15*a^2*x^2-12*a*x+8)*(-a*c*x+c)^(1/2)/a^3/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.33, size = 106, normalized size = 0.99 \[ \frac {2 \, {\left (5 \, a^{4} \sqrt {c} x^{4} - a^{3} \sqrt {c} x^{3} + 2 \, a^{2} \sqrt {c} x^{2} - 8 \, a \sqrt {c} x - 16 \, \sqrt {c}\right )}}{35 \, \sqrt {a x + 1} a^{3}} + \frac {2 \, {\left (3 \, a^{3} \sqrt {c} x^{3} - a^{2} \sqrt {c} x^{2} + 4 \, a \sqrt {c} x + 8 \, \sqrt {c}\right )}}{15 \, \sqrt {a x + 1} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*a^4*sqrt(c)*x^4 - a^3*sqrt(c)*x^3 + 2*a^2*sqrt(c)*x^2 - 8*a*sqrt(c)*x - 16*sqrt(c))/(sqrt(a*x + 1)*a^3
) + 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^3)

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mupad [B]  time = 0.95, size = 53, normalized size = 0.50 \[ \frac {\sqrt {c-a\,c\,x}\,\left (\frac {8\,x}{105\,a^2}+\frac {2\,a\,x^4}{7}+\frac {16}{105\,a^3}+\frac {12\,x^3}{35}-\frac {2\,x^2}{105\,a}\right )}{\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a*c*x)^(1/2)*((8*x)/(105*a^2) + (2*a*x^4)/7 + 16/(105*a^3) + (12*x^3)/35 - (2*x^2)/(105*a)))/(1 - a^2*x^
2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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