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

Optimal. Leaf size=72 \[ -\frac {5 \sin ^{-1}(a x)}{2 a^3 c}+\frac {(a x+1)^2}{a^3 c \sqrt {1-a^2 x^2}}+\frac {(a x+6) \sqrt {1-a^2 x^2}}{2 a^3 c} \]

[Out]

-5/2*arcsin(a*x)/a^3/c+(a*x+1)^2/a^3/c/(-a^2*x^2+1)^(1/2)+1/2*(a*x+6)*(-a^2*x^2+1)^(1/2)/a^3/c

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6128, 852, 1635, 780, 216} \[ \frac {(a x+1)^2}{a^3 c \sqrt {1-a^2 x^2}}+\frac {(a x+6) \sqrt {1-a^2 x^2}}{2 a^3 c}-\frac {5 \sin ^{-1}(a x)}{2 a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^3*c*Sqrt[1 - a^2*x^2]) + ((6 + a*x)*Sqrt[1 - a^2*x^2])/(2*a^3*c) - (5*ArcSin[a*x])/(2*a^3*c)

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 780

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

Rule 852

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

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 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 \frac {e^{\tanh ^{-1}(a x)} x^2}{c-a c x} \, dx &=c \int \frac {x^2 \sqrt {1-a^2 x^2}}{(c-a c x)^2} \, dx\\ &=\frac {\int \frac {x^2 (c+a c x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {(1+a x)^2}{a^3 c \sqrt {1-a^2 x^2}}-\frac {\int \frac {\left (\frac {2}{a^2}+\frac {x}{a}\right ) (c+a c x)}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=\frac {(1+a x)^2}{a^3 c \sqrt {1-a^2 x^2}}+\frac {(6+a x) \sqrt {1-a^2 x^2}}{2 a^3 c}-\frac {5 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^2 c}\\ &=\frac {(1+a x)^2}{a^3 c \sqrt {1-a^2 x^2}}+\frac {(6+a x) \sqrt {1-a^2 x^2}}{2 a^3 c}-\frac {5 \sin ^{-1}(a x)}{2 a^3 c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 64, normalized size = 0.89 \[ \frac {10 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-\frac {\sqrt {a x+1} \left (a^2 x^2+3 a x-8\right )}{\sqrt {1-a x}}}{2 a^3 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((Sqrt[1 + a*x]*(-8 + 3*a*x + a^2*x^2))/Sqrt[1 - a*x]) + 10*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(2*a^3*c)

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 78, normalized size = 1.08 \[ \frac {8 \, a x + 10 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a^{2} x^{2} + 3 \, a x - 8\right )} \sqrt {-a^{2} x^{2} + 1} - 8}{2 \, {\left (a^{4} c x - a^{3} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*a*x + 10*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a^2*x^2 + 3*a*x - 8)*sqrt(-a^2*x^2 + 1) -
8)/(a^4*c*x - a^3*c)

________________________________________________________________________________________

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),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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 120, normalized size = 1.67 \[ \frac {x \sqrt {-a^{2} x^{2}+1}}{2 c \,a^{2}}-\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 c \,a^{2} \sqrt {a^{2}}}+\frac {2 \sqrt {-a^{2} x^{2}+1}}{a^{3} c}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{c \,a^{4} \left (x -\frac {1}{a}\right )} \]

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),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 83, normalized size = 1.15 \[ -\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4} c x - a^{3} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{2} c} - \frac {5 \, \arcsin \left (a x\right )}{2 \, a^{3} c} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3} c} \]

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),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 129, normalized size = 1.79 \[ \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2\,\sqrt {-a^2}}{a^3\,c}+\frac {x\,\sqrt {-a^2}}{2\,a^2\,c}\right )}{\sqrt {-a^2}}-\frac {5\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^2\,c\,\sqrt {-a^2}}+\frac {2\,\sqrt {1-a^2\,x^2}}{a^2\,c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{2}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{3}}{a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c} \]

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),x)

[Out]

-(Integral(x**2/(a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**3/(a*x*sqrt(-a**2*x**2 +
1) - sqrt(-a**2*x**2 + 1)), x))/c

________________________________________________________________________________________