3.347 \(\int \frac {e^{\tanh ^{-1}(a x)} x^2}{(c-a c x)^3} \, dx\)

Optimal. Leaf size=107 \[ -\frac {\sin ^{-1}(a x)}{a^3 c^3}-\frac {3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}+\frac {2 \sqrt {1-a^2 x^2}}{a^3 c^3 (1-a x)} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6128, 1637, 659, 651, 663, 216} \[ -\frac {3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}+\frac {2 \sqrt {1-a^2 x^2}}{a^3 c^3 (1-a x)}-\frac {\sin ^{-1}(a x)}{a^3 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^2)/(c - a*c*x)^3,x]

[Out]

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

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 651

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

Rule 659

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

Rule 663

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

Rule 1637

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p,
(d + e*x)^m*Pq, x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq
, x] + 2*p + 1, 0] && ILtQ[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)^3} \, dx &=c \int \frac {x^2 \sqrt {1-a^2 x^2}}{(c-a c x)^4} \, dx\\ &=c \int \left (\frac {\sqrt {1-a^2 x^2}}{a^2 c^4 (-1+a x)^4}+\frac {2 \sqrt {1-a^2 x^2}}{a^2 c^4 (-1+a x)^3}+\frac {\sqrt {1-a^2 x^2}}{a^2 c^4 (-1+a x)^2}\right ) \, dx\\ &=\frac {\int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^4} \, dx}{a^2 c^3}+\frac {\int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^2} \, dx}{a^2 c^3}+\frac {2 \int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^3} \, dx}{a^2 c^3}\\ &=\frac {2 \sqrt {1-a^2 x^2}}{a^3 c^3 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}-\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^3 c^3 (1-a x)^3}-\frac {\int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^3} \, dx}{5 a^2 c^3}-\frac {\int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^2 c^3}\\ &=\frac {2 \sqrt {1-a^2 x^2}}{a^3 c^3 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^4}-\frac {3 \left (1-a^2 x^2\right )^{3/2}}{5 a^3 c^3 (1-a x)^3}-\frac {\sin ^{-1}(a x)}{a^3 c^3}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 77, normalized size = 0.72 \[ \frac {\sqrt {a x+1} \left (-a^2 x^2+3 a x+4\right )+20 \sqrt {2} (a x-1) \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {1}{2} (1-a x)\right )}{15 a^3 c^3 (1-a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*x^2)/(c - a*c*x)^3,x]

[Out]

(Sqrt[1 + a*x]*(4 + 3*a*x - a^2*x^2) + 20*Sqrt[2]*(-1 + a*x)*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 - a*x)/2])
/(15*a^3*c^3*(1 - a*x)^(5/2))

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fricas [A]  time = 0.49, size = 138, normalized size = 1.29 \[ \frac {8 \, a^{3} x^{3} - 24 \, a^{2} x^{2} + 24 \, a x + 10 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (13 \, a^{2} x^{2} - 19 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} - 8}{5 \, {\left (a^{6} c^{3} x^{3} - 3 \, a^{5} c^{3} x^{2} + 3 \, a^{4} c^{3} x - a^{3} c^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

1/5*(8*a^3*x^3 - 24*a^2*x^2 + 24*a*x + 10*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a
*x)) - (13*a^2*x^2 - 19*a*x + 8)*sqrt(-a^2*x^2 + 1) - 8)/(a^6*c^3*x^3 - 3*a^5*c^3*x^2 + 3*a^4*c^3*x - a^3*c^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)^3,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 = 0.05, size = 167, normalized size = 1.56 \[ -\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{3} a^{2} \sqrt {a^{2}}}-\frac {7 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 c^{3} a^{5} \left (x -\frac {1}{a}\right )^{2}}-\frac {13 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 c^{3} a^{4} \left (x -\frac {1}{a}\right )}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 c^{3} a^{6} \left (x -\frac {1}{a}\right )^{3}} \]

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)^3,x)

[Out]

-1/c^3/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-7/5/c^3/a^5/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1
/a))^(1/2)-13/5/c^3/a^4/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5/c^3/a^6/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(
x-1/a))^(1/2)

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maxima [A]  time = 0.41, size = 144, normalized size = 1.35 \[ -\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{6} c^{3} x^{3} - 3 \, a^{5} c^{3} x^{2} + 3 \, a^{4} c^{3} x - a^{3} c^{3}\right )}} - \frac {7 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{5} c^{3} x^{2} - 2 \, a^{4} c^{3} x + a^{3} c^{3}\right )}} - \frac {13 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{4} c^{3} x - a^{3} c^{3}\right )}} - \frac {\arcsin \left (a x\right )}{a^{3} c^{3}} \]

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)^3,x, algorithm="maxima")

[Out]

-2/5*sqrt(-a^2*x^2 + 1)/(a^6*c^3*x^3 - 3*a^5*c^3*x^2 + 3*a^4*c^3*x - a^3*c^3) - 7/5*sqrt(-a^2*x^2 + 1)/(a^5*c^
3*x^2 - 2*a^4*c^3*x + a^3*c^3) - 13/5*sqrt(-a^2*x^2 + 1)/(a^4*c^3*x - a^3*c^3) - arcsin(a*x)/(a^3*c^3)

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mupad [B]  time = 0.82, size = 259, normalized size = 2.42 \[ \frac {4\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^7\,c^3\,x^2-2\,a^6\,c^3\,x+a^5\,c^3\right )}-\frac {13\,\sqrt {1-a^2\,x^2}}{5\,\left (a\,c^3\,\sqrt {-a^2}-a^2\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {5\,\sqrt {1-a^2\,x^2}}{3\,\left (a^5\,c^3\,x^2-2\,a^4\,c^3\,x+a^3\,c^3\right )}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (a\,c^3\,\sqrt {-a^2}+3\,a^3\,c^3\,x^2\,\sqrt {-a^2}-a^4\,c^3\,x^3\,\sqrt {-a^2}-3\,a^2\,c^3\,x\,\sqrt {-a^2}\right )}-\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^2\,c^3\,\sqrt {-a^2}} \]

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)^3),x)

[Out]

(4*a^2*(1 - a^2*x^2)^(1/2))/(15*(a^5*c^3 - 2*a^6*c^3*x + a^7*c^3*x^2)) - (13*(1 - a^2*x^2)^(1/2))/(5*(a*c^3*(-
a^2)^(1/2) - a^2*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (5*(1 - a^2*x^2)^(1/2))/(3*(a^3*c^3 - 2*a^4*c^3*x + a^5*c
^3*x^2)) - (2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(a*c^3*(-a^2)^(1/2) + 3*a^3*c^3*x^2*(-a^2)^(1/2) - a^4*c^3*
x^3*(-a^2)^(1/2) - 3*a^2*c^3*x*(-a^2)^(1/2))) - asinh(x*(-a^2)^(1/2))/(a^2*c^3*(-a^2)^(1/2))

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

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)**3,x)

[Out]

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

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