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

Optimal. Leaf size=104 \[ \frac {11 \sin ^{-1}(a x)}{2 a^4 c^2}+\frac {(a x+1)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 c^2 \sqrt {1-a^2 x^2}}-\frac {(a x+12) \sqrt {1-a^2 x^2}}{2 a^4 c^2} \]

[Out]

1/3*(a*x+1)^3/a^4/c^2/(-a^2*x^2+1)^(3/2)+11/2*arcsin(a*x)/a^4/c^2-3*(a*x+1)^2/a^4/c^2/(-a^2*x^2+1)^(1/2)-1/2*(
a*x+12)*(-a^2*x^2+1)^(1/2)/a^4/c^2

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Rubi [A]  time = 0.30, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {(a x+1)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 c^2 \sqrt {1-a^2 x^2}}-\frac {(a x+12) \sqrt {1-a^2 x^2}}{2 a^4 c^2}+\frac {11 \sin ^{-1}(a x)}{2 a^4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^3/(3*a^4*c^2*(1 - a^2*x^2)^(3/2)) - (3*(1 + a*x)^2)/(a^4*c^2*Sqrt[1 - a^2*x^2]) - ((12 + a*x)*Sqrt[1
- a^2*x^2])/(2*a^4*c^2) + (11*ArcSin[a*x])/(2*a^4*c^2)

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^3}{(c-a c x)^2} \, dx &=c \int \frac {x^3 \sqrt {1-a^2 x^2}}{(c-a c x)^3} \, dx\\ &=\frac {\int \frac {x^3 (c+a c x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^5}\\ &=\frac {(1+a x)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {\int \frac {(c+a c x)^2 \left (\frac {3}{a^3}+\frac {3 x}{a^2}+\frac {3 x^2}{a}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^4}\\ &=\frac {(1+a x)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (1+a x)^2}{a^4 c^2 \sqrt {1-a^2 x^2}}+\frac {\int \frac {\left (\frac {15}{a^3}+\frac {3 x}{a^2}\right ) (c+a c x)}{\sqrt {1-a^2 x^2}} \, dx}{3 c^3}\\ &=\frac {(1+a x)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (1+a x)^2}{a^4 c^2 \sqrt {1-a^2 x^2}}-\frac {(12+a x) \sqrt {1-a^2 x^2}}{2 a^4 c^2}+\frac {11 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{2 a^3 c^2}\\ &=\frac {(1+a x)^3}{3 a^4 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (1+a x)^2}{a^4 c^2 \sqrt {1-a^2 x^2}}-\frac {(12+a x) \sqrt {1-a^2 x^2}}{2 a^4 c^2}+\frac {11 \sin ^{-1}(a x)}{2 a^4 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 72, normalized size = 0.69 \[ -\frac {\frac {\sqrt {a x+1} \left (3 a^3 x^3+12 a^2 x^2-71 a x+52\right )}{(1-a x)^{3/2}}+66 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{6 a^4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*((Sqrt[1 + a*x]*(52 - 71*a*x + 12*a^2*x^2 + 3*a^3*x^3))/(1 - a*x)^(3/2) + 66*ArcSin[Sqrt[1 - a*x]/Sqrt[2]
])/(a^4*c^2)

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fricas [A]  time = 0.43, size = 117, normalized size = 1.12 \[ -\frac {52 \, a^{2} x^{2} - 104 \, a x + 66 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 71 \, a x + 52\right )} \sqrt {-a^{2} x^{2} + 1} + 52}{6 \, {\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a*c*x+c)^2,x, algorithm="fricas")

[Out]

-1/6*(52*a^2*x^2 - 104*a*x + 66*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^3*x^3 + 12
*a^2*x^2 - 71*a*x + 52)*sqrt(-a^2*x^2 + 1) + 52)/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2)

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giac [A]  time = 0.24, size = 176, normalized size = 1.69 \[ -\frac {4 \, a^{3} c^{6} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} + 132 \, a^{3} c^{6} \arctan \left (\sqrt {-\frac {2 \, c}{a c x - c} - 1}\right ) - 72 \, a^{3} c^{6} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - \frac {3 \, {\left (7 \, a^{3} c^{6} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} + 5 \, a^{3} c^{6} \sqrt {-\frac {2 \, c}{a c x - c} - 1}\right )} {\left (a c x - c\right )}^{2}}{c^{2}}}{12 \, a^{6} c^{8} {\left | a \right |} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\relax (a) \mathrm {sgn}\relax (c)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a*c*x+c)^2,x, algorithm="giac")

[Out]

-1/12*(4*a^3*c^6*(-2*c/(a*c*x - c) - 1)^(3/2) + 132*a^3*c^6*arctan(sqrt(-2*c/(a*c*x - c) - 1)) - 72*a^3*c^6*sq
rt(-2*c/(a*c*x - c) - 1) - 3*(7*a^3*c^6*(-2*c/(a*c*x - c) - 1)^(3/2) + 5*a^3*c^6*sqrt(-2*c/(a*c*x - c) - 1))*(
a*c*x - c)^2/c^2)/(a^6*c^8*abs(a)*sgn(1/(a*c*x - c))*sgn(a)*sgn(c))

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maple [A]  time = 0.04, size = 164, normalized size = 1.58 \[ -\frac {x \sqrt {-a^{2} x^{2}+1}}{2 c^{2} a^{3}}+\frac {11 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 c^{2} a^{3} \sqrt {a^{2}}}-\frac {3 \sqrt {-a^{2} x^{2}+1}}{c^{2} a^{4}}+\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 c^{2} a^{6} \left (x -\frac {1}{a}\right )^{2}}+\frac {19 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 c^{2} a^{5} \left (x -\frac {1}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2/a^3*x*(-a^2*x^2+1)^(1/2)+11/2/c^2/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-3/c^2/a^4*
(-a^2*x^2+1)^(1/2)+2/3/c^2/a^6/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+19/3/c^2/a^5/(x-1/a)*(-a^2*(x-1/a)
^2-2*a*(x-1/a))^(1/2)

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maxima [A]  time = 0.48, size = 130, normalized size = 1.25 \[ \frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}} + \frac {19 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{5} c^{2} x - a^{4} c^{2}\right )}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{3} c^{2}} + \frac {11 \, \arcsin \left (a x\right )}{2 \, a^{4} c^{2}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a*c*x+c)^2,x, algorithm="maxima")

[Out]

2/3*sqrt(-a^2*x^2 + 1)/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2) + 19/3*sqrt(-a^2*x^2 + 1)/(a^5*c^2*x - a^4*c^2) -
1/2*sqrt(-a^2*x^2 + 1)*x/(a^3*c^2) + 11/2*arcsin(a*x)/(a^4*c^2) - 3*sqrt(-a^2*x^2 + 1)/(a^4*c^2)

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mupad [B]  time = 0.05, size = 166, normalized size = 1.60 \[ \frac {2\,\sqrt {1-a^2\,x^2}}{3\,\left (a^6\,c^2\,x^2-2\,a^5\,c^2\,x+a^4\,c^2\right )}+\frac {19\,\sqrt {1-a^2\,x^2}}{3\,\left (a^2\,c^2\,\sqrt {-a^2}-a^3\,c^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {3\,\sqrt {1-a^2\,x^2}}{a^4\,c^2}-\frac {x\,\sqrt {1-a^2\,x^2}}{2\,a^3\,c^2}+\frac {11\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^3\,c^2\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(1 - a^2*x^2)^(1/2))/(3*(a^4*c^2 - 2*a^5*c^2*x + a^6*c^2*x^2)) + (19*(1 - a^2*x^2)^(1/2))/(3*(a^2*c^2*(-a^2
)^(1/2) - a^3*c^2*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (3*(1 - a^2*x^2)^(1/2))/(a^4*c^2) - (x*(1 - a^2*x^2)^(1/2))/
(2*a^3*c^2) + (11*asinh(x*(-a^2)^(1/2)))/(2*a^3*c^2*(-a^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{3}}{a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a x^{4}}{a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(x**3/(a**2*x**2*sqrt(-a**2*x**2 + 1) - 2*a*x*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Inte
gral(a*x**4/(a**2*x**2*sqrt(-a**2*x**2 + 1) - 2*a*x*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x))/c**2

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