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

Optimal. Leaf size=137 \[ -\frac {4 \sin ^{-1}(a x)}{a^4 c^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}-\frac {14 \left (1-a^2 x^2\right )^{3/2}}{15 a^4 c^3 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^4 c^3 (1-a x)^4}+\frac {8 \sqrt {1-a^2 x^2}}{a^4 c^3 (1-a x)} \]

[Out]

1/5*(-a^2*x^2+1)^(3/2)/a^4/c^3/(-a*x+1)^4-14/15*(-a^2*x^2+1)^(3/2)/a^4/c^3/(-a*x+1)^3-(-a^2*x^2+1)^(3/2)/a^4/c
^3/(-a*x+1)^2-4*arcsin(a*x)/a^4/c^3+8*(-a^2*x^2+1)^(1/2)/a^4/c^3/(-a*x+1)

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Rubi [A]  time = 0.33, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6128, 1639, 1637, 659, 651, 663, 216} \[ -\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}-\frac {14 \left (1-a^2 x^2\right )^{3/2}}{15 a^4 c^3 (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^4 c^3 (1-a x)^4}+\frac {8 \sqrt {1-a^2 x^2}}{a^4 c^3 (1-a x)}-\frac {4 \sin ^{-1}(a x)}{a^4 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[1 - a^2*x^2])/(a^4*c^3*(1 - a*x)) + (1 - a^2*x^2)^(3/2)/(5*a^4*c^3*(1 - a*x)^4) - (14*(1 - a^2*x^2)^(3
/2))/(15*a^4*c^3*(1 - a*x)^3) - (1 - a^2*x^2)^(3/2)/(a^4*c^3*(1 - a*x)^2) - (4*ArcSin[a*x])/(a^4*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 1639

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[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)^3} \, dx &=c \int \frac {x^3 \sqrt {1-a^2 x^2}}{(c-a c x)^4} \, dx\\ &=-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}+\frac {\int \frac {\sqrt {1-a^2 x^2} \left (2 a^2 c^3-5 a^3 c^3 x+4 a^4 c^3 x^2\right )}{(c-a c x)^4} \, dx}{a^5 c^2}\\ &=-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}+\frac {\int \left (\frac {a^2 \sqrt {1-a^2 x^2}}{c (-1+a x)^4}+\frac {3 a^2 \sqrt {1-a^2 x^2}}{c (-1+a x)^3}+\frac {4 a^2 \sqrt {1-a^2 x^2}}{c (-1+a x)^2}\right ) \, dx}{a^5 c^2}\\ &=-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}+\frac {\int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^4} \, dx}{a^3 c^3}+\frac {3 \int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^3} \, dx}{a^3 c^3}+\frac {4 \int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^2} \, dx}{a^3 c^3}\\ &=\frac {8 \sqrt {1-a^2 x^2}}{a^4 c^3 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^4 c^3 (1-a x)^4}-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}-\frac {\int \frac {\sqrt {1-a^2 x^2}}{(-1+a x)^3} \, dx}{5 a^3 c^3}-\frac {4 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^3 c^3}\\ &=\frac {8 \sqrt {1-a^2 x^2}}{a^4 c^3 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^4 c^3 (1-a x)^4}-\frac {14 \left (1-a^2 x^2\right )^{3/2}}{15 a^4 c^3 (1-a x)^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}-\frac {4 \sin ^{-1}(a x)}{a^4 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 72, normalized size = 0.53 \[ \frac {\frac {\sqrt {a x+1} \left (-15 a^3 x^3+149 a^2 x^2-222 a x+94\right )}{(1-a x)^{5/2}}+120 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{15 a^4 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((Sqrt[1 + a*x]*(94 - 222*a*x + 149*a^2*x^2 - 15*a^3*x^3))/(1 - a*x)^(5/2) + 120*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]
)/(15*a^4*c^3)

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fricas [A]  time = 0.49, size = 145, normalized size = 1.06 \[ \frac {94 \, a^{3} x^{3} - 282 \, a^{2} x^{2} + 282 \, a x + 120 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{3} x^{3} - 149 \, a^{2} x^{2} + 222 \, a x - 94\right )} \sqrt {-a^{2} x^{2} + 1} - 94}{15 \, {\left (a^{7} c^{3} x^{3} - 3 \, a^{6} c^{3} x^{2} + 3 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(94*a^3*x^3 - 282*a^2*x^2 + 282*a*x + 120*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) -
1)/(a*x)) + (15*a^3*x^3 - 149*a^2*x^2 + 222*a*x - 94)*sqrt(-a^2*x^2 + 1) - 94)/(a^7*c^3*x^3 - 3*a^6*c^3*x^2 +
3*a^5*c^3*x - a^4*c^3)

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giac [A]  time = 0.19, size = 186, normalized size = 1.36 \[ -\frac {4 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{a^{3} c^{3} {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{3}} - \frac {2 \, {\left (\frac {335 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {505 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {285 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {60 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 79\right )}}{15 \, a^{3} c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \]

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)^3,x, algorithm="giac")

[Out]

-4*arcsin(a*x)*sgn(a)/(a^3*c^3*abs(a)) + sqrt(-a^2*x^2 + 1)/(a^4*c^3) - 2/15*(335*(sqrt(-a^2*x^2 + 1)*abs(a) +
a)/(a^2*x) - 505*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 285*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3
) - 60*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 79)/(a^3*c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1
)^5*abs(a))

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maple [A]  time = 0.05, size = 186, normalized size = 1.36 \[ \frac {\sqrt {-a^{2} x^{2}+1}}{c^{3} a^{4}}-\frac {4 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{3} a^{3} \sqrt {a^{2}}}-\frac {31 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 c^{3} a^{6} \left (x -\frac {1}{a}\right )^{2}}-\frac {104 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 c^{3} a^{5} \left (x -\frac {1}{a}\right )}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 c^{3} a^{7} \left (x -\frac {1}{a}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3/a^4*(-a^2*x^2+1)^(1/2)-4/c^3/a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-31/15/c^3/a^6/(x-1
/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-104/15/c^3/a^5/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5/c^3/a^7
/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)

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maxima [A]  time = 0.46, size = 163, normalized size = 1.19 \[ -\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{7} c^{3} x^{3} - 3 \, a^{6} c^{3} x^{2} + 3 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} - \frac {31 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{6} c^{3} x^{2} - 2 \, a^{5} c^{3} x + a^{4} c^{3}\right )}} - \frac {104 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{5} c^{3} x - a^{4} c^{3}\right )}} - \frac {4 \, \arcsin \left (a x\right )}{a^{4} c^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^3 - 3*a^6*c^3*x^2 + 3*a^5*c^3*x - a^4*c^3) - 31/15*sqrt(-a^2*x^2 + 1)/(a^6*
c^3*x^2 - 2*a^5*c^3*x + a^4*c^3) - 104/15*sqrt(-a^2*x^2 + 1)/(a^5*c^3*x - a^4*c^3) - 4*arcsin(a*x)/(a^4*c^3) +
sqrt(-a^2*x^2 + 1)/(a^4*c^3)

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mupad [B]  time = 0.05, size = 234, normalized size = 1.71 \[ \frac {\sqrt {1-a^2\,x^2}}{a^4\,c^3}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (a^2\,c^3\,\sqrt {-a^2}+3\,a^4\,c^3\,x^2\,\sqrt {-a^2}-a^5\,c^3\,x^3\,\sqrt {-a^2}-3\,a^3\,c^3\,x\,\sqrt {-a^2}\right )}-\frac {104\,\sqrt {1-a^2\,x^2}}{15\,\left (a^2\,c^3\,\sqrt {-a^2}-a^3\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {31\,\sqrt {1-a^2\,x^2}}{15\,\left (a^6\,c^3\,x^2-2\,a^5\,c^3\,x+a^4\,c^3\right )}-\frac {4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^3\,c^3\,\sqrt {-a^2}} \]

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)^3),x)

[Out]

(1 - a^2*x^2)^(1/2)/(a^4*c^3) - (2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(a^2*c^3*(-a^2)^(1/2) + 3*a^4*c^3*x^2*
(-a^2)^(1/2) - a^5*c^3*x^3*(-a^2)^(1/2) - 3*a^3*c^3*x*(-a^2)^(1/2))) - (104*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3*
(-a^2)^(1/2) - a^3*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (31*(1 - a^2*x^2)^(1/2))/(15*(a^4*c^3 - 2*a^5*c^3*x + a
^6*c^3*x^2)) - (4*asinh(x*(-a^2)^(1/2)))/(a^3*c^3*(-a^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(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) + Integral(a*x**4/(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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