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

Optimal. Leaf size=114 \[ -\frac {3 \sin ^{-1}(a x)}{a^4 c}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}+\frac {11 \sqrt {1-a^2 x^2}}{3 a^4 c}+\frac {(a x+1)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6128, 852, 1635, 1815, 641, 216} \[ \frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c}+\frac {11 \sqrt {1-a^2 x^2}}{3 a^4 c}+\frac {(a x+1)^2}{a^4 c \sqrt {1-a^2 x^2}}-\frac {3 \sin ^{-1}(a x)}{a^4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^4*c*Sqrt[1 - a^2*x^2]) + (11*Sqrt[1 - a^2*x^2])/(3*a^4*c) + (x*Sqrt[1 - a^2*x^2])/(a^3*c) + (x^
2*Sqrt[1 - a^2*x^2])/(3*a^2*c) - (3*ArcSin[a*x])/(a^4*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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

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} \, dx &=c \int \frac {x^3 \sqrt {1-a^2 x^2}}{(c-a c x)^2} \, dx\\ &=\frac {\int \frac {x^3 (c+a c x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}-\frac {\int \frac {(c+a c x) \left (\frac {2}{a^3}+\frac {x}{a^2}+\frac {x^2}{a}\right )}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}+\frac {\int \frac {-\frac {6 c}{a}-11 c x-6 a c x^2}{\sqrt {1-a^2 x^2}} \, dx}{3 a^2 c^2}\\ &=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}-\frac {\int \frac {18 a c+22 a^2 c x}{\sqrt {1-a^2 x^2}} \, dx}{6 a^4 c^2}\\ &=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {11 \sqrt {1-a^2 x^2}}{3 a^4 c}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}-\frac {3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^3 c}\\ &=\frac {(1+a x)^2}{a^4 c \sqrt {1-a^2 x^2}}+\frac {11 \sqrt {1-a^2 x^2}}{3 a^4 c}+\frac {x \sqrt {1-a^2 x^2}}{a^3 c}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a^2 c}-\frac {3 \sin ^{-1}(a x)}{a^4 c}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 72, normalized size = 0.63 \[ \frac {18 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-\frac {\sqrt {a x+1} \left (a^3 x^3+2 a^2 x^2+5 a x-14\right )}{\sqrt {1-a x}}}{3 a^4 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((Sqrt[1 + a*x]*(-14 + 5*a*x + 2*a^2*x^2 + a^3*x^3))/Sqrt[1 - a*x]) + 18*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(3*a
^4*c)

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fricas [A]  time = 0.60, size = 86, normalized size = 0.75 \[ \frac {14 \, a x + 18 \, {\left (a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (a^{3} x^{3} + 2 \, a^{2} x^{2} + 5 \, a x - 14\right )} \sqrt {-a^{2} x^{2} + 1} - 14}{3 \, {\left (a^{5} c x - a^{4} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(14*a*x + 18*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (a^3*x^3 + 2*a^2*x^2 + 5*a*x - 14)*sqrt(-a
^2*x^2 + 1) - 14)/(a^5*c*x - a^4*c)

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giac [A]  time = 0.21, size = 101, normalized size = 0.89 \[ \frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} {\left (x {\left (\frac {x}{a^{2} c} + \frac {3}{a^{3} c}\right )} + \frac {8}{a^{4} c}\right )} - \frac {3 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{a^{3} c {\left | a \right |}} + \frac {4}{a^{3} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} \]

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),x, algorithm="giac")

[Out]

1/3*sqrt(-a^2*x^2 + 1)*(x*(x/(a^2*c) + 3/(a^3*c)) + 8/(a^4*c)) - 3*arcsin(a*x)*sgn(a)/(a^3*c*abs(a)) + 4/(a^3*
c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))

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maple [A]  time = 0.04, size = 142, normalized size = 1.25 \[ \frac {x^{2} \sqrt {-a^{2} x^{2}+1}}{3 a^{2} c}+\frac {8 \sqrt {-a^{2} x^{2}+1}}{3 a^{4} c}+\frac {x \sqrt {-a^{2} x^{2}+1}}{a^{3} c}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c \,a^{3} \sqrt {a^{2}}}-\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{c \,a^{5} \left (x -\frac {1}{a}\right )} \]

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),x)

[Out]

1/3*x^2*(-a^2*x^2+1)^(1/2)/a^2/c+8/3*(-a^2*x^2+1)^(1/2)/a^4/c+x*(-a^2*x^2+1)^(1/2)/a^3/c-3/c/a^3/(a^2)^(1/2)*a
rctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-2/c/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 = 105, normalized size = 0.92 \[ -\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{a^{5} c x - a^{4} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{2}}{3 \, a^{2} c} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{3} c} - \frac {3 \, \arcsin \left (a x\right )}{a^{4} c} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, a^{4} c} \]

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.82, size = 158, normalized size = 1.39 \[ \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{3\,c\,{\left (-a^2\right )}^{3/2}}-\frac {2}{a^2\,c\,\sqrt {-a^2}}+\frac {a^2\,x^2}{3\,c\,{\left (-a^2\right )}^{3/2}}+\frac {x\,\sqrt {-a^2}}{a^3\,c}\right )}{\sqrt {-a^2}}-\frac {3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^3\,c\,\sqrt {-a^2}}+\frac {2\,\sqrt {1-a^2\,x^2}}{a^3\,c\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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