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

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

[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)

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Rubi [A]  time = 0.285799, antiderivative size = 114, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0509299, 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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Maple [A]  time = 0.042, size = 142, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}}{3\,{a}^{2}c}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{8}{3\,{a}^{4}c}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{x}{{a}^{3}c}\sqrt{-{a}^{2}{x}^{2}+1}}-3\,{\frac{1}{{a}^{3}c\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }-2\,{\frac{1}{c{a}^{5}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.5502, size = 198, normalized size = 1.74 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \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} \end{align*}

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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Giac [A]  time = 1.3224, size = 136, normalized size = 1.19 \begin{align*} \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}\left (a\right )}{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 |}} \end{align*}

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))