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

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

[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)

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Rubi [A]  time = 0.304152, antiderivative size = 104, normalized size of antiderivative = 1., 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 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 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 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)^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*}

Mathematica [A]  time = 0.0834507, 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]

-((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]])/(
6*a^4*c^2)

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Maple [A]  time = 0.044, size = 164, normalized size = 1.6 \begin{align*} -{\frac{x}{2\,{c}^{2}{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{11}{2\,{c}^{2}{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-3\,{\frac{\sqrt{-{a}^{2}{x}^{2}+1}}{{a}^{4}{c}^{2}}}+{\frac{2}{3\,{c}^{2}{a}^{6}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{19}{3\,{c}^{2}{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)^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 [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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60952, size = 267, normalized size = 2.57 \begin{align*} -\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 )}} \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)^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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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}} \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)**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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{3}}{\sqrt{-a^{2} x^{2} + 1}{\left (a c x - c\right )}^{2}}\,{d x} \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)^2,x, algorithm="giac")

[Out]

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