3.336 \(\int \frac{e^{\tanh ^{-1}(a x)} x^4}{(c-a c x)^2} \, dx\)

Optimal. Leaf size=159 \[ -\frac{2 (a x+1)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}+\frac{(a x+1)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{(a x+5)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}-\frac{(a x+5) \sqrt{1-a^2 x^2}}{6 a^5 c^2}-\frac{5 \sqrt{1-a^2 x^2}}{2 a^5 c^2}+\frac{17 \sin ^{-1}(a x)}{2 a^5 c^2} \]

[Out]

(1 + a*x)^3/(3*a^5*c^2*(1 - a^2*x^2)^(3/2)) - (2*(1 + a*x)^3)/(a^5*c^2*Sqrt[1 - a^2*x^2]) - (5*Sqrt[1 - a^2*x^
2])/(2*a^5*c^2) - ((5 + a*x)*Sqrt[1 - a^2*x^2])/(6*a^5*c^2) - ((5 + a*x)^2*Sqrt[1 - a^2*x^2])/(3*a^5*c^2) + (1
7*ArcSin[a*x])/(2*a^5*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.517402, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {6128, 852, 1635, 1625, 1654, 21, 743, 641, 216} \[ -\frac{2 (a x+1)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}+\frac{(a x+1)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{(a x+5)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}-\frac{(a x+5) \sqrt{1-a^2 x^2}}{6 a^5 c^2}-\frac{5 \sqrt{1-a^2 x^2}}{2 a^5 c^2}+\frac{17 \sin ^{-1}(a x)}{2 a^5 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x^4)/(c - a*c*x)^2,x]

[Out]

(1 + a*x)^3/(3*a^5*c^2*(1 - a^2*x^2)^(3/2)) - (2*(1 + a*x)^3)/(a^5*c^2*Sqrt[1 - a^2*x^2]) - (5*Sqrt[1 - a^2*x^
2])/(2*a^5*c^2) - ((5 + a*x)*Sqrt[1 - a^2*x^2])/(6*a^5*c^2) - ((5 + a*x)^2*Sqrt[1 - a^2*x^2])/(3*a^5*c^2) + (1
7*ArcSin[a*x])/(2*a^5*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 1625

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + 1)*Polynomial
Quotient[Pq, d + e*x, x]*(a + c*x^2)^p, x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[PolynomialRe
mainder[Pq, d + e*x, x], 0]

Rule 1654

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 - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

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^4}{(c-a c x)^2} \, dx &=c \int \frac{x^4 \sqrt{1-a^2 x^2}}{(c-a c x)^3} \, dx\\ &=\frac{\int \frac{x^4 (c+a c x)^3}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^5}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{\int \frac{(c+a c x)^2 \left (\frac{3}{a^4}+\frac{3 x}{a^3}+\frac{3 x^2}{a^2}+\frac{3 x^3}{a}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^4}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{\int \frac{(c+a c x)^3 \left (\frac{3}{a^4 c}+\frac{3 x^2}{a^2 c}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^4}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{\left (\frac{15}{a^4 c}+\frac{3 x}{a^3 c}\right ) (c+a c x)^2}{\sqrt{1-a^2 x^2}} \, dx}{3 c^3}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}-\frac{(5+a x)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}-\frac{a^4 \int \frac{\left (-\frac{45}{a^4}-\frac{9 x}{a^3}\right ) \left (\frac{15}{a^4 c}+\frac{3 x}{a^3 c}\right )}{\sqrt{1-a^2 x^2}} \, dx}{81 c}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}-\frac{(5+a x)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}+\frac{a^4 \int \frac{\left (-\frac{45}{a^4}-\frac{9 x}{a^3}\right )^2}{\sqrt{1-a^2 x^2}} \, dx}{243 c^2}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}-\frac{(5+a x) \sqrt{1-a^2 x^2}}{6 a^5 c^2}-\frac{(5+a x)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}-\frac{a^2 \int \frac{-\frac{4131}{a^6}-\frac{1215 x}{a^5}}{\sqrt{1-a^2 x^2}} \, dx}{486 c^2}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 a^5 c^2}-\frac{(5+a x) \sqrt{1-a^2 x^2}}{6 a^5 c^2}-\frac{(5+a x)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}+\frac{17 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{2 a^4 c^2}\\ &=\frac{(1+a x)^3}{3 a^5 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{2 (1+a x)^3}{a^5 c^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 a^5 c^2}-\frac{(5+a x) \sqrt{1-a^2 x^2}}{6 a^5 c^2}-\frac{(5+a x)^2 \sqrt{1-a^2 x^2}}{3 a^5 c^2}+\frac{17 \sin ^{-1}(a x)}{2 a^5 c^2}\\ \end{align*}

Mathematica [A]  time = 0.094149, size = 80, normalized size = 0.5 \[ -\frac{\frac{\sqrt{a x+1} \left (2 a^4 x^4+5 a^3 x^3+18 a^2 x^2-109 a x+80\right )}{(1-a x)^{3/2}}+102 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{6 a^5 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*x^4)/(c - a*c*x)^2,x]

[Out]

-((Sqrt[1 + a*x]*(80 - 109*a*x + 18*a^2*x^2 + 5*a^3*x^3 + 2*a^4*x^4))/(1 - a*x)^(3/2) + 102*ArcSin[Sqrt[1 - a*
x]/Sqrt[2]])/(6*a^5*c^2)

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 187, normalized size = 1.2 \begin{align*} -{\frac{{x}^{2}}{3\,{c}^{2}{a}^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{17}{3\,{a}^{5}{c}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3\,x}{2\,{a}^{4}{c}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{17}{2\,{a}^{4}{c}^{2}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{2}{3\,{c}^{2}{a}^{7}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{25}{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 ) ^{-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^4/(-a*c*x+c)^2,x)

[Out]

-1/3/c^2/a^3*x^2*(-a^2*x^2+1)^(1/2)-17/3*(-a^2*x^2+1)^(1/2)/a^5/c^2-3/2/c^2/a^4*x*(-a^2*x^2+1)^(1/2)+17/2/c^2/
a^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/3/c^2/a^7/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1
/2)+25/3/c^2/a^6/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)

________________________________________________________________________________________

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^4/(-a*c*x+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.63265, size = 286, normalized size = 1.8 \begin{align*} -\frac{80 \, a^{2} x^{2} - 160 \, a x + 102 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (2 \, a^{4} x^{4} + 5 \, a^{3} x^{3} + 18 \, a^{2} x^{2} - 109 \, a x + 80\right )} \sqrt{-a^{2} x^{2} + 1} + 80}{6 \,{\left (a^{7} c^{2} x^{2} - 2 \, a^{6} c^{2} x + a^{5} 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^4/(-a*c*x+c)^2,x, algorithm="fricas")

[Out]

-1/6*(80*a^2*x^2 - 160*a*x + 102*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^4*x^4 + 5
*a^3*x^3 + 18*a^2*x^2 - 109*a*x + 80)*sqrt(-a^2*x^2 + 1) + 80)/(a^7*c^2*x^2 - 2*a^6*c^2*x + a^5*c^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{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 + \int \frac{a x^{5}}{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**4/(-a*c*x+c)**2,x)

[Out]

(Integral(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) + Inte
gral(a*x**5/(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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )} x^{4}}{\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^4/(-a*c*x+c)^2,x, algorithm="giac")

[Out]

integrate((a*x + 1)*x^4/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)^2), x)