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

Optimal. Leaf size=101 \[ \frac{x^3 (a x+1)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{1-a^2 x^2}}{3 a^3 c}+\frac{(9 a x+16) \sqrt{1-a^2 x^2}}{6 a^5 c}-\frac{3 \sin ^{-1}(a x)}{2 a^5 c} \]

[Out]

(x^3*(1 + a*x))/(a^2*c*Sqrt[1 - a^2*x^2]) + (4*x^2*Sqrt[1 - a^2*x^2])/(3*a^3*c) + ((16 + 9*a*x)*Sqrt[1 - a^2*x
^2])/(6*a^5*c) - (3*ArcSin[a*x])/(2*a^5*c)

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Rubi [A]  time = 0.132048, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {6148, 819, 833, 780, 216} \[ \frac{x^3 (a x+1)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{1-a^2 x^2}}{3 a^3 c}+\frac{(9 a x+16) \sqrt{1-a^2 x^2}}{6 a^5 c}-\frac{3 \sin ^{-1}(a x)}{2 a^5 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(1 + a*x))/(a^2*c*Sqrt[1 - a^2*x^2]) + (4*x^2*Sqrt[1 - a^2*x^2])/(3*a^3*c) + ((16 + 9*a*x)*Sqrt[1 - a^2*x
^2])/(6*a^5*c) - (3*ArcSin[a*x])/(2*a^5*c)

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 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^4}{c-a^2 c x^2} \, dx &=\frac{\int \frac{x^4 (1+a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=\frac{x^3 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}-\frac{\int \frac{x^2 (3+4 a x)}{\sqrt{1-a^2 x^2}} \, dx}{a^2 c}\\ &=\frac{x^3 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{1-a^2 x^2}}{3 a^3 c}+\frac{\int \frac{x \left (-8 a-9 a^2 x\right )}{\sqrt{1-a^2 x^2}} \, dx}{3 a^4 c}\\ &=\frac{x^3 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{1-a^2 x^2}}{3 a^3 c}+\frac{(16+9 a x) \sqrt{1-a^2 x^2}}{6 a^5 c}-\frac{3 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{2 a^4 c}\\ &=\frac{x^3 (1+a x)}{a^2 c \sqrt{1-a^2 x^2}}+\frac{4 x^2 \sqrt{1-a^2 x^2}}{3 a^3 c}+\frac{(16+9 a x) \sqrt{1-a^2 x^2}}{6 a^5 c}-\frac{3 \sin ^{-1}(a x)}{2 a^5 c}\\ \end{align*}

Mathematica [A]  time = 0.0473442, size = 74, normalized size = 0.73 \[ -\frac{2 a^4 x^4+3 a^3 x^3+8 a^2 x^2+9 \sqrt{1-a^2 x^2} \sin ^{-1}(a x)-9 a x-16}{6 a^5 c \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-16 - 9*a*x + 8*a^2*x^2 + 3*a^3*x^3 + 2*a^4*x^4 + 9*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(6*a^5*c*Sqrt[1 - a^2*x^2
])

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Maple [A]  time = 0.04, size = 143, normalized size = 1.4 \begin{align*}{\frac{{x}^{2}}{3\,{a}^{3}c}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{5}{3\,{a}^{5}c}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{x}{2\,{a}^{4}c}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{3}{2\,{a}^{4}c}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-{\frac{1}{c{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^2*c*x^2+c),x)

[Out]

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

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Maxima [B]  time = 1.84586, size = 502, normalized size = 4.97 \begin{align*} -\frac{a^{2} c{\left (\frac{6 \, \sqrt{-a^{2} x^{2} + 1} c}{\sqrt{a^{2} c^{2}} a^{6} c x + a^{6} c^{2}} + \frac{6 \, \sqrt{-a^{2} x^{2} + 1} c}{\sqrt{a^{2} c^{2}} a^{6} c x - a^{6} c^{2}} - \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{7} c x + \sqrt{a^{2} c^{2}} a^{5}} + \frac{6 \, \sqrt{-a^{2} x^{2} + 1}}{a^{7} c x - \sqrt{a^{2} c^{2}} a^{5}} - \frac{4 \, \sqrt{a^{2} c^{2}} \sqrt{-a^{2} x^{2} + 1} x^{2}}{a^{5} c^{2}} - \frac{6 \, \sqrt{a^{2} c^{2}} \sqrt{-a^{2} x^{2} + 1} x}{a^{6} c^{2}} + \frac{25 \, \sqrt{a^{2} c^{2}} \sqrt{-a^{2} x^{2} + 1}}{a^{7} c^{2}} - \frac{30 \, \sqrt{a^{2} c^{2}} \arcsin \left (\frac{x}{c \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}{a^{8} c^{3} \sqrt{\frac{1}{a^{2} c^{2}}}} - \frac{45 \, \left (a^{2} c^{2}\right )^{\frac{3}{2}} \sqrt{-a^{2} x^{2} + 1}}{a^{9} c^{4}} + \frac{48 \, \left (a^{2} c^{2}\right )^{\frac{3}{2}} \arcsin \left (\frac{x}{c \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}{a^{10} c^{5} \sqrt{\frac{1}{a^{2} c^{2}}}}\right )}}{12 \, \sqrt{a^{2} 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^2*c*x^2+c),x, algorithm="maxima")

[Out]

-1/12*a^2*c*(6*sqrt(-a^2*x^2 + 1)*c/(sqrt(a^2*c^2)*a^6*c*x + a^6*c^2) + 6*sqrt(-a^2*x^2 + 1)*c/(sqrt(a^2*c^2)*
a^6*c*x - a^6*c^2) - 6*sqrt(-a^2*x^2 + 1)/(a^7*c*x + sqrt(a^2*c^2)*a^5) + 6*sqrt(-a^2*x^2 + 1)/(a^7*c*x - sqrt
(a^2*c^2)*a^5) - 4*sqrt(a^2*c^2)*sqrt(-a^2*x^2 + 1)*x^2/(a^5*c^2) - 6*sqrt(a^2*c^2)*sqrt(-a^2*x^2 + 1)*x/(a^6*
c^2) + 25*sqrt(a^2*c^2)*sqrt(-a^2*x^2 + 1)/(a^7*c^2) - 30*sqrt(a^2*c^2)*arcsin(x/(c*sqrt(1/(a^2*c^2))))/(a^8*c
^3*sqrt(1/(a^2*c^2))) - 45*(a^2*c^2)^(3/2)*sqrt(-a^2*x^2 + 1)/(a^9*c^4) + 48*(a^2*c^2)^(3/2)*arcsin(x/(c*sqrt(
1/(a^2*c^2))))/(a^10*c^5*sqrt(1/(a^2*c^2))))/sqrt(a^2*c^2)

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Fricas [A]  time = 1.59177, size = 198, normalized size = 1.96 \begin{align*} \frac{16 \, a x + 18 \,{\left (a x - 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (2 \, a^{3} x^{3} + a^{2} x^{2} + 7 \, a x - 16\right )} \sqrt{-a^{2} x^{2} + 1} - 16}{6 \,{\left (a^{6} c x - a^{5} 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^4/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

1/6*(16*a*x + 18*(a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^3*x^3 + a^2*x^2 + 7*a*x - 16)*sqrt(-a
^2*x^2 + 1) - 16)/(a^6*c*x - a^5*c)

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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} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x^{5}}{- a^{2} x^{2} \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**4/(-a**2*c*x**2+c),x)

[Out]

(Integral(x**4/(-a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**5/(-a**2*x**2*sqrt
(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x))/c

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Giac [A]  time = 1.19979, size = 138, normalized size = 1.37 \begin{align*} \frac{1}{6} \, \sqrt{-a^{2} x^{2} + 1}{\left (x{\left (\frac{2 \, x}{a^{3} c} + \frac{3}{a^{4} c}\right )} + \frac{10}{a^{5} c}\right )} - \frac{3 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \, a^{4} c{\left | a \right |}} + \frac{2}{a^{4} 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^4/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

1/6*sqrt(-a^2*x^2 + 1)*(x*(2*x/(a^3*c) + 3/(a^4*c)) + 10/(a^5*c)) - 3/2*arcsin(a*x)*sgn(a)/(a^4*c*abs(a)) + 2/
(a^4*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)*abs(a))