3.1116 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{\sqrt{c-a^2 c x^2}}{a^4 c^2}+\frac{2 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^4 c^{3/2}}+\frac{(a x+1)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (a x+1)}{3 a^4 c \sqrt{c-a^2 c x^2}} \]

[Out]

(1 + a*x)^2/(3*a^4*(c - a^2*c*x^2)^(3/2)) - (8*(1 + a*x))/(3*a^4*c*Sqrt[c - a^2*c*x^2]) - Sqrt[c - a^2*c*x^2]/
(a^4*c^2) + (2*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(a^4*c^(3/2))

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Rubi [A]  time = 0.320685, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6151, 1635, 641, 217, 203} \[ -\frac{\sqrt{c-a^2 c x^2}}{a^4 c^2}+\frac{2 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^4 c^{3/2}}+\frac{(a x+1)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (a x+1)}{3 a^4 c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*x^3)/(c - a^2*c*x^2)^(3/2),x]

[Out]

(1 + a*x)^2/(3*a^4*(c - a^2*c*x^2)^(3/2)) - (8*(1 + a*x))/(3*a^4*c*Sqrt[c - a^2*c*x^2]) - Sqrt[c - a^2*c*x^2]/
(a^4*c^2) + (2*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(a^4*c^(3/2))

Rule 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[x^m*(c
+ d*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] ||
GtQ[c, 0]) && IGtQ[n/2, 0]

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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{2 \tanh ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx &=c \int \frac{x^3 (1+a x)^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\\ &=\frac{(1+a x)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{1}{3} \int \frac{(1+a x) \left (\frac{2}{a^3}+\frac{3 x}{a^2}+\frac{3 x^2}{a}\right )}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac{(1+a x)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (1+a x)}{3 a^4 c \sqrt{c-a^2 c x^2}}+\frac{\int \frac{\frac{6}{a^3}+\frac{3 x}{a^2}}{\sqrt{c-a^2 c x^2}} \, dx}{3 c}\\ &=\frac{(1+a x)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (1+a x)}{3 a^4 c \sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{a^4 c^2}+\frac{2 \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{a^3 c}\\ &=\frac{(1+a x)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (1+a x)}{3 a^4 c \sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{a^4 c^2}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )}{a^3 c}\\ &=\frac{(1+a x)^2}{3 a^4 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8 (1+a x)}{3 a^4 c \sqrt{c-a^2 c x^2}}-\frac{\sqrt{c-a^2 c x^2}}{a^4 c^2}+\frac{2 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^4 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.166011, size = 90, normalized size = 0.77 \[ \frac{\frac{\left (-3 a^2 x^2+14 a x-10\right ) \sqrt{c-a^2 c x^2}}{(a x-1)^2}-6 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{3 a^4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*x^3)/(c - a^2*c*x^2)^(3/2),x]

[Out]

(((-10 + 14*a*x - 3*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(-1 + a*x)^2 - 6*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(S
qrt[c]*(-1 + a^2*x^2))])/(3*a^4*c^2)

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Maple [A]  time = 0.04, size = 190, normalized size = 1.6 \begin{align*}{\frac{{x}^{2}}{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}}-4\,{\frac{1}{{a}^{4}\sqrt{-{a}^{2}c{x}^{2}+c}c}}-4\,{\frac{x}{{a}^{3}\sqrt{-{a}^{2}c{x}^{2}+c}c}}+2\,{\frac{1}{{a}^{3}c\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-{a}^{2}c{x}^{2}+c}}} \right ) }-{\frac{2}{3\,{a}^{5}c} \left ( x-{a}^{-1} \right ) ^{-1}{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}}+{\frac{4\,x}{3\,{a}^{3}c}{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.79853, size = 506, normalized size = 4.32 \begin{align*} \left [-\frac{3 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt{-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (3 \, a^{2} x^{2} - 14 \, a x + 10\right )}}{3 \,{\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}}, -\frac{6 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (3 \, a^{2} x^{2} - 14 \, a x + 10\right )}}{3 \,{\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) + sqrt(-a^2
*c*x^2 + c)*(3*a^2*x^2 - 14*a*x + 10))/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2), -1/3*(6*(a^2*x^2 - 2*a*x + 1)*sq
rt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + sqrt(-a^2*c*x^2 + c)*(3*a^2*x^2 - 14*a*x + 10
))/(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*} - \int \frac{x^{3}}{- a^{3} c x^{3} \sqrt{- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt{- a^{2} c x^{2} + c} + a c x \sqrt{- a^{2} c x^{2} + c} - c \sqrt{- a^{2} c x^{2} + c}}\, dx - \int \frac{a x^{4}}{- a^{3} c x^{3} \sqrt{- a^{2} c x^{2} + c} + a^{2} c x^{2} \sqrt{- a^{2} c x^{2} + c} + a c x \sqrt{- a^{2} c x^{2} + c} - c \sqrt{- a^{2} c x^{2} + c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21464, size = 89, normalized size = 0.76 \begin{align*} \frac{2 \, \sqrt{-c} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{a^{3} c^{2}{\left | a \right |}} - \frac{\sqrt{-a^{2} c x^{2} + c}}{a^{4} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(-c)*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a^3*c^2*abs(a)) - sqrt(-a^2*c*x^2 + c)/(a^4*c^2)