3.1117 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^2}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ \frac{\tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^3 c^{3/2}}+\frac{(a x+1)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{5 (a x+1)}{3 a^3 c \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.262683, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6151, 1635, 778, 217, 203} \[ \frac{\tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^3 c^{3/2}}+\frac{(a x+1)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{5 (a x+1)}{3 a^3 c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(3*a^3*(c - a^2*c*x^2)^(3/2)) - (5*(1 + a*x))/(3*a^3*c*Sqrt[c - a^2*c*x^2]) + ArcTan[(a*Sqrt[c]*x)
/Sqrt[c - a^2*c*x^2]]/(a^3*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 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[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^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx &=c \int \frac{x^2 (1+a x)^2}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\\ &=\frac{(1+a x)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{1}{3} \int \frac{\left (\frac{2}{a^2}+\frac{3 x}{a}\right ) (1+a x)}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac{(1+a x)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{5 (1+a x)}{3 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{a^2 c}\\ &=\frac{(1+a x)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{5 (1+a x)}{3 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\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^2 c}\\ &=\frac{(1+a x)^2}{3 a^3 \left (c-a^2 c x^2\right )^{3/2}}-\frac{5 (1+a x)}{3 a^3 c \sqrt{c-a^2 c x^2}}+\frac{\tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^3 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.116325, size = 82, normalized size = 0.88 \[ \frac{\frac{(5 a x-4) \sqrt{c-a^2 c x^2}}{(a x-1)^2}-3 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{3 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(-1 + a*x)^2 - 3*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a
^2*x^2))])/(3*a^3*c^2)

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Maple [B]  time = 0.039, size = 166, normalized size = 1.8 \begin{align*} -3\,{\frac{x}{{a}^{2}\sqrt{-{a}^{2}c{x}^{2}+c}c}}+{\frac{1}{{a}^{2}c}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-2\,{\frac{1}{{a}^{3}\sqrt{-{a}^{2}c{x}^{2}+c}c}}-{\frac{2}{3\,{a}^{4}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}^{2}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^2/(-a^2*c*x^2+c)^(3/2),x)

[Out]

-3*x/a^2/c/(-a^2*c*x^2+c)^(1/2)+1/a^2/c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/a^3/c/(-a
^2*c*x^2+c)^(1/2)-2/3/a^4/c/(x-1/a)/(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)+4/3/a^2/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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.75023, size = 471, normalized size = 5.06 \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 ) - 2 \, \sqrt{-a^{2} c x^{2} + c}{\left (5 \, a x - 4\right )}}{6 \,{\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} c^{2}\right )}}, -\frac{3 \,{\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 (5 \, a x - 4\right )}}{3 \,{\left (a^{5} c^{2} x^{2} - 2 \, a^{4} c^{2} x + a^{3} 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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/6*(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) - 2*sqrt(-a
^2*c*x^2 + c)*(5*a*x - 4))/(a^5*c^2*x^2 - 2*a^4*c^2*x + a^3*c^2), -1/3*(3*(a^2*x^2 - 2*a*x + 1)*sqrt(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)*(5*a*x - 4))/(a^5*c^2*x^2 - 2*a^4*c^
2*x + a^3*c^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2}}{- 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^{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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x**2/(-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**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)

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Giac [A]  time = 1.21104, size = 58, normalized size = 0.62 \begin{align*} \frac{\sqrt{-c} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{a^{2} c^{2}{\left | a \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^2/(-a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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