3.1087 \(\int e^{2 \tanh ^{-1}(a x)} x^2 (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=136 \[ \frac{3 c^{3/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{16 a^3}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}+\frac{3 c x \sqrt{c-a^2 c x^2}}{16 a^2}-\frac{(45 a x+32) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3} \]

[Out]

(3*c*x*Sqrt[c - a^2*c*x^2])/(16*a^2) - (2*x^2*(c - a^2*c*x^2)^(3/2))/(5*a) - (x^3*(c - a^2*c*x^2)^(3/2))/6 - (
(32 + 45*a*x)*(c - a^2*c*x^2)^(3/2))/(120*a^3) + (3*c^(3/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(16*a^3
)

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Rubi [A]  time = 0.308353, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {6151, 1809, 833, 780, 195, 217, 203} \[ \frac{3 c^{3/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{16 a^3}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}+\frac{3 c x \sqrt{c-a^2 c x^2}}{16 a^2}-\frac{(45 a x+32) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c*x*Sqrt[c - a^2*c*x^2])/(16*a^2) - (2*x^2*(c - a^2*c*x^2)^(3/2))/(5*a) - (x^3*(c - a^2*c*x^2)^(3/2))/6 - (
(32 + 45*a*x)*(c - a^2*c*x^2)^(3/2))/(120*a^3) + (3*c^(3/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(16*a^3
)

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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 e^{2 \tanh ^{-1}(a x)} x^2 \left (c-a^2 c x^2\right )^{3/2} \, dx &=c \int x^2 (1+a x)^2 \sqrt{c-a^2 c x^2} \, dx\\ &=-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{\int x^2 \left (-9 a^2 c-12 a^3 c x\right ) \sqrt{c-a^2 c x^2} \, dx}{6 a^2}\\ &=-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}+\frac{\int x \left (24 a^3 c^2+45 a^4 c^2 x\right ) \sqrt{c-a^2 c x^2} \, dx}{30 a^4 c}\\ &=-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac{(3 c) \int \sqrt{c-a^2 c x^2} \, dx}{8 a^2}\\ &=\frac{3 c x \sqrt{c-a^2 c x^2}}{16 a^2}-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac{\left (3 c^2\right ) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{16 a^2}\\ &=\frac{3 c x \sqrt{c-a^2 c x^2}}{16 a^2}-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac{\left (3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )}{16 a^2}\\ &=\frac{3 c x \sqrt{c-a^2 c x^2}}{16 a^2}-\frac{2 x^2 \left (c-a^2 c x^2\right )^{3/2}}{5 a}-\frac{1}{6} x^3 \left (c-a^2 c x^2\right )^{3/2}-\frac{(32+45 a x) \left (c-a^2 c x^2\right )^{3/2}}{120 a^3}+\frac{3 c^{3/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{16 a^3}\\ \end{align*}

Mathematica [A]  time = 0.118549, size = 105, normalized size = 0.77 \[ \frac{c \left (40 a^5 x^5+96 a^4 x^4+50 a^3 x^3-32 a^2 x^2-45 a x-64\right ) \sqrt{c-a^2 c x^2}-45 c^{3/2} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{240 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c - a^2*c*x^2]*(-64 - 45*a*x - 32*a^2*x^2 + 50*a^3*x^3 + 96*a^4*x^4 + 40*a^5*x^5) - 45*c^(3/2)*ArcTan[
(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(240*a^3)

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Maple [B]  time = 0.04, size = 244, normalized size = 1.8 \begin{align*}{\frac{x}{6\,{a}^{2}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{13\,x}{24\,{a}^{2}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{13\,cx}{16\,{a}^{2}}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{13\,{c}^{2}}{16\,{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}+{\frac{2}{5\,{a}^{3}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{2}{3\,{a}^{3}} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{cx}{{a}^{2}}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}+{\frac{{c}^{2}}{{a}^{2}}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}} \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]

1/6*x*(-a^2*c*x^2+c)^(5/2)/a^2/c-13/24/a^2*x*(-a^2*c*x^2+c)^(3/2)-13/16*c*x*(-a^2*c*x^2+c)^(1/2)/a^2-13/16/a^2
*c^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/5/a^3*(-a^2*c*x^2+c)^(5/2)/c-2/3/a^3*(-c*a^2
*(x-1/a)^2-2*a*c*(x-1/a))^(3/2)+1/a^2*c*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)*x+1/a^2*c^2/(a^2*c)^(1/2)*arcta
n((a^2*c)^(1/2)*x/(-c*a^2*(x-1/a)^2-2*a*c*(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)^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 = 2.66154, size = 508, normalized size = 3.74 \begin{align*} \left [\frac{45 \, \sqrt{-c} c \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) + 2 \,{\left (40 \, a^{5} c x^{5} + 96 \, a^{4} c x^{4} + 50 \, a^{3} c x^{3} - 32 \, a^{2} c x^{2} - 45 \, a c x - 64 \, c\right )} \sqrt{-a^{2} c x^{2} + c}}{480 \, a^{3}}, -\frac{45 \, c^{\frac{3}{2}} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) -{\left (40 \, a^{5} c x^{5} + 96 \, a^{4} c x^{4} + 50 \, a^{3} c x^{3} - 32 \, a^{2} c x^{2} - 45 \, a c x - 64 \, c\right )} \sqrt{-a^{2} c x^{2} + c}}{240 \, a^{3}}\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]

[1/480*(45*sqrt(-c)*c*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) + 2*(40*a^5*c*x^5 + 96*a^4*c*
x^4 + 50*a^3*c*x^3 - 32*a^2*c*x^2 - 45*a*c*x - 64*c)*sqrt(-a^2*c*x^2 + c))/a^3, -1/240*(45*c^(3/2)*arctan(sqrt
(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) - (40*a^5*c*x^5 + 96*a^4*c*x^4 + 50*a^3*c*x^3 - 32*a^2*c*x^2 - 4
5*a*c*x - 64*c)*sqrt(-a^2*c*x^2 + c))/a^3]

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Sympy [C]  time = 13.3175, size = 515, normalized size = 3.79 \begin{align*} a^{2} c \left (\begin{cases} \frac{i a^{2} \sqrt{c} x^{7}}{6 \sqrt{a^{2} x^{2} - 1}} - \frac{5 i \sqrt{c} x^{5}}{24 \sqrt{a^{2} x^{2} - 1}} - \frac{i \sqrt{c} x^{3}}{48 a^{2} \sqrt{a^{2} x^{2} - 1}} + \frac{i \sqrt{c} x}{16 a^{4} \sqrt{a^{2} x^{2} - 1}} - \frac{i \sqrt{c} \operatorname{acosh}{\left (a x \right )}}{16 a^{5}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{a^{2} \sqrt{c} x^{7}}{6 \sqrt{- a^{2} x^{2} + 1}} + \frac{5 \sqrt{c} x^{5}}{24 \sqrt{- a^{2} x^{2} + 1}} + \frac{\sqrt{c} x^{3}}{48 a^{2} \sqrt{- a^{2} x^{2} + 1}} - \frac{\sqrt{c} x}{16 a^{4} \sqrt{- a^{2} x^{2} + 1}} + \frac{\sqrt{c} \operatorname{asin}{\left (a x \right )}}{16 a^{5}} & \text{otherwise} \end{cases}\right ) + 2 a c \left (\begin{cases} \frac{x^{4} \sqrt{- a^{2} c x^{2} + c}}{5} - \frac{x^{2} \sqrt{- a^{2} c x^{2} + c}}{15 a^{2}} - \frac{2 \sqrt{- a^{2} c x^{2} + c}}{15 a^{4}} & \text{for}\: a \neq 0 \\\frac{\sqrt{c} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + c \left (\begin{cases} \frac{i a^{2} \sqrt{c} x^{5}}{4 \sqrt{a^{2} x^{2} - 1}} - \frac{3 i \sqrt{c} x^{3}}{8 \sqrt{a^{2} x^{2} - 1}} + \frac{i \sqrt{c} x}{8 a^{2} \sqrt{a^{2} x^{2} - 1}} - \frac{i \sqrt{c} \operatorname{acosh}{\left (a x \right )}}{8 a^{3}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{a^{2} \sqrt{c} x^{5}}{4 \sqrt{- a^{2} x^{2} + 1}} + \frac{3 \sqrt{c} x^{3}}{8 \sqrt{- a^{2} x^{2} + 1}} - \frac{\sqrt{c} x}{8 a^{2} \sqrt{- a^{2} x^{2} + 1}} + \frac{\sqrt{c} \operatorname{asin}{\left (a x \right )}}{8 a^{3}} & \text{otherwise} \end{cases}\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)

[Out]

a**2*c*Piecewise((I*a**2*sqrt(c)*x**7/(6*sqrt(a**2*x**2 - 1)) - 5*I*sqrt(c)*x**5/(24*sqrt(a**2*x**2 - 1)) - I*
sqrt(c)*x**3/(48*a**2*sqrt(a**2*x**2 - 1)) + I*sqrt(c)*x/(16*a**4*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*acosh(a*x)/
(16*a**5), Abs(a**2*x**2) > 1), (-a**2*sqrt(c)*x**7/(6*sqrt(-a**2*x**2 + 1)) + 5*sqrt(c)*x**5/(24*sqrt(-a**2*x
**2 + 1)) + sqrt(c)*x**3/(48*a**2*sqrt(-a**2*x**2 + 1)) - sqrt(c)*x/(16*a**4*sqrt(-a**2*x**2 + 1)) + sqrt(c)*a
sin(a*x)/(16*a**5), True)) + 2*a*c*Piecewise((x**4*sqrt(-a**2*c*x**2 + c)/5 - x**2*sqrt(-a**2*c*x**2 + c)/(15*
a**2) - 2*sqrt(-a**2*c*x**2 + c)/(15*a**4), Ne(a, 0)), (sqrt(c)*x**4/4, True)) + c*Piecewise((I*a**2*sqrt(c)*x
**5/(4*sqrt(a**2*x**2 - 1)) - 3*I*sqrt(c)*x**3/(8*sqrt(a**2*x**2 - 1)) + I*sqrt(c)*x/(8*a**2*sqrt(a**2*x**2 -
1)) - I*sqrt(c)*acosh(a*x)/(8*a**3), Abs(a**2*x**2) > 1), (-a**2*sqrt(c)*x**5/(4*sqrt(-a**2*x**2 + 1)) + 3*sqr
t(c)*x**3/(8*sqrt(-a**2*x**2 + 1)) - sqrt(c)*x/(8*a**2*sqrt(-a**2*x**2 + 1)) + sqrt(c)*asin(a*x)/(8*a**3), Tru
e))

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Giac [A]  time = 1.16868, size = 144, normalized size = 1.06 \begin{align*} \frac{1}{240} \, \sqrt{-a^{2} c x^{2} + c}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, a^{2} c x + 12 \, a c\right )} x + 25 \, c\right )} x - \frac{16 \, c}{a}\right )} x - \frac{45 \, c}{a^{2}}\right )} x - \frac{64 \, c}{a^{3}}\right )} - \frac{3 \, c^{2} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{16 \, a^{2} \sqrt{-c}{\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]

1/240*sqrt(-a^2*c*x^2 + c)*((2*((4*(5*a^2*c*x + 12*a*c)*x + 25*c)*x - 16*c/a)*x - 45*c/a^2)*x - 64*c/a^3) - 3/
16*c^2*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a^2*sqrt(-c)*abs(a))