3.1138 \(\int e^{3 \tanh ^{-1}(a x)} x^2 (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=111 \[ \frac {13 c \sin ^{-1}(a x)}{8 a^3}-\frac {19 c x^2 \sqrt {1-a^2 x^2}}{15 a}-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {c (195 a x+304) \sqrt {1-a^2 x^2}}{120 a^3} \]

[Out]

13/8*c*arcsin(a*x)/a^3-19/15*c*x^2*(-a^2*x^2+1)^(1/2)/a-3/4*c*x^3*(-a^2*x^2+1)^(1/2)-1/5*a*c*x^4*(-a^2*x^2+1)^
(1/2)-1/120*c*(195*a*x+304)*(-a^2*x^2+1)^(1/2)/a^3

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Rubi [A]  time = 0.25, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6148, 1809, 833, 780, 216} \[ -\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {19 c x^2 \sqrt {1-a^2 x^2}}{15 a}-\frac {c (195 a x+304) \sqrt {1-a^2 x^2}}{120 a^3}+\frac {13 c \sin ^{-1}(a x)}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-19*c*x^2*Sqrt[1 - a^2*x^2])/(15*a) - (3*c*x^3*Sqrt[1 - a^2*x^2])/4 - (a*c*x^4*Sqrt[1 - a^2*x^2])/5 - (c*(304
+ 195*a*x)*Sqrt[1 - a^2*x^2])/(120*a^3) + (13*c*ArcSin[a*x])/(8*a^3)

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]

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 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 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 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]

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} x^2 \left (c-a^2 c x^2\right ) \, dx &=c \int \frac {x^2 (1+a x)^3}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {c \int \frac {x^2 \left (-5 a^2-19 a^3 x-15 a^4 x^2\right )}{\sqrt {1-a^2 x^2}} \, dx}{5 a^2}\\ &=-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}+\frac {c \int \frac {x^2 \left (65 a^4+76 a^5 x\right )}{\sqrt {1-a^2 x^2}} \, dx}{20 a^4}\\ &=-\frac {19 c x^2 \sqrt {1-a^2 x^2}}{15 a}-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {c \int \frac {x \left (-152 a^5-195 a^6 x\right )}{\sqrt {1-a^2 x^2}} \, dx}{60 a^6}\\ &=-\frac {19 c x^2 \sqrt {1-a^2 x^2}}{15 a}-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {c (304+195 a x) \sqrt {1-a^2 x^2}}{120 a^3}+\frac {(13 c) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}\\ &=-\frac {19 c x^2 \sqrt {1-a^2 x^2}}{15 a}-\frac {3}{4} c x^3 \sqrt {1-a^2 x^2}-\frac {1}{5} a c x^4 \sqrt {1-a^2 x^2}-\frac {c (304+195 a x) \sqrt {1-a^2 x^2}}{120 a^3}+\frac {13 c \sin ^{-1}(a x)}{8 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 62, normalized size = 0.56 \[ \frac {195 c \sin ^{-1}(a x)-c \sqrt {1-a^2 x^2} \left (24 a^4 x^4+90 a^3 x^3+152 a^2 x^2+195 a x+304\right )}{120 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(c*Sqrt[1 - a^2*x^2]*(304 + 195*a*x + 152*a^2*x^2 + 90*a^3*x^3 + 24*a^4*x^4)) + 195*c*ArcSin[a*x])/(120*a^3)

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fricas [A]  time = 0.61, size = 80, normalized size = 0.72 \[ -\frac {390 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (24 \, a^{4} c x^{4} + 90 \, a^{3} c x^{3} + 152 \, a^{2} c x^{2} + 195 \, a c x + 304 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{120 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(390*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (24*a^4*c*x^4 + 90*a^3*c*x^3 + 152*a^2*c*x^2 + 195*a*c*
x + 304*c)*sqrt(-a^2*x^2 + 1))/a^3

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giac [A]  time = 0.23, size = 69, normalized size = 0.62 \[ -\frac {1}{120} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (3 \, {\left (4 \, a c x + 15 \, c\right )} x + \frac {76 \, c}{a}\right )} x + \frac {195 \, c}{a^{2}}\right )} x + \frac {304 \, c}{a^{3}}\right )} + \frac {13 \, c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, a^{2} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*sqrt(-a^2*x^2 + 1)*((2*(3*(4*a*c*x + 15*c)*x + 76*c/a)*x + 195*c/a^2)*x + 304*c/a^3) + 13/8*c*arcsin(a*
x)*sgn(a)/(a^2*abs(a))

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maple [A]  time = 0.07, size = 170, normalized size = 1.53 \[ \frac {c \,a^{3} x^{6}}{5 \sqrt {-a^{2} x^{2}+1}}+\frac {16 c a \,x^{4}}{15 \sqrt {-a^{2} x^{2}+1}}+\frac {19 c \,x^{2}}{15 a \sqrt {-a^{2} x^{2}+1}}-\frac {38 c}{15 a^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {3 c \,a^{2} x^{5}}{4 \sqrt {-a^{2} x^{2}+1}}+\frac {7 c \,x^{3}}{8 \sqrt {-a^{2} x^{2}+1}}-\frac {13 c x}{8 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {13 c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{2} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c*a^3*x^6/(-a^2*x^2+1)^(1/2)+16/15*c*a*x^4/(-a^2*x^2+1)^(1/2)+19/15*c/a*x^2/(-a^2*x^2+1)^(1/2)-38/15*c/a^3
/(-a^2*x^2+1)^(1/2)+3/4*c*a^2*x^5/(-a^2*x^2+1)^(1/2)+7/8*c*x^3/(-a^2*x^2+1)^(1/2)-13/8*c*x/a^2/(-a^2*x^2+1)^(1
/2)+13/8*c/a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))

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maxima [A]  time = 0.42, size = 148, normalized size = 1.33 \[ \frac {a^{3} c x^{6}}{5 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {3 \, a^{2} c x^{5}}{4 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {16 \, a c x^{4}}{15 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {7 \, c x^{3}}{8 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {19 \, c x^{2}}{15 \, \sqrt {-a^{2} x^{2} + 1} a} - \frac {13 \, c x}{8 \, \sqrt {-a^{2} x^{2} + 1} a^{2}} + \frac {13 \, c \arcsin \left (a x\right )}{8 \, a^{3}} - \frac {38 \, c}{15 \, \sqrt {-a^{2} x^{2} + 1} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^3*c*x^6/sqrt(-a^2*x^2 + 1) + 3/4*a^2*c*x^5/sqrt(-a^2*x^2 + 1) + 16/15*a*c*x^4/sqrt(-a^2*x^2 + 1) + 7/8*c
*x^3/sqrt(-a^2*x^2 + 1) + 19/15*c*x^2/(sqrt(-a^2*x^2 + 1)*a) - 13/8*c*x/(sqrt(-a^2*x^2 + 1)*a^2) + 13/8*c*arcs
in(a*x)/a^3 - 38/15*c/(sqrt(-a^2*x^2 + 1)*a^3)

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mupad [B]  time = 0.03, size = 119, normalized size = 1.07 \[ \frac {13\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a^2\,\sqrt {-a^2}}-\frac {3\,c\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {19\,c\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a}-\frac {13\,c\,x\,\sqrt {1-a^2\,x^2}}{8\,a^2}-\frac {a\,c\,x^4\,\sqrt {1-a^2\,x^2}}{5}-\frac {38\,c\,\sqrt {1-a^2\,x^2}}{15\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*c*asinh(x*(-a^2)^(1/2)))/(8*a^2*(-a^2)^(1/2)) - (3*c*x^3*(1 - a^2*x^2)^(1/2))/4 - (19*c*x^2*(1 - a^2*x^2)^
(1/2))/(15*a) - (13*c*x*(1 - a^2*x^2)^(1/2))/(8*a^2) - (a*c*x^4*(1 - a^2*x^2)^(1/2))/5 - (38*c*(1 - a^2*x^2)^(
1/2))/(15*a^3)

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sympy [C]  time = 18.45, size = 371, normalized size = 3.34 \[ a^{3} c \left (\begin {cases} - \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} - \frac {8 \sqrt {- a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \frac {i x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {3 i x}{8 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \operatorname {acosh}{\left (a x \right )}}{8 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8 a^{5}} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c*Piecewise((-x**4*sqrt(-a**2*x**2 + 1)/(5*a**2) - 4*x**2*sqrt(-a**2*x**2 + 1)/(15*a**4) - 8*sqrt(-a**2*x
**2 + 1)/(15*a**6), Ne(a, 0)), (x**6/6, True)) + 3*a**2*c*Piecewise((-I*x**5/(4*sqrt(a**2*x**2 - 1)) - I*x**3/
(8*a**2*sqrt(a**2*x**2 - 1)) + 3*I*x/(8*a**4*sqrt(a**2*x**2 - 1)) - 3*I*acosh(a*x)/(8*a**5), Abs(a**2*x**2) >
1), (x**5/(4*sqrt(-a**2*x**2 + 1)) + x**3/(8*a**2*sqrt(-a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 + 1)) +
3*asin(a*x)/(8*a**5), True)) + 3*a*c*Piecewise((-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 + 1)/(
3*a**4), Ne(a, 0)), (x**4/4, True)) + c*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2*a**3),
Abs(a**2*x**2) > 1), (x**3/(2*sqrt(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a**3), Tr
ue))

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