3.1109 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x^2}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=93 \[ \frac {(a x+1)^2}{a^3 \sqrt {c-a^2 c x^2}}+\frac {(a x+6) \sqrt {c-a^2 c x^2}}{2 a^3 c}-\frac {5 \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a^3 \sqrt {c}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 93, normalized size of antiderivative = 1.00, 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, 780, 217, 203} \[ \frac {(a x+1)^2}{a^3 \sqrt {c-a^2 c x^2}}+\frac {(a x+6) \sqrt {c-a^2 c x^2}}{2 a^3 c}-\frac {5 \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a^3 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^3*Sqrt[c - a^2*c*x^2]) + ((6 + a*x)*Sqrt[c - a^2*c*x^2])/(2*a^3*c) - (5*ArcTan[(a*Sqrt[c]*x)/Sq
rt[c - a^2*c*x^2]])/(2*a^3*Sqrt[c])

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])

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 94, normalized size = 1.01 \[ \frac {\left (a^2 x^2+3 a x-8\right ) \sqrt {c-a^2 c x^2}+5 \sqrt {c} (a x-1) \tan ^{-1}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (a^2 x^2-1\right )}\right )}{2 a^3 c (a x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8 + 3*a*x + a^2*x^2)*Sqrt[c - a^2*c*x^2] + 5*Sqrt[c]*(-1 + a*x)*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(
-1 + a^2*x^2))])/(2*a^3*c*(-1 + a*x))

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fricas [A]  time = 0.72, size = 184, normalized size = 1.98 \[ \left [-\frac {5 \, {\left (a x - 1\right )} \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 \, a x - 8\right )}}{4 \, {\left (a^{4} c x - a^{3} c\right )}}, \frac {5 \, {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} + 3 \, a x - 8\right )}}{2 \, {\left (a^{4} c x - a^{3} c\right )}}\right ] \]

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)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(5*(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
)*(a^2*x^2 + 3*a*x - 8))/(a^4*c*x - a^3*c), 1/2*(5*(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)*(a^2*x^2 + 3*a*x - 8))/(a^4*c*x - a^3*c)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 126, normalized size = 1.35 \[ \frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2 a^{2} c}-\frac {5 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 a^{2} \sqrt {a^{2} c}}+\frac {2 \sqrt {-a^{2} c \,x^{2}+c}}{a^{3} c}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{a^{4} c \left (x -\frac {1}{a}\right )} \]

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)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.50, size = 89, normalized size = 0.96 \[ -\frac {1}{2} \, a {\left (\frac {4 \, \sqrt {-a^{2} c x^{2} + c}}{a^{5} c x - a^{4} c} - \frac {\sqrt {-a^{2} c x^{2} + c} x}{a^{3} c} + \frac {5 \, \arcsin \left (a x\right )}{a^{4} \sqrt {c}} - \frac {4 \, \sqrt {-a^{2} c x^{2} + c}}{a^{4} c}\right )} \]

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)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {x^2\,{\left (a\,x+1\right )}^2}{\sqrt {c-a^2\,c\,x^2}\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{2}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x^{3}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx \]

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)**(1/2),x)

[Out]

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

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