3.12.10 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x}{\sqrt {c-a^2 c x^2}} \, dx\) [1110]

Optimal. Leaf size=84 \[ \frac {(1+a x)^2}{a^2 \sqrt {c-a^2 c x^2}}+\frac {2 \sqrt {c-a^2 c x^2}}{a^2 c}-\frac {2 \text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2 \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6286, 803, 655, 223, 209} \begin {gather*} -\frac {2 \text {ArcTan}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a^2 \sqrt {c}}+\frac {(a x+1)^2}{a^2 \sqrt {c-a^2 c x^2}}+\frac {2 \sqrt {c-a^2 c x^2}}{a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 223

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 655

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 803

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g + e*f)*(
d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(p + 1))), x] - Dist[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

Rule 6286

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

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Mathematica [A]
time = 0.08, size = 78, normalized size = 0.93 \begin {gather*} \frac {\frac {(-3+a x) \sqrt {c-a^2 c x^2}}{-1+a x}+2 \sqrt {c} \text {ArcTan}\left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 103, normalized size = 1.23

method	result	size
default	\(\frac {\sqrt {-a^{2} c \,x^{2}+c}}{a^{2} c}-\frac {2 \arctan \left (\frac {\sqrt {c \,a^{2}}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{a \sqrt {c \,a^{2}}}-\frac {2 \sqrt {-c \,a^{2} \left (x -\frac {1}{a}\right )^{2}-2 c a \left (x -\frac {1}{a}\right )}}{a^{3} c \left (x -\frac {1}{a}\right )}\)	\(103\)
risch	\(-\frac {a^{2} x^{2}-1}{a^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {2 \arctan \left (\frac {\sqrt {c \,a^{2}}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{a \sqrt {c \,a^{2}}}-\frac {2 \sqrt {-c \,a^{2} \left (x -\frac {1}{a}\right )^{2}-2 c a \left (x -\frac {1}{a}\right )}}{a^{3} c \left (x -\frac {1}{a}\right )}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 67, normalized size = 0.80 \begin {gather*} -a {\left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{4} c x - a^{3} c} + \frac {2 \, \arcsin \left (a x\right )}{a^{3} \sqrt {c}} - \frac {\sqrt {-a^{2} c x^{2} + c}}{a^{3} c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 166, normalized size = 1.98 \begin {gather*} \left [-\frac {{\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 ) - \sqrt {-a^{2} c x^{2} + c} {\left (a x - 3\right )}}{a^{3} c x - a^{2} c}, \frac {2 \, {\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 x - 3\right )}}{a^{3} c x - a^{2} c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx - \int \frac {a x^{2}}{a x \sqrt {- a^{2} c x^{2} + c} - \sqrt {- a^{2} c x^{2} + c}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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