3.1080 \(\int e^{2 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=86 \[ -\frac {(a x+1) \sqrt {c-a^2 c x^2}}{2 a}-\frac {3 \sqrt {c-a^2 c x^2}}{2 a}+\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6141, 671, 641, 217, 203} \[ -\frac {(a x+1) \sqrt {c-a^2 c x^2}}{2 a}-\frac {3 \sqrt {c-a^2 c x^2}}{2 a}+\frac {3 \sqrt {c} \tan ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 641

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 671

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

Rule 6141

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^(n/2), Int[(c + d*x^2)^(p -
n/2)*(1 + a*x)^n, x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) && IGt
Q[n/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.88 \[ -\frac {\sqrt {c-a^2 c x^2} \left (\sqrt {1-a^2 x^2} (a x+4)+6 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{2 a \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(Sqrt[c - a^2*c*x^2]*((4 + a*x)*Sqrt[1 - a^2*x^2] + 6*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(a*Sqrt[1 - a^2*x^2
])

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fricas [A]  time = 0.70, size = 134, normalized size = 1.56 \[ \left [-\frac {2 \, \sqrt {-a^{2} c x^{2} + c} {\left (a x + 4\right )} - 3 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{4 \, a}, -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (a x + 4\right )} + 3 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{2 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 62, normalized size = 0.72 \[ -\frac {1}{2} \, \sqrt {-a^{2} c x^{2} + c} {\left (x + \frac {4}{a}\right )} - \frac {3 \, c \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{2 \, \sqrt {-c} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(-a^2*c*x^2 + c)*(x + 4/a) - 3/2*c*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(sqrt(-c)*abs(a))

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maple [A]  time = 0.04, size = 134, normalized size = 1.56 \[ -\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}-\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}{a}+\frac {2 c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 52, normalized size = 0.60 \[ -\frac {1}{2} \, a {\left (\frac {\sqrt {-a^{2} c x^{2} + c} x}{a} - \frac {3 \, \sqrt {c} \arcsin \left (a x\right )}{a^{2}} + \frac {4 \, \sqrt {-a^{2} c x^{2} + c}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {- a^{2} c x^{2} + c}}{a x - 1}\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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