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

Optimal. Leaf size=235 \[ -\frac {i c \sqrt {1-a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a \sqrt {c-a^2 c x^2}}+\frac {i c \sqrt {1-a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a \sqrt {c-a^2 c x^2}}+\frac {\sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x)-\frac {c \sqrt {1-a^2 x^2} \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a \sqrt {c-a^2 c x^2}} \]

[Out]

-c*arctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a/(-a^2*c*x^2+c)^(1/2)-1/2*I*c*polylog
(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))*(-a^2*x^2+1)^(1/2)/a/(-a^2*c*x^2+c)^(1/2)+1/2*I*c*polylog(2,I*(-a*x+1)^(1/
2)/(a*x+1)^(1/2))*(-a^2*x^2+1)^(1/2)/a/(-a^2*c*x^2+c)^(1/2)+1/2*(-a^2*c*x^2+c)^(1/2)/a+1/2*x*arctanh(a*x)*(-a^
2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5942, 5954, 5950} \[ -\frac {i c \sqrt {1-a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a \sqrt {c-a^2 c x^2}}+\frac {i c \sqrt {1-a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{2 a \sqrt {c-a^2 c x^2}}+\frac {\sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x)-\frac {c \sqrt {1-a^2 x^2} \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{a \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5942

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(b*(d + e*x^2)^q)/(2*c*
q*(2*q + 1)), x] + (Dist[(2*d*q)/(2*q + 1), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x]), x], x] + Simp[(x*(d
+ e*x^2)^q*(a + b*ArcTanh[c*x]))/(2*q + 1), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 0]

Rule 5950

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-2*(a + b*ArcTanh[c*x])*
ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(c*Sqrt[d]), x] + (-Simp[(I*b*PolyLog[2, -((I*Sqrt[1 - c*x])/Sqrt[1 + c*x
])])/(c*Sqrt[d]), x] + Simp[(I*b*PolyLog[2, (I*Sqrt[1 - c*x])/Sqrt[1 + c*x]])/(c*Sqrt[d]), x]) /; FreeQ[{a, b,
c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]

Rule 5954

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcTanh[c*x])^p/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d
+ e, 0] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x) \, dx &=\frac {\sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x)+\frac {1}{2} c \int \frac {\tanh ^{-1}(a x)}{\sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x)+\frac {\left (c \sqrt {1-a^2 x^2}\right ) \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{2} x \sqrt {c-a^2 c x^2} \tanh ^{-1}(a x)-\frac {c \sqrt {1-a^2 x^2} \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{a \sqrt {c-a^2 c x^2}}-\frac {i c \sqrt {1-a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a \sqrt {c-a^2 c x^2}}+\frac {i c \sqrt {1-a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{2 a \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 119, normalized size = 0.51 \[ \frac {\sqrt {c \left (1-a^2 x^2\right )} \left (-\frac {i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )+\tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )\right )}{\sqrt {1-a^2 x^2}}+a x \tanh ^{-1}(a x)+1\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c*(1 - a^2*x^2)]*(1 + a*x*ArcTanh[a*x] - (I*(ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log[1 + I/E^ArcTa
nh[a*x]]) + PolyLog[2, (-I)/E^ArcTanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(2*a)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-a^{2} c x^{2} + c} \operatorname {artanh}\left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*arctanh(a*x), x)

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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^2*c*x^2+c)^(1/2)*arctanh(a*x),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 = 0.55, size = 319, normalized size = 1.36 \[ \frac {\left (a x \arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{2 a}+\frac {i \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a \left (a x +1\right ) \left (a x -1\right )}-\frac {i \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a \left (a x +1\right ) \left (a x -1\right )}+\frac {i \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a \left (a x +1\right ) \left (a x -1\right )}-\frac {i \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a \left (a x +1\right ) \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x*arctanh(a*x)+1)*(-(a*x-1)*(a*x+1)*c)^(1/2)/a+1/2*I/a*(-(a*x-1)*(a*x+1)*c)^(1/2)/(a*x+1)*(-a^2*x^2+1)^
(1/2)/(a*x-1)*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*I/a*(-(a*x-1)*(a*x+1)*c)^(1/2)/(a*x+1)*(-a^2
*x^2+1)^(1/2)/(a*x-1)*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*I/a*(-(a*x-1)*(a*x+1)*c)^(1/2)/(a*x+
1)*(-a^2*x^2+1)^(1/2)/(a*x-1)*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*I/a*(-(a*x-1)*(a*x+1)*c)^(1/2)/(a*x+1)
*(-a^2*x^2+1)^(1/2)/(a*x-1)*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} c x^{2} + c} \operatorname {artanh}\left (a x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*arctanh(a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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