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

Optimal. Leaf size=118 \[ -x \sqrt {c-\frac {c}{a^2 x^2}}+\frac {2 x \sqrt {c-\frac {c}{a^2 x^2}} \sin ^{-1}(a x)}{\sqrt {1-a x} \sqrt {a x+1}}-\frac {x \sqrt {c-\frac {c}{a^2 x^2}} \tanh ^{-1}\left (\sqrt {1-a x} \sqrt {a x+1}\right )}{\sqrt {1-a x} \sqrt {a x+1}} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6159, 6129, 102, 157, 41, 216, 92, 208} \[ -x \sqrt {c-\frac {c}{a^2 x^2}}+\frac {2 x \sqrt {c-\frac {c}{a^2 x^2}} \sin ^{-1}(a x)}{\sqrt {1-a x} \sqrt {a x+1}}-\frac {x \sqrt {c-\frac {c}{a^2 x^2}} \tanh ^{-1}\left (\sqrt {1-a x} \sqrt {a x+1}\right )}{\sqrt {1-a x} \sqrt {a x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 208

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

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 6129

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

Rule 6159

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.69 \[ -\frac {x \sqrt {c-\frac {c}{a^2 x^2}} \left (\sqrt {a^2 x^2-1}+2 \log \left (\sqrt {a^2 x^2-1}+a x\right )-\tan ^{-1}\left (\frac {1}{\sqrt {a^2 x^2-1}}\right )\right )}{\sqrt {a^2 x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 270, normalized size = 2.29 \[ \left [-\frac {2 \, a x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 4 \, \sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) - \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} - 2 \, a \sqrt {-c} x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 2 \, c}{x^{2}}\right )}{2 \, a}, -\frac {a x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - \sqrt {c} \arctan \left (\frac {a \sqrt {c} x \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) - \sqrt {c} \log \left (2 \, a^{2} c x^{2} - 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right )}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [A]  time = 0.04, size = 198, normalized size = 1.68 \[ -\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (2 \sqrt {c}\, \ln \left (\frac {\sqrt {c}\, \sqrt {\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}}+c x}{\sqrt {c}}\right ) a \sqrt {-\frac {c}{a^{2}}}+2 \sqrt {\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}}\, a^{2} \sqrt {-\frac {c}{a^{2}}}-\sqrt {-\frac {c}{a^{2}}}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2}-c \ln \left (\frac {2 \sqrt {-\frac {c}{a^{2}}}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2}-2 c}{a^{2} x}\right )\right )}{\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2} \sqrt {-\frac {c}{a^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (a x + 1\right )}^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(-((c - c/(a^2*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 \[ - \int \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}}}{a x - 1}\, dx - \int \frac {a x \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a x - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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