3.9.88 \(\int e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}} x \, dx\) [888]

Optimal. Leaf size=98 \[ \frac {3 \sqrt {c-\frac {c}{a^2 x^2}} x}{2 a}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)}{2 a}-\frac {3 \sqrt {c-\frac {c}{a^2 x^2}} x \text {ArcSin}(a x)}{2 a \sqrt {1-a x} \sqrt {1+a x}} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6302, 6294, 6264, 52, 41, 222} \begin {gather*} -\frac {3 x \text {ArcSin}(a x) \sqrt {c-\frac {c}{a^2 x^2}}}{2 a \sqrt {1-a x} \sqrt {a x+1}}+\frac {x (a x+1) \sqrt {c-\frac {c}{a^2 x^2}}}{2 a}+\frac {3 x \sqrt {c-\frac {c}{a^2 x^2}}}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[c - c/(a^2*x^2)]*x)/(2*a) + (Sqrt[c - c/(a^2*x^2)]*x*(1 + a*x))/(2*a) - (3*Sqrt[c - c/(a^2*x^2)]*x*Arc
Sin[a*x])/(2*a*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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 222

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 6264

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 6294

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/x^(2*p))*(1 - a*x)^p*(1 + a*x)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d
, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &&  !GtQ[c, 0]

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 77, normalized size = 0.79 \begin {gather*} \frac {\sqrt {c-\frac {c}{a^2 x^2}} x \left ((4+a x) \sqrt {1-a^2 x^2}+6 \text {ArcSin}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{2 a \sqrt {1-a^2 x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 147, normalized size = 1.50

method	result	size
risch	\(\frac {\left (a x +4\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x}{2 a}+\frac {3 \ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {c \left (a^{2} x^{2}-1\right )}\, x}{2 \sqrt {a^{2} c}\, \left (a^{2} x^{2}-1\right )}\)	\(114\)
default	\(\frac {\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \left (x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2}-\sqrt {c}\, \ln \left (\sqrt {c}\, x +\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right )+4 \sqrt {c}\, \ln \left (\frac {\sqrt {\frac {c \left (a x +1\right ) \left (a x -1\right )}{a^{2}}}\, \sqrt {c}+c x}{\sqrt {c}}\right )+4 \sqrt {\frac {c \left (a x +1\right ) \left (a x -1\right )}{a^{2}}}\, a \right )}{2 \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\, a^{2}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 188, normalized size = 1.92 \begin {gather*} \left [\frac {2 \, {\left (a^{2} x^{2} + 4 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} + 3 \, \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 )}{4 \, a^{2}}, \frac {{\left (a^{2} x^{2} + 4 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \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 )}{2 \, a^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{a x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.41, size = 106, normalized size = 1.08 \begin {gather*} \frac {1}{4} \, {\left (2 \, \sqrt {a^{2} c x^{2} - c} {\left (\frac {x \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {4 \, \mathrm {sgn}\left (x\right )}{a^{3}}\right )} - \frac {6 \, \sqrt {c} \log \left ({\left | -\sqrt {a^{2} c} x + \sqrt {a^{2} c x^{2} - c} \right |}\right ) \mathrm {sgn}\left (x\right )}{a^{2} {\left | a \right |}} + \frac {{\left (3 \, a \sqrt {c} \log \left ({\left | c \right |}\right ) - 8 \, \sqrt {-c} {\left | a \right |}\right )} \mathrm {sgn}\left (x\right )}{a^{3} {\left | a \right |}}\right )} {\left | a \right |} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(a^2*c*x^2 - c)*(x*sgn(x)/a^2 + 4*sgn(x)/a^3) - 6*sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 -
c)))*sgn(x)/(a^2*abs(a)) + (3*a*sqrt(c)*log(abs(c)) - 8*sqrt(-c)*abs(a))*sgn(x)/(a^3*abs(a)))*abs(a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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