3.7.65 \(\int e^{-\tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [665]

Optimal. Leaf size=58 \[ \frac {c (1+a x) \sqrt {1-a^2 x^2}}{a^2 x}+\frac {c \text {ArcSin}(a x)}{a}-\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6292, 6284, 827, 858, 222, 272, 65, 214} \begin {gather*} \frac {c \sqrt {1-a^2 x^2} (a x+1)}{a^2 x}-\frac {c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{a}+\frac {c \text {ArcSin}(a x)}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
+ 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 858

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

Rule 6284

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*((1 -
a^2*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] || G
tQ[c, 0]) && ILtQ[(n - 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 6292

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 55, normalized size = 0.95 \begin {gather*} \frac {c \left ((1+a x) \sqrt {1-a^2 x^2}+a x \text {ArcSin}(a x)-a x \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\right )}{a^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.74, size = 101, normalized size = 1.74

method	result	size
risch	\(-\frac {\left (a^{2} x^{2}-1\right ) c}{x \sqrt {-a^{2} x^{2}+1}\, a^{2}}+\frac {\left (a \sqrt {-a^{2} x^{2}+1}+\frac {a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}-a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right ) c}{a^{2}}\)	\(97\)
default	\(\frac {c \left (a \left (\sqrt {-a^{2} x^{2}+1}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+2 a^{2} \left (\frac {x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}\right )\right )}{a^{2}}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.41, size = 85, normalized size = 1.47 \begin {gather*} -\frac {2 \, a c x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - a c x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - a c x - \sqrt {-a^{2} x^{2} + 1} {\left (a c x + c\right )}}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 2.98, size = 177, normalized size = 3.05 \begin {gather*} \frac {c \left (\begin {cases} i \sqrt {a^{2} x^{2} - 1} - \log {\left (a x \right )} + \frac {\log {\left (a^{2} x^{2} \right )}}{2} + i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt {- a^{2} x^{2} + 1} + \frac {\log {\left (a^{2} x^{2} \right )}}{2} - \log {\left (\sqrt {- a^{2} x^{2} + 1} + 1 \right )} & \text {otherwise} \end {cases}\right )}{a} - \frac {c \left (\begin {cases} - \frac {i a^{2} x}{\sqrt {a^{2} x^{2} - 1}} + i a \operatorname {acosh}{\left (a x \right )} + \frac {i}{x \sqrt {a^{2} x^{2} - 1}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {a^{2} x}{\sqrt {- a^{2} x^{2} + 1}} - a \operatorname {asin}{\left (a x \right )} - \frac {1}{x \sqrt {- a^{2} x^{2} + 1}} & \text {otherwise} \end {cases}\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*Piecewise((I*sqrt(a**2*x**2 - 1) - log(a*x) + log(a**2*x**2)/2 + I*asin(1/(a*x)), Abs(a**2*x**2) > 1), (sqrt
(-a**2*x**2 + 1) + log(a**2*x**2)/2 - log(sqrt(-a**2*x**2 + 1) + 1), True))/a - c*Piecewise((-I*a**2*x/sqrt(a*
*2*x**2 - 1) + I*a*acosh(a*x) + I/(x*sqrt(a**2*x**2 - 1)), Abs(a**2*x**2) > 1), (a**2*x/sqrt(-a**2*x**2 + 1) -
a*asin(a*x) - 1/(x*sqrt(-a**2*x**2 + 1)), True))/a**2

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (54) = 108\).
time = 0.42, size = 128, normalized size = 2.21 \begin {gather*} -\frac {a^{2} c x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} - \frac {c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c}{a} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c}{2 \, a^{2} x {\left | a \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*c*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) + c*arcsin(a*x)*sgn(a)/abs(a) - c*log(1/2*abs(-2*sqrt(-a
^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + sqrt(-a^2*x^2 + 1)*c/a + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*
c/(a^2*x*abs(a))

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Mupad [B]
time = 0.04, size = 76, normalized size = 1.31 \begin {gather*} \frac {c\,\sqrt {1-a^2\,x^2}}{a}-\frac {c\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )}{a}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c\,\sqrt {1-a^2\,x^2}}{a^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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