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

Optimal. Leaf size=79 \[ -\frac {\sqrt {c-a c x}}{x}+5 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6130, 21, 98, 156, 63, 208, 206} \[ -\frac {\sqrt {c-a c x}}{x}+5 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c - a*c*x]/x) + 5*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] - 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c
*x]/(Sqrt[2]*Sqrt[c])]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 63

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 98

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

Rule 156

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 6130

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

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^2} \, dx &=\int \frac {(1-a x) \sqrt {c-a c x}}{x^2 (1+a x)} \, dx\\ &=\frac {\int \frac {(c-a c x)^{3/2}}{x^2 (1+a x)} \, dx}{c}\\ &=-\frac {\sqrt {c-a c x}}{x}-\frac {\int \frac {\frac {5 a c^2}{2}-\frac {3}{2} a^2 c^2 x}{x (1+a x) \sqrt {c-a c x}} \, dx}{c}\\ &=-\frac {\sqrt {c-a c x}}{x}-\frac {1}{2} (5 a c) \int \frac {1}{x \sqrt {c-a c x}} \, dx+\left (4 a^2 c\right ) \int \frac {1}{(1+a x) \sqrt {c-a c x}} \, dx\\ &=-\frac {\sqrt {c-a c x}}{x}+5 \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a}-\frac {x^2}{a c}} \, dx,x,\sqrt {c-a c x}\right )-(8 a) \operatorname {Subst}\left (\int \frac {1}{2-\frac {x^2}{c}} \, dx,x,\sqrt {c-a c x}\right )\\ &=-\frac {\sqrt {c-a c x}}{x}+5 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 79, normalized size = 1.00 \[ -\frac {\sqrt {c-a c x}}{x}+5 a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {c}}\right )-4 \sqrt {2} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a c x}}{\sqrt {2} \sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c - a*c*x]/x) + 5*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c*x]/Sqrt[c]] - 4*Sqrt[2]*a*Sqrt[c]*ArcTanh[Sqrt[c - a*c
*x]/(Sqrt[2]*Sqrt[c])]

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fricas [A]  time = 0.63, size = 175, normalized size = 2.22 \[ \left [\frac {4 \, \sqrt {2} a \sqrt {c} x \log \left (\frac {a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} \sqrt {c} - 3 \, c}{a x + 1}\right ) + 5 \, a \sqrt {c} x \log \left (\frac {a c x - 2 \, \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{x}\right ) - 2 \, \sqrt {-a c x + c}}{2 \, x}, \frac {4 \, \sqrt {2} a \sqrt {-c} x \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {-c}}{2 \, c}\right ) - 5 \, a \sqrt {-c} x \arctan \left (\frac {\sqrt {-a c x + c} \sqrt {-c}}{c}\right ) - \sqrt {-a c x + c}}{x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*sqrt(2)*a*sqrt(c)*x*log((a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a*x + 1)) + 5*a*sqrt(c)*x*
log((a*c*x - 2*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/x) - 2*sqrt(-a*c*x + c))/x, (4*sqrt(2)*a*sqrt(-c)*x*arctan(1/2*
sqrt(2)*sqrt(-a*c*x + c)*sqrt(-c)/c) - 5*a*sqrt(-c)*x*arctan(sqrt(-a*c*x + c)*sqrt(-c)/c) - sqrt(-a*c*x + c))/
x]

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giac [A]  time = 0.18, size = 72, normalized size = 0.91 \[ \frac {4 \, \sqrt {2} a c \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} - \frac {5 \, a c \arctan \left (\frac {\sqrt {-a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {\sqrt {-a c x + c}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2)*a*c*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) - 5*a*c*arctan(sqrt(-a*c*x + c)/sqrt(-c))
/sqrt(-c) - sqrt(-a*c*x + c)/x

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maple [A]  time = 0.04, size = 71, normalized size = 0.90 \[ 2 a c \left (-\frac {2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a c x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {-a c x +c}}{2 x a c}+\frac {5 \arctanh \left (\frac {\sqrt {-a c x +c}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*c*(-2*2^(1/2)/c^(1/2)*arctanh(1/2*(-a*c*x+c)^(1/2)*2^(1/2)/c^(1/2))-1/2*(-a*c*x+c)^(1/2)/x/a/c+5/2/c^(1/2)
*arctanh((-a*c*x+c)^(1/2)/c^(1/2)))

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maxima [A]  time = 0.44, size = 111, normalized size = 1.41 \[ \frac {1}{2} \, a c {\left (\frac {4 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-a c x + c}}{\sqrt {2} \sqrt {c} + \sqrt {-a c x + c}}\right )}{\sqrt {c}} - \frac {5 \, \log \left (\frac {\sqrt {-a c x + c} - \sqrt {c}}{\sqrt {-a c x + c} + \sqrt {c}}\right )}{\sqrt {c}} - \frac {2 \, \sqrt {-a c x + c}}{a c x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*c*(4*sqrt(2)*log(-(sqrt(2)*sqrt(c) - sqrt(-a*c*x + c))/(sqrt(2)*sqrt(c) + sqrt(-a*c*x + c)))/sqrt(c) - 5
*log((sqrt(-a*c*x + c) - sqrt(c))/(sqrt(-a*c*x + c) + sqrt(c)))/sqrt(c) - 2*sqrt(-a*c*x + c)/(a*c*x))

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mupad [B]  time = 0.86, size = 62, normalized size = 0.78 \[ 5\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c-a\,c\,x}}{\sqrt {c}}\right )-\frac {\sqrt {c-a\,c\,x}}{x}-4\,\sqrt {2}\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-a\,c\,x}}{2\,\sqrt {c}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a*c^(1/2)*atanh((c - a*c*x)^(1/2)/c^(1/2)) - (c - a*c*x)^(1/2)/x - 4*2^(1/2)*a*c^(1/2)*atanh((2^(1/2)*(c - a
*c*x)^(1/2))/(2*c^(1/2)))

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sympy [B]  time = 14.77, size = 162, normalized size = 2.05 \[ \frac {a c^{2} \sqrt {\frac {1}{c^{3}}} \log {\left (- c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {- a c x + c} \right )}}{2} - \frac {a c^{2} \sqrt {\frac {1}{c^{3}}} \log {\left (c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {- a c x + c} \right )}}{2} - \frac {6 a c \operatorname {atan}{\left (\frac {\sqrt {- a c x + c}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + \frac {4 \sqrt {2} a c \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {- a c x + c}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - \frac {\sqrt {- a c x + c}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**2*sqrt(c**(-3))*log(-c**2*sqrt(c**(-3)) + sqrt(-a*c*x + c))/2 - a*c**2*sqrt(c**(-3))*log(c**2*sqrt(c**(-3
)) + sqrt(-a*c*x + c))/2 - 6*a*c*atan(sqrt(-a*c*x + c)/sqrt(-c))/sqrt(-c) + 4*sqrt(2)*a*c*atan(sqrt(2)*sqrt(-a
*c*x + c)/(2*sqrt(-c)))/sqrt(-c) - sqrt(-a*c*x + c)/x

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