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

Optimal. Leaf size=86 \[ -2 \sqrt {c-\frac {c}{a x}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6133, 25, 434, 446, 84, 156, 63, 208} \[ -2 \sqrt {c-\frac {c}{a x}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 25

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

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 84

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

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 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 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6133

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, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 86, normalized size = 1.00 \[ -2 \sqrt {c-\frac {c}{a x}}-2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )+4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 206, normalized size = 2.40 \[ \left [2 \, \sqrt {2} \sqrt {c} \log \left (-\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + 3 \, a c x - c}{a x + 1}\right ) + \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) - 2 \, \sqrt {\frac {a c x - c}{a x}}, -4 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - 2 \, \sqrt {\frac {a c x - c}{a x}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2*sqrt(2)*sqrt(c)*log(-(2*sqrt(2)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + 3*a*c*x - c)/(a*x + 1)) + sqrt(c)*log
(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c) - 2*sqrt((a*c*x - c)/(a*x)), -4*sqrt(2)*sqrt(-c)*arctan
(1/2*sqrt(2)*sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) + 2*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/c) - 2*s
qrt((a*c*x - c)/(a*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((c-c/a/x)^(1/2)/(a*x+1)^2*(-a^2*x^2+1)/x,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)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.05, size = 227, normalized size = 2.64 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-4 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}\, x^{2}+2 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} x^{2} \sqrt {\frac {1}{a}}+2 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\, \sqrt {\frac {1}{a}}+2 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{2} a -2 \sqrt {a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{2}-3 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, x^{2} a \right )}{x \sqrt {\left (a x -1\right ) x}\, \sqrt {a}\, \sqrt {\frac {1}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(a*x-1)/a/x)^(1/2)/x*(-4*(a*x^2-x)^(1/2)*a^(3/2)*(1/a)^(1/2)*x^2+2*((a*x-1)*x)^(1/2)*a^(3/2)*x^2*(1/a)^(1/2
)+2*(a*x^2-x)^(3/2)*a^(1/2)*(1/a)^(1/2)+2*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^2*
a-2*a^(1/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x^2-3*ln(1/2*(2*((a*x-1)*x
)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*x^2*a)/((a*x-1)*x)^(1/2)/a^(1/2)/(1/a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 12.33, size = 80, normalized size = 0.93 \[ \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c - \frac {c}{a x}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - \frac {4 \sqrt {2} c \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {c - \frac {c}{a x}}}{2 \sqrt {- c}} \right )}}{\sqrt {- c}} - 2 \sqrt {c - \frac {c}{a x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c*atan(sqrt(c - c/(a*x))/sqrt(-c))/sqrt(-c) - 4*sqrt(2)*c*atan(sqrt(2)*sqrt(c - c/(a*x))/(2*sqrt(-c)))/sqrt(
-c) - 2*sqrt(c - c/(a*x))

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