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

Optimal. Leaf size=93 \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )+\frac {5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6133, 25, 514, 375, 98, 156, 63, 208} \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )+\frac {5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 514

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

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

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Mathematica [A]  time = 0.05, size = 93, normalized size = 1.00 \[ x \left (-\sqrt {c-\frac {c}{a x}}\right )+\frac {5 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a}-\frac {4 \sqrt {2} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 217, normalized size = 2.33 \[ \left [-\frac {2 \, a x \sqrt {\frac {a c x - c}{a x}} - 4 \, \sqrt {2} \sqrt {c} \log \left (\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} - 3 \, a c x + c}{a x + 1}\right ) - 5 \, \sqrt {c} \log \left (-2 \, a c x - 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{2 \, a}, -\frac {a x \sqrt {\frac {a c x - c}{a x}} - 4 \, \sqrt {2} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + 5 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right )}{a}\right ] \]

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, algorithm="fricas")

[Out]

[-1/2*(2*a*x*sqrt((a*c*x - c)/(a*x)) - 4*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)) - 5*sqrt(c)*log(-2*a*c*x - 2*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x)) + c))/a, -(a*x*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) + 5*sqrt(-c)*arc
tan(sqrt(-c)*sqrt((a*c*x - c)/(a*x))/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((c-c/a/x)^(1/2)/(a*x+1)^2*(-a^2*x^2+1),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)]Error: Bad Argument Type

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maple [B]  time = 0.05, size = 190, normalized size = 2.04 \[ \frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}-4 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}+4 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) \sqrt {a}-\ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \sqrt {\frac {1}{a}}+6 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \sqrt {\frac {1}{a}}\right )}{2 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}} \]

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)

[Out]

1/2*(c*(a*x-1)/a/x)^(1/2)*x*(2*(a*x^2-x)^(1/2)*a^(3/2)*(1/a)^(1/2)-4*((a*x-1)*x)^(1/2)*a^(3/2)*(1/a)^(1/2)+4*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))*a^(1/2)-ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/
2)+2*a*x-1)/a^(1/2))*a*(1/a)^(1/2)+6*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a*(1/a)^(1/2))/((a*
x-1)*x)^(1/2)/a^(3/2)/(1/a)^(1/2)

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

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, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

[Out]

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

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