3.6.49 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{\sqrt {c-\frac {c}{a x}}} \, dx\) [549]

Optimal. Leaf size=96 \[ -\frac {\sqrt {c-\frac {c}{a x}} x}{c}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 96, 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 = {6268, 25, 528, 382, 101, 162, 65, 214} \begin {gather*} -\frac {x \sqrt {c-\frac {c}{a x}}}{c}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]),x]

[Out]

-((Sqrt[c - c/(a*x)]*x)/c) + (3*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*Sqrt[c]) - (2*Sqrt[2]*ArcTanh[Sqrt[c -
c/(a*x)]/(Sqrt[2]*Sqrt[c])])/(a*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 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 101

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 162

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 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 382

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 528

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 6268

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

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Mathematica [A]
time = 0.03, size = 96, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {c-\frac {c}{a x}} x}{c}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a \sqrt {c}}-\frac {2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{a \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*Sqrt[c - c/(a*x)]),x]

[Out]

-((Sqrt[c - c/(a*x)]*x)/c) + (3*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(a*Sqrt[c]) - (2*Sqrt[2]*ArcTanh[Sqrt[c -
c/(a*x)]/(Sqrt[2]*Sqrt[c])])/(a*Sqrt[c])

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Maple [A]
time = 0.39, size = 136, normalized size = 1.42

method	result	size
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (2 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}-2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) \sqrt {a}-3 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \sqrt {\frac {1}{a}}\right )}{2 \sqrt {\left (a x -1\right ) x}\, c \,a^{\frac {3}{2}} \sqrt {\frac {1}{a}}}\)	\(136\)
risch	\(-\frac {a x -1}{a \sqrt {\frac {c \left (a x -1\right )}{a x}}}-\frac {\left (-\frac {3 \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c x a}\right )}{2 a \sqrt {a^{2} c}}-\frac {\sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right )}{a^{2} \sqrt {c}}\right ) \sqrt {a c x \left (a x -1\right )}}{\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a*x-1)/a/x)^(1/2)*x*(2*((a*x-1)*x)^(1/2)*a^(3/2)*(1/a)^(1/2)-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))*a^(1/2)-3*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)/c/a^(3/2)/(1/a)^(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(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a/x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 232, normalized size = 2.42 \begin {gather*} \left [-\frac {2 \, a x \sqrt {\frac {a c x - c}{a x}} - 2 \, \sqrt {2} \sqrt {c} \log \left (\frac {\frac {2 \, \sqrt {2} a x \sqrt {\frac {a c x - c}{a x}}}{\sqrt {c}} - 3 \, a x + 1}{a x + 1}\right ) - 3 \, \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 c}, \frac {2 \, \sqrt {2} c \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {2} a x \sqrt {-\frac {1}{c}} \sqrt {\frac {a c x - c}{a x}}}{a x - 1}\right ) - a x \sqrt {\frac {a c x - c}{a x}} - 3 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right )}{a c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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