3.6.75 \(\int e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx\) [575]

Optimal. Leaf size=80 \[ -\frac {7 \sqrt {c-\frac {c}{a x}} x}{4 a}-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2-\frac {7 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{4 a^2} \]

[Out]

-7/4*arctanh((c-c/a/x)^(1/2)/c^(1/2))*c^(1/2)/a^2-7/4*x*(c-c/a/x)^(1/2)/a-1/2*x^2*(c-c/a/x)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6268, 25, 445, 457, 79, 44, 65, 214} \begin {gather*} -\frac {7 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{4 a^2}-\frac {1}{2} x^2 \sqrt {c-\frac {c}{a x}}-\frac {7 x \sqrt {c-\frac {c}{a x}}}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Sqrt[c - c/(a*x)]*x)/(4*a) - (Sqrt[c - c/(a*x)]*x^2)/2 - (7*Sqrt[c]*ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]])/(4
*a^2)

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 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 79

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

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 445

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 457

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 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 e^{2 \tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} x \, dx &=\int \frac {\sqrt {c-\frac {c}{a x}} x (1+a x)}{1-a x} \, dx\\ &=-\frac {c \int \frac {1+a x}{\sqrt {c-\frac {c}{a x}}} \, dx}{a}\\ &=-\frac {c \int \frac {\left (a+\frac {1}{x}\right ) x}{\sqrt {c-\frac {c}{a x}}} \, dx}{a}\\ &=\frac {c \text {Subst}\left (\int \frac {a+x}{x^3 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2+\frac {(7 c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{4 a}\\ &=-\frac {7 \sqrt {c-\frac {c}{a x}} x}{4 a}-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2+\frac {(7 c) \text {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{8 a^2}\\ &=-\frac {7 \sqrt {c-\frac {c}{a x}} x}{4 a}-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2-\frac {7 \text {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )}{4 a}\\ &=-\frac {7 \sqrt {c-\frac {c}{a x}} x}{4 a}-\frac {1}{2} \sqrt {c-\frac {c}{a x}} x^2-\frac {7 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{4 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 77, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {c-\frac {c}{a x}} \left (a \sqrt {1-\frac {1}{a x}} x (7+2 a x)+7 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}}\right )\right )}{4 a^2 \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[c - c/(a*x)]*(a*Sqrt[1 - 1/(a*x)]*x*(7 + 2*a*x) + 7*ArcTanh[Sqrt[1 - 1/(a*x)]]))/(a^2*Sqrt[1 - 1/(a
*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(138\) vs. \(2(64)=128\).
time = 0.75, size = 139, normalized size = 1.74

method	result	size
risch	\(-\frac {\left (2 a x +7\right ) x \sqrt {\frac {c \left (a x -1\right )}{a x}}}{4 a}-\frac {7 \ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c x a}\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {a c x \left (a x -1\right )}}{8 a \sqrt {a^{2} c}\, \left (a x -1\right )}\)	\(111\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (4 a^{\frac {5}{2}} \sqrt {a \,x^{2}-x}\, x +16 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}}-2 \sqrt {a \,x^{2}-x}\, a^{\frac {3}{2}}+8 a \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right )-\ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \right )}{8 \sqrt {\left (a x -1\right ) x}\, a^{\frac {5}{2}}}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*(a*x-1)/a/x)^(1/2)*x*(4*a^(5/2)*(a*x^2-x)^(1/2)*x+16*((a*x-1)*x)^(1/2)*a^(3/2)-2*(a*x^2-x)^(1/2)*a^(3/
2)+8*a*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*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)/((a*x-1)*x)^(1/2)/a^(5/2)

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

[Out]

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

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Fricas [A]
time = 0.36, size = 147, normalized size = 1.84 \begin {gather*} \left [-\frac {2 \, {\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 7 \, \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right )}{8 \, a^{2}}, -\frac {{\left (2 \, a^{2} x^{2} + 7 \, a x\right )} \sqrt {\frac {a c x - c}{a x}} - 7 \, \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right )}{4 \, a^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(2*(2*a^2*x^2 + 7*a*x)*sqrt((a*c*x - c)/(a*x)) - 7*sqrt(c)*log(-2*a*c*x + 2*a*sqrt(c)*x*sqrt((a*c*x - c)
/(a*x)) + c))/a^2, -1/4*((2*a^2*x^2 + 7*a*x)*sqrt((a*c*x - c)/(a*x)) - 7*sqrt(-c)*arctan(sqrt(-c)*sqrt((a*c*x
- c)/(a*x))/c))/a^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx - \int \frac {a x^{2} \sqrt {c - \frac {c}{a x}}}{a x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(x*sqrt(c - c/(a*x))/(a*x - 1), x) - Integral(a*x**2*sqrt(c - c/(a*x))/(a*x - 1), x)

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Giac [A]
time = 0.44, size = 112, normalized size = 1.40 \begin {gather*} -\frac {1}{4} \, \sqrt {a^{2} c x^{2} - a c x} {\left (\frac {2 \, x {\left | a \right |}}{a^{2} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left | a \right |}}{a^{3} \mathrm {sgn}\left (x\right )}\right )} - \frac {7 \, \sqrt {c} \log \left ({\left | a \right |} {\left | c \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, a^{2}} + \frac {7 \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {a^{2} c} x - \sqrt {a^{2} c x^{2} - a c x}\right )} \sqrt {c} {\left | a \right |} + a c \right |}\right )}{8 \, a^{2} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(a^2*c*x^2 - a*c*x)*(2*x*abs(a)/(a^2*sgn(x)) + 7*abs(a)/(a^3*sgn(x))) - 7/8*sqrt(c)*log(abs(a)*abs(c)
)*sgn(x)/a^2 + 7/8*sqrt(c)*log(abs(-2*(sqrt(a^2*c)*x - sqrt(a^2*c*x^2 - a*c*x))*sqrt(c)*abs(a) + a*c))/(a^2*sg
n(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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