3.572 \(\int \frac {e^{\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx\)

Optimal. Leaf size=172 \[ \frac {32 a^3 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{315 x \sqrt {1-a x}}-\frac {16 a^2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{105 x^2 \sqrt {1-a x}}-\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{9 x^4 \sqrt {1-a x}}+\frac {4 a (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{21 x^3 \sqrt {1-a x}} \]

[Out]

-2/9*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^4/(-a*x+1)^(1/2)+4/21*a*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^3/(-a*x+1)^(1/2)-
16/105*a^2*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^2/(-a*x+1)^(1/2)+32/315*a^3*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x/(-a*x+1
)^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6134, 6128, 848, 45, 37} \[ -\frac {16 a^2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{105 x^2 \sqrt {1-a x}}+\frac {32 a^3 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{315 x \sqrt {1-a x}}+\frac {4 a (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{21 x^3 \sqrt {1-a x}}-\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{9 x^4 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^5,x]

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(9*x^4*Sqrt[1 - a*x]) + (4*a*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(21*x^3
*Sqrt[1 - a*x]) - (16*a^2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(105*x^2*Sqrt[1 - a*x]) + (32*a^3*Sqrt[c - c/(a*x
)]*(1 + a*x)^(3/2))/(315*x*Sqrt[1 - a*x])

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 6128

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

Rule 6134

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {e^{\tanh ^{-1}(a x)} \sqrt {1-a x}}{x^{11/2}} \, dx}{\sqrt {1-a x}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1-a^2 x^2}}{x^{11/2} \sqrt {1-a x}} \, dx}{\sqrt {1-a x}}\\ &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{11/2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{9 x^4 \sqrt {1-a x}}-\frac {\left (2 a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{9/2}} \, dx}{3 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{9 x^4 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{21 x^3 \sqrt {1-a x}}+\frac {\left (8 a^2 \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{7/2}} \, dx}{21 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{9 x^4 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{21 x^3 \sqrt {1-a x}}-\frac {16 a^2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{105 x^2 \sqrt {1-a x}}-\frac {\left (16 a^3 \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{5/2}} \, dx}{105 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{9 x^4 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{21 x^3 \sqrt {1-a x}}-\frac {16 a^2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{105 x^2 \sqrt {1-a x}}+\frac {32 a^3 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{315 x \sqrt {1-a x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 63, normalized size = 0.37 \[ \frac {2 (a x+1)^{3/2} \left (16 a^3 x^3-24 a^2 x^2+30 a x-35\right ) \sqrt {c-\frac {c}{a x}}}{315 x^4 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*Sqrt[c - c/(a*x)])/x^5,x]

[Out]

(2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(-35 + 30*a*x - 24*a^2*x^2 + 16*a^3*x^3))/(315*x^4*Sqrt[1 - a*x])

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fricas [A]  time = 0.40, size = 74, normalized size = 0.43 \[ -\frac {2 \, {\left (16 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 5 \, a x - 35\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{315 \, {\left (a x^{5} - x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-2/315*(16*a^4*x^4 - 8*a^3*x^3 + 6*a^2*x^2 - 5*a*x - 35)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))/(a*x^5 - x
^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^5,x, algorithm="giac")

[Out]

integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^5), x)

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maple [A]  time = 0.03, size = 62, normalized size = 0.36 \[ \frac {2 \left (a x +1\right )^{2} \left (16 x^{3} a^{3}-24 a^{2} x^{2}+30 a x -35\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{315 x^{4} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(a*x+1)^2*(16*a^3*x^3-24*a^2*x^2+30*a*x-35)*(c*(a*x-1)/a/x)^(1/2)/x^4/(-a^2*x^2+1)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a/x)^(1/2)/x^5,x, algorithm="maxima")

[Out]

integrate((a*x + 1)*sqrt(c - c/(a*x))/(sqrt(-a^2*x^2 + 1)*x^5), x)

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mupad [B]  time = 1.13, size = 68, normalized size = 0.40 \[ -\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (-\frac {32\,a^5\,x^5}{315}-\frac {16\,a^4\,x^4}{315}+\frac {4\,a^3\,x^3}{315}-\frac {2\,a^2\,x^2}{315}+\frac {16\,a\,x}{63}+\frac {2}{9}\right )}{x^4\,\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - c/(a*x))^(1/2)*((16*a*x)/63 - (2*a^2*x^2)/315 + (4*a^3*x^3)/315 - (16*a^4*x^4)/315 - (32*a^5*x^5)/315 +
2/9))/(x^4*(1 - a^2*x^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (-1 + \frac {1}{a x}\right )} \left (a x + 1\right )}{x^{5} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(c-c/a/x)**(1/2)/x**5,x)

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**5*sqrt(-(a*x - 1)*(a*x + 1))), x)

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