3.523 \(\int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac {26 a c \sqrt {1-\frac {1}{a^2 x^2}}}{35 x^2 \sqrt {c-\frac {c}{a x}}}+\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 x^3 \sqrt {c-\frac {c}{a x}}}-\frac {104}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {104 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}} \]

[Out]

-104/105*a^3*c*(1-1/a^2/x^2)^(1/2)/(c-c/a/x)^(1/2)+2/7*c*(1-1/a^2/x^2)^(1/2)/x^3/(c-c/a/x)^(1/2)-26/35*a*c*(1-
1/a^2/x^2)^(1/2)/x^2/(c-c/a/x)^(1/2)-104/105*a^3*(1-1/a^2/x^2)^(1/2)*(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6178, 881, 871, 795, 649} \[ -\frac {104}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}-\frac {104 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}}-\frac {26 a c \sqrt {1-\frac {1}{a^2 x^2}}}{35 x^2 \sqrt {c-\frac {c}{a x}}}+\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 x^3 \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a*x)]/(E^ArcCoth[a*x]*x^4),x]

[Out]

(-104*a^3*c*Sqrt[1 - 1/(a^2*x^2)])/(105*Sqrt[c - c/(a*x)]) - (104*a^3*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)])
/105 + (2*c*Sqrt[1 - 1/(a^2*x^2)])/(7*Sqrt[c - c/(a*x)]*x^3) - (26*a*c*Sqrt[1 - 1/(a^2*x^2)])/(35*Sqrt[c - c/(
a*x)]*x^2)

Rule 649

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 795

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(g*(d + e*x)^m
*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2)), Int[(d +
e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && NeQ[m + 2*p +
2, 0] && NeQ[m, 2]

Rule 871

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

Rule 881

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(d +
e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + p + 2)), x] - Dist[(e*f*(p + 1) - d*g*(2*n + p
+ 3))/(g*(n + p + 2)), Int[(d + e*x)^(m - 1)*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m,
n, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1]
&& IntegerQ[2*p]

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {e^{-\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (c-\frac {c x}{a}\right )^{3/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 \sqrt {c-\frac {c}{a x}} x^3}-\frac {13}{7} \operatorname {Subst}\left (\int \frac {x^2 \sqrt {c-\frac {c x}{a}}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 \sqrt {c-\frac {c}{a x}} x^3}-\frac {26 a c \sqrt {1-\frac {1}{a^2 x^2}}}{35 \sqrt {c-\frac {c}{a x}} x^2}+\frac {1}{35} (52 a) \operatorname {Subst}\left (\int \frac {x \sqrt {c-\frac {c x}{a}}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {104}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 \sqrt {c-\frac {c}{a x}} x^3}-\frac {26 a c \sqrt {1-\frac {1}{a^2 x^2}}}{35 \sqrt {c-\frac {c}{a x}} x^2}-\frac {1}{105} \left (52 a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-\frac {c x}{a}}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {104 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}}-\frac {104}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {2 c \sqrt {1-\frac {1}{a^2 x^2}}}{7 \sqrt {c-\frac {c}{a x}} x^3}-\frac {26 a c \sqrt {1-\frac {1}{a^2 x^2}}}{35 \sqrt {c-\frac {c}{a x}} x^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 66, normalized size = 0.44 \[ -\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \left (104 a^3 x^3-52 a^2 x^2+39 a x-15\right ) \sqrt {c-\frac {c}{a x}}}{105 x^2 (a x-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^ArcCoth[a*x]*x^4),x]

[Out]

(-2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(-15 + 39*a*x - 52*a^2*x^2 + 104*a^3*x^3))/(105*x^2*(-1 + a*x))

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fricas [A]  time = 0.51, size = 77, normalized size = 0.52 \[ -\frac {2 \, {\left (104 \, a^{4} x^{4} + 52 \, a^{3} x^{3} - 13 \, a^{2} x^{2} + 24 \, a x - 15\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(1/2)/x^4,x, algorithm="fricas")

[Out]

-2/105*(104*a^4*x^4 + 52*a^3*x^3 - 13*a^2*x^2 + 24*a*x - 15)*sqrt((a*x - 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))
/(a*x^4 - x^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} \sqrt {\frac {a x - 1}{a x + 1}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(sqrt(c - c/(a*x))*sqrt((a*x - 1)/(a*x + 1))/x^4, x)

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maple [A]  time = 0.04, size = 70, normalized size = 0.47 \[ -\frac {2 \left (a x +1\right ) \left (104 x^{3} a^{3}-52 a^{2} x^{2}+39 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {a x -1}{a x +1}}}{105 x^{3} \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(a*x+1)*(104*a^3*x^3-52*a^2*x^2+39*a*x-15)*(c*(a*x-1)/a/x)^(1/2)*((a*x-1)/(a*x+1))^(1/2)/x^3/(a*x-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} \sqrt {\frac {a x - 1}{a x + 1}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(1/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(c - c/(a*x))*sqrt((a*x - 1)/(a*x + 1))/x^4, x)

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mupad [B]  time = 1.41, size = 100, normalized size = 0.67 \[ -\frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (104\,a^3\,x^3+156\,a^2\,x^2+143\,a\,x+167\right )\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3}-\frac {304\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*((a*x - 1)/(a*x + 1))^(1/2)*(143*a*x + 156*a^2*x^2 + 104*a^3*x^3 + 167)*((c*(a*x - 1))/(a*x))^(1/2))/(105
*x^3) - (304*((a*x - 1)/(a*x + 1))^(1/2)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3*(a*x - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)**(1/2)*((a*x-1)/(a*x+1))**(1/2)/x**4,x)

[Out]

Timed out

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