3.540 \(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\)

Optimal. Leaf size=188 \[ \frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {472}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {1888 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}} \]

[Out]

a^3*(c-c/a/x)^(7/2)/c^3/(1-1/a^2/x^2)^(1/2)+59/35*a^3*(c-c/a/x)^(3/2)*(1-1/a^2/x^2)^(1/2)/c+2/7*a^3*(c-c/a/x)^
(5/2)*(1-1/a^2/x^2)^(1/2)/c^2+1888/105*a^3*c*(1-1/a^2/x^2)^(1/2)/(c-c/a/x)^(1/2)+472/105*a^3*(1-1/a^2/x^2)^(1/
2)*(c-c/a/x)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6178, 1635, 795, 657, 649} \[ \frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {472}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {1888 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^4),x]

[Out]

(1888*a^3*c*Sqrt[1 - 1/(a^2*x^2)])/(105*Sqrt[c - c/(a*x)]) + (472*a^3*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)])
/105 + (59*a^3*Sqrt[1 - 1/(a^2*x^2)]*(c - c/(a*x))^(3/2))/(35*c) + (2*a^3*Sqrt[1 - 1/(a^2*x^2)]*(c - c/(a*x))^
(5/2))/(7*c^2) + (a^3*(c - c/(a*x))^(7/2))/(c^3*Sqrt[1 - 1/(a^2*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 657

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*(m + 2*p + 1)), x] + Dist[(2*c*d*Simplify[m + p])/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^
2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

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^{-3 \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 )^{7/2}}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {7 a^2}{2}-a x\right ) \left (c-\frac {c x}{a}\right )^{5/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^2}\\ &=\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\left (59 a^2\right ) \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{5/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{14 c^2}\\ &=\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\left (236 a^2\right ) \operatorname {Subst}\left (\int \frac {\left (c-\frac {c x}{a}\right )^{3/2}}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{35 c}\\ &=\frac {472}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {1}{105} \left (944 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 {1888 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}}+\frac {472}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 78, normalized size = 0.41 \[ \frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \left (1336 a^4 x^4+668 a^3 x^3-167 a^2 x^2+66 a x-15\right ) \sqrt {c-\frac {c}{a x}}}{105 x^2 \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^4),x]

[Out]

(2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(-15 + 66*a*x - 167*a^2*x^2 + 668*a^3*x^3 + 1336*a^4*x^4))/(105*x
^2*(-1 + a^2*x^2))

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fricas [A]  time = 0.82, size = 77, normalized size = 0.41 \[ \frac {2 \, {\left (1336 \, a^{4} x^{4} + 668 \, a^{3} x^{3} - 167 \, a^{2} x^{2} + 66 \, a x - 15\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="fricas")

[Out]

2/105*(1336*a^4*x^4 + 668*a^3*x^3 - 167*a^2*x^2 + 66*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(-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)/(a*x+1))^(3/2)/x^4,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),abs(a*x+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 78, normalized size = 0.41 \[ \frac {2 \left (a x +1\right ) \left (1336 x^{4} a^{4}+668 x^{3} a^{3}-167 a^{2} x^{2}+66 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{105 x^{3} \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x)

[Out]

2/105*(a*x+1)*(1336*a^4*x^4+668*a^3*x^3-167*a^2*x^2+66*a*x-15)*(c*(a*x-1)/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x
^3/(a*x-1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a x}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate(sqrt(c - c/(a*x))*((a*x - 1)/(a*x + 1))^(3/2)/x^4, x)

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mupad [B]  time = 1.41, size = 100, normalized size = 0.53 \[ \frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (1336\,a^3\,x^3+2004\,a^2\,x^2+1837\,a\,x+1903\right )\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3}+\frac {3776\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - c/(a*x))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^4,x)

[Out]

(2*((a*x - 1)/(a*x + 1))^(1/2)*(1837*a*x + 2004*a^2*x^2 + 1336*a^3*x^3 + 1903)*((c*(a*x - 1))/(a*x))^(1/2))/(1
05*x^3) + (3776*((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))**(3/2)/x**4,x)

[Out]

Timed out

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