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

Optimal. Leaf size=148 \[ -\frac {11 a^2 c \sqrt {1-a^2 x^2}}{8 x \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right ) \]

[Out]

11/8*a^3*arctanh(c^(1/2)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(1/2))*c^(1/2)-1/3*c*(-a^2*x^2+1)^(1/2)/x^3/(-a*c*x+c)^
(1/2)+11/12*a*c*(-a^2*x^2+1)^(1/2)/x^2/(-a*c*x+c)^(1/2)-11/8*a^2*c*(-a^2*x^2+1)^(1/2)/x/(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6128, 879, 873, 875, 208} \[ -\frac {11 a^2 c \sqrt {1-a^2 x^2}}{8 x \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Sqrt[1 - a^2*x^2])/(3*x^3*Sqrt[c - a*c*x]) + (11*a*c*Sqrt[1 - a^2*x^2])/(12*x^2*Sqrt[c - a*c*x]) - (11*a^2
*c*Sqrt[1 - a^2*x^2])/(8*x*Sqrt[c - a*c*x]) + (11*a^3*Sqrt[c]*ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c
*x]])/8

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 873

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

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 879

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

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]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^4} \, dx &=\frac {\int \frac {(c-a c x)^{3/2}}{x^4 \sqrt {1-a^2 x^2}} \, dx}{c}\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}-\frac {1}{6} (11 a) \int \frac {\sqrt {c-a c x}}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}+\frac {1}{8} \left (11 a^2\right ) \int \frac {\sqrt {c-a c x}}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{8 x \sqrt {c-a c x}}-\frac {1}{16} \left (11 a^3\right ) \int \frac {\sqrt {c-a c x}}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{8 x \sqrt {c-a c x}}-\frac {1}{8} \left (11 a^5 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a^2 c+a^2 c^2 x^2} \, dx,x,\frac {\sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a c x}}+\frac {11 a c \sqrt {1-a^2 x^2}}{12 x^2 \sqrt {c-a c x}}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{8 x \sqrt {c-a c x}}+\frac {11}{8} a^3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-a^2 x^2}}{\sqrt {c-a c x}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 56, normalized size = 0.38 \[ \frac {c \sqrt {1-a^2 x^2} \left (11 a^3 x^3 \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};a x+1\right )-1\right )}{3 x^3 \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*Sqrt[1 - a^2*x^2]*(-1 + 11*a^3*x^3*Hypergeometric2F1[1/2, 3, 3/2, 1 + a*x]))/(3*x^3*Sqrt[c - a*c*x])

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fricas [A]  time = 0.60, size = 248, normalized size = 1.68 \[ \left [\frac {33 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) + 2 \, {\left (33 \, a^{2} x^{2} - 22 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{48 \, {\left (a x^{4} - x^{3}\right )}}, \frac {33 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + {\left (33 \, a^{2} x^{2} - 22 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{24 \, {\left (a x^{4} - x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(33*(a^4*x^4 - a^3*x^3)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c)
- 2*c)/(a*x^2 - x)) + 2*(33*a^2*x^2 - 22*a*x + 8)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a*x^4 - x^3), 1/24*(33
*(a^4*x^4 - a^3*x^3)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + (33*a^2*x
^2 - 22*a*x + 8)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a*x^4 - x^3)]

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giac [A]  time = 0.26, size = 132, normalized size = 0.89 \[ -\frac {1}{24} \, a^{3} c^{2} {\left (\frac {33 \, \arctan \left (\frac {\sqrt {a c x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c^{2}} + \frac {33 \, {\left (a c x + c\right )}^{\frac {5}{2}} - 88 \, {\left (a c x + c\right )}^{\frac {3}{2}} c + 63 \, \sqrt {a c x + c} c^{2}}{a^{3} c^{5} x^{3}}\right )} {\left | c \right |} + \frac {33 \, a^{3} c {\left | c \right |} \arctan \left (\frac {\sqrt {2} \sqrt {c}}{\sqrt {-c}}\right ) + 19 \, \sqrt {2} a^{3} \sqrt {-c} \sqrt {c} {\left | c \right |}}{24 \, \sqrt {-c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*a^3*c^2*(33*arctan(sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c^2) + (33*(a*c*x + c)^(5/2) - 88*(a*c*x + c)^(3/
2)*c + 63*sqrt(a*c*x + c)*c^2)/(a^3*c^5*x^3))*abs(c) + 1/24*(33*a^3*c*abs(c)*arctan(sqrt(2)*sqrt(c)/sqrt(-c))
+ 19*sqrt(2)*a^3*sqrt(-c)*sqrt(c)*abs(c))/(sqrt(-c)*c)

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maple [A]  time = 0.05, size = 121, normalized size = 0.82 \[ -\frac {\sqrt {-c \left (a x -1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (33 c \arctanh \left (\frac {\sqrt {c \left (a x +1\right )}}{\sqrt {c}}\right ) x^{3} a^{3}-33 x^{2} a^{2} \sqrt {c \left (a x +1\right )}\, \sqrt {c}+22 x a \sqrt {c \left (a x +1\right )}\, \sqrt {c}-8 \sqrt {c \left (a x +1\right )}\, \sqrt {c}\right )}{24 \sqrt {c}\, \left (a x -1\right ) \sqrt {c \left (a x +1\right )}\, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(-c*(a*x-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(33*c*arctanh((c*(a*x+1))^(1/2)/c^(1/2))*x^3*a^3-33*x^2*a^2*(c*(a*
x+1))^(1/2)*c^(1/2)+22*x*a*(c*(a*x+1))^(1/2)*c^(1/2)-8*(c*(a*x+1))^(1/2)*c^(1/2))/c^(1/2)/(a*x-1)/(c*(a*x+1))^
(1/2)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{x^4\,\left (a\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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