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

Optimal. Leaf size=128 \[ -\frac {16 a^2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{105 x \sqrt {1-a x}}-\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{7 x^3 \sqrt {1-a x}}+\frac {8 a (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{35 x^2 \sqrt {1-a x}} \]

[Out]

-2/7*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^3/(-a*x+1)^(1/2)+8/35*a*(a*x+1)^(3/2)*(c-c/a/x)^(1/2)/x^2/(-a*x+1)^(1/2)-
16/105*a^2*(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 = 128, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ -\frac {16 a^2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{105 x \sqrt {1-a x}}+\frac {8 a (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{35 x^2 \sqrt {1-a x}}-\frac {2 (a x+1)^{3/2} \sqrt {c-\frac {c}{a x}}}{7 x^3 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(7*x^3*Sqrt[1 - a*x]) + (8*a*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(35*x^2
*Sqrt[1 - a*x]) - (16*a^2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2))/(105*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^4} \, dx &=\frac {\left (\sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {e^{\tanh ^{-1}(a x)} \sqrt {1-a x}}{x^{9/2}} \, 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^{9/2} \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^{9/2}} \, dx}{\sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{7 x^3 \sqrt {1-a x}}-\frac {\left (4 a \sqrt {c-\frac {c}{a x}} \sqrt {x}\right ) \int \frac {\sqrt {1+a x}}{x^{7/2}} \, dx}{7 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{7 x^3 \sqrt {1-a x}}+\frac {8 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{35 x^2 \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^{5/2}} \, dx}{35 \sqrt {1-a x}}\\ &=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{7 x^3 \sqrt {1-a x}}+\frac {8 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{35 x^2 \sqrt {1-a x}}-\frac {16 a^2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{105 x \sqrt {1-a x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 55, normalized size = 0.43 \[ -\frac {2 (a x+1)^{3/2} \left (8 a^2 x^2-12 a x+15\right ) \sqrt {c-\frac {c}{a x}}}{105 x^3 \sqrt {1-a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(15 - 12*a*x + 8*a^2*x^2))/(105*x^3*Sqrt[1 - a*x])

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fricas [A]  time = 1.12, size = 66, normalized size = 0.52 \[ \frac {2 \, {\left (8 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + 3 \, a x + 15\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(8*a^3*x^3 - 4*a^2*x^2 + 3*a*x + 15)*sqrt(-a^2*x^2 + 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 \[ \int \frac {{\left (a x + 1\right )} \sqrt {c - \frac {c}{a x}}}{\sqrt {-a^{2} x^{2} + 1} x^{4}}\,{d x} \]

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^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 54, normalized size = 0.42 \[ -\frac {2 \left (a x +1\right )^{2} \left (8 a^{2} x^{2}-12 a x +15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{105 x^{3} \sqrt {-a^{2} x^{2}+1}} \]

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^4,x)

[Out]

-2/105*(a*x+1)^2*(8*a^2*x^2-12*a*x+15)*(c*(a*x-1)/a/x)^(1/2)/x^3/(-a^2*x^2+1)^(1/2)

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

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^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.14, size = 60, normalized size = 0.47 \[ -\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (\frac {16\,a^4\,x^4}{105}+\frac {8\,a^3\,x^3}{105}-\frac {2\,a^2\,x^2}{105}+\frac {12\,a\,x}{35}+\frac {2}{7}\right )}{x^3\,\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c - c/(a*x))^(1/2)*((12*a*x)/35 - (2*a^2*x^2)/105 + (8*a^3*x^3)/105 + (16*a^4*x^4)/105 + 2/7))/(x^3*(1 - a^
2*x^2)^(1/2))

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

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**4,x)

[Out]

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

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