3.5.34 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^3} \, dx\) [434]

Optimal. Leaf size=163 \[ \frac {47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {1+a x}\right )}{4 (c-a c x)^{3/2}} \]

[Out]

-47/4*a^2*c^2*(-a*x+1)^(3/2)*arctanh((a*x+1)^(1/2))/(-a*c*x+c)^(3/2)+47/4*a^2*c^2*(-a*x+1)^(3/2)/(-a*c*x+c)^(3
/2)/(a*x+1)^(1/2)-1/2*c^2*(-a*x+1)^(3/2)/x^2/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)+13/4*a*c^2*(-a*x+1)^(3/2)/x/(-a*c*
x+c)^(3/2)/(a*x+1)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6265, 23, 91, 79, 53, 65, 214} \begin {gather*} \frac {47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt {a x+1} (c-a c x)^{3/2}}-\frac {47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {a x+1}\right )}{4 (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {a x+1} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {a x+1} (c-a c x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(47*a^2*c^2*(1 - a*x)^(3/2))/(4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (c^2*(1 - a*x)^(3/2))/(2*x^2*Sqrt[1 + a*x]*
(c - a*c*x)^(3/2)) + (13*a*c^2*(1 - a*x)^(3/2))/(4*x*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) - (47*a^2*c^2*(1 - a*x)^
(3/2)*ArcTanh[Sqrt[1 + a*x]])/(4*(c - a*c*x)^(3/2))

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 53

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

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

Rule 91

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

Rule 214

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

Rule 6265

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

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x^3} \, dx &=\int \frac {(1-a x)^{3/2} \sqrt {c-a c x}}{x^3 (1+a x)^{3/2}} \, dx\\ &=\frac {(1-a x)^{3/2} \int \frac {(c-a c x)^2}{x^3 (1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {(1-a x)^{3/2} \int \frac {-\frac {13 a c^2}{2}+2 a^2 c^2 x}{x^2 (1+a x)^{3/2}} \, dx}{2 (c-a c x)^{3/2}}\\ &=-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {\left (47 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x (1+a x)^{3/2}} \, dx}{8 (c-a c x)^{3/2}}\\ &=\frac {47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {\left (47 a^2 c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x \sqrt {1+a x}} \, dx}{8 (c-a c x)^{3/2}}\\ &=\frac {47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {\left (47 a c^2 (1-a x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{a}+\frac {x^2}{a}} \, dx,x,\sqrt {1+a x}\right )}{4 (c-a c x)^{3/2}}\\ &=\frac {47 a^2 c^2 (1-a x)^{3/2}}{4 \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {c^2 (1-a x)^{3/2}}{2 x^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {13 a c^2 (1-a x)^{3/2}}{4 x \sqrt {1+a x} (c-a c x)^{3/2}}-\frac {47 a^2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {1+a x}\right )}{4 (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 65, normalized size = 0.40 \begin {gather*} \frac {c \sqrt {1-a x} \left (-2+13 a x+47 a^2 x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+a x\right )\right )}{4 x^2 \sqrt {1+a x} \sqrt {c-a c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[1 - a*x]*(-2 + 13*a*x + 47*a^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + a*x]))/(4*x^2*Sqrt[1 + a*x]*Sqr
t[c - a*c*x])

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Maple [A]
time = 1.13, size = 100, normalized size = 0.61

method	result	size
default	\(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (47 \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}}{\sqrt {c}}\right ) a^{2} x^{2} \sqrt {\left (a x +1\right ) c}-47 a^{2} x^{2} \sqrt {c}-13 a x \sqrt {c}+2 \sqrt {c}\right )}{4 \sqrt {c}\, \left (a x -1\right ) \left (a x +1\right ) x^{2}}\)	\(100\)
risch	\(-\frac {\left (15 a^{2} x^{2}+13 a x -2\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{4 x^{2} \sqrt {\left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {a^{2} \left (-\frac {94 \arctanh \left (\frac {\sqrt {c x a +c}}{\sqrt {c}}\right )}{\sqrt {c}}+\frac {64}{\sqrt {c x a +c}}\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{8 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a*c*x+c)^(1/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

1/4*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(47*arctanh(((a*x+1)*c)^(1/2)/c^(1/2))*a^2*x^2*((a*x+1)*c)^(1/2)-47*
a^2*x^2*c^(1/2)-13*a*x*c^(1/2)+2*c^(1/2))/c^(1/2)/(a*x-1)/(a*x+1)/x^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 252, normalized size = 1.55 \begin {gather*} \left [\frac {47 \, {\left (a^{4} x^{4} - a^{2} x^{2}\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 (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{8 \, {\left (a^{2} x^{4} - x^{2}\right )}}, -\frac {47 \, {\left (a^{4} x^{4} - a^{2} x^{2}\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 (47 \, a^{2} x^{2} + 13 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{4 \, {\left (a^{2} x^{4} - x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(47*(a^4*x^4 - a^2*x^2)*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*(47*a^2*x^2 + 13*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^2*x^4 - x^2), -1/4*(4
7*(a^4*x^4 - a^2*x^2)*sqrt(-c)*arctan(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + (47*a^2*
x^2 + 13*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^2*x^4 - x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{x^{3} \left (a x + 1\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(a*x - 1))*(-(a*x - 1)*(a*x + 1))**(3/2)/(x**3*(a*x + 1)**3), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}\,\sqrt {c-a\,c\,x}}{x^3\,{\left (a\,x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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