3.277 \(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac {a x^2 \left (1-\frac {1}{a x}\right )^{5/2}}{\sqrt {\frac {1}{a x}+1} (c-a c x)^{5/2}}-\frac {a^{3/2} \left (1-\frac {1}{a x}\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )}{\sqrt {2} \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}} \]

[Out]

-1/2*a^(3/2)*(1-1/a/x)^(5/2)*arctanh(2^(1/2)*(1/x)^(1/2)/a^(1/2)/(1+1/a/x)^(1/2))/(1/x)^(5/2)/(-a*c*x+c)^(5/2)
*2^(1/2)+a*(1-1/a/x)^(5/2)*x^2/(-a*c*x+c)^(5/2)/(1+1/a/x)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6176, 6181, 94, 93, 206} \[ \frac {a x^2 \left (1-\frac {1}{a x}\right )^{5/2}}{\sqrt {\frac {1}{a x}+1} (c-a c x)^{5/2}}-\frac {a^{3/2} \left (1-\frac {1}{a x}\right )^{5/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )}{\sqrt {2} \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2)),x]

[Out]

(a*(1 - 1/(a*x))^(5/2)*x^2)/(Sqrt[1 + 1/(a*x)]*(c - a*c*x)^(5/2)) - (a^(3/2)*(1 - 1/(a*x))^(5/2)*ArcTanh[(Sqrt
[2]*Sqrt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/(a*x)])])/(Sqrt[2]*(x^(-1))^(5/2)*(c - a*c*x)^(5/2))

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6176

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

Rule 6181

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

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 122, normalized size = 1.02 \[ \frac {\sqrt {1-\frac {1}{a x}} \left (2 \sqrt {\frac {1}{x}}-\sqrt {2} \sqrt {a} \sqrt {\frac {1}{a x}+1} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )\right )}{2 a c^2 \sqrt {\frac {1}{x}} \sqrt {\frac {1}{a x}+1} \sqrt {c-a c x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2)),x]

[Out]

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

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fricas [A]  time = 0.58, size = 235, normalized size = 1.96 \[ \left [-\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {-c} \log \left (-\frac {a^{2} c x^{2} + 2 \, a c x + 2 \, \sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 4 \, \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{4 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}, \frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}}}{a c x - c}\right ) - 2 \, \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/4*(sqrt(2)*(a*x - 1)*sqrt(-c)*log(-(a^2*c*x^2 + 2*a*c*x + 2*sqrt(2)*sqrt(-a*c*x + c)*(a*x + 1)*sqrt(-c)*sq
rt((a*x - 1)/(a*x + 1)) - 3*c)/(a^2*x^2 - 2*a*x + 1)) + 4*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^3
*x - a*c^3), 1/2*(sqrt(2)*(a*x - 1)*sqrt(c)*arctan(sqrt(2)*sqrt(-a*c*x + c)*sqrt(c)*sqrt((a*x - 1)/(a*x + 1))/
(a*c*x - c)) - 2*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c^3*x - a*c^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(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(5/2),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(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 = 0.06, size = 85, normalized size = 0.71 \[ -\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}\, \left (\arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sqrt {-c \left (a x +1\right )}+2 \sqrt {c}\right )}{2 \left (a x -1\right )^{2} c^{\frac {7}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(5/2),x)

[Out]

-1/2*((a*x-1)/(a*x+1))^(3/2)*(a*x+1)/(a*x-1)^2*(-c*(a*x-1))^(1/2)/c^(7/2)*(arctan(1/2*(-c*(a*x+1))^(1/2)*2^(1/
2)/c^(1/2))*2^(1/2)*(-c*(a*x+1))^(1/2)+2*c^(1/2))/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (-a c x + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(((a*x - 1)/(a*x + 1))^(3/2)/(-a*c*x + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{{\left (c-a\,c\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a*c*x)^(5/2),x)

[Out]

int(((a*x - 1)/(a*x + 1))^(3/2)/(c - a*c*x)^(5/2), x)

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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(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**(5/2),x)

[Out]

Timed out

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