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

Optimal. Leaf size=95 \[ \frac {64 c^2 x \sqrt {1-\frac {1}{a^2 x^2}}}{15 \sqrt {c-a c x}}+\frac {16}{15} c x \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-a c x}+\frac {2}{5} x \sqrt {1-\frac {1}{a^2 x^2}} (c-a c x)^{3/2} \]

[Out]

2/5*x*(-a*c*x+c)^(3/2)*(1-1/a^2/x^2)^(1/2)+64/15*c^2*x*(1-1/a^2/x^2)^(1/2)/(-a*c*x+c)^(1/2)+16/15*c*x*(1-1/a^2
/x^2)^(1/2)*(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 137, normalized size of antiderivative = 1.44, 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, 89, 78, 37} \[ \frac {86 \sqrt {\frac {1}{a x}+1} (c-a c x)^{3/2}}{15 a^2 x \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {2 x \sqrt {\frac {1}{a x}+1} (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}-\frac {28 \sqrt {\frac {1}{a x}+1} (c-a c x)^{3/2}}{15 a \left (1-\frac {1}{a x}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-28*Sqrt[1 + 1/(a*x)]*(c - a*c*x)^(3/2))/(15*a*(1 - 1/(a*x))^(3/2)) + (86*Sqrt[1 + 1/(a*x)]*(c - a*c*x)^(3/2)
)/(15*a^2*(1 - 1/(a*x))^(3/2)*x) + (2*Sqrt[1 + 1/(a*x)]*x*(c - a*c*x)^(3/2))/(5*(1 - 1/(a*x))^(3/2))

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 78

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] || LtQ
[p, n]))))

Rule 89

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 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 e^{-\coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx &=\frac {(c-a c x)^{3/2} \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{3/2} x^{3/2} \, dx}{\left (1-\frac {1}{a x}\right )^{3/2} x^{3/2}}\\ &=-\frac {\left (\left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2}{x^{7/2} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {2 \sqrt {1+\frac {1}{a x}} x (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}-\frac {\left (2 \left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {-\frac {7}{a}+\frac {5 x}{2 a^2}}{x^{5/2} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{5 \left (1-\frac {1}{a x}\right )^{3/2}}\\ &=-\frac {28 \sqrt {1+\frac {1}{a x}} (c-a c x)^{3/2}}{15 a \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {2 \sqrt {1+\frac {1}{a x}} x (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}-\frac {\left (43 \left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x^{3/2} \sqrt {1+\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )}{15 a^2 \left (1-\frac {1}{a x}\right )^{3/2}}\\ &=-\frac {28 \sqrt {1+\frac {1}{a x}} (c-a c x)^{3/2}}{15 a \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {86 \sqrt {1+\frac {1}{a x}} (c-a c x)^{3/2}}{15 a^2 \left (1-\frac {1}{a x}\right )^{3/2} x}+\frac {2 \sqrt {1+\frac {1}{a x}} x (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.63 \[ -\frac {2 c \sqrt {\frac {1}{a x}+1} \left (3 a^2 x^2-14 a x+43\right ) \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*c*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x]*(43 - 14*a*x + 3*a^2*x^2))/(15*a*Sqrt[1 - 1/(a*x)])

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fricas [A]  time = 0.60, size = 64, normalized size = 0.67 \[ -\frac {2 \, {\left (3 \, a^{3} c x^{3} - 11 \, a^{2} c x^{2} + 29 \, a c x + 43 \, c\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*c*x^3 - 11*a^2*c*x^2 + 29*a*c*x + 43*c)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)

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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*c*x+c)^(3/2)*((a*x-1)/(a*x+1))^(1/2),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 = 0.04, size = 56, normalized size = 0.59 \[ \frac {2 \left (a x +1\right ) \left (3 a^{2} x^{2}-14 a x +43\right ) \left (-a c x +c \right )^{\frac {3}{2}} \sqrt {\frac {a x -1}{a x +1}}}{15 a \left (a x -1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+1)*(3*a^2*x^2-14*a*x+43)*(-a*c*x+c)^(3/2)*((a*x-1)/(a*x+1))^(1/2)/a/(a*x-1)^2

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maxima [A]  time = 0.33, size = 72, normalized size = 0.76 \[ -\frac {2 \, {\left (3 \, a^{3} \sqrt {-c} c x^{3} - 11 \, a^{2} \sqrt {-c} c x^{2} + 29 \, a \sqrt {-c} c x + 43 \, \sqrt {-c} c\right )} {\left (a x - 1\right )}}{15 \, {\left (a^{2} x - a\right )} \sqrt {a x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*sqrt(-c)*c*x^3 - 11*a^2*sqrt(-c)*c*x^2 + 29*a*sqrt(-c)*c*x + 43*sqrt(-c)*c)*(a*x - 1)/((a^2*x - a
)*sqrt(a*x + 1))

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mupad [B]  time = 1.34, size = 82, normalized size = 0.86 \[ -\frac {2\,c\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (3\,a^2\,x^2-8\,a\,x+21\right )}{15\,a}-\frac {128\,c\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{15\,a\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (2*c*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(3*a^2*x^2 - 8*a*x + 21))/(15*a) - (128*c*(c - a*c*x)^(1/
2)*((a*x - 1)/(a*x + 1))^(1/2))/(15*a*(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((-a*c*x+c)**(3/2)*((a*x-1)/(a*x+1))**(1/2),x)

[Out]

Timed out

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