3.5.41 \(\int e^{\coth ^{-1}(a x)} (c-\frac {c}{a x})^{3/2} \, dx\) [441]

Optimal. Leaf size=117 \[ \frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a \sqrt {c-\frac {c}{a x}}}+\frac {c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x}{\left (c-\frac {c}{a x}\right )^{3/2}}-\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6312, 893, 879, 889, 214} \begin {gather*} -\frac {c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a x}}}\right )}{a}+\frac {c^3 x \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{\left (c-\frac {c}{a x}\right )^{3/2}}+\frac {c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{a \sqrt {c-\frac {c}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 879

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

Rule 889

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 893

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 6312

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 70, normalized size = 0.60 \begin {gather*} \frac {c \sqrt {c-\frac {c}{a x}} \left (\sqrt {1+\frac {1}{a x}} (2+a x)-\tanh ^{-1}\left (\sqrt {1+\frac {1}{a x}}\right )\right )}{a \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 106, normalized size = 0.91

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c \left (2 a^{\frac {3}{2}} x \sqrt {x \left (a x +1\right )}-\ln \left (\frac {2 \sqrt {x \left (a x +1\right )}\, \sqrt {a}+2 a x +1}{2 \sqrt {a}}\right ) a x +4 \sqrt {x \left (a x +1\right )}\, \sqrt {a}\right )}{2 \sqrt {\frac {a x -1}{a x +1}}\, a^{\frac {3}{2}} \sqrt {x \left (a x +1\right )}}\)	\(106\)
risch	\(\frac {\left (a^{2} x^{2}+3 a x +2\right ) c \sqrt {\frac {c \left (a x -1\right )}{a x}}}{a \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}-\frac {\ln \left (\frac {\frac {1}{2} a c +c \,a^{2} x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}+a c x}\right ) c \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {c a x \left (a x +1\right )}}{2 \sqrt {a^{2} c}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/((a*x-1)/(a*x+1))^(1/2)*(c*(a*x-1)/a/x)^(1/2)*c*(2*a^(3/2)*x*(x*(a*x+1))^(1/2)-ln(1/2*(2*(x*(a*x+1))^(1/2)
*a^(1/2)+2*a*x+1)/a^(1/2))*a*x+4*(x*(a*x+1))^(1/2)*a^(1/2))/a^(3/2)/(x*(a*x+1))^(1/2)

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

[Out]

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

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Fricas [A]
time = 0.40, size = 313, normalized size = 2.68 \begin {gather*} \left [\frac {{\left (a c x - c\right )} \sqrt {c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt {c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (a^{2} c x^{2} + 3 \, a c x + 2 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{4 \, {\left (a^{2} x - a\right )}}, \frac {{\left (a c x - c\right )} \sqrt {-c} \arctan \left (\frac {2 \, {\left (a^{2} x^{2} + a x\right )} \sqrt {-c} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \, {\left (a^{2} c x^{2} + 3 \, a c x + 2 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} x - a\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((a*c*x - c)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x - 1)
/(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(a^2*c*x^2 + 3*a*c*x + 2*c)*sqrt((a*x - 1)/(a*x + 1))*
sqrt((a*c*x - c)/(a*x)))/(a^2*x - a), 1/2*((a*c*x - c)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x -
1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(a^2*c*x^2 + 3*a*c*x + 2*c)*sqrt((a*x - 1
)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a^2*x - a)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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