3.3.29 \(\int e^{\coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx\) [229]

Optimal. Leaf size=77 \[ \frac {8 a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{15 (c-a c x)^{3/2}}+\frac {2 a^2 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{5 \sqrt {c-a c x}} \]

[Out]

8/15*a^2*c^3*(1-1/a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(3/2)+2/5*a^2*c^2*(1-1/a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6311, 6316, 79, 37} \begin {gather*} \frac {2 x \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}-\frac {14 \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^{3/2}}{15 a \left (1-\frac {1}{a x}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^(3/2))/(15*a*(1 - 1/(a*x))^(3/2)) + (2*(1 + 1/(a*x))^(3/2)*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 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 6311

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 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[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 ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right ) \sqrt {1+\frac {x}{a}}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{3/2}}\\ &=\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^{3/2}}{5 \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {\left (7 \left (\frac {1}{x}\right )^{3/2} (c-a c x)^{3/2}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 a \left (1-\frac {1}{a x}\right )^{3/2}}\\ &=-\frac {14 \left (1+\frac {1}{a x}\right )^{3/2} (c-a c x)^{3/2}}{15 a \left (1-\frac {1}{a x}\right )^{3/2}}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} 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 = 0.02, size = 57, normalized size = 0.74 \begin {gather*} -\frac {2 c \sqrt {1+\frac {1}{a x}} (1+a x) (-7+3 a x) \sqrt {c-a c x}}{15 a \sqrt {1-\frac {1}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Sqrt[1 + 1/(a*x)]*(1 + a*x)*(-7 + 3*a*x)*Sqrt[c - a*c*x])/(15*a*Sqrt[1 - 1/(a*x)])

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Maple [A]
time = 0.12, size = 43, normalized size = 0.56

method	result	size
default	\(-\frac {2 \sqrt {-c \left (a x -1\right )}\, c \left (a x +1\right ) \left (3 a x -7\right )}{15 \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(43\)
gosper	\(\frac {2 \left (a x +1\right ) \left (3 a x -7\right ) \left (-a c x +c \right )^{\frac {3}{2}}}{15 a \left (a x -1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(48\)
risch	\(\frac {2 c^{2} \left (a x -1\right ) \left (3 a^{2} x^{2}-4 a x -7\right )}{15 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/((a*x-1)/(a*x+1))^(1/2)*(-c*(a*x-1))^(1/2)*c*(a*x+1)*(3*a*x-7)/a

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Maxima [A]
time = 0.26, size = 45, normalized size = 0.58 \begin {gather*} -\frac {2 \, {\left (3 \, a^{2} \sqrt {-c} c x^{2} - 4 \, a \sqrt {-c} c x - 7 \, \sqrt {-c} c\right )} \sqrt {a x + 1}}{15 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^2*sqrt(-c)*c*x^2 - 4*a*sqrt(-c)*c*x - 7*sqrt(-c)*c)*sqrt(a*x + 1)/a

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Fricas [A]
time = 0.38, size = 64, normalized size = 0.83 \begin {gather*} -\frac {2 \, {\left (3 \, a^{3} c x^{3} - a^{2} c x^{2} - 11 \, a c x - 7 \, c\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{15 \, {\left (a^{2} x - a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*c*x^3 - a^2*c*x^2 - 11*a*c*x - 7*c)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^2*x - a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-c*(a*x - 1))**(3/2)/sqrt((a*x - 1)/(a*x + 1)), 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(1/((a*x-1)/(a*x+1))^(1/2)*(-a*c*x+c)^(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 [B]
time = 1.37, size = 50, normalized size = 0.65 \begin {gather*} -\frac {2\,c\,\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^2\,\left (3\,a\,x-7\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{15\,a\,\left (a\,x-1\right )} \end {gather*}

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)^2*(3*a*x - 7)*((a*x - 1)/(a*x + 1))^(1/2))/(15*a*(a*x - 1))

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