3.6.3 \(\int \frac {e^{2 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^2} \, dx\) [503]

Optimal. Leaf size=42 \[ 4 a \sqrt {c-\frac {c}{a x}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6302, 6268, 25, 528, 455, 45} \begin {gather*} 4 a \sqrt {c-\frac {c}{a x}}-\frac {2 a \left (c-\frac {c}{a x}\right )^{3/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*a*Sqrt[c - c/(a*x)] - (2*a*(c - c/(a*x))^(3/2))/(3*c)

Rule 25

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 6268

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, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[p] && IntegerQ[n/2] &
&  !GtQ[c, 0]

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 28, normalized size = 0.67 \begin {gather*} \frac {2 \sqrt {c-\frac {c}{a x}} (1+5 a x)}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.12, size = 173, normalized size = 4.12

method	result	size
gosper	\(\frac {2 \left (5 a x +1\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{3 x}\)	\(27\)
trager	\(\frac {2 \left (5 a x +1\right ) \sqrt {-\frac {-a c x +c}{a x}}}{3 x}\)	\(29\)
risch	\(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (5 a^{2} x^{2}-4 a x -1\right )}{3 \left (a x -1\right ) x}\)	\(42\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (-6 \sqrt {a \,x^{2}-x}\, a^{\frac {5}{2}} x^{3}-6 a^{\frac {5}{2}} \sqrt {\left (a x -1\right ) x}\, x^{3}+12 a^{\frac {3}{2}} \left (a \,x^{2}-x \right )^{\frac {3}{2}} x +3 \ln \left (\frac {2 \sqrt {a \,x^{2}-x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x^{3}-3 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a^{2} x^{3}+2 \left (a \,x^{2}-x \right )^{\frac {3}{2}} \sqrt {a}\right )}{3 x^{2} \sqrt {\left (a x -1\right ) x}\, \sqrt {a}}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(c*(a*x-1)/a/x)^(1/2)/x^2*(-6*(a*x^2-x)^(1/2)*a^(5/2)*x^3-6*a^(5/2)*((a*x-1)*x)^(1/2)*x^3+12*a^(3/2)*(a*x
^2-x)^(3/2)*x+3*ln(1/2*(2*(a*x^2-x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a^2*x^3-3*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1
/2)+2*a*x-1)/a^(1/2))*a^2*x^3+2*(a*x^2-x)^(3/2)*a^(1/2))/((a*x-1)*x)^(1/2)/a^(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)*(c-c/a/x)^(1/2)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.32, size = 28, normalized size = 0.67 \begin {gather*} \frac {2 \, {\left (5 \, a x + 1\right )} \sqrt {\frac {a c x - c}{a x}}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(-1 + 1/(a*x)))*(a*x + 1)/(x**2*(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)*(c-c/a/x)^(1/2)/x^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 = 1.26, size = 24, normalized size = 0.57 \begin {gather*} \frac {2\,\sqrt {c-\frac {c}{a\,x}}\,\left (5\,a\,x+1\right )}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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