3.6.51 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^{5/2}} \, dx\) [551]

Optimal. Leaf size=119 \[ \frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6268, 25, 528, 382, 105, 157, 162, 65, 214} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}+\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(a*c^2*Sqrt[c - c/(a*x)]) - x/(c^2*Sqrt[c - c/(a*x)]) - ArcTanh[Sqrt[c - c/(a*x)]/Sqrt[c]]/(a*c^(5/2)) - Arc
Tanh[Sqrt[c - c/(a*x)]/(Sqrt[2]*Sqrt[c])]/(Sqrt[2]*a*c^(5/2))

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[(a + b/x^n)^p*((c +
d/x^n)^q/x^2), x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 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]

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx &=\int \frac {1-a x}{\left (c-\frac {c}{a x}\right )^{5/2} (1+a x)} \, dx\\ &=-\frac {a \int \frac {x}{\left (c-\frac {c}{a x}\right )^{3/2} (1+a x)} \, dx}{c}\\ &=-\frac {a \int \frac {1}{\left (a+\frac {1}{x}\right ) \left (c-\frac {c}{a x}\right )^{3/2}} \, dx}{c}\\ &=\frac {a \text {Subst}\left (\int \frac {1}{x^2 (a+x) \left (c-\frac {c x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}-\frac {\text {Subst}\left (\int \frac {-\frac {c}{2}-\frac {3 c x}{2 a}}{x (a+x) \left (c-\frac {c x}{a}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^2}\\ &=\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}+\frac {\text {Subst}\left (\int \frac {\frac {c^2}{2}+\frac {c^2 x}{a}}{x (a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{c^4}\\ &=\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a c^2}+\frac {\text {Subst}\left (\int \frac {1}{(a+x) \sqrt {c-\frac {c x}{a}}} \, dx,x,\frac {1}{x}\right )}{2 a c^2}\\ &=\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}-\frac {\text {Subst}\left (\int \frac {1}{a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )}{c^3}-\frac {\text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-\frac {c}{a x}}\right )}{c^3}\\ &=\frac {2}{a c^2 \sqrt {c-\frac {c}{a x}}}-\frac {x}{c^2 \sqrt {c-\frac {c}{a x}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {c}}\right )}{a c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a c^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.04, size = 67, normalized size = 0.56 \begin {gather*} \frac {-a x+\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {a-\frac {1}{x}}{2 a}\right )+\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1-\frac {1}{a x}\right )}{a c^2 \sqrt {c-\frac {c}{a x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x) + Hypergeometric2F1[-1/2, 1, 1/2, (a - x^(-1))/(2*a)] + Hypergeometric2F1[-1/2, 1, 1/2, 1 - 1/(a*x)])/
(a*c^2*Sqrt[c - c/(a*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(369\) vs. \(2(100)=200\).
time = 0.48, size = 370, normalized size = 3.11

method	result	size
risch	\(-\frac {a x -1}{a \,c^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}}-\frac {\left (\frac {\ln \left (\frac {-\frac {1}{2} a c +a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c x a}\right )}{2 a^{3} \sqrt {a^{2} c}}-\frac {\sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+\left (x -\frac {1}{a}\right ) a c}}{a^{5} c \left (x -\frac {1}{a}\right )}-\frac {\sqrt {2}\, \ln \left (\frac {4 c -3 \left (x +\frac {1}{a}\right ) a c +2 \sqrt {2}\, \sqrt {c}\, \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-3 \left (x +\frac {1}{a}\right ) a c +2 c}}{x +\frac {1}{a}}\right )}{4 a^{4} \sqrt {c}}\right ) a^{2} \sqrt {a c x \left (a x -1\right )}}{c^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, x}\)	\(229\)
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \left (8 a^{\frac {7}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, x^{2}-a^{\frac {5}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x^{2}-4 a^{\frac {5}{2}} \left (\left (a x -1\right ) x \right )^{\frac {3}{2}} \sqrt {\frac {1}{a}}-16 a^{\frac {5}{2}} \sqrt {\left (a x -1\right ) x}\, \sqrt {\frac {1}{a}}\, x +2 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{3} x^{2}+2 a^{\frac {3}{2}} \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) x +8 \sqrt {\left (a x -1\right ) x}\, a^{\frac {3}{2}} \sqrt {\frac {1}{a}}-4 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) \sqrt {\frac {1}{a}}\, a^{2} x -\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {\frac {1}{a}}\, \sqrt {\left (a x -1\right ) x}\, a -3 a x +1}{a x +1}\right ) \sqrt {a}+2 \ln \left (\frac {2 \sqrt {\left (a x -1\right ) x}\, \sqrt {a}+2 a x -1}{2 \sqrt {a}}\right ) a \sqrt {\frac {1}{a}}\right )}{4 a^{\frac {3}{2}} \sqrt {\left (a x -1\right ) x}\, c^{3} \left (a x -1\right )^{2} \sqrt {\frac {1}{a}}}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)^2*(-a^2*x^2+1)/(c-c/a/x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(c*(a*x-1)/a/x)^(1/2)*x*(8*a^(7/2)*((a*x-1)*x)^(1/2)*(1/a)^(1/2)*x^2-a^(5/2)*2^(1/2)*ln((2*2^(1/2)*(1/a)^
(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x^2-4*a^(5/2)*((a*x-1)*x)^(3/2)*(1/a)^(1/2)-16*a^(5/2)*((a*x-1)*x)
^(1/2)*(1/a)^(1/2)*x+2*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^3*x^2+2*a^(3/2)*2^(
1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*((a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*x+8*((a*x-1)*x)^(1/2)*a^(3/2)*(1/a)^(1/2)
-4*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*(1/a)^(1/2)*a^2*x-2^(1/2)*ln((2*2^(1/2)*(1/a)^(1/2)*(
(a*x-1)*x)^(1/2)*a-3*a*x+1)/(a*x+1))*a^(1/2)+2*ln(1/2*(2*((a*x-1)*x)^(1/2)*a^(1/2)+2*a*x-1)/a^(1/2))*a*(1/a)^(
1/2))/a^(3/2)/((a*x-1)*x)^(1/2)/c^3/(a*x-1)^2/(1/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)^2*(-a^2*x^2+1)/(c-c/a/x)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 284, normalized size = 2.39 \begin {gather*} \left [\frac {\sqrt {2} {\left (a x - 1\right )} \sqrt {c} \log \left (\frac {2 \, \sqrt {2} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} - 3 \, a c x + c}{a x + 1}\right ) + 2 \, {\left (a x - 1\right )} \sqrt {c} \log \left (-2 \, a c x + 2 \, a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}} + c\right ) - 4 \, {\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{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 {-c} \sqrt {\frac {a c x - c}{a x}}}{2 \, c}\right ) + 2 \, {\left (a x - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} \sqrt {\frac {a c x - c}{a x}}}{c}\right ) - 2 \, {\left (a^{2} x^{2} - 2 \, a x\right )} \sqrt {\frac {a c x - c}{a x}}}{2 \, {\left (a^{2} c^{3} x - a c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x/(a*c**2*x*sqrt(c - c/(a*x)) - c**2*sqrt(c - c/(a*x)) - c**2*sqrt(c - c/(a*x))/(a*x) + c**2*sqrt(
c - c/(a*x))/(a**2*x**2)), x) - Integral(-1/(a*c**2*x*sqrt(c - c/(a*x)) - c**2*sqrt(c - c/(a*x)) - c**2*sqrt(c
- c/(a*x))/(a*x) + c**2*sqrt(c - c/(a*x))/(a**2*x**2)), 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)^2*(-a^2*x^2+1)/(c-c/a/x)^(5/2),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 {a^2\,x^2-1}{{\left (c-\frac {c}{a\,x}\right )}^{5/2}\,{\left (a\,x+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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