3.2.99 \(\int e^{-\coth ^{-1}(a x)} (c-a c x)^2 \, dx\) [199]

Optimal. Leaf size=100 \[ \frac {11}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

[Out]

-5/2*c^2*arctanh((1-1/a^2/x^2)^(1/2))/a+11/3*c^2*x*(1-1/a^2/x^2)^(1/2)-3/2*a*c^2*x^2*(1-1/a^2/x^2)^(1/2)+1/3*a
^2*c^2*x^3*(1-1/a^2/x^2)^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6310, 6313, 1821, 821, 272, 65, 214} \begin {gather*} -\frac {3}{2} a c^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {11}{3} c^2 x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}+\frac {1}{3} a^2 c^2 x^3 \sqrt {1-\frac {1}{a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^2/E^ArcCoth[a*x],x]

[Out]

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

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 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 1821

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6310

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int e^{-\coth ^{-1}(a x)} (c-a c x)^2 \, dx &=\left (a^2 c^2\right ) \int e^{-\coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^2 x^2 \, dx\\ &=-\left (\left (a^2 c^2\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^3}{x^4 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\right )\\ &=\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {1}{3} \left (a^2 c^2\right ) \text {Subst}\left (\int \frac {\frac {9}{a}-\frac {11 x}{a^2}+\frac {3 x^2}{a^3}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{6} \left (a^2 c^2\right ) \text {Subst}\left (\int \frac {\frac {22}{a^2}-\frac {15 x}{a^3}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {11}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=\frac {11}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3+\frac {\left (5 c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a}\\ &=\frac {11}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {1}{2} \left (5 a c^2\right ) \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )\\ &=\frac {11}{3} c^2 \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3}{2} a c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{3} a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x^3-\frac {5 c^2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 64, normalized size = 0.64 \begin {gather*} \frac {c^2 \left (a \sqrt {1-\frac {1}{a^2 x^2}} x \left (22-9 a x+2 a^2 x^2\right )-15 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{6 a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^2*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(22 - 9*a*x + 2*a^2*x^2) - 15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/(6*a)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(84)=168\).
time = 0.12, size = 176, normalized size = 1.76

method	result	size
risch	\(\frac {\left (2 a^{2} x^{2}-9 a x +22\right ) \left (a x +1\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}}{6 a}-\frac {5 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c^{2} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{2 \sqrt {a^{2}}\, \left (a x -1\right )}\)	\(112\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c^{2} \left (9 \sqrt {a^{2}}\, \sqrt {a^{2} x^{2}-1}\, a x -2 \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-9 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -24 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}+24 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )\right )}{6 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, a \sqrt {a^{2}}}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*c^2*(9*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*a*x-2*((a*x+1)*(a*x-1))^(3/2)*(a^2)^
(1/2)-9*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a-24*(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2)+24*a*ln
((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2)))/((a*x+1)*(a*x-1))^(1/2)/a/(a^2)^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (84) = 168\).
time = 0.26, size = 181, normalized size = 1.81 \begin {gather*} -\frac {1}{6} \, a {\left (\frac {15 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {15 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {2 \, {\left (33 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 40 \, c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 15 \, c^{2} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {3 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {3 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - a^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a*(15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + 2*(33*
c^2*((a*x - 1)/(a*x + 1))^(5/2) - 40*c^2*((a*x - 1)/(a*x + 1))^(3/2) + 15*c^2*sqrt((a*x - 1)/(a*x + 1)))/(3*(a
*x - 1)*a^2/(a*x + 1) - 3*(a*x - 1)^2*a^2/(a*x + 1)^2 + (a*x - 1)^3*a^2/(a*x + 1)^3 - a^2))

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Fricas [A]
time = 0.38, size = 104, normalized size = 1.04 \begin {gather*} -\frac {15 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 15 \, c^{2} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (2 \, a^{3} c^{2} x^{3} - 7 \, a^{2} c^{2} x^{2} + 13 \, a c^{2} x + 22 \, c^{2}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c^2*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (2*a^3*c^2*x^3 -
7*a^2*c^2*x^2 + 13*a*c^2*x + 22*c^2)*sqrt((a*x - 1)/(a*x + 1)))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 90, normalized size = 0.90 \begin {gather*} \frac {5 \, c^{2} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{2 \, {\left | a \right |}} + \frac {1}{6} \, \sqrt {a^{2} x^{2} - 1} {\left ({\left (2 \, a c^{2} x \mathrm {sgn}\left (a x + 1\right ) - 9 \, c^{2} \mathrm {sgn}\left (a x + 1\right )\right )} x + \frac {22 \, c^{2} \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*c^2*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/abs(a) + 1/6*sqrt(a^2*x^2 - 1)*((2*a*c^2*x*sgn(a*
x + 1) - 9*c^2*sgn(a*x + 1))*x + 22*c^2*sgn(a*x + 1)/a)

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Mupad [B]
time = 1.21, size = 140, normalized size = 1.40 \begin {gather*} \frac {5\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}-\frac {40\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+11\,c^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{a-\frac {3\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {3\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}}-\frac {5\,c^2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*c^2*((a*x - 1)/(a*x + 1))^(1/2) - (40*c^2*((a*x - 1)/(a*x + 1))^(3/2))/3 + 11*c^2*((a*x - 1)/(a*x + 1))^(5/
2))/(a - (3*a*(a*x - 1))/(a*x + 1) + (3*a*(a*x - 1)^2)/(a*x + 1)^2 - (a*(a*x - 1)^3)/(a*x + 1)^3) - (5*c^2*ata
nh(((a*x - 1)/(a*x + 1))^(1/2)))/a

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