3.300 \(\int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)^2}{x^2} \, dx\)

Optimal. Leaf size=58 \[ -\frac {c^2 (a x+1) \sqrt {1-a^2 x^2}}{x}+a c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-a c^2 \sin ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6128, 813, 844, 216, 266, 63, 208} \[ -\frac {c^2 (a x+1) \sqrt {1-a^2 x^2}}{x}+a c^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-a c^2 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 266

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 6128

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

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 84, normalized size = 1.45 \[ \frac {1}{2} c^2 \left (\frac {2 (a x-1) (a x+1)^2}{x \sqrt {1-a^2 x^2}}+2 a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )-a \sin ^{-1}(a x)+2 a \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 91, normalized size = 1.57 \[ \frac {2 \, a c^{2} x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - a c^{2} x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - a c^{2} x - {\left (a c^{2} x + c^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*c^2*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - a*c^2*x*log((sqrt(-a^2*x^2 + 1) - 1)/x) - a*c^2*x - (a*c^2
*x + c^2)*sqrt(-a^2*x^2 + 1))/x

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giac [B]  time = 0.23, size = 140, normalized size = 2.41 \[ \frac {a^{4} c^{2} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} - \frac {a^{2} c^{2} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{{\left | a \right |}} + \frac {a^{2} c^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \sqrt {-a^{2} x^{2} + 1} a c^{2} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{2}}{2 \, x {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^4*c^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) - a^2*c^2*arcsin(a*x)*sgn(a)/abs(a) + a^2*c^2*log(1/2*a
bs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - sqrt(-a^2*x^2 + 1)*a*c^2 - 1/2*(sqrt(-a^2*x^2 +
1)*abs(a) + a)*c^2/(x*abs(a))

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maple [A]  time = 0.04, size = 91, normalized size = 1.57 \[ -c^{2} a \sqrt {-a^{2} x^{2}+1}-\frac {c^{2} a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+c^{2} a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {c^{2} \sqrt {-a^{2} x^{2}+1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 80, normalized size = 1.38 \[ -a c^{2} \arcsin \left (a x\right ) + a c^{2} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \sqrt {-a^{2} x^{2} + 1} a c^{2} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c^2*arcsin(a*x) + a*c^2*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - sqrt(-a^2*x^2 + 1)*a*c^2 - sqrt(-a^2*
x^2 + 1)*c^2/x

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mupad [B]  time = 0.04, size = 87, normalized size = 1.50 \[ -a\,c^2\,\sqrt {1-a^2\,x^2}-\frac {c^2\,\sqrt {1-a^2\,x^2}}{x}-\frac {a^2\,c^2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}-a\,c^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- a*c^2*(1 - a^2*x^2)^(1/2) - (c^2*(1 - a^2*x^2)^(1/2))/x - a*c^2*atan((1 - a^2*x^2)^(1/2)*1i)*1i - (a^2*c^2*a
sinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2)

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sympy [C]  time = 4.60, size = 153, normalized size = 2.64 \[ a^{3} c^{2} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) - a^{2} c^{2} \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) - a c^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + c^{2} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c**2*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) - a**2*c**2*Piecewise((sqrt(a**
(-2))*asin(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) - a*c**2*Piecewise((-acos
h(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + c**2*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a*
*2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))

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