3.219 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a c x) \, dx\)

Optimal. Leaf size=91 \[ -\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{5 c \sqrt{1-a^2 x^2} (1-a x)}{2 a}-\frac{15 c \sqrt{1-a^2 x^2}}{2 a}-\frac{15 c \sin ^{-1}(a x)}{2 a} \]

[Out]

(-2*c*(1 - a*x)^3)/(a*Sqrt[1 - a^2*x^2]) - (15*c*Sqrt[1 - a^2*x^2])/(2*a) - (5*c*(1 - a*x)*Sqrt[1 - a^2*x^2])/
(2*a) - (15*c*ArcSin[a*x])/(2*a)

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Rubi [A]  time = 0.0641629, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6127, 669, 671, 641, 216} \[ -\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{5 c \sqrt{1-a^2 x^2} (1-a x)}{2 a}-\frac{15 c \sqrt{1-a^2 x^2}}{2 a}-\frac{15 c \sin ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)/E^(3*ArcTanh[a*x]),x]

[Out]

(-2*c*(1 - a*x)^3)/(a*Sqrt[1 - a^2*x^2]) - (15*c*Sqrt[1 - a^2*x^2])/(2*a) - (5*c*(1 - a*x)*Sqrt[1 - a^2*x^2])/
(2*a) - (15*c*ArcSin[a*x])/(2*a)

Rule 6127

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

Rule 669

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

Rule 671

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

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]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} (c-a c x) \, dx &=\frac{\int \frac{(c-a c x)^4}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=-\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{5 \int \frac{(c-a c x)^2}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=-\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{5 c (1-a x) \sqrt{1-a^2 x^2}}{2 a}-\frac{15}{2} \int \frac{c-a c x}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{15 c \sqrt{1-a^2 x^2}}{2 a}-\frac{5 c (1-a x) \sqrt{1-a^2 x^2}}{2 a}-\frac{1}{2} (15 c) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{2 c (1-a x)^3}{a \sqrt{1-a^2 x^2}}-\frac{15 c \sqrt{1-a^2 x^2}}{2 a}-\frac{5 c (1-a x) \sqrt{1-a^2 x^2}}{2 a}-\frac{15 c \sin ^{-1}(a x)}{2 a}\\ \end{align*}

Mathematica [C]  time = 0.0148091, size = 43, normalized size = 0.47 \[ -\frac{c (1-a x)^{7/2} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{7}{2},\frac{9}{2},\frac{1}{2} (1-a x)\right )}{7 \sqrt{2} a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(c*(1 - a*x)^(7/2)*Hypergeometric2F1[3/2, 7/2, 9/2, (1 - a*x)/2])/(7*Sqrt[2]*a)

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Maple [B]  time = 0.043, size = 169, normalized size = 1.9 \begin{align*} -5\,{\frac{c \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{3} \left ( x+{a}^{-1} \right ) ^{2}}}-5\,{\frac{c \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{3/2}}{a}}-{\frac{15\,cx}{2}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{15\,c}{2}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}-2\,{\frac{c \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{{a}^{4} \left ( x+{a}^{-1} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-5*c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-15/2*c*(-a^2*(x
+1/a)^2+2*a*(x+1/a))^(1/2)*x-15/2*c/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))-2*c/a
^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)

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Maxima [A]  time = 1.44679, size = 147, normalized size = 1.62 \begin{align*} \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} c}{a^{3} x^{2} + 2 \, a^{2} x + a} - \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} c}{2 \,{\left (a^{2} x + a\right )}} - \frac{15 \, c \arcsin \left (a x\right )}{2 \, a} - \frac{12 \, \sqrt{-a^{2} x^{2} + 1} c}{a^{2} x + a} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} c}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-a^2*x^2 + 1)^(3/2)*c/(a^3*x^2 + 2*a^2*x + a) - 1/2*(-a^2*x^2 + 1)^(3/2)*c/(a^2*x + a) - 15/2*c*arcsin(a*x)
/a - 12*sqrt(-a^2*x^2 + 1)*c/(a^2*x + a) - 3/2*sqrt(-a^2*x^2 + 1)*c/a

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Fricas [A]  time = 1.64923, size = 192, normalized size = 2.11 \begin{align*} -\frac{24 \, a c x - 30 \,{\left (a c x + c\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) -{\left (a^{2} c x^{2} - 7 \, a c x - 24 \, c\right )} \sqrt{-a^{2} x^{2} + 1} + 24 \, c}{2 \,{\left (a^{2} x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(24*a*c*x - 30*(a*c*x + c)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (a^2*c*x^2 - 7*a*c*x - 24*c)*sqrt(-a^
2*x^2 + 1) + 24*c)/(a^2*x + a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24556, size = 99, normalized size = 1.09 \begin{align*} -\frac{15 \, c \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{2 \,{\left | a \right |}} + \frac{1}{2} \, \sqrt{-a^{2} x^{2} + 1}{\left (c x - \frac{8 \, c}{a}\right )} + \frac{16 \, c}{{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/2*c*arcsin(a*x)*sgn(a)/abs(a) + 1/2*sqrt(-a^2*x^2 + 1)*(c*x - 8*c/a) + 16*c/(((sqrt(-a^2*x^2 + 1)*abs(a) +
a)/(a^2*x) + 1)*abs(a))