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]

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

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Rubi [A]  time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.02, size = 43, normalized size = 0.47 \[ -\frac {c (1-a x)^{7/2} \, _2F_1\left (\frac {3}{2},\frac {7}{2};\frac {9}{2};\frac {1}{2} (1-a x)\right )}{7 \sqrt {2} a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 81, normalized size = 0.89 \[ -\frac {24 \, a c x - 30 \, {\left (a c x + c\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (a^{2} c x^{2} - 7 \, a c x - 24 \, c\right )} \sqrt {-a^{2} x^{2} + 1} + 24 \, c}{2 \, {\left (a^{2} x + a\right )}} \]

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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giac [A]  time = 0.20, size = 73, normalized size = 0.80 \[ -\frac {15 \, c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, {\left | a \right |}} + \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} {\left (c x - \frac {8 \, c}{a}\right )} + \frac {16 \, c}{{\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \]

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))

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maple [B]  time = 0.04, size = 169, normalized size = 1.86 \[ -\frac {5 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{3} \left (x +\frac {1}{a}\right )^{2}}-\frac {5 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a}-\frac {15 c \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{2}-\frac {15 c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}-\frac {2 c \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{4} \left (x +\frac {1}{a}\right )^{3}} \]

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 = 0.45, size = 109, normalized size = 1.20 \[ \frac {2 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{a^{3} x^{2} + 2 \, a^{2} x + a} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c}{2 \, {\left (a^{2} x + a\right )}} - \frac {15 \, c \arcsin \left (a x\right )}{2 \, a} - \frac {12 \, \sqrt {-a^{2} x^{2} + 1} c}{a^{2} x + a} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} c}{2 \, a} \]

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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mupad [B]  time = 0.80, size = 96, normalized size = 1.05 \[ \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {4\,a\,c}{\sqrt {-a^2}}+\frac {c\,x\,\sqrt {-a^2}}{2}\right )-\frac {15\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2}+\frac {8\,c\,\sqrt {1-a^2\,x^2}}{x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}}}{\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1 - a^2*x^2)^(1/2)*((4*a*c)/(-a^2)^(1/2) + (c*x*(-a^2)^(1/2))/2) - (15*c*asinh(x*(-a^2)^(1/2)))/2 + (8*c*(1
- a^2*x^2)^(1/2))/(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a))/(-a^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - c \left (\int \left (- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\right )\, dx + \int \frac {a x \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx + \int \frac {a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\, dx + \int \left (- \frac {a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1}}{a^{3} x^{3} + 3 a^{2} x^{2} + 3 a x + 1}\right )\, dx\right ) \]

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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