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

Optimal. Leaf size=133 \[ \frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {35 c^3 \sqrt {1-a^2 x^2}}{8 a}+\frac {35 c^3 \sin ^{-1}(a x)}{8 a} \]

[Out]

35/8*c^3*arcsin(a*x)/a+35/8*c^3*(-a^2*x^2+1)^(1/2)/a+35/24*c^3*(-a*x+1)*(-a^2*x^2+1)^(1/2)/a+7/12*c^3*(-a*x+1)
^2*(-a^2*x^2+1)^(1/2)/a+1/4*c^3*(-a*x+1)^3*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.10, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6127, 671, 641, 216} \[ \frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {35 c^3 \sqrt {1-a^2 x^2}}{8 a}+\frac {35 c^3 \sin ^{-1}(a x)}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)^3/E^ArcTanh[a*x],x]

[Out]

(35*c^3*Sqrt[1 - a^2*x^2])/(8*a) + (35*c^3*(1 - a*x)*Sqrt[1 - a^2*x^2])/(24*a) + (7*c^3*(1 - a*x)^2*Sqrt[1 - a
^2*x^2])/(12*a) + (c^3*(1 - a*x)^3*Sqrt[1 - a^2*x^2])/(4*a) + (35*c^3*ArcSin[a*x])/(8*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 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^{-\tanh ^{-1}(a x)} (c-a c x)^3 \, dx &=\frac {\int \frac {(c-a c x)^4}{\sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {7}{4} \int \frac {(c-a c x)^3}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {1}{12} (35 c) \int \frac {(c-a c x)^2}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {1}{8} \left (35 c^2\right ) \int \frac {c-a c x}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {35 c^3 \sqrt {1-a^2 x^2}}{8 a}+\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {1}{8} \left (35 c^3\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {35 c^3 \sqrt {1-a^2 x^2}}{8 a}+\frac {35 c^3 (1-a x) \sqrt {1-a^2 x^2}}{24 a}+\frac {7 c^3 (1-a x)^2 \sqrt {1-a^2 x^2}}{12 a}+\frac {c^3 (1-a x)^3 \sqrt {1-a^2 x^2}}{4 a}+\frac {35 c^3 \sin ^{-1}(a x)}{8 a}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 80, normalized size = 0.60 \[ \frac {c^3 \left (\frac {\sqrt {a x+1} \left (6 a^4 x^4-38 a^3 x^3+113 a^2 x^2-241 a x+160\right )}{\sqrt {1-a x}}-210 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{24 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*((Sqrt[1 + a*x]*(160 - 241*a*x + 113*a^2*x^2 - 38*a^3*x^3 + 6*a^4*x^4))/Sqrt[1 - a*x] - 210*ArcSin[Sqrt[1
- a*x]/Sqrt[2]]))/(24*a)

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fricas [A]  time = 0.75, size = 81, normalized size = 0.61 \[ -\frac {210 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (6 \, a^{3} c^{3} x^{3} - 32 \, a^{2} c^{3} x^{2} + 81 \, a c^{3} x - 160 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(210*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (6*a^3*c^3*x^3 - 32*a^2*c^3*x^2 + 81*a*c^3*x - 160*c^3
)*sqrt(-a^2*x^2 + 1))/a

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giac [A]  time = 0.78, size = 67, normalized size = 0.50 \[ \frac {35 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, {\left | a \right |}} + \frac {1}{24} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {160 \, c^{3}}{a} - {\left (81 \, c^{3} + 2 \, {\left (3 \, a^{2} c^{3} x - 16 \, a c^{3}\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1/24*sqrt(-a^2*x^2 + 1)*(160*c^3/a - (81*c^3 + 2*(3*a^2*c^3*x - 16*a*c^3)
*x)*x)

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maple [A]  time = 0.04, size = 160, normalized size = 1.20 \[ \frac {c^{3} x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}-\frac {29 c^{3} x \sqrt {-a^{2} x^{2}+1}}{8}-\frac {29 c^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 \sqrt {a^{2}}}-\frac {4 c^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 a}+\frac {8 c^{3} \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a}+\frac {8 c^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^3*x*(-a^2*x^2+1)^(3/2)-29/8*c^3*x*(-a^2*x^2+1)^(1/2)-29/8*c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2
+1)^(1/2))-4/3*c^3*(-a^2*x^2+1)^(3/2)/a+8*c^3/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+8*c^3/(a^2)^(1/2)*arctan((a
^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))

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maxima [A]  time = 0.43, size = 89, normalized size = 0.67 \[ \frac {1}{4} \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{3} x - \frac {29}{8} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x - \frac {4 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} c^{3}}{3 \, a} + \frac {35 \, c^{3} \arcsin \left (a x\right )}{8 \, a} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-a^2*x^2 + 1)^(3/2)*c^3*x - 29/8*sqrt(-a^2*x^2 + 1)*c^3*x - 4/3*(-a^2*x^2 + 1)^(3/2)*c^3/a + 35/8*c^3*arc
sin(a*x)/a + 8*sqrt(-a^2*x^2 + 1)*c^3/a

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mupad [B]  time = 0.03, size = 105, normalized size = 0.79 \[ \frac {35\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,\sqrt {-a^2}}-\frac {27\,c^3\,x\,\sqrt {1-a^2\,x^2}}{8}+\frac {20\,c^3\,\sqrt {1-a^2\,x^2}}{3\,a}+\frac {4\,a\,c^3\,x^2\,\sqrt {1-a^2\,x^2}}{3}-\frac {a^2\,c^3\,x^3\,\sqrt {1-a^2\,x^2}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(35*c^3*asinh(x*(-a^2)^(1/2)))/(8*(-a^2)^(1/2)) - (27*c^3*x*(1 - a^2*x^2)^(1/2))/8 + (20*c^3*(1 - a^2*x^2)^(1/
2))/(3*a) + (4*a*c^3*x^2*(1 - a^2*x^2)^(1/2))/3 - (a^2*c^3*x^3*(1 - a^2*x^2)^(1/2))/4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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