3.286 \(\int e^{\tanh ^{-1}(a x)} x^4 (c-a c x) \, dx\)

Optimal. Leaf size=83 \[ \frac {c \sin ^{-1}(a x)}{16 a^5}+\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}-\frac {c x \sqrt {1-a^2 x^2}}{16 a^4} \]

[Out]

1/16*c*arcsin(a*x)/a^5-1/16*c*x*(-a^2*x^2+1)^(1/2)/a^4-1/24*c*x^3*(-a^2*x^2+1)^(1/2)/a^2+1/6*c*x^5*(-a^2*x^2+1
)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {6128, 279, 321, 216} \[ \frac {1}{6} c x^5 \sqrt {1-a^2 x^2}-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}-\frac {c x \sqrt {1-a^2 x^2}}{16 a^4}+\frac {c \sin ^{-1}(a x)}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^4*(c - a*c*x),x]

[Out]

-(c*x*Sqrt[1 - a^2*x^2])/(16*a^4) - (c*x^3*Sqrt[1 - a^2*x^2])/(24*a^2) + (c*x^5*Sqrt[1 - a^2*x^2])/6 + (c*ArcS
in[a*x])/(16*a^5)

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 e^{\tanh ^{-1}(a x)} x^4 (c-a c x) \, dx &=c \int x^4 \sqrt {1-a^2 x^2} \, dx\\ &=\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}+\frac {1}{6} c \int \frac {x^4}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}+\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}+\frac {c \int \frac {x^2}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}\\ &=-\frac {c x \sqrt {1-a^2 x^2}}{16 a^4}-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}+\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}+\frac {c \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{16 a^4}\\ &=-\frac {c x \sqrt {1-a^2 x^2}}{16 a^4}-\frac {c x^3 \sqrt {1-a^2 x^2}}{24 a^2}+\frac {1}{6} c x^5 \sqrt {1-a^2 x^2}+\frac {c \sin ^{-1}(a x)}{16 a^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 50, normalized size = 0.60 \[ \frac {c \left (a x \sqrt {1-a^2 x^2} \left (8 a^4 x^4-2 a^2 x^2-3\right )+3 \sin ^{-1}(a x)\right )}{48 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^4*(c - a*c*x),x]

[Out]

(c*(a*x*Sqrt[1 - a^2*x^2]*(-3 - 2*a^2*x^2 + 8*a^4*x^4) + 3*ArcSin[a*x]))/(48*a^5)

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 69, normalized size = 0.83 \[ -\frac {6 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (8 \, a^{5} c x^{5} - 2 \, a^{3} c x^{3} - 3 \, a c x\right )} \sqrt {-a^{2} x^{2} + 1}}{48 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="fricas")

[Out]

-1/48*(6*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (8*a^5*c*x^5 - 2*a^3*c*x^3 - 3*a*c*x)*sqrt(-a^2*x^2 + 1))/
a^5

________________________________________________________________________________________

giac [A]  time = 0.22, size = 57, normalized size = 0.69 \[ \frac {1}{48} \, \sqrt {-a^{2} x^{2} + 1} {\left (2 \, {\left (4 \, c x^{2} - \frac {c}{a^{2}}\right )} x^{2} - \frac {3 \, c}{a^{4}}\right )} x + \frac {c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{16 \, a^{4} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="giac")

[Out]

1/48*sqrt(-a^2*x^2 + 1)*(2*(4*c*x^2 - c/a^2)*x^2 - 3*c/a^4)*x + 1/16*c*arcsin(a*x)*sgn(a)/(a^4*abs(a))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 91, normalized size = 1.10 \[ \frac {c \,x^{5} \sqrt {-a^{2} x^{2}+1}}{6}-\frac {c \,x^{3} \sqrt {-a^{2} x^{2}+1}}{24 a^{2}}-\frac {c x \sqrt {-a^{2} x^{2}+1}}{16 a^{4}}+\frac {c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 a^{4} \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x)

[Out]

1/6*c*x^5*(-a^2*x^2+1)^(1/2)-1/24*c*x^3*(-a^2*x^2+1)^(1/2)/a^2-1/16*c*x*(-a^2*x^2+1)^(1/2)/a^4+1/16*c/a^4/(a^2
)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 69, normalized size = 0.83 \[ \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} c x^{5} - \frac {\sqrt {-a^{2} x^{2} + 1} c x^{3}}{24 \, a^{2}} - \frac {\sqrt {-a^{2} x^{2} + 1} c x}{16 \, a^{4}} + \frac {c \arcsin \left (a x\right )}{16 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*x^4*(-a*c*x+c),x, algorithm="maxima")

[Out]

1/6*sqrt(-a^2*x^2 + 1)*c*x^5 - 1/24*sqrt(-a^2*x^2 + 1)*c*x^3/a^2 - 1/16*sqrt(-a^2*x^2 + 1)*c*x/a^4 + 1/16*c*ar
csin(a*x)/a^5

________________________________________________________________________________________

mupad [B]  time = 0.78, size = 82, normalized size = 0.99 \[ \frac {c\,x^5\,\sqrt {1-a^2\,x^2}}{6}-\frac {c\,x^3\,\sqrt {1-a^2\,x^2}}{24\,a^2}-\frac {c\,x\,\sqrt {1-a^2\,x^2}}{16\,a^4}+\frac {c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{16\,a^4\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(c - a*c*x)*(a*x + 1))/(1 - a^2*x^2)^(1/2),x)

[Out]

(c*x^5*(1 - a^2*x^2)^(1/2))/6 - (c*x^3*(1 - a^2*x^2)^(1/2))/(24*a^2) - (c*x*(1 - a^2*x^2)^(1/2))/(16*a^4) + (c
*asinh(x*(-a^2)^(1/2)))/(16*a^4*(-a^2)^(1/2))

________________________________________________________________________________________

sympy [A]  time = 6.53, size = 192, normalized size = 2.31 \[ c \left (\begin {cases} \frac {i a^{2} x^{7}}{6 \sqrt {a^{2} x^{2} - 1}} - \frac {5 i x^{5}}{24 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{48 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {i x}{16 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{16 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {a^{2} x^{7}}{6 \sqrt {- a^{2} x^{2} + 1}} + \frac {5 x^{5}}{24 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{48 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{16 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{16 a^{5}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*x**4*(-a*c*x+c),x)

[Out]

c*Piecewise((I*a**2*x**7/(6*sqrt(a**2*x**2 - 1)) - 5*I*x**5/(24*sqrt(a**2*x**2 - 1)) - I*x**3/(48*a**2*sqrt(a*
*2*x**2 - 1)) + I*x/(16*a**4*sqrt(a**2*x**2 - 1)) - I*acosh(a*x)/(16*a**5), Abs(a**2*x**2) > 1), (-a**2*x**7/(
6*sqrt(-a**2*x**2 + 1)) + 5*x**5/(24*sqrt(-a**2*x**2 + 1)) + x**3/(48*a**2*sqrt(-a**2*x**2 + 1)) - x/(16*a**4*
sqrt(-a**2*x**2 + 1)) + asin(a*x)/(16*a**5), True))

________________________________________________________________________________________