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

Optimal. Leaf size=94 \[ -\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 (14-15 a x) \left (1-a^2 x^2\right )^{3/2}}{30 a^2}-\frac {c^3 x \sqrt {1-a^2 x^2}}{4 a}-\frac {c^3 \sin ^{-1}(a x)}{4 a^2} \]

[Out]

-1/5*c^3*x^2*(-a^2*x^2+1)^(3/2)-1/30*c^3*(-15*a*x+14)*(-a^2*x^2+1)^(3/2)/a^2-1/4*c^3*arcsin(a*x)/a^2-1/4*c^3*x
*(-a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6128, 1809, 780, 195, 216} \[ -\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 (14-15 a x) \left (1-a^2 x^2\right )^{3/2}}{30 a^2}-\frac {c^3 x \sqrt {1-a^2 x^2}}{4 a}-\frac {c^3 \sin ^{-1}(a x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x*(c - a*c*x)^3,x]

[Out]

-(c^3*x*Sqrt[1 - a^2*x^2])/(4*a) - (c^3*x^2*(1 - a^2*x^2)^(3/2))/5 - (c^3*(14 - 15*a*x)*(1 - a^2*x^2)^(3/2))/(
30*a^2) - (c^3*ArcSin[a*x])/(4*a^2)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

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 (c-a c x)^3 \, dx &=c \int x (c-a c x)^2 \sqrt {1-a^2 x^2} \, dx\\ &=-\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c \int x \left (-7 a^2 c^2+10 a^3 c^2 x\right ) \sqrt {1-a^2 x^2} \, dx}{5 a^2}\\ &=-\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 (14-15 a x) \left (1-a^2 x^2\right )^{3/2}}{30 a^2}-\frac {c^3 \int \sqrt {1-a^2 x^2} \, dx}{2 a}\\ &=-\frac {c^3 x \sqrt {1-a^2 x^2}}{4 a}-\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 (14-15 a x) \left (1-a^2 x^2\right )^{3/2}}{30 a^2}-\frac {c^3 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac {c^3 x \sqrt {1-a^2 x^2}}{4 a}-\frac {1}{5} c^3 x^2 \left (1-a^2 x^2\right )^{3/2}-\frac {c^3 (14-15 a x) \left (1-a^2 x^2\right )^{3/2}}{30 a^2}-\frac {c^3 \sin ^{-1}(a x)}{4 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 75, normalized size = 0.80 \[ \frac {c^3 \left (\sqrt {1-a^2 x^2} \left (12 a^4 x^4-30 a^3 x^3+16 a^2 x^2+15 a x-28\right )+30 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{60 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(Sqrt[1 - a^2*x^2]*(-28 + 15*a*x + 16*a^2*x^2 - 30*a^3*x^3 + 12*a^4*x^4) + 30*ArcSin[Sqrt[1 - a*x]/Sqrt[2
]]))/(60*a^2)

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fricas [A]  time = 0.48, size = 92, normalized size = 0.98 \[ \frac {30 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (12 \, a^{4} c^{3} x^{4} - 30 \, a^{3} c^{3} x^{3} + 16 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x - 28 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{60 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(30*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (12*a^4*c^3*x^4 - 30*a^3*c^3*x^3 + 16*a^2*c^3*x^2 + 15*a
*c^3*x - 28*c^3)*sqrt(-a^2*x^2 + 1))/a^2

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giac [A]  time = 0.37, size = 81, normalized size = 0.86 \[ -\frac {c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{4 \, a {\left | a \right |}} + \frac {1}{60} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (\frac {15 \, c^{3}}{a} + 2 \, {\left (8 \, c^{3} + 3 \, {\left (2 \, a^{2} c^{3} x - 5 \, a c^{3}\right )} x\right )} x\right )} x - \frac {28 \, c^{3}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c^3*arcsin(a*x)*sgn(a)/(a*abs(a)) + 1/60*sqrt(-a^2*x^2 + 1)*((15*c^3/a + 2*(8*c^3 + 3*(2*a^2*c^3*x - 5*a*
c^3)*x)*x)*x - 28*c^3/a^2)

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maple [A]  time = 0.04, size = 140, normalized size = 1.49 \[ \frac {c^{3} a^{2} x^{4} \sqrt {-a^{2} x^{2}+1}}{5}+\frac {4 c^{3} x^{2} \sqrt {-a^{2} x^{2}+1}}{15}-\frac {7 c^{3} \sqrt {-a^{2} x^{2}+1}}{15 a^{2}}-\frac {c^{3} a \,x^{3} \sqrt {-a^{2} x^{2}+1}}{2}+\frac {c^{3} x \sqrt {-a^{2} x^{2}+1}}{4 a}-\frac {c^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{4 a \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c^3*a^2*x^4*(-a^2*x^2+1)^(1/2)+4/15*c^3*x^2*(-a^2*x^2+1)^(1/2)-7/15*c^3/a^2*(-a^2*x^2+1)^(1/2)-1/2*c^3*a*x
^3*(-a^2*x^2+1)^(1/2)+1/4*c^3*x*(-a^2*x^2+1)^(1/2)/a-1/4*c^3/a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(
1/2))

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maxima [A]  time = 0.41, size = 118, normalized size = 1.26 \[ \frac {1}{5} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{4} - \frac {1}{2} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{3} + \frac {4}{15} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x^{2} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{3} x}{4 \, a} - \frac {c^{3} \arcsin \left (a x\right )}{4 \, a^{2}} - \frac {7 \, \sqrt {-a^{2} x^{2} + 1} c^{3}}{15 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^4 - 1/2*sqrt(-a^2*x^2 + 1)*a*c^3*x^3 + 4/15*sqrt(-a^2*x^2 + 1)*c^3*x^2 + 1/4*
sqrt(-a^2*x^2 + 1)*c^3*x/a - 1/4*c^3*arcsin(a*x)/a^2 - 7/15*sqrt(-a^2*x^2 + 1)*c^3/a^2

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mupad [B]  time = 0.06, size = 108, normalized size = 1.15 \[ \frac {c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}{4\,a^3}-\frac {\frac {2\,c^3\,{\left (1-a^2\,x^2\right )}^{3/2}}{3}-\frac {c^3\,{\left (1-a^2\,x^2\right )}^{5/2}}{5}}{a^2}-\frac {\frac {c^3\,x\,\sqrt {1-a^2\,x^2}}{4}-\frac {c^3\,x\,{\left (1-a^2\,x^2\right )}^{3/2}}{2}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2))/(4*a^3) - ((2*c^3*(1 - a^2*x^2)^(3/2))/3 - (c^3*(1 - a^2*x^2)^(5/2))/
5)/a^2 - ((c^3*x*(1 - a^2*x^2)^(1/2))/4 - (c^3*x*(1 - a^2*x^2)^(3/2))/2)/a

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sympy [A]  time = 7.71, size = 355, normalized size = 3.78 \[ - a^{4} c^{3} \left (\begin {cases} - \frac {x^{4} \sqrt {- a^{2} x^{2} + 1}}{5 a^{2}} - \frac {4 x^{2} \sqrt {- a^{2} x^{2} + 1}}{15 a^{4}} - \frac {8 \sqrt {- a^{2} x^{2} + 1}}{15 a^{6}} & \text {for}\: a \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{3} \left (\begin {cases} - \frac {i x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {3 i x}{8 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \operatorname {acosh}{\left (a x \right )}}{8 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8 a^{5}} & \text {otherwise} \end {cases}\right ) - 2 a c^{3} \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) + c^{3} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**4*c**3*Piecewise((-x**4*sqrt(-a**2*x**2 + 1)/(5*a**2) - 4*x**2*sqrt(-a**2*x**2 + 1)/(15*a**4) - 8*sqrt(-a*
*2*x**2 + 1)/(15*a**6), Ne(a, 0)), (x**6/6, True)) + 2*a**3*c**3*Piecewise((-I*x**5/(4*sqrt(a**2*x**2 - 1)) -
I*x**3/(8*a**2*sqrt(a**2*x**2 - 1)) + 3*I*x/(8*a**4*sqrt(a**2*x**2 - 1)) - 3*I*acosh(a*x)/(8*a**5), Abs(a**2*x
**2) > 1), (x**5/(4*sqrt(-a**2*x**2 + 1)) + x**3/(8*a**2*sqrt(-a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 +
1)) + 3*asin(a*x)/(8*a**5), True)) - 2*a*c**3*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2*
a**3), Abs(a**2*x**2) > 1), (x**3/(2*sqrt(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a*
*3), True)) + c**3*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True))

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