3.1139 \(\int e^{3 \tanh ^{-1}(a x)} x (c-a^2 c x^2) \, dx\)

Optimal. Leaf size=120 \[ -\frac {c \sqrt {1-a^2 x^2} (a x+1)^3}{4 a^2}-\frac {c \sqrt {1-a^2 x^2} (a x+1)^2}{4 a^2}-\frac {5 c \sqrt {1-a^2 x^2} (a x+1)}{8 a^2}-\frac {15 c \sqrt {1-a^2 x^2}}{8 a^2}+\frac {15 c \sin ^{-1}(a x)}{8 a^2} \]

[Out]

15/8*c*arcsin(a*x)/a^2-15/8*c*(-a^2*x^2+1)^(1/2)/a^2-5/8*c*(a*x+1)*(-a^2*x^2+1)^(1/2)/a^2-1/4*c*(a*x+1)^2*(-a^
2*x^2+1)^(1/2)/a^2-1/4*c*(a*x+1)^3*(-a^2*x^2+1)^(1/2)/a^2

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6148, 795, 671, 641, 216} \[ -\frac {c \sqrt {1-a^2 x^2} (a x+1)^3}{4 a^2}-\frac {c \sqrt {1-a^2 x^2} (a x+1)^2}{4 a^2}-\frac {5 c \sqrt {1-a^2 x^2} (a x+1)}{8 a^2}-\frac {15 c \sqrt {1-a^2 x^2}}{8 a^2}+\frac {15 c \sin ^{-1}(a x)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])*x*(c - a^2*c*x^2),x]

[Out]

(-15*c*Sqrt[1 - a^2*x^2])/(8*a^2) - (5*c*(1 + a*x)*Sqrt[1 - a^2*x^2])/(8*a^2) - (c*(1 + a*x)^2*Sqrt[1 - a^2*x^
2])/(4*a^2) - (c*(1 + a*x)^3*Sqrt[1 - a^2*x^2])/(4*a^2) + (15*c*ArcSin[a*x])/(8*a^2)

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 795

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

Rule 6148

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

Rubi steps

\begin {align*} \int e^{3 \tanh ^{-1}(a x)} x \left (c-a^2 c x^2\right ) \, dx &=c \int \frac {x (1+a x)^3}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c (1+a x)^3 \sqrt {1-a^2 x^2}}{4 a^2}+\frac {(3 c) \int \frac {(1+a x)^3}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac {c (1+a x)^2 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {c (1+a x)^3 \sqrt {1-a^2 x^2}}{4 a^2}+\frac {(5 c) \int \frac {(1+a x)^2}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac {5 c (1+a x) \sqrt {1-a^2 x^2}}{8 a^2}-\frac {c (1+a x)^2 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {c (1+a x)^3 \sqrt {1-a^2 x^2}}{4 a^2}+\frac {(15 c) \int \frac {1+a x}{\sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac {15 c \sqrt {1-a^2 x^2}}{8 a^2}-\frac {5 c (1+a x) \sqrt {1-a^2 x^2}}{8 a^2}-\frac {c (1+a x)^2 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {c (1+a x)^3 \sqrt {1-a^2 x^2}}{4 a^2}+\frac {(15 c) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{8 a}\\ &=-\frac {15 c \sqrt {1-a^2 x^2}}{8 a^2}-\frac {5 c (1+a x) \sqrt {1-a^2 x^2}}{8 a^2}-\frac {c (1+a x)^2 \sqrt {1-a^2 x^2}}{4 a^2}-\frac {c (1+a x)^3 \sqrt {1-a^2 x^2}}{4 a^2}+\frac {15 c \sin ^{-1}(a x)}{8 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 54, normalized size = 0.45 \[ \frac {15 c \sin ^{-1}(a x)-c \sqrt {1-a^2 x^2} \left (2 a^3 x^3+8 a^2 x^2+15 a x+24\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcTanh[a*x])*x*(c - a^2*c*x^2),x]

[Out]

(-(c*Sqrt[1 - a^2*x^2]*(24 + 15*a*x + 8*a^2*x^2 + 2*a^3*x^3)) + 15*c*ArcSin[a*x])/(8*a^2)

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fricas [A]  time = 0.46, size = 71, normalized size = 0.59 \[ -\frac {30 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (2 \, a^{3} c x^{3} + 8 \, a^{2} c x^{2} + 15 \, a c x + 24 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(30*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^3*c*x^3 + 8*a^2*c*x^2 + 15*a*c*x + 24*c)*sqrt(-a^2*x^
2 + 1))/a^2

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giac [A]  time = 1.61, size = 58, normalized size = 0.48 \[ -\frac {1}{8} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (a c x + 4 \, c\right )} x + \frac {15 \, c}{a}\right )} x + \frac {24 \, c}{a^{2}}\right )} + \frac {15 \, c \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{8 \, a {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(-a^2*x^2 + 1)*((2*(a*c*x + 4*c)*x + 15*c/a)*x + 24*c/a^2) + 15/8*c*arcsin(a*x)*sgn(a)/(a*abs(a))

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maple [A]  time = 0.05, size = 148, normalized size = 1.23 \[ \frac {c \,a^{3} x^{5}}{4 \sqrt {-a^{2} x^{2}+1}}+\frac {13 c a \,x^{3}}{8 \sqrt {-a^{2} x^{2}+1}}-\frac {15 c x}{8 a \sqrt {-a^{2} x^{2}+1}}+\frac {15 c \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a \sqrt {a^{2}}}+\frac {c \,a^{2} x^{4}}{\sqrt {-a^{2} x^{2}+1}}+\frac {2 c \,x^{2}}{\sqrt {-a^{2} x^{2}+1}}-\frac {3 c}{a^{2} \sqrt {-a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c*a^3*x^5/(-a^2*x^2+1)^(1/2)+13/8*c*a*x^3/(-a^2*x^2+1)^(1/2)-15/8*c/a*x/(-a^2*x^2+1)^(1/2)+15/8*c/a/(a^2)^
(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+c*a^2*x^4/(-a^2*x^2+1)^(1/2)+2*c*x^2/(-a^2*x^2+1)^(1/2)-3*c/a^2
/(-a^2*x^2+1)^(1/2)

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maxima [A]  time = 0.43, size = 126, normalized size = 1.05 \[ \frac {a^{3} c x^{5}}{4 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {a^{2} c x^{4}}{\sqrt {-a^{2} x^{2} + 1}} + \frac {13 \, a c x^{3}}{8 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {2 \, c x^{2}}{\sqrt {-a^{2} x^{2} + 1}} - \frac {15 \, c x}{8 \, \sqrt {-a^{2} x^{2} + 1} a} + \frac {15 \, c \arcsin \left (a x\right )}{8 \, a^{2}} - \frac {3 \, c}{\sqrt {-a^{2} x^{2} + 1} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^3*c*x^5/sqrt(-a^2*x^2 + 1) + a^2*c*x^4/sqrt(-a^2*x^2 + 1) + 13/8*a*c*x^3/sqrt(-a^2*x^2 + 1) + 2*c*x^2/sq
rt(-a^2*x^2 + 1) - 15/8*c*x/(sqrt(-a^2*x^2 + 1)*a) + 15/8*c*arcsin(a*x)/a^2 - 3*c/(sqrt(-a^2*x^2 + 1)*a^2)

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mupad [B]  time = 0.03, size = 98, normalized size = 0.82 \[ \frac {15\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a\,\sqrt {-a^2}}-c\,x^2\,\sqrt {1-a^2\,x^2}-\frac {15\,c\,x\,\sqrt {1-a^2\,x^2}}{8\,a}-\frac {a\,c\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {3\,c\,\sqrt {1-a^2\,x^2}}{a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*c*asinh(x*(-a^2)^(1/2)))/(8*a*(-a^2)^(1/2)) - c*x^2*(1 - a^2*x^2)^(1/2) - (15*c*x*(1 - a^2*x^2)^(1/2))/(8*
a) - (a*c*x^3*(1 - a^2*x^2)^(1/2))/4 - (3*c*(1 - a^2*x^2)^(1/2))/a^2

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sympy [A]  time = 16.33, size = 326, normalized size = 2.72 \[ a^{3} c \left (\begin {cases} - \frac {i x^{5}}{4 \sqrt {a^{2} x^{2} - 1}} - \frac {i x^{3}}{8 a^{2} \sqrt {a^{2} x^{2} - 1}} + \frac {3 i x}{8 a^{4} \sqrt {a^{2} x^{2} - 1}} - \frac {3 i \operatorname {acosh}{\left (a x \right )}}{8 a^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{5}}{4 \sqrt {- a^{2} x^{2} + 1}} + \frac {x^{3}}{8 a^{2} \sqrt {- a^{2} x^{2} + 1}} - \frac {3 x}{8 a^{4} \sqrt {- a^{2} x^{2} + 1}} + \frac {3 \operatorname {asin}{\left (a x \right )}}{8 a^{5}} & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c*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*s
qrt(-a**2*x**2 + 1)) - 3*x/(8*a**4*sqrt(-a**2*x**2 + 1)) + 3*asin(a*x)/(8*a**5), True)) + 3*a**2*c*Piecewise((
-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 + 1)/(3*a**4), Ne(a, 0)), (x**4/4, True)) + 3*a*c*Piec
ewise((-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*Piecewise((x**2/2, Eq(a**2, 0)), (-
sqrt(-a**2*x**2 + 1)/a**2, True))

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