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

Optimal. Leaf size=105 \[ -\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}+\frac {5}{24} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {5}{16} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \sin ^{-1}(a x)}{16 a} \]

[Out]

5/24*c^3*x*(-a^2*x^2+1)^(3/2)+1/6*c^3*x*(-a^2*x^2+1)^(5/2)-1/7*c^3*(-a^2*x^2+1)^(7/2)/a+5/16*c^3*arcsin(a*x)/a
+5/16*c^3*x*(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6138, 641, 195, 216} \[ -\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}+\frac {5}{24} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {5}{16} c^3 x \sqrt {1-a^2 x^2}+\frac {5 c^3 \sin ^{-1}(a x)}{16 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*c^3*x*Sqrt[1 - a^2*x^2])/16 + (5*c^3*x*(1 - a^2*x^2)^(3/2))/24 + (c^3*x*(1 - a^2*x^2)^(5/2))/6 - (c^3*(1 -
a^2*x^2)^(7/2))/(7*a) + (5*c^3*ArcSin[a*x])/(16*a)

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 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 6138

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

Rubi steps

\begin {align*} \int e^{\tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^3 \, dx &=c^3 \int (1+a x) \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+c^3 \int \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{6} \left (5 c^3\right ) \int \left (1-a^2 x^2\right )^{3/2} \, dx\\ &=\frac {5}{24} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{8} \left (5 c^3\right ) \int \sqrt {1-a^2 x^2} \, dx\\ &=\frac {5}{16} c^3 x \sqrt {1-a^2 x^2}+\frac {5}{24} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {1}{16} \left (5 c^3\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=\frac {5}{16} c^3 x \sqrt {1-a^2 x^2}+\frac {5}{24} c^3 x \left (1-a^2 x^2\right )^{3/2}+\frac {1}{6} c^3 x \left (1-a^2 x^2\right )^{5/2}-\frac {c^3 \left (1-a^2 x^2\right )^{7/2}}{7 a}+\frac {5 c^3 \sin ^{-1}(a x)}{16 a}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 91, normalized size = 0.87 \[ \frac {c^3 \left (\sqrt {1-a^2 x^2} \left (48 a^6 x^6+56 a^5 x^5-144 a^4 x^4-182 a^3 x^3+144 a^2 x^2+231 a x-48\right )-210 \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{336 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c^3*(Sqrt[1 - a^2*x^2]*(-48 + 231*a*x + 144*a^2*x^2 - 182*a^3*x^3 - 144*a^4*x^4 + 56*a^5*x^5 + 48*a^6*x^6) -
210*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(336*a)

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fricas [A]  time = 0.65, size = 115, normalized size = 1.10 \[ -\frac {210 \, c^{3} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) - {\left (48 \, a^{6} c^{3} x^{6} + 56 \, a^{5} c^{3} x^{5} - 144 \, a^{4} c^{3} x^{4} - 182 \, a^{3} c^{3} x^{3} + 144 \, a^{2} c^{3} x^{2} + 231 \, a c^{3} x - 48 \, c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{336 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/336*(210*c^3*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) - (48*a^6*c^3*x^6 + 56*a^5*c^3*x^5 - 144*a^4*c^3*x^4 -
182*a^3*c^3*x^3 + 144*a^2*c^3*x^2 + 231*a*c^3*x - 48*c^3)*sqrt(-a^2*x^2 + 1))/a

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giac [A]  time = 0.19, size = 103, normalized size = 0.98 \[ \frac {5 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{16 \, {\left | a \right |}} - \frac {1}{336} \, \sqrt {-a^{2} x^{2} + 1} {\left (\frac {48 \, c^{3}}{a} - {\left (231 \, c^{3} + 2 \, {\left (72 \, a c^{3} - {\left (91 \, a^{2} c^{3} + 4 \, {\left (18 \, a^{3} c^{3} - {\left (6 \, a^{5} c^{3} x + 7 \, a^{4} c^{3}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*c^3*arcsin(a*x)*sgn(a)/abs(a) - 1/336*sqrt(-a^2*x^2 + 1)*(48*c^3/a - (231*c^3 + 2*(72*a*c^3 - (91*a^2*c^3
+ 4*(18*a^3*c^3 - (6*a^5*c^3*x + 7*a^4*c^3)*x)*x)*x)*x)*x)

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maple [B]  time = 0.06, size = 183, normalized size = 1.74 \[ \frac {c^{3} a^{5} x^{6} \sqrt {-a^{2} x^{2}+1}}{7}-\frac {3 c^{3} a^{3} x^{4} \sqrt {-a^{2} x^{2}+1}}{7}+\frac {3 c^{3} a \,x^{2} \sqrt {-a^{2} x^{2}+1}}{7}-\frac {c^{3} \sqrt {-a^{2} x^{2}+1}}{7 a}+\frac {c^{3} a^{4} x^{5} \sqrt {-a^{2} x^{2}+1}}{6}-\frac {13 c^{3} a^{2} x^{3} \sqrt {-a^{2} x^{2}+1}}{24}+\frac {11 c^{3} x \sqrt {-a^{2} x^{2}+1}}{16}+\frac {5 c^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{16 \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*c^3*a^5*x^6*(-a^2*x^2+1)^(1/2)-3/7*c^3*a^3*x^4*(-a^2*x^2+1)^(1/2)+3/7*c^3*a*x^2*(-a^2*x^2+1)^(1/2)-1/7*c^3
*(-a^2*x^2+1)^(1/2)/a+1/6*c^3*a^4*x^5*(-a^2*x^2+1)^(1/2)-13/24*c^3*a^2*x^3*(-a^2*x^2+1)^(1/2)+11/16*c^3*x*(-a^
2*x^2+1)^(1/2)+5/16*c^3/(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 = 164, normalized size = 1.56 \[ \frac {1}{7} \, \sqrt {-a^{2} x^{2} + 1} a^{5} c^{3} x^{6} + \frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} a^{4} c^{3} x^{5} - \frac {3}{7} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{3} x^{4} - \frac {13}{24} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{3} x^{3} + \frac {3}{7} \, \sqrt {-a^{2} x^{2} + 1} a c^{3} x^{2} + \frac {11}{16} \, \sqrt {-a^{2} x^{2} + 1} c^{3} x + \frac {5 \, c^{3} \arcsin \left (a x\right )}{16 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{3}}{7 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*sqrt(-a^2*x^2 + 1)*a^5*c^3*x^6 + 1/6*sqrt(-a^2*x^2 + 1)*a^4*c^3*x^5 - 3/7*sqrt(-a^2*x^2 + 1)*a^3*c^3*x^4 -
13/24*sqrt(-a^2*x^2 + 1)*a^2*c^3*x^3 + 3/7*sqrt(-a^2*x^2 + 1)*a*c^3*x^2 + 11/16*sqrt(-a^2*x^2 + 1)*c^3*x + 5/
16*c^3*arcsin(a*x)/a - 1/7*sqrt(-a^2*x^2 + 1)*c^3/a

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mupad [B]  time = 0.90, size = 100, normalized size = 0.95 \[ \frac {5\,c^3\,x\,\sqrt {1-a^2\,x^2}}{16}+\frac {5\,c^3\,x\,{\left (1-a^2\,x^2\right )}^{3/2}}{24}+\frac {c^3\,x\,{\left (1-a^2\,x^2\right )}^{5/2}}{6}-\frac {c^3\,{\left (1-a^2\,x^2\right )}^{7/2}}{7\,a}-\frac {5\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}{16\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*c^3*x*(1 - a^2*x^2)^(1/2))/16 + (5*c^3*x*(1 - a^2*x^2)^(3/2))/24 + (c^3*x*(1 - a^2*x^2)^(5/2))/6 - (c^3*(1
- a^2*x^2)^(7/2))/(7*a) - (5*c^3*asinh(x*(-a^2)^(1/2))*(-a^2)^(1/2))/(16*a^2)

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sympy [A]  time = 24.98, size = 267, normalized size = 2.54 \[ \begin {cases} \frac {- \frac {c^{3} \left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} + c^{3} \left (\begin {cases} \frac {a x \sqrt {- a^{2} x^{2} + 1}}{2} + \frac {\operatorname {asin}{\left (a x \right )}}{2} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) - 2 c^{3} \left (\begin {cases} - \frac {a x \left (- 2 a^{2} x^{2} + 1\right ) \sqrt {- a^{2} x^{2} + 1}}{8} + \frac {\operatorname {asin}{\left (a x \right )}}{8} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) - 2 c^{3} \left (\begin {cases} \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {5}{2}}}{5} - \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) + c^{3} \left (\begin {cases} - \frac {a^{3} x^{3} \left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{6} - \frac {a x \left (- 2 a^{2} x^{2} + 1\right ) \sqrt {- a^{2} x^{2} + 1}}{16} + \frac {\operatorname {asin}{\left (a x \right )}}{16} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right ) + c^{3} \left (\begin {cases} - \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {7}{2}}}{7} + \frac {2 \left (- a^{2} x^{2} + 1\right )^{\frac {5}{2}}}{5} - \frac {\left (- a^{2} x^{2} + 1\right )^{\frac {3}{2}}}{3} & \text {for}\: a x > -1 \wedge a x < 1 \end {cases}\right )}{a} & \text {for}\: a \neq 0 \\c^{3} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-c**3*(-a**2*x**2 + 1)**(3/2)/3 + c**3*Piecewise((a*x*sqrt(-a**2*x**2 + 1)/2 + asin(a*x)/2, (a*x >
-1) & (a*x < 1))) - 2*c**3*Piecewise((-a*x*(-2*a**2*x**2 + 1)*sqrt(-a**2*x**2 + 1)/8 + asin(a*x)/8, (a*x > -1
) & (a*x < 1))) - 2*c**3*Piecewise(((-a**2*x**2 + 1)**(5/2)/5 - (-a**2*x**2 + 1)**(3/2)/3, (a*x > -1) & (a*x <
1))) + c**3*Piecewise((-a**3*x**3*(-a**2*x**2 + 1)**(3/2)/6 - a*x*(-2*a**2*x**2 + 1)*sqrt(-a**2*x**2 + 1)/16
+ asin(a*x)/16, (a*x > -1) & (a*x < 1))) + c**3*Piecewise((-(-a**2*x**2 + 1)**(7/2)/7 + 2*(-a**2*x**2 + 1)**(5
/2)/5 - (-a**2*x**2 + 1)**(3/2)/3, (a*x > -1) & (a*x < 1))))/a, Ne(a, 0)), (c**3*x, True))

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