3.1254 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^4 \, dx\)

Optimal. Leaf size=167 \[ \frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac{55}{128} c^4 x \sqrt{1-a^2 x^2}+\frac{55 c^4 \sin ^{-1}(a x)}{128 a} \]

[Out]

(55*c^4*x*Sqrt[1 - a^2*x^2])/128 + (55*c^4*x*(1 - a^2*x^2)^(3/2))/192 + (11*c^4*x*(1 - a^2*x^2)^(5/2))/48 + (1
1*c^4*(1 - a^2*x^2)^(7/2))/(56*a) + (11*c^4*(1 - a*x)*(1 - a^2*x^2)^(7/2))/(72*a) + (c^4*(1 - a*x)^2*(1 - a^2*
x^2)^(7/2))/(9*a) + (55*c^4*ArcSin[a*x])/(128*a)

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Rubi [A]  time = 0.0943462, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6139, 671, 641, 195, 216} \[ \frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac{55}{128} c^4 x \sqrt{1-a^2 x^2}+\frac{55 c^4 \sin ^{-1}(a x)}{128 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^4/E^(3*ArcTanh[a*x]),x]

[Out]

(55*c^4*x*Sqrt[1 - a^2*x^2])/128 + (55*c^4*x*(1 - a^2*x^2)^(3/2))/192 + (11*c^4*x*(1 - a^2*x^2)^(5/2))/48 + (1
1*c^4*(1 - a^2*x^2)^(7/2))/(56*a) + (11*c^4*(1 - a*x)*(1 - a^2*x^2)^(7/2))/(72*a) + (c^4*(1 - a*x)^2*(1 - a^2*
x^2)^(7/2))/(9*a) + (55*c^4*ArcSin[a*x])/(128*a)

Rule 6139

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] && ILtQ[(n - 1)/2, 0] &&
!IntegerQ[p - n/2]

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 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 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]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^4 \, dx &=c^4 \int (1-a x)^3 \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{9} \left (11 c^4\right ) \int (1-a x)^2 \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{8} \left (11 c^4\right ) \int (1-a x) \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{8} \left (11 c^4\right ) \int \left (1-a^2 x^2\right )^{5/2} \, dx\\ &=\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{48} \left (55 c^4\right ) \int \left (1-a^2 x^2\right )^{3/2} \, dx\\ &=\frac{55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{64} \left (55 c^4\right ) \int \sqrt{1-a^2 x^2} \, dx\\ &=\frac{55}{128} c^4 x \sqrt{1-a^2 x^2}+\frac{55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{1}{128} \left (55 c^4\right ) \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx\\ &=\frac{55}{128} c^4 x \sqrt{1-a^2 x^2}+\frac{55}{192} c^4 x \left (1-a^2 x^2\right )^{3/2}+\frac{11}{48} c^4 x \left (1-a^2 x^2\right )^{5/2}+\frac{11 c^4 \left (1-a^2 x^2\right )^{7/2}}{56 a}+\frac{11 c^4 (1-a x) \left (1-a^2 x^2\right )^{7/2}}{72 a}+\frac{c^4 (1-a x)^2 \left (1-a^2 x^2\right )^{7/2}}{9 a}+\frac{55 c^4 \sin ^{-1}(a x)}{128 a}\\ \end{align*}

Mathematica [A]  time = 0.147153, size = 107, normalized size = 0.64 \[ -\frac{c^4 \left (\sqrt{1-a^2 x^2} \left (896 a^8 x^8-3024 a^7 x^7+1024 a^6 x^6+7224 a^5 x^5-8448 a^4 x^4-3066 a^3 x^3+10240 a^2 x^2-4599 a x-3712\right )+6930 \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{8064 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a^2*c*x^2)^4/E^(3*ArcTanh[a*x]),x]

[Out]

-(c^4*(Sqrt[1 - a^2*x^2]*(-3712 - 4599*a*x + 10240*a^2*x^2 - 3066*a^3*x^3 - 8448*a^4*x^4 + 7224*a^5*x^5 + 1024
*a^6*x^6 - 3024*a^7*x^7 + 896*a^8*x^8) + 6930*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/(8064*a)

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Maple [A]  time = 0.05, size = 173, normalized size = 1. \begin{align*} -{\frac{{c}^{4}{a}^{3}{x}^{4}}{9} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{22\,{c}^{4}a{x}^{2}}{63} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{29\,{c}^{4}}{63\,a} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{3\,{a}^{2}{c}^{4}{x}^{3}}{8} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{7\,{c}^{4}x}{48} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}+{\frac{55\,{c}^{4}x}{192} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{c}^{4}x}{128}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{55\,{c}^{4}}{128}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

-1/9*c^4*a^3*x^4*(-a^2*x^2+1)^(5/2)-22/63*c^4*a*x^2*(-a^2*x^2+1)^(5/2)+29/63*c^4*(-a^2*x^2+1)^(5/2)/a+3/8*c^4*
a^2*x^3*(-a^2*x^2+1)^(5/2)-7/48*c^4*x*(-a^2*x^2+1)^(5/2)+55/192*c^4*x*(-a^2*x^2+1)^(3/2)+55/128*c^4*x*(-a^2*x^
2+1)^(1/2)+55/128*c^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.4669, size = 208, normalized size = 1.25 \begin{align*} -\frac{1}{9} \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} a^{3} c^{4} x^{4} + \frac{3}{8} \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} a^{2} c^{4} x^{3} - \frac{22}{63} \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} a c^{4} x^{2} - \frac{7}{48} \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} c^{4} x + \frac{55}{192} \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} c^{4} x + \frac{29 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} c^{4}}{63 \, a} + \frac{55}{128} \, \sqrt{-a^{2} x^{2} + 1} c^{4} x + \frac{55 \, c^{4} \arcsin \left (a x\right )}{128 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

-1/9*(-a^2*x^2 + 1)^(5/2)*a^3*c^4*x^4 + 3/8*(-a^2*x^2 + 1)^(5/2)*a^2*c^4*x^3 - 22/63*(-a^2*x^2 + 1)^(5/2)*a*c^
4*x^2 - 7/48*(-a^2*x^2 + 1)^(5/2)*c^4*x + 55/192*(-a^2*x^2 + 1)^(3/2)*c^4*x + 29/63*(-a^2*x^2 + 1)^(5/2)*c^4/a
+ 55/128*sqrt(-a^2*x^2 + 1)*c^4*x + 55/128*c^4*arcsin(a*x)/a

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Fricas [A]  time = 2.65421, size = 325, normalized size = 1.95 \begin{align*} -\frac{6930 \, c^{4} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (896 \, a^{8} c^{4} x^{8} - 3024 \, a^{7} c^{4} x^{7} + 1024 \, a^{6} c^{4} x^{6} + 7224 \, a^{5} c^{4} x^{5} - 8448 \, a^{4} c^{4} x^{4} - 3066 \, a^{3} c^{4} x^{3} + 10240 \, a^{2} c^{4} x^{2} - 4599 \, a c^{4} x - 3712 \, c^{4}\right )} \sqrt{-a^{2} x^{2} + 1}}{8064 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-1/8064*(6930*c^4*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (896*a^8*c^4*x^8 - 3024*a^7*c^4*x^7 + 1024*a^6*c^4*
x^6 + 7224*a^5*c^4*x^5 - 8448*a^4*c^4*x^4 - 3066*a^3*c^4*x^3 + 10240*a^2*c^4*x^2 - 4599*a*c^4*x - 3712*c^4)*sq
rt(-a^2*x^2 + 1))/a

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**4/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17329, size = 170, normalized size = 1.02 \begin{align*} \frac{55 \, c^{4} \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{128 \,{\left | a \right |}} + \frac{1}{8064} \, \sqrt{-a^{2} x^{2} + 1}{\left (\frac{3712 \, c^{4}}{a} +{\left (4599 \, c^{4} - 2 \,{\left (5120 \, a c^{4} -{\left (1533 \, a^{2} c^{4} + 4 \,{\left (1056 \, a^{3} c^{4} -{\left (903 \, a^{4} c^{4} + 2 \,{\left (64 \, a^{5} c^{4} + 7 \,{\left (8 \, a^{7} c^{4} x - 27 \, a^{6} c^{4}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^4/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

55/128*c^4*arcsin(a*x)*sgn(a)/abs(a) + 1/8064*sqrt(-a^2*x^2 + 1)*(3712*c^4/a + (4599*c^4 - 2*(5120*a*c^4 - (15
33*a^2*c^4 + 4*(1056*a^3*c^4 - (903*a^4*c^4 + 2*(64*a^5*c^4 + 7*(8*a^7*c^4*x - 27*a^6*c^4)*x)*x)*x)*x)*x)*x)*x
)