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

Optimal. Leaf size=235 \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{9/2}}{9 a^4 (c-a c x)^{3/2}}-\frac{2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{a^4 (c-a c x)^{3/2}}+\frac{38 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^4 (c-a c x)^{3/2}}-\frac{50 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^4 (c-a c x)^{3/2}}+\frac{32 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^4 (c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{a^4 \sqrt{a x+1} (c-a c x)^{3/2}} \]

[Out]

(8*c^2*(1 - a*x)^(3/2))/(a^4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (32*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^4*(c
- a*c*x)^(3/2)) - (50*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^4*(c - a*c*x)^(3/2)) + (38*c^2*(1 - a*x)^(3/2
)*(1 + a*x)^(5/2))/(5*a^4*(c - a*c*x)^(3/2)) - (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(7/2))/(a^4*(c - a*c*x)^(3/2))
+ (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(9/2))/(9*a^4*(c - a*c*x)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.153605, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6130, 23, 88} \[ \frac{2 c^2 (1-a x)^{3/2} (a x+1)^{9/2}}{9 a^4 (c-a c x)^{3/2}}-\frac{2 c^2 (1-a x)^{3/2} (a x+1)^{7/2}}{a^4 (c-a c x)^{3/2}}+\frac{38 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^4 (c-a c x)^{3/2}}-\frac{50 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^4 (c-a c x)^{3/2}}+\frac{32 c^2 (1-a x)^{3/2} \sqrt{a x+1}}{a^4 (c-a c x)^{3/2}}+\frac{8 c^2 (1-a x)^{3/2}}{a^4 \sqrt{a x+1} (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^2*(1 - a*x)^(3/2))/(a^4*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (32*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^4*(c
- a*c*x)^(3/2)) - (50*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^4*(c - a*c*x)^(3/2)) + (38*c^2*(1 - a*x)^(3/2
)*(1 + a*x)^(5/2))/(5*a^4*(c - a*c*x)^(3/2)) - (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(7/2))/(a^4*(c - a*c*x)^(3/2))
+ (2*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(9/2))/(9*a^4*(c - a*c*x)^(3/2))

Rule 6130

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

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} x^3 \sqrt{c-a c x} \, dx &=\int \frac{x^3 (1-a x)^{3/2} \sqrt{c-a c x}}{(1+a x)^{3/2}} \, dx\\ &=\frac{(1-a x)^{3/2} \int \frac{x^3 (c-a c x)^2}{(1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=\frac{(1-a x)^{3/2} \int \left (-\frac{4 c^2}{a^3 (1+a x)^{3/2}}+\frac{16 c^2}{a^3 \sqrt{1+a x}}-\frac{25 c^2 \sqrt{1+a x}}{a^3}+\frac{19 c^2 (1+a x)^{3/2}}{a^3}-\frac{7 c^2 (1+a x)^{5/2}}{a^3}+\frac{c^2 (1+a x)^{7/2}}{a^3}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=\frac{8 c^2 (1-a x)^{3/2}}{a^4 \sqrt{1+a x} (c-a c x)^{3/2}}+\frac{32 c^2 (1-a x)^{3/2} \sqrt{1+a x}}{a^4 (c-a c x)^{3/2}}-\frac{50 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^4 (c-a c x)^{3/2}}+\frac{38 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^4 (c-a c x)^{3/2}}-\frac{2 c^2 (1-a x)^{3/2} (1+a x)^{7/2}}{a^4 (c-a c x)^{3/2}}+\frac{2 c^2 (1-a x)^{3/2} (1+a x)^{9/2}}{9 a^4 (c-a c x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.048824, size = 76, normalized size = 0.32 \[ \frac{2 c \sqrt{1-a x} \left (5 a^5 x^5-20 a^4 x^4+41 a^3 x^3-82 a^2 x^2+328 a x+656\right )}{45 a^4 \sqrt{a x+1} \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(656 + 328*a*x - 82*a^2*x^2 + 41*a^3*x^3 - 20*a^4*x^4 + 5*a^5*x^5))/(45*a^4*Sqrt[1 + a*x]*S
qrt[c - a*c*x])

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 79, normalized size = 0.3 \begin{align*}{\frac{10\,{x}^{5}{a}^{5}-40\,{x}^{4}{a}^{4}+82\,{x}^{3}{a}^{3}-164\,{a}^{2}{x}^{2}+656\,ax+1312}{45\, \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{2}{a}^{4}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}\sqrt{-acx+c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(5*a^5*x^5-20*a^4*x^4+41*a^3*x^3-82*a^2*x^2+328*a*x+656)/(a*x+1)^2/(a
*x-1)^2/a^4

________________________________________________________________________________________

Maxima [A]  time = 1.04195, size = 116, normalized size = 0.49 \begin{align*} \frac{2 \,{\left (5 \, a^{5} \sqrt{c} x^{5} - 20 \, a^{4} \sqrt{c} x^{4} + 41 \, a^{3} \sqrt{c} x^{3} - 82 \, a^{2} \sqrt{c} x^{2} + 328 \, a \sqrt{c} x + 656 \, \sqrt{c}\right )} \sqrt{a x + 1}{\left (a x - 1\right )}}{45 \,{\left (a^{6} x^{2} - a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*(5*a^5*sqrt(c)*x^5 - 20*a^4*sqrt(c)*x^4 + 41*a^3*sqrt(c)*x^3 - 82*a^2*sqrt(c)*x^2 + 328*a*sqrt(c)*x + 656
*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^6*x^2 - a^4)

________________________________________________________________________________________

Fricas [A]  time = 1.53572, size = 170, normalized size = 0.72 \begin{align*} -\frac{2 \,{\left (5 \, a^{5} x^{5} - 20 \, a^{4} x^{4} + 41 \, a^{3} x^{3} - 82 \, a^{2} x^{2} + 328 \, a x + 656\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{45 \,{\left (a^{6} x^{2} - a^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45*(5*a^5*x^5 - 20*a^4*x^4 + 41*a^3*x^3 - 82*a^2*x^2 + 328*a*x + 656)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(
a^6*x^2 - a^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.28954, size = 146, normalized size = 0.62 \begin{align*} -\frac{928 \, \sqrt{2}{\left | c \right |}}{45 \, a^{4} \sqrt{c}} + \frac{2 \,{\left (5 \,{\left (a c x + c\right )}^{\frac{9}{2}}{\left | c \right |} - 45 \,{\left (a c x + c\right )}^{\frac{7}{2}} c{\left | c \right |} + 171 \,{\left (a c x + c\right )}^{\frac{5}{2}} c^{2}{\left | c \right |} - 375 \,{\left (a c x + c\right )}^{\frac{3}{2}} c^{3}{\left | c \right |} + 720 \, \sqrt{a c x + c} c^{4}{\left | c \right |} + \frac{180 \, c^{5}{\left | c \right |}}{\sqrt{a c x + c}}\right )}}{45 \, a^{4} c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-928/45*sqrt(2)*abs(c)/(a^4*sqrt(c)) + 2/45*(5*(a*c*x + c)^(9/2)*abs(c) - 45*(a*c*x + c)^(7/2)*c*abs(c) + 171*
(a*c*x + c)^(5/2)*c^2*abs(c) - 375*(a*c*x + c)^(3/2)*c^3*abs(c) + 720*sqrt(a*c*x + c)*c^4*abs(c) + 180*c^5*abs
(c)/sqrt(a*c*x + c))/(a^4*c^5)