3.296 \(\int e^{\coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx\)

Optimal. Leaf size=140 \[ \frac {16 x \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{105 a^2 \sqrt {1-\frac {1}{a x}}}+\frac {2 x^3 \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{7 \sqrt {1-\frac {1}{a x}}}-\frac {8 x^2 \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{35 a \sqrt {1-\frac {1}{a x}}} \]

[Out]

16/105*(1+1/a/x)^(3/2)*x*(-a*c*x+c)^(1/2)/a^2/(1-1/a/x)^(1/2)-8/35*(1+1/a/x)^(3/2)*x^2*(-a*c*x+c)^(1/2)/a/(1-1
/a/x)^(1/2)+2/7*(1+1/a/x)^(3/2)*x^3*(-a*c*x+c)^(1/2)/(1-1/a/x)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6176, 6181, 45, 37} \[ \frac {16 x \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{105 a^2 \sqrt {1-\frac {1}{a x}}}+\frac {2 x^3 \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{7 \sqrt {1-\frac {1}{a x}}}-\frac {8 x^2 \left (\frac {1}{a x}+1\right )^{3/2} \sqrt {c-a c x}}{35 a \sqrt {1-\frac {1}{a x}}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*x^2*Sqrt[c - a*c*x],x]

[Out]

(16*(1 + 1/(a*x))^(3/2)*x*Sqrt[c - a*c*x])/(105*a^2*Sqrt[1 - 1/(a*x)]) - (8*(1 + 1/(a*x))^(3/2)*x^2*Sqrt[c - a
*c*x])/(35*a*Sqrt[1 - 1/(a*x)]) + (2*(1 + 1/(a*x))^(3/2)*x^3*Sqrt[c - a*c*x])/(7*Sqrt[1 - 1/(a*x)])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 6176

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

Rule 6181

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

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(a x)} x^2 \sqrt {c-a c x} \, dx &=\frac {\sqrt {c-a c x} \int e^{\coth ^{-1}(a x)} \sqrt {1-\frac {1}{a x}} x^{5/2} \, dx}{\sqrt {1-\frac {1}{a x}} \sqrt {x}}\\ &=-\frac {\left (\sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{9/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a x}}}\\ &=\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^3 \sqrt {c-a c x}}{7 \sqrt {1-\frac {1}{a x}}}+\frac {\left (4 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{7 a \sqrt {1-\frac {1}{a x}}}\\ &=-\frac {8 \left (1+\frac {1}{a x}\right )^{3/2} x^2 \sqrt {c-a c x}}{35 a \sqrt {1-\frac {1}{a x}}}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^3 \sqrt {c-a c x}}{7 \sqrt {1-\frac {1}{a x}}}-\frac {\left (8 \sqrt {\frac {1}{x}} \sqrt {c-a c x}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{35 a^2 \sqrt {1-\frac {1}{a x}}}\\ &=\frac {16 \left (1+\frac {1}{a x}\right )^{3/2} x \sqrt {c-a c x}}{105 a^2 \sqrt {1-\frac {1}{a x}}}-\frac {8 \left (1+\frac {1}{a x}\right )^{3/2} x^2 \sqrt {c-a c x}}{35 a \sqrt {1-\frac {1}{a x}}}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x^3 \sqrt {c-a c x}}{7 \sqrt {1-\frac {1}{a x}}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.46 \[ \frac {2 \sqrt {\frac {1}{a x}+1} (a x+1) \left (15 a^2 x^2-12 a x+8\right ) \sqrt {c-a c x}}{105 a^3 \sqrt {1-\frac {1}{a x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]*x^2*Sqrt[c - a*c*x],x]

[Out]

(2*Sqrt[1 + 1/(a*x)]*(1 + a*x)*Sqrt[c - a*c*x]*(8 - 12*a*x + 15*a^2*x^2))/(105*a^3*Sqrt[1 - 1/(a*x)])

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fricas [A]  time = 0.51, size = 69, normalized size = 0.49 \[ \frac {2 \, {\left (15 \, a^{4} x^{4} + 18 \, a^{3} x^{3} - a^{2} x^{2} + 4 \, a x + 8\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{105 \, {\left (a^{4} x - a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^4*x^4 + 18*a^3*x^3 - a^2*x^2 + 4*a*x + 8)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*x + 1))/(a^4*x - a^3)

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giac [A]  time = 0.17, size = 102, normalized size = 0.73 \[ \frac {2 \, {\left (\frac {22 \, \sqrt {2} \sqrt {-c}}{a^{2} \mathrm {sgn}\relax (c)} + \frac {15 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} - 42 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c - 35 \, {\left (-a c x - c\right )}^{\frac {3}{2}} c^{2}}{a^{2} c^{3} \mathrm {sgn}\left (-a c x - c\right )}\right )}}{105 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(22*sqrt(2)*sqrt(-c)/(a^2*sgn(c)) + (15*(a*c*x + c)^3*sqrt(-a*c*x - c) - 42*(a*c*x + c)^2*sqrt(-a*c*x -
c)*c - 35*(-a*c*x - c)^(3/2)*c^2)/(a^2*c^3*sgn(-a*c*x - c)))/a

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maple [A]  time = 0.04, size = 49, normalized size = 0.35 \[ \frac {2 \left (a x +1\right ) \left (15 a^{2} x^{2}-12 a x +8\right ) \sqrt {-a c x +c}}{105 a^{3} \sqrt {\frac {a x -1}{a x +1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(a*x+1)*(15*a^2*x^2-12*a*x+8)*(-a*c*x+c)^(1/2)/a^3/((a*x-1)/(a*x+1))^(1/2)

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maxima [A]  time = 0.32, size = 55, normalized size = 0.39 \[ \frac {2 \, {\left (15 \, a^{3} \sqrt {-c} x^{3} + 3 \, a^{2} \sqrt {-c} x^{2} - 4 \, a \sqrt {-c} x + 8 \, \sqrt {-c}\right )} \sqrt {a x + 1}}{105 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*a^3*sqrt(-c)*x^3 + 3*a^2*sqrt(-c)*x^2 - 4*a*sqrt(-c)*x + 8*sqrt(-c))*sqrt(a*x + 1)/a^3

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mupad [B]  time = 1.38, size = 57, normalized size = 0.41 \[ \frac {2\,\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (15\,a^2\,x^2-12\,a\,x+8\right )}{105\,a^3\,\left (a\,x-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(c - a*c*x)^(1/2)*(a*x + 1)^2*((a*x - 1)/(a*x + 1))^(1/2)*(15*a^2*x^2 - 12*a*x + 8))/(105*a^3*(a*x - 1))

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sympy [C]  time = 50.08, size = 105, normalized size = 0.75 \[ - \frac {94 i x}{105 a^{2} \sqrt {\frac {1}{a c x + c}}} + \frac {10 i}{21 a^{3} \sqrt {\frac {1}{a c x + c}}} - \frac {32 i \left (- a c x + c\right )^{2}}{35 a^{3} c^{2} \sqrt {\frac {1}{a c x + c}}} + \frac {2 i \left (- a c x + c\right )^{3}}{7 a^{3} c^{3} \sqrt {\frac {1}{a c x + c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-94*I*x/(105*a**2*sqrt(1/(a*c*x + c))) + 10*I/(21*a**3*sqrt(1/(a*c*x + c))) - 32*I*(-a*c*x + c)**2/(35*a**3*c*
*2*sqrt(1/(a*c*x + c))) + 2*I*(-a*c*x + c)**3/(7*a**3*c**3*sqrt(1/(a*c*x + c)))

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