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

Optimal. Leaf size=103 \[ \frac {2 (c-a c x)^{5/2}}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {16 (c-a c x)^{3/2}}{3 a c \sqrt {1-a^2 x^2}}-\frac {64 \sqrt {c-a c x}}{3 a \sqrt {1-a^2 x^2}} \]

[Out]

16/3*(-a*c*x+c)^(3/2)/a/c/(-a^2*x^2+1)^(1/2)+2/3*(-a*c*x+c)^(5/2)/a/c^2/(-a^2*x^2+1)^(1/2)-64/3*(-a*c*x+c)^(1/
2)/a/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6127, 657, 649} \[ \frac {2 (c-a c x)^{5/2}}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {16 (c-a c x)^{3/2}}{3 a c \sqrt {1-a^2 x^2}}-\frac {64 \sqrt {c-a c x}}{3 a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*Sqrt[c - a*c*x])/(3*a*Sqrt[1 - a^2*x^2]) + (16*(c - a*c*x)^(3/2))/(3*a*c*Sqrt[1 - a^2*x^2]) + (2*(c - a*c
*x)^(5/2))/(3*a*c^2*Sqrt[1 - a^2*x^2])

Rule 649

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*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p,
0]

Rule 657

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*Simplify[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, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p]
, 0]

Rule 6127

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

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x} \, dx &=\frac {\int \frac {(c-a c x)^{7/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {2 (c-a c x)^{5/2}}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {8 \int \frac {(c-a c x)^{5/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac {16 (c-a c x)^{3/2}}{3 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{5/2}}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {32 \int \frac {(c-a c x)^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c}\\ &=-\frac {64 \sqrt {c-a c x}}{3 a \sqrt {1-a^2 x^2}}+\frac {16 (c-a c x)^{3/2}}{3 a c \sqrt {1-a^2 x^2}}+\frac {2 (c-a c x)^{5/2}}{3 a c^2 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.50 \[ \frac {2 c \sqrt {1-a x} \left (a^2 x^2-10 a x-23\right )}{3 a \sqrt {a x+1} \sqrt {c-a c x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(-23 - 10*a*x + a^2*x^2))/(3*a*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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fricas [A]  time = 0.47, size = 49, normalized size = 0.48 \[ -\frac {2 \, {\left (a^{2} x^{2} - 10 \, a x - 23\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{3 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(a^2*x^2 - 10*a*x - 23)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^3*x^2 - a)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.03, size = 54, normalized size = 0.52 \[ \frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \sqrt {-a c x +c}\, \left (a^{2} x^{2}-10 a x -23\right )}{3 \left (a x +1\right )^{2} \left (a x -1\right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(1/2)*(a^2*x^2-10*a*x-23)/(a*x+1)^2/(a*x-1)^2/a

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maxima [A]  time = 0.35, size = 50, normalized size = 0.49 \[ \frac {2 \, {\left (a^{2} \sqrt {c} x^{2} - 10 \, a \sqrt {c} x - 23 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{3 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*sqrt(c)*x^2 - 10*a*sqrt(c)*x - 23*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^3*x^2 - a)

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mupad [B]  time = 0.97, size = 40, normalized size = 0.39 \[ -\frac {2\,\sqrt {c-a\,c\,x}\,\left (-a^2\,x^2+10\,a\,x+23\right )}{3\,a\,\sqrt {1-a^2\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(c - a*c*x)^(1/2)*(10*a*x - a^2*x^2 + 23))/(3*a*(1 - a^2*x^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (a x - 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}{\left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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