3.5.30 \(\int e^{-3 \tanh ^{-1}(a x)} x \sqrt {c-a c x} \, dx\) [430]

Optimal. Leaf size=157 \[ \frac {8 c^2 (1-a x)^{3/2}}{a^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {16 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{a^2 (c-a c x)^{3/2}}-\frac {10 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^2 (c-a c x)^{3/2}} \]

[Out]

-10/3*c^2*(-a*x+1)^(3/2)*(a*x+1)^(3/2)/a^2/(-a*c*x+c)^(3/2)+2/5*c^2*(-a*x+1)^(3/2)*(a*x+1)^(5/2)/a^2/(-a*c*x+c
)^(3/2)+8*c^2*(-a*x+1)^(3/2)/a^2/(-a*c*x+c)^(3/2)/(a*x+1)^(1/2)+16*c^2*(-a*x+1)^(3/2)*(a*x+1)^(1/2)/a^2/(-a*c*
x+c)^(3/2)

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Rubi [A]
time = 0.07, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6265, 23, 78} \begin {gather*} \frac {2 c^2 (1-a x)^{3/2} (a x+1)^{5/2}}{5 a^2 (c-a c x)^{3/2}}-\frac {10 c^2 (1-a x)^{3/2} (a x+1)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac {16 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{a^2 (c-a c x)^{3/2}}+\frac {8 c^2 (1-a x)^{3/2}}{a^2 \sqrt {a x+1} (c-a c x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^2*(1 - a*x)^(3/2))/(a^2*Sqrt[1 + a*x]*(c - a*c*x)^(3/2)) + (16*c^2*(1 - a*x)^(3/2)*Sqrt[1 + a*x])/(a^2*(c
- a*c*x)^(3/2)) - (10*c^2*(1 - a*x)^(3/2)*(1 + a*x)^(3/2))/(3*a^2*(c - a*c*x)^(3/2)) + (2*c^2*(1 - a*x)^(3/2)
*(1 + a*x)^(5/2))/(5*a^2*(c - a*c*x)^(3/2))

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 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6265

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]
)

Rubi steps

\begin {align*} \int e^{-3 \tanh ^{-1}(a x)} x \sqrt {c-a c x} \, dx &=\int \frac {x (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 (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 (1+a x)^{3/2}}+\frac {8 c^2}{a \sqrt {1+a x}}-\frac {5 c^2 \sqrt {1+a x}}{a}+\frac {c^2 (1+a x)^{3/2}}{a}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=\frac {8 c^2 (1-a x)^{3/2}}{a^2 \sqrt {1+a x} (c-a c x)^{3/2}}+\frac {16 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{a^2 (c-a c x)^{3/2}}-\frac {10 c^2 (1-a x)^{3/2} (1+a x)^{3/2}}{3 a^2 (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} (1+a x)^{5/2}}{5 a^2 (c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 60, normalized size = 0.38 \begin {gather*} \frac {2 c \sqrt {1-a x} \left (158+79 a x-16 a^2 x^2+3 a^3 x^3\right )}{15 a^2 \sqrt {1+a x} \sqrt {c-a c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[1 - a*x]*(158 + 79*a*x - 16*a^2*x^2 + 3*a^3*x^3))/(15*a^2*Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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Maple [A]
time = 1.13, size = 64, normalized size = 0.41

method	result	size
gosper	\(\frac {2 \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}} \sqrt {-c x a +c}\, \left (3 a^{3} x^{3}-16 a^{2} x^{2}+79 a x +158\right )}{15 \left (a x +1\right )^{2} a^{2} \left (a x -1\right )^{2}}\)	\(63\)
default	\(-\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (3 a^{3} x^{3}-16 a^{2} x^{2}+79 a x +158\right )}{15 \left (a x -1\right ) \left (a x +1\right ) a^{2}}\)	\(64\)
risch	\(-\frac {2 \left (3 a^{2} x^{2}-19 a x +98\right ) \left (a x +1\right ) \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{15 a^{2} \sqrt {\left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}-\frac {8 \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c}{a^{2} \sqrt {\left (a x +1\right ) c}\, \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(3*a^3*x^3-16*a^2*x^2+79*a*x+158)/(a*x-1)/(a*x+1)/a^2

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Maxima [A]
time = 0.29, size = 64, normalized size = 0.41 \begin {gather*} \frac {2 \, {\left (3 \, a^{3} \sqrt {c} x^{3} - 16 \, a^{2} \sqrt {c} x^{2} + 79 \, a \sqrt {c} x + 158 \, \sqrt {c}\right )} \sqrt {a x + 1} {\left (a x - 1\right )}}{15 \, {\left (a^{4} x^{2} - a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^3*sqrt(c)*x^3 - 16*a^2*sqrt(c)*x^2 + 79*a*sqrt(c)*x + 158*sqrt(c))*sqrt(a*x + 1)*(a*x - 1)/(a^4*x^2
- a^2)

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Fricas [A]
time = 0.41, size = 60, normalized size = 0.38 \begin {gather*} -\frac {2 \, {\left (3 \, a^{3} x^{3} - 16 \, a^{2} x^{2} + 79 \, a x + 158\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{15 \, {\left (a^{4} x^{2} - a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*a^3*x^3 - 16*a^2*x^2 + 79*a*x + 158)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^4*x^2 - a^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-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,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [B]
time = 1.00, size = 97, normalized size = 0.62 \begin {gather*} \frac {\sqrt {c-a\,c\,x}\,\left (\frac {316\,\sqrt {1-a^2\,x^2}}{15\,a^4}+\frac {158\,x\,\sqrt {1-a^2\,x^2}}{15\,a^3}+\frac {2\,x^3\,\sqrt {1-a^2\,x^2}}{5\,a}-\frac {32\,x^2\,\sqrt {1-a^2\,x^2}}{15\,a^2}\right )}{\frac {1}{a^2}-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a*c*x)^(1/2)*((316*(1 - a^2*x^2)^(1/2))/(15*a^4) + (158*x*(1 - a^2*x^2)^(1/2))/(15*a^3) + (2*x^3*(1 - a^
2*x^2)^(1/2))/(5*a) - (32*x^2*(1 - a^2*x^2)^(1/2))/(15*a^2)))/(1/a^2 - x^2)

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