∫ e−3 tanh−1(ax) √ ____ c − acx x dx [432]

Optimal. Leaf size=107 \[ \frac {8 c^2 (1-a x)^{3/2}}{\sqrt {1+a x} (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{(c-a c x)^{3/2}}-\frac {2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {1+a x}\right )}{(c-a c x)^{3/2}} \]

-2*c^2*(-a*x+1)^(3/2)*arctanh((a*x+1)^(1/2))/(-a*c*x+c)^(3/2)+8*c^2*(-a*x+1)^(3/2)/(-a*c*x+c)^(3/2)/(a*x+1)^(1

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Rubi [A]
time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6265, 23, 89, 65, 214} \begin {gather*} \frac {2 c^2 (1-a x)^{3/2} \sqrt {a x+1}}{(c-a c x)^{3/2}}+\frac {8 c^2 (1-a x)^{3/2}}{\sqrt {a x+1} (c-a c x)^{3/2}}-\frac {2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {a x+1}\right )}{(c-a c x)^{3/2}} \end {gather*}

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

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

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

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub

st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &

& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr

and[(e + f*x)^FractionalPart[p], (c + d*x)^n*((e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},

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]

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx &=\int \frac {(1-a x)^{3/2} \sqrt {c-a c x}}{x (1+a x)^{3/2}} \, dx\\ &=\frac {(1-a x)^{3/2} \int \frac {(c-a c x)^2}{x (1+a x)^{3/2}} \, dx}{(c-a c x)^{3/2}}\\ &=\frac {(1-a x)^{3/2} \int \left (-\frac {4 a c^2}{(1+a x)^{3/2}}+\frac {a c^2}{\sqrt {1+a x}}+\frac {c^2}{x \sqrt {1+a x}}\right ) \, dx}{(c-a c x)^{3/2}}\\ &=\frac {8 c^2 (1-a x)^{3/2}}{\sqrt {1+a x} (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{(c-a c x)^{3/2}}+\frac {\left (c^2 (1-a x)^{3/2}\right ) \int \frac {1}{x \sqrt {1+a x}} \, dx}{(c-a c x)^{3/2}}\\ &=\frac {8 c^2 (1-a x)^{3/2}}{\sqrt {1+a x} (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{(c-a c x)^{3/2}}+\frac {\left (2 c^2 (1-a x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{a}+\frac {x^2}{a}} \, dx,x,\sqrt {1+a x}\right )}{a (c-a c x)^{3/2}}\\ &=\frac {8 c^2 (1-a x)^{3/2}}{\sqrt {1+a x} (c-a c x)^{3/2}}+\frac {2 c^2 (1-a x)^{3/2} \sqrt {1+a x}}{(c-a c x)^{3/2}}-\frac {2 c^2 (1-a x)^{3/2} \tanh ^{-1}\left (\sqrt {1+a x}\right )}{(c-a c x)^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 51, normalized size = 0.48 \begin {gather*} \frac {2 c \sqrt {1-a x} \left (4+a x+\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+a x\right )\right )}{\sqrt {1+a x} \sqrt {c-a c x}} \end {gather*}

(2*c*Sqrt[1 - a*x]*(4 + a*x + Hypergeometric2F1[-1/2, 1, 1/2, 1 + a*x]))/(Sqrt[1 + a*x]*Sqrt[c - a*c*x])

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method	result	size

default	\(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, \left (\sqrt {c}\, \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}}{\sqrt {c}}\right ) \sqrt {\left (a x +1\right ) c}-c x a -5 c \right )}{\left (a x -1\right ) \left (a x +1\right ) c}\)	\(78\)

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

2*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)*(c^(1/2)*arctanh(((a*x+1)*c)^(1/2)/c^(1/2))*((a*x+1)*c)^(1/2)-c*x*a-5*

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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

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Fricas [A]
time = 0.38, size = 209, normalized size = 1.95 \begin {gather*} \left [\frac {{\left (a^{2} x^{2} - 1\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 5\right )}}{a^{2} x^{2} - 1}, -\frac {2 \, {\left ({\left (a^{2} x^{2} - 1\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (a x + 5\right )}\right )}}{a^{2} x^{2} - 1}\right ] \end {gather*}

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

[((a^2*x^2 - 1)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/(a*x^2

- x)) - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 5))/(a^2*x^2 - 1), -2*((a^2*x^2 - 1)*sqrt(-c)*arctan(sqrt

(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(a*x + 5))/(a^

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

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

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

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

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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}\,\sqrt {c-a\,c\,x}}{x\,{\left (a\,x+1\right )}^3} \,d x \end {gather*}

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

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

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3.5.32 \(\int \frac {e^{-3 \tanh ^{-1}(a x)} \sqrt {c-a c x}}{x} \, dx\) [432]