∫ e−3 tanh−1(ax)(c − c __ a2x2 )3∕2 dx [735]

Optimal. Leaf size=146 \[ -\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x}{2 \left (1-a^2 x^2\right )^{3/2}}+\frac {3 a \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^2}{\left (1-a^2 x^2\right )^{3/2}}-\frac {a^3 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^4}{\left (1-a^2 x^2\right )^{3/2}}+\frac {3 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^3 \log (x)}{\left (1-a^2 x^2\right )^{3/2}} \]

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

^(3/2)*x^4/(-a^2*x^2+1)^(3/2)+3*a^2*(c-c/a^2/x^2)^(3/2)*x^3*ln(x)/(-a^2*x^2+1)^(3/2)

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Rubi [A]
time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6295, 6285, 45} \begin {gather*} \frac {3 a x^2 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac {x \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{2 \left (1-a^2 x^2\right )^{3/2}}+\frac {3 a^2 x^3 \log (x) \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}}-\frac {a^3 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{3/2}}{\left (1-a^2 x^2\right )^{3/2}} \end {gather*}

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

(a^3*(c - c/(a^2*x^2))^(3/2)*x^4)/(1 - a^2*x^2)^(3/2) + (3*a^2*(c - c/(a^2*x^2))^(3/2)*x^3*Log[x])/(1 - a^2*x

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d

*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]

&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -

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

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[x^(2*p)*((c + d/x^2)^p/(

1 + c*(x^2/d))^p), Int[(u/x^(2*p))*(1 + c*(x^2/d))^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] &

& EqQ[c + a^2*d, 0] && !IntegerQ[p] && !IntegerQ[n/2]

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

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Mathematica [A]
time = 0.03, size = 64, normalized size = 0.44 \begin {gather*} \frac {c \sqrt {c-\frac {c}{a^2 x^2}} \left (1-6 a x+2 a^3 x^3-6 a^2 x^2 \log (x)\right )}{2 a^2 x \sqrt {1-a^2 x^2}} \end {gather*}

Integrate[(c - c/(a^2*x^2))^(3/2)/E^(3*ArcTanh[a*x]),x]

(c*Sqrt[c - c/(a^2*x^2)]*(1 - 6*a*x + 2*a^3*x^3 - 6*a^2*x^2*Log[x]))/(2*a^2*x*Sqrt[1 - a^2*x^2])

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method	result	size

default	\(\frac {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {3}{2}} x \sqrt {-a^{2} x^{2}+1}\, \left (-2 a^{3} x^{3}+6 a^{2} \ln \left (x \right ) x^{2}+6 a x -1\right )}{2 \left (a^{2} x^{2}-1\right )^{2}}\)	\(70\)

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

1/2*(c*(a^2*x^2-1)/a^2/x^2)^(3/2)*x/(a^2*x^2-1)^2*(-a^2*x^2+1)^(1/2)*(-2*a^3*x^3+6*a^2*ln(x)*x^2+6*a*x-1)

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

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

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

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Fricas [A]
time = 0.40, size = 376, normalized size = 2.58 \begin {gather*} \left [\frac {3 \, {\left (a^{3} c x^{3} - a c x\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} - {\left (a x^{5} - a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - 6 \, a + 1\right )} c x^{2} - 6 \, a c x + c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}, \frac {6 \, {\left (a^{3} c x^{3} - a c x\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} + a x\right )} \sqrt {c} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} - {\left (a^{2} + 1\right )} c x^{2} + c}\right ) - {\left (2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - 6 \, a + 1\right )} c x^{2} - 6 \, a c x + c\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{4} x^{3} - a^{2} x\right )}}\right ] \end {gather*}

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

[1/2*(3*(a^3*c*x^3 - a*c*x)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 - (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqr

t(-c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c)/(a^2*x^4 - x^2)) - (2*a^3*c*x^3 - (2*a^3 - 6*a + 1)*c*x^2 - 6*a*c*x

+ c)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^4*x^3 - a^2*x), 1/2*(6*(a^3*c*x^3 - a*c*x)*sqrt(c

)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 + a*x)*sqrt(c)*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 - (a^2 + 1)*c*x^2

+ c)) - (2*a^3*c*x^3 - (2*a^3 - 6*a + 1)*c*x^2 - 6*a*c*x + c)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 - c)/(a^2*x^

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

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

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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

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

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

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3.8.35 \(\int e^{-3 \tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\) [735]