3.7.89 \(\int e^{\tanh ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} \, dx\) [689]

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

[Out]

-1/2*(c-c/a^2/x^2)^(3/2)*x/(-a^2*x^2+1)^(3/2)-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)-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 = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6295, 6285, 76} \begin {gather*} -\frac {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 {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*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 6285

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
] || GtQ[c, 0])

Rule 6295

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]

Rubi steps

\begin {align*} \int e^{\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^{\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) (1+a x)^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 \left (-a^3+\frac {1}{x^3}+\frac {a}{x^2}-\frac {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 {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 {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 = 0.02, size = 72, normalized size = 0.49 \begin {gather*} \frac {c \sqrt {c-\frac {c}{a^2 x^2}} \left (1+2 a x+3 a^2 x^2+2 a^3 x^3+2 a^2 x^2 \log (x)\right )}{2 a^2 x \sqrt {1-a^2 x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.39, size = 70, normalized size = 0.48

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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+2*ln(x)*a^2*x^2+2*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*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.45, size = 373, normalized size = 2.55 \begin {gather*} \left [\frac {{\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} + 2 \, a + 1\right )} c x^{2} + 2 \, 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 {2 \, {\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} + 2 \, a + 1\right )} c x^{2} + 2 \, 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*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + 1)*(c - c/(a^2*x^2))^(3/2)/sqrt(-a^2*x^2 + 1), 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 (a\,x+1\right )}{\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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