3.10.85 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^3 (c-a^2 c x^2)^{5/2}} \, dx\) [985]

Optimal. Leaf size=345 \[ -\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

-1/2*(-a^2*x^2+1)^(1/2)/c^2/x^2/(-a^2*c*x^2+c)^(1/2)-a*(-a^2*x^2+1)^(1/2)/c^2/x/(-a^2*c*x^2+c)^(1/2)+1/8*a^2*(
-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)+a^2*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)
+1/8*a^2*(-a^2*x^2+1)^(1/2)/c^2/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+3*a^2*ln(x)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)
^(1/2)-39/16*a^2*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)^(1/2)-9/16*a^2*ln(a*x+1)*(-a^2*x^2+1)^(1/2)/
c^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6288, 6285, 90} \begin {gather*} \frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)^(5/2)),x]

[Out]

-1/2*Sqrt[1 - a^2*x^2]/(c^2*x^2*Sqrt[c - a^2*c*x^2]) - (a*Sqrt[1 - a^2*x^2])/(c^2*x*Sqrt[c - a^2*c*x^2]) + (a^
2*Sqrt[1 - a^2*x^2])/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + (a^2*Sqrt[1 - a^2*x^2])/(c^2*(1 - a*x)*Sqrt[c -
a^2*c*x^2]) + (a^2*Sqrt[1 - a^2*x^2])/(8*c^2*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (3*a^2*Sqrt[1 - a^2*x^2]*Log[x]
)/(c^2*Sqrt[c - a^2*c*x^2]) - (39*a^2*Sqrt[1 - a^2*x^2]*Log[1 - a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) - (9*a^2*Sq
rt[1 - a^2*x^2]*Log[1 + a*x])/(16*c^2*Sqrt[c - a^2*c*x^2])

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 6288

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{x^3 \left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{\tanh ^{-1}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {1}{x^3 (1-a x)^3 (1+a x)^2} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{x^3}+\frac {a}{x^2}+\frac {3 a^2}{x}-\frac {a^3}{4 (-1+a x)^3}+\frac {a^3}{(-1+a x)^2}-\frac {39 a^3}{16 (-1+a x)}-\frac {a^3}{8 (1+a x)^2}-\frac {9 a^3}{16 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=-\frac {\sqrt {1-a^2 x^2}}{2 c^2 x^2 \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {a^2 \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {3 a^2 \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {39 a^2 \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}-\frac {9 a^2 \sqrt {1-a^2 x^2} \log (1+a x)}{16 c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 115, normalized size = 0.33 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-\frac {8}{x^2}-\frac {16 a}{x}+\frac {16 a^2}{1-a x}+\frac {2 a^2}{(-1+a x)^2}+\frac {2 a^2}{1+a x}+48 a^2 \log (x)-39 a^2 \log (1-a x)-9 a^2 \log (1+a x)\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x^3*(c - a^2*c*x^2)^(5/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-8/x^2 - (16*a)/x + (16*a^2)/(1 - a*x) + (2*a^2)/(-1 + a*x)^2 + (2*a^2)/(1 + a*x) + 48*a^2
*Log[x] - 39*a^2*Log[1 - a*x] - 9*a^2*Log[1 + a*x]))/(16*c^2*Sqrt[c - a^2*c*x^2])

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Maple [A]
time = 0.06, size = 242, normalized size = 0.70

method	result	size
default	\(\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (9 \ln \left (a x +1\right ) a^{5} x^{5}-48 a^{5} \ln \left (x \right ) x^{5}+39 \ln \left (a x -1\right ) a^{5} x^{5}-9 \ln \left (a x +1\right ) a^{4} x^{4}+48 a^{4} \ln \left (x \right ) x^{4}-39 \ln \left (a x -1\right ) a^{4} x^{4}+30 a^{4} x^{4}-9 \ln \left (a x +1\right ) a^{3} x^{3}+48 a^{3} \ln \left (x \right ) x^{3}-39 \ln \left (a x -1\right ) a^{3} x^{3}-6 a^{3} x^{3}+9 \ln \left (a x +1\right ) a^{2} x^{2}-48 a^{2} \ln \left (x \right ) x^{2}+39 \ln \left (a x -1\right ) a^{2} x^{2}-44 a^{2} x^{2}+8 a x +8\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} \left (a x -1\right )^{2} \left (a x +1\right ) x^{2}}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^3/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(9*ln(a*x+1)*a^5*x^5-48*a^5*ln(x)*x^5+39*ln(a*x-1)*a^5*x^5-9*ln
(a*x+1)*a^4*x^4+48*a^4*ln(x)*x^4-39*ln(a*x-1)*a^4*x^4+30*a^4*x^4-9*ln(a*x+1)*a^3*x^3+48*a^3*ln(x)*x^3-39*ln(a*
x-1)*a^3*x^3-6*a^3*x^3+9*ln(a*x+1)*a^2*x^2-48*a^2*ln(x)*x^2+39*ln(a*x-1)*a^2*x^2-44*a^2*x^2+8*a*x+8)/(a^2*x^2-
1)/c^3/(a*x-1)^2/(a*x+1)/x^2

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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)/x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)*x^3), x)

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Fricas [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)/x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^10 - a^6*c^3*x^9 - 3*a^5*c^3*x^8 + 3*a^4*c^3*x^7 +
3*a^3*c^3*x^6 - 3*a^2*c^3*x^5 - a*c^3*x^4 + c^3*x^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x + 1}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, 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)/x**3/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Integral((a*x + 1)/(x**3*sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), 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)/x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a\,x+1}{x^3\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x^3*(c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int((a*x + 1)/(x^3*(c - a^2*c*x^2)^(5/2)*(1 - a^2*x^2)^(1/2)), x)

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