3.965 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^4 \sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=187 \[ -\frac {a^2 \sqrt {1-a^2 x^2}}{x \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{2 x^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a^2 c x^2}}+\frac {a^3 \sqrt {1-a^2 x^2} \log (x)}{\sqrt {c-a^2 c x^2}}-\frac {a^3 \sqrt {1-a^2 x^2} \log (1-a x)}{\sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 187, 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 = {6153, 6150, 44} \[ -\frac {a^2 \sqrt {1-a^2 x^2}}{x \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{2 x^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 x^3 \sqrt {c-a^2 c x^2}}+\frac {a^3 \sqrt {1-a^2 x^2} \log (x)}{\sqrt {c-a^2 c x^2}}-\frac {a^3 \sqrt {1-a^2 x^2} \log (1-a x)}{\sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x^4*Sqrt[c - a^2*c*x^2]),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 6150

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 6153

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

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Mathematica [A]  time = 0.04, size = 72, normalized size = 0.39 \[ \frac {\sqrt {1-a^2 x^2} \left (a^3 \log (x)-a^3 \log (1-a x)-\frac {a^2}{x}-\frac {a}{2 x^2}-\frac {1}{3 x^3}\right )}{\sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x^4*Sqrt[c - a^2*c*x^2]),x]

[Out]

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

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fricas [A]  time = 0.89, size = 482, normalized size = 2.58 \[ \left [\frac {3 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {c} \log \left (-\frac {4 \, a^{5} c x^{5} - {\left (2 \, a^{6} - 4 \, a^{5} + 6 \, a^{4} - 4 \, a^{3} + a^{2}\right )} c x^{6} - {\left (4 \, a^{4} + 4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} c x^{4} + 5 \, a^{2} c x^{2} - 4 \, a c x + {\left (4 \, a^{3} x^{3} - {\left (4 \, a^{3} - 6 \, a^{2} + 4 \, a - 1\right )} x^{4} - 6 \, a^{2} x^{2} + 4 \, a x - 1\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} + c}{a^{4} x^{6} - 2 \, a^{3} x^{5} + 2 \, a x^{3} - x^{2}}\right ) + \sqrt {-a^{2} c x^{2} + c} {\left (6 \, a^{2} x^{2} - {\left (6 \, a^{2} + 3 \, a + 2\right )} x^{3} + 3 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{2} c x^{5} - c x^{3}\right )}}, -\frac {6 \, {\left (a^{5} x^{5} - a^{3} x^{3}\right )} \sqrt {-c} \arctan \left (-\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, a^{2} - 2 \, a + 1\right )} x^{2} - 2 \, a x + 1\right )} \sqrt {-c}}{2 \, a^{3} c x^{3} - {\left (2 \, a^{3} - a^{2}\right )} c x^{4} - {\left (a^{2} - 2 \, a + 1\right )} c x^{2} - 2 \, a c x + c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (6 \, a^{2} x^{2} - {\left (6 \, a^{2} + 3 \, a + 2\right )} x^{3} + 3 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, {\left (a^{2} c x^{5} - c x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 84, normalized size = 0.45 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (6 a^{3} \ln \relax (x ) x^{3}-6 \ln \left (a x -1\right ) x^{3} a^{3}-6 a^{2} x^{2}-3 a x -2\right )}{6 \left (a^{2} x^{2}-1\right ) c \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(1/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a\,x+1}{x^4\,\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x^4*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)),x)

[Out]

int((a*x + 1)/(x^4*(c - a^2*c*x^2)^(1/2)*(1 - a^2*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{x^{4} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**4/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Integral((a*x + 1)/(x**4*sqrt(-(a*x - 1)*(a*x + 1))*sqrt(-c*(a*x - 1)*(a*x + 1))), x)

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