3.984 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^2 (c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=295 \[ \frac {3 a \sqrt {1-a^2 x^2}}{4 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {23 a \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}+\frac {7 a \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

-(-a^2*x^2+1)^(1/2)/c^2/x/(-a^2*c*x^2+c)^(1/2)+1/8*a*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)^2/(-a^2*c*x^2+c)^(1/2)+3/
4*a*(-a^2*x^2+1)^(1/2)/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/8*a*(-a^2*x^2+1)^(1/2)/c^2/(a*x+1)/(-a^2*c*x^2+c)^(
1/2)+a*ln(x)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2+c)^(1/2)-23/16*a*ln(-a*x+1)*(-a^2*x^2+1)^(1/2)/c^2/(-a^2*c*x^2
+c)^(1/2)+7/16*a*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 = 0.25, antiderivative size = 295, 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, 88} \[ \frac {3 a \sqrt {1-a^2 x^2}}{4 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {23 a \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}+\frac {7 a \sqrt {1-a^2 x^2} \log (a x+1)}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - a^2*x^2]/(c^2*x*Sqrt[c - a^2*c*x^2])) + (a*Sqrt[1 - a^2*x^2])/(8*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2
]) + (3*a*Sqrt[1 - a^2*x^2])/(4*c^2*(1 - a*x)*Sqrt[c - a^2*c*x^2]) - (a*Sqrt[1 - a^2*x^2])/(8*c^2*(1 + a*x)*Sq
rt[c - a^2*c*x^2]) + (a*Sqrt[1 - a^2*x^2]*Log[x])/(c^2*Sqrt[c - a^2*c*x^2]) - (23*a*Sqrt[1 - a^2*x^2]*Log[1 -
a*x])/(16*c^2*Sqrt[c - a^2*c*x^2]) + (7*a*Sqrt[1 - a^2*x^2]*Log[1 + a*x])/(16*c^2*Sqrt[c - a^2*c*x^2])

Rule 88

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 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^2 \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^2 \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^2 (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^2}+\frac {a}{x}-\frac {a^2}{4 (-1+a x)^3}+\frac {3 a^2}{4 (-1+a x)^2}-\frac {23 a^2}{16 (-1+a x)}+\frac {a^2}{8 (1+a x)^2}+\frac {7 a^2}{16 (1+a x)}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=-\frac {\sqrt {1-a^2 x^2}}{c^2 x \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {3 a \sqrt {1-a^2 x^2}}{4 c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {a \sqrt {1-a^2 x^2}}{8 c^2 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {a \sqrt {1-a^2 x^2} \log (x)}{c^2 \sqrt {c-a^2 c x^2}}-\frac {23 a \sqrt {1-a^2 x^2} \log (1-a x)}{16 c^2 \sqrt {c-a^2 c x^2}}+\frac {7 a \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 = 0.09, size = 97, normalized size = 0.33 \[ \frac {\sqrt {1-a^2 x^2} \left (\frac {12 a}{1-a x}-\frac {2 a}{a x+1}+\frac {2 a}{(a x-1)^2}+16 a \log (x)-23 a \log (1-a x)+7 a \log (a x+1)-\frac {16}{x}\right )}{16 c^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-16/x + (12*a)/(1 - a*x) + (2*a)/(-1 + a*x)^2 - (2*a)/(1 + a*x) + 16*a*Log[x] - 23*a*Log[1
 - a*x] + 7*a*Log[1 + a*x]))/(16*c^2*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{a^{7} c^{3} x^{9} - a^{6} c^{3} x^{8} - 3 \, a^{5} c^{3} x^{7} + 3 \, a^{4} c^{3} x^{6} + 3 \, a^{3} c^{3} x^{5} - 3 \, a^{2} c^{3} x^{4} - a c^{3} x^{3} + c^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.05, size = 222, normalized size = 0.75 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (16 a^{4} \ln \relax (x ) x^{4}-23 \ln \left (a x -1\right ) x^{4} a^{4}+7 \ln \left (a x +1\right ) x^{4} a^{4}-16 a^{3} \ln \relax (x ) x^{3}+23 \ln \left (a x -1\right ) x^{3} a^{3}-7 a^{3} x^{3} \ln \left (a x +1\right )-30 x^{3} a^{3}-16 a^{2} \ln \relax (x ) x^{2}+23 \ln \left (a x -1\right ) x^{2} a^{2}-7 \ln \left (a x +1\right ) x^{2} a^{2}+22 a^{2} x^{2}+16 a \ln \relax (x ) x -23 \ln \left (a x -1\right ) x a +7 a x \ln \left (a x +1\right )+28 a x -16\right )}{16 \left (a^{2} x^{2}-1\right ) c^{3} \left (a x -1\right )^{2} \left (a x +1\right ) x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(16*a^4*ln(x)*x^4-23*ln(a*x-1)*x^4*a^4+7*ln(a*x+1)*x^4*a^4-16*
a^3*ln(x)*x^3+23*ln(a*x-1)*x^3*a^3-7*a^3*x^3*ln(a*x+1)-30*x^3*a^3-16*a^2*ln(x)*x^2+23*ln(a*x-1)*x^2*a^2-7*ln(a
*x+1)*x^2*a^2+22*a^2*x^2+16*a*ln(x)*x-23*ln(a*x-1)*x*a+7*a*x*ln(a*x+1)+28*a*x-16)/(a^2*x^2-1)/c^3/(a*x-1)^2/(a
*x+1)/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \sqrt {-a^{2} x^{2} + 1} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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