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

Optimal. Leaf size=269 \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (1-a x)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

[Out]

(-a^2*x^2+1)^(5/2)/a^5/(c-c/a^2/x^2)^(5/2)/x^4+1/6*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(-a*x+1)^3-9
/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x^5/(-a*x+1)^2+31/8*(-a^2*x^2+1)^(5/2)/a^6/(c-c/a^2/x^2)^(5/2)/x
^5/(-a*x+1)+49/16*(-a^2*x^2+1)^(5/2)*ln(-a*x+1)/a^6/(c-c/a^2/x^2)^(5/2)/x^5-1/16*(-a^2*x^2+1)^(5/2)*ln(a*x+1)/
a^6/(c-c/a^2/x^2)^(5/2)/x^5

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Rubi [A]  time = 0.22, antiderivative size = 269, 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 = {6160, 6150, 88} \[ \frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 x^5 (1-a x)^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 x^5 (1-a x)^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 x^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (a x+1)}{16 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - a^2*x^2)^(5/2)/(a^5*(c - c/(a^2*x^2))^(5/2)*x^4) + (1 - a^2*x^2)^(5/2)/(6*a^6*(c - c/(a^2*x^2))^(5/2)*x^5
*(1 - a*x)^3) - (9*(1 - a^2*x^2)^(5/2))/(8*a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 - a*x)^2) + (31*(1 - a^2*x^2)^(5
/2))/(8*a^6*(c - c/(a^2*x^2))^(5/2)*x^5*(1 - a*x)) + (49*(1 - a^2*x^2)^(5/2)*Log[1 - a*x])/(16*a^6*(c - c/(a^2
*x^2))^(5/2)*x^5) - ((1 - a^2*x^2)^(5/2)*Log[1 + a*x])/(16*a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

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 6160

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*(1 + (c*x^2)/d)^p*E^(n*ArcTanh[a*x]))/x^(2*p), 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 \frac {e^{3 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {e^{3 \tanh ^{-1}(a x)} x^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5}{(1-a x)^4 (1+a x)} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2} \int \left (\frac {1}{a^5}+\frac {1}{2 a^5 (-1+a x)^4}+\frac {9}{4 a^5 (-1+a x)^3}+\frac {31}{8 a^5 (-1+a x)^2}+\frac {49}{16 a^5 (-1+a x)}-\frac {1}{16 a^5 (1+a x)}\right ) \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left (1-a^2 x^2\right )^{5/2}}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^4}+\frac {\left (1-a^2 x^2\right )^{5/2}}{6 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)^3}-\frac {9 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)^2}+\frac {31 \left (1-a^2 x^2\right )^{5/2}}{8 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5 (1-a x)}+\frac {49 \left (1-a^2 x^2\right )^{5/2} \log (1-a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}-\frac {\left (1-a^2 x^2\right )^{5/2} \log (1+a x)}{16 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 113, normalized size = 0.42 \[ \frac {\sqrt {1-a^2 x^2} \left (2 \left (24 a^4 x^4-72 a^3 x^3-21 a^2 x^2+135 a x-70\right )+147 (a x-1)^3 \log (1-a x)-3 (a x-1)^3 \log (a x+1)\right )}{48 a^2 c^2 x (a x-1)^3 \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(2*(-70 + 135*a*x - 21*a^2*x^2 - 72*a^3*x^3 + 24*a^4*x^4) + 147*(-1 + a*x)^3*Log[1 - a*x] -
3*(-1 + a*x)^3*Log[1 + a*x]))/(48*a^2*c^2*Sqrt[c - c/(a^2*x^2)]*x*(-1 + a*x)^3)

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fricas [F]  time = 2.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} a^{6} x^{6} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{7} c^{3} x^{7} - 3 \, a^{6} c^{3} x^{6} + a^{5} c^{3} x^{5} + 5 \, a^{4} c^{3} x^{4} - 5 \, a^{3} c^{3} x^{3} - a^{2} c^{3} x^{2} + 3 \, a c^{3} x - c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 176, normalized size = 0.65 \[ \frac {\left (48 x^{4} a^{4}+147 \ln \left (a x -1\right ) x^{3} a^{3}-3 a^{3} x^{3} \ln \left (a x +1\right )-144 x^{3} a^{3}-441 \ln \left (a x -1\right ) x^{2} a^{2}+9 \ln \left (a x +1\right ) x^{2} a^{2}-42 a^{2} x^{2}+441 \ln \left (a x -1\right ) x a -9 a x \ln \left (a x +1\right )+270 a x -147 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )-140\right ) \left (a x +1\right )^{2} \sqrt {-a^{2} x^{2}+1}}{48 \left (a x -1\right ) a^{6} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*x^4*a^4+147*ln(a*x-1)*x^3*a^3-3*a^3*x^3*ln(a*x+1)-144*x^3*a^3-441*ln(a*x-1)*x^2*a^2+9*ln(a*x+1)*x^2*a
^2-42*a^2*x^2+441*ln(a*x-1)*x*a-9*a*x*ln(a*x+1)+270*a*x-147*ln(a*x-1)+3*ln(a*x+1)-140)*(a*x+1)^2*(-a^2*x^2+1)^
(1/2)/(a*x-1)/a^6/x^5/(c*(a^2*x^2-1)/a^2/x^2)^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(c-c/a^2/x^2)^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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