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

Optimal. Leaf size=194 \[ \frac {(1-a x)^2}{a^2 x \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (a x+1)^2 (13 a x+28) (1-a x)^3}{15 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (a x+1)^{5/2} (1-a x)^{5/2} \sin ^{-1}(a x)}{a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {2 (a x+1) (1-a x)^3}{15 a^4 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (1-a x)^3}{5 a^3 x^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

[Out]

(-a*x+1)^2/a^2/(c-c/a^2/x^2)^(5/2)/x+2/5*(-a*x+1)^3/a^3/(c-c/a^2/x^2)^(5/2)/x^2-2/15*(-a*x+1)^3*(a*x+1)/a^4/(c
-c/a^2/x^2)^(5/2)/x^3+2/15*(-a*x+1)^3*(a*x+1)^2*(13*a*x+28)/a^6/(c-c/a^2/x^2)^(5/2)/x^5+2*(-a*x+1)^(5/2)*(a*x+
1)^(5/2)*arcsin(a*x)/a^6/(c-c/a^2/x^2)^(5/2)/x^5

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {6159, 6129, 98, 150, 143, 41, 216} \[ -\frac {2 (a x+1) (1-a x)^3}{15 a^4 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (a x+1)^2 (13 a x+28) (1-a x)^3}{15 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (1-a x)^3}{5 a^3 x^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {(1-a x)^2}{a^2 x \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (a x+1)^{5/2} (1-a x)^{5/2} \sin ^{-1}(a x)}{a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

(1 - a*x)^2/(a^2*(c - c/(a^2*x^2))^(5/2)*x) + (2*(1 - a*x)^3)/(5*a^3*(c - c/(a^2*x^2))^(5/2)*x^2) - (2*(1 - a*
x)^3*(1 + a*x))/(15*a^4*(c - c/(a^2*x^2))^(5/2)*x^3) + (2*(1 - a*x)^3*(1 + a*x)^2*(28 + 13*a*x))/(15*a^6*(c -
c/(a^2*x^2))^(5/2)*x^5) + (2*(1 - a*x)^(5/2)*(1 + a*x)^(5/2)*ArcSin[a*x])/(a^6*(c - c/(a^2*x^2))^(5/2)*x^5)

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 6129

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

Rule 6159

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(x^(2*p)*(c + d/x^2)^p)/(
(1 - a*x)^p*(1 + a*x)^p), Int[(u*(1 - a*x)^p*(1 + a*x)^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] &&  !GtQ[c, 0]

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx &=\frac {\left ((1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {e^{-2 \tanh ^{-1}(a x)} x^5}{(1-a x)^{5/2} (1+a x)^{5/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {\left ((1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {x^5}{(1-a x)^{3/2} (1+a x)^{7/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {\left ((1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {x^3 (4+2 a x)}{\sqrt {1-a x} (1+a x)^{7/2}} \, dx}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}+\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}-\frac {\left ((1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {x^2 \left (6 a+8 a^2 x\right )}{\sqrt {1-a x} (1+a x)^{5/2}} \, dx}{5 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}+\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}-\frac {2 (1-a x)^3 (1+a x)}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}-\frac {\left ((1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {x \left (-4 a^2+26 a^3 x\right )}{\sqrt {1-a x} (1+a x)^{3/2}} \, dx}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}+\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}-\frac {2 (1-a x)^3 (1+a x)}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}+\frac {2 (1-a x)^3 (1+a x)^2 (28+13 a x)}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {\left (2 (1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {1}{\sqrt {1-a x} \sqrt {1+a x}} \, dx}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}+\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}-\frac {2 (1-a x)^3 (1+a x)}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}+\frac {2 (1-a x)^3 (1+a x)^2 (28+13 a x)}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {\left (2 (1-a x)^{5/2} (1+a x)^{5/2}\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{a^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}+\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}-\frac {2 (1-a x)^3 (1+a x)}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}+\frac {2 (1-a x)^3 (1+a x)^2 (28+13 a x)}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}+\frac {2 (1-a x)^{5/2} (1+a x)^{5/2} \sin ^{-1}(a x)}{a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 105, normalized size = 0.54 \[ \frac {-15 a^4 x^4-76 a^3 x^3-32 a^2 x^2+30 (a x+1)^2 \sqrt {a^2 x^2-1} \log \left (\sqrt {a^2 x^2-1}+a x\right )+82 a x+56}{15 a^2 c^2 x (a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a^2*x^2))^(5/2)),x]

[Out]

(56 + 82*a*x - 32*a^2*x^2 - 76*a^3*x^3 - 15*a^4*x^4 + 30*(1 + a*x)^2*Sqrt[-1 + a^2*x^2]*Log[a*x + Sqrt[-1 + a^
2*x^2]])/(15*a^2*c^2*Sqrt[c - c/(a^2*x^2)]*x*(1 + a*x)^2)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 352, normalized size = 1.81 \[ \left [\frac {15 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt {c} \log \left (2 \, a^{2} c x^{2} + 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right ) - {\left (15 \, a^{5} x^{5} + 76 \, a^{4} x^{4} + 32 \, a^{3} x^{3} - 82 \, a^{2} x^{2} - 56 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{15 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}}, -\frac {30 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) + {\left (15 \, a^{5} x^{5} + 76 \, a^{4} x^{4} + 32 \, a^{3} x^{3} - 82 \, a^{2} x^{2} - 56 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{15 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*sqrt(c)*log(2*a^2*c*x^2 + 2*a^2*sqrt(c)*x^2*sqrt((a^2*c*x^2 - c)/(
a^2*x^2)) - c) - (15*a^5*x^5 + 76*a^4*x^4 + 32*a^3*x^3 - 82*a^2*x^2 - 56*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))
/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3), -1/15*(30*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*sqrt(-c)*arc
tan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^2 - c)) + (15*a^5*x^5 + 76*a^4*x^4 + 32*a^3*x^3
- 82*a^2*x^2 - 56*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3)]

________________________________________________________________________________________

giac [A]  time = 0.25, size = 208, normalized size = 1.07 \[ -\frac {4 \, \arctan \left (\frac {\sqrt {c - \frac {2 \, c}{a x + 1}}}{\sqrt {-c}}\right )}{a \sqrt {-c} c^{2} \mathrm {sgn}\left (-\frac {1}{a x + 1} + 1\right )} + \frac {8 \, c - \frac {17 \, c}{a x + 1}}{4 \, {\left ({\left (c - \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} - \sqrt {c - \frac {2 \, c}{a x + 1}} c\right )} a c^{2} \mathrm {sgn}\left (-\frac {1}{a x + 1} + 1\right )} - \frac {3 \, a^{4} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{\frac {5}{2}} c^{20} + 35 \, a^{4} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} c^{21} + 345 \, a^{4} \sqrt {c - \frac {2 \, c}{a x + 1}} c^{22}}{120 \, a^{5} c^{25} \mathrm {sgn}\left (-\frac {1}{a x + 1} + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*arctan(sqrt(c - 2*c/(a*x + 1))/sqrt(-c))/(a*sqrt(-c)*c^2*sgn(-1/(a*x + 1) + 1)) + 1/4*(8*c - 17*c/(a*x + 1)
)/(((c - 2*c/(a*x + 1))^(3/2) - sqrt(c - 2*c/(a*x + 1))*c)*a*c^2*sgn(-1/(a*x + 1) + 1)) - 1/120*(3*a^4*(c - 2*
c/(a*x + 1))^(5/2)*c^20 + 35*a^4*(c - 2*c/(a*x + 1))^(3/2)*c^21 + 345*a^4*sqrt(c - 2*c/(a*x + 1))*c^22)/(a^5*c
^25*sgn(-1/(a*x + 1) + 1))

________________________________________________________________________________________

maple [B]  time = 0.05, size = 462, normalized size = 2.38 \[ -\frac {\left (15 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} x^{5} a^{5}+45 x^{4} c^{\frac {5}{2}} a^{4} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}}+16 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} x^{4} a^{4}-60 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} x^{3} a^{3}+16 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} x^{3} a^{3}-30 \ln \left (x \sqrt {c}+\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right ) \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} x \,a^{4} c -90 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} x^{2} a^{2}-24 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} x^{2} a^{2}-30 \ln \left (x \sqrt {c}+\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right ) \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} a^{3} c +50 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} x a -24 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} x a +50 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}}+6 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}}\right ) \left (a x -1\right )}{15 \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} x^{5} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {5}{2}} a^{6} c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*c^(5/2)*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*x^5*a^5+45*x^4*c^(5/2)*a^4*((a*x-1)*(a*x+1)*c/a^2)^(3/2)+16*c^
(5/2)*(c*(a^2*x^2-1)/a^2)^(3/2)*x^4*a^4-60*c^(5/2)*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*x^3*a^3+16*c^(5/2)*(c*(a^2*x^
2-1)/a^2)^(3/2)*x^3*a^3-30*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2)^(1/2))*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*(c*(a^2*x^2-1
)/a^2)^(3/2)*x*a^4*c-90*c^(5/2)*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*x^2*a^2-24*c^(5/2)*(c*(a^2*x^2-1)/a^2)^(3/2)*x^2
*a^2-30*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2)^(1/2))*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*(c*(a^2*x^2-1)/a^2)^(3/2)*a^3*c+
50*c^(5/2)*((a*x-1)*(a*x+1)*c/a^2)^(3/2)*x*a-24*c^(5/2)*(c*(a^2*x^2-1)/a^2)^(3/2)*x*a+50*c^(5/2)*((a*x-1)*(a*x
+1)*c/a^2)^(3/2)+6*c^(5/2)*(c*(a^2*x^2-1)/a^2)^(3/2))*(a*x-1)/((a*x-1)*(a*x+1)*c/a^2)^(3/2)/x^5/(c*(a^2*x^2-1)
/a^2/x^2)^(5/2)/a^6/c^(5/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {a^2\,x^2-1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,{\left (a\,x+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a x}{a c^{2} x \sqrt {c - \frac {c}{a^{2} x^{2}}} + c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}} - \frac {2 c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a x} - \frac {2 c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{2} x^{2}} + \frac {c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{3} x^{3}} + \frac {c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{4} x^{4}}}\, dx - \int \left (- \frac {1}{a c^{2} x \sqrt {c - \frac {c}{a^{2} x^{2}}} + c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}} - \frac {2 c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a x} - \frac {2 c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{2} x^{2}} + \frac {c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{3} x^{3}} + \frac {c^{2} \sqrt {c - \frac {c}{a^{2} x^{2}}}}{a^{4} x^{4}}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a*x/(a*c**2*x*sqrt(c - c/(a**2*x**2)) + c**2*sqrt(c - c/(a**2*x**2)) - 2*c**2*sqrt(c - c/(a**2*x**2)
)/(a*x) - 2*c**2*sqrt(c - c/(a**2*x**2))/(a**2*x**2) + c**2*sqrt(c - c/(a**2*x**2))/(a**3*x**3) + c**2*sqrt(c
- c/(a**2*x**2))/(a**4*x**4)), x) - Integral(-1/(a*c**2*x*sqrt(c - c/(a**2*x**2)) + c**2*sqrt(c - c/(a**2*x**2
)) - 2*c**2*sqrt(c - c/(a**2*x**2))/(a*x) - 2*c**2*sqrt(c - c/(a**2*x**2))/(a**2*x**2) + c**2*sqrt(c - c/(a**2
*x**2))/(a**3*x**3) + c**2*sqrt(c - c/(a**2*x**2))/(a**4*x**4)), x)

________________________________________________________________________________________