3.8.2 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{(c-\frac {c}{a^2 x^2})^{5/2}} \, dx\) [702]

Optimal. Leaf size=203 \[ \frac {(1+a x)^2}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {58 (1-a x)^2 (1+a x)^2}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}+\frac {2 (1-a x)^3 (1+a x)^2 (28+43 a x)}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}-\frac {2 (1-a x)^{5/2} (1+a x)^{5/2} \text {ArcSin}(a x)}{a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5} \]

[Out]

1/5*(a*x+1)^2/a^2/(c-c/a^2/x^2)^(5/2)/x-2/3*(-a*x+1)*(a*x+1)^2/a^3/(c-c/a^2/x^2)^(5/2)/x^2+58/15*(-a*x+1)^2*(a
*x+1)^2/a^4/(c-c/a^2/x^2)^(5/2)/x^3+2/15*(-a*x+1)^3*(a*x+1)^2*(43*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

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Rubi [A]
time = 0.26, antiderivative size = 203, 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 = {6294, 6264, 100, 155, 148, 41, 222} \begin {gather*} \frac {(a x+1)^2}{5 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} \text {ArcSin}(a x)}{a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {2 (a x+1)^2 (43 a x+28) (1-a x)^3}{15 a^6 x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}+\frac {58 (a x+1)^2 (1-a x)^2}{15 a^4 x^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}-\frac {2 (a x+1)^2 (1-a x)}{3 a^3 x^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(5*a^2*(c - c/(a^2*x^2))^(5/2)*x) - (2*(1 - a*x)*(1 + a*x)^2)/(3*a^3*(c - c/(a^2*x^2))^(5/2)*x^2)
+ (58*(1 - a*x)^2*(1 + a*x)^2)/(15*a^4*(c - c/(a^2*x^2))^(5/2)*x^3) + (2*(1 - a*x)^3*(1 + a*x)^2*(28 + 43*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 100

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 148

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[m, 1] ||  !SumSimplerQ[n, 1])

Rule 155

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 222

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 6264

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 6294

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/x^(2*p))*(1 - a*x)^p*(1 + a*x)^p*E^(n*ArcTanh[a*x]), 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)^{7/2} (1+a x)^{3/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1+a x)^2}{5 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+6 a x)}{(1-a x)^{5/2} (1+a x)^{3/2}} \, dx}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1+a x)^2}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 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 (-30 a-28 a^2 x\right )}{(1-a x)^{3/2} (1+a x)^{3/2}} \, dx}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac {(1+a x)^2}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {58 (1-a x)^2 (1+a x)^2}{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 (116 a^2+86 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}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {58 (1-a x)^2 (1+a x)^2}{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+43 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}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {58 (1-a x)^2 (1+a x)^2}{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+43 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}{5 a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x) (1+a x)^2}{3 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {58 (1-a x)^2 (1+a x)^2}{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+43 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*}

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Mathematica [A]
time = 0.07, size = 105, normalized size = 0.52 \begin {gather*} \frac {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 \left (a x+\sqrt {-1+a^2 x^2}\right )}{15 a^2 c^2 \sqrt {c-\frac {c}{a^2 x^2}} x (-1+a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(461\) vs. \(2(181)=362\).
time = 0.88, size = 462, normalized size = 2.28

method	result	size
risch	\(-\frac {a^{2} x^{2}-1}{a^{2} c^{2} \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x}-\frac {\left (\frac {2 \ln \left (\frac {x \,a^{2} c}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right )}{a^{5} \sqrt {a^{2} c}}-\frac {\sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+2 \left (x -\frac {1}{a}\right ) a c}}{10 a^{9} c \left (x -\frac {1}{a}\right )^{3}}-\frac {41 \sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+2 \left (x -\frac {1}{a}\right ) a c}}{60 a^{8} c \left (x -\frac {1}{a}\right )^{2}}-\frac {383 \sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+2 \left (x -\frac {1}{a}\right ) a c}}{120 a^{7} c \left (x -\frac {1}{a}\right )}+\frac {\sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-2 \left (x +\frac {1}{a}\right ) a c}}{8 a^{7} c \left (x +\frac {1}{a}\right )}\right ) a^{4} \sqrt {c \left (a^{2} x^{2}-1\right )}}{c^{2} \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x}\)	\(300\)
default	\(\frac {\left (-15 x^{5} c^{\frac {5}{2}} a^{5} \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}} a^{4} x^{4}+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}} a^{3} x^{3}+60 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} a^{3} x^{3}-30 \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} \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}} a^{4} c x -24 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} a^{2} x^{2}-90 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} a^{2} x^{2}+30 \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} \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}} a^{3} c +24 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}} a x -50 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{a^{2}}\right )^{\frac {3}{2}} a x +6 c^{\frac {5}{2}} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )^{\frac {3}{2}}+50 c^{\frac {5}{2}} \left (\frac {\left (a x -1\right ) \left (a x +1\right ) c}{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}}}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.37, size = 353, normalized size = 1.74 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)*arct
an(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)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 \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}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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