3.1095 \(\int \frac {e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=131 \[ -\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}+\frac {1}{4} a^5 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )-\frac {a^3 c \sqrt {c-a^2 c x^2}}{4 x^2} \]

[Out]

-1/5*(-a^2*c*x^2+c)^(3/2)/x^5-1/2*a*(-a^2*c*x^2+c)^(3/2)/x^4-7/15*a^2*(-a^2*c*x^2+c)^(3/2)/x^3+1/4*a^5*c^(3/2)
*arctanh((-a^2*c*x^2+c)^(1/2)/c^(1/2))-1/4*a^3*c*(-a^2*c*x^2+c)^(1/2)/x^2

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Rubi [A]  time = 0.29, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6151, 1807, 835, 807, 266, 47, 63, 208} \[ \frac {1}{4} a^5 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )-\frac {a^3 c \sqrt {c-a^2 c x^2}}{4 x^2}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2))/x^6,x]

[Out]

-(a^3*c*Sqrt[c - a^2*c*x^2])/(4*x^2) - (c - a^2*c*x^2)^(3/2)/(5*x^5) - (a*(c - a^2*c*x^2)^(3/2))/(2*x^4) - (7*
a^2*(c - a^2*c*x^2)^(3/2))/(15*x^3) + (a^5*c^(3/2)*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]])/4

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 6151

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

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2}}{x^6} \, dx &=c \int \frac {(1+a x)^2 \sqrt {c-a^2 c x^2}}{x^6} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {1}{5} \int \frac {\left (-10 a c-7 a^2 c x\right ) \sqrt {c-a^2 c x^2}}{x^5} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}+\frac {\int \frac {\left (28 a^2 c^2+10 a^3 c^2 x\right ) \sqrt {c-a^2 c x^2}}{x^4} \, dx}{20 c}\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}+\frac {1}{2} \left (a^3 c\right ) \int \frac {\sqrt {c-a^2 c x^2}}{x^3} \, dx\\ &=-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}+\frac {1}{4} \left (a^3 c\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-a^2 c x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {a^3 c \sqrt {c-a^2 c x^2}}{4 x^2}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}-\frac {1}{8} \left (a^5 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )\\ &=-\frac {a^3 c \sqrt {c-a^2 c x^2}}{4 x^2}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}+\frac {1}{4} \left (a^3 c\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c-a^2 c x^2}\right )\\ &=-\frac {a^3 c \sqrt {c-a^2 c x^2}}{4 x^2}-\frac {\left (c-a^2 c x^2\right )^{3/2}}{5 x^5}-\frac {a \left (c-a^2 c x^2\right )^{3/2}}{2 x^4}-\frac {7 a^2 \left (c-a^2 c x^2\right )^{3/2}}{15 x^3}+\frac {1}{4} a^5 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 104, normalized size = 0.79 \[ -\frac {1}{4} a^5 c^{3/2} \log (x)+\frac {1}{4} a^5 c^{3/2} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )+\frac {c \left (28 a^4 x^4+15 a^3 x^3-16 a^2 x^2-30 a x-12\right ) \sqrt {c-a^2 c x^2}}{60 x^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(3/2))/x^6,x]

[Out]

(c*Sqrt[c - a^2*c*x^2]*(-12 - 30*a*x - 16*a^2*x^2 + 15*a^3*x^3 + 28*a^4*x^4))/(60*x^5) - (a^5*c^(3/2)*Log[x])/
4 + (a^5*c^(3/2)*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x^2]])/4

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fricas [A]  time = 0.65, size = 209, normalized size = 1.60 \[ \left [\frac {15 \, a^{5} c^{\frac {3}{2}} x^{5} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, {\left (28 \, a^{4} c x^{4} + 15 \, a^{3} c x^{3} - 16 \, a^{2} c x^{2} - 30 \, a c x - 12 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{120 \, x^{5}}, \frac {15 \, a^{5} \sqrt {-c} c x^{5} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + {\left (28 \, a^{4} c x^{4} + 15 \, a^{3} c x^{3} - 16 \, a^{2} c x^{2} - 30 \, a c x - 12 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{60 \, x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[1/120*(15*a^5*c^(3/2)*x^5*log(-(a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) + 2*(28*a^4*c*x^4 + 15
*a^3*c*x^3 - 16*a^2*c*x^2 - 30*a*c*x - 12*c)*sqrt(-a^2*c*x^2 + c))/x^5, 1/60*(15*a^5*sqrt(-c)*c*x^5*arctan(sqr
t(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) + (28*a^4*c*x^4 + 15*a^3*c*x^3 - 16*a^2*c*x^2 - 30*a*c*x - 12*c)*s
qrt(-a^2*c*x^2 + c))/x^5]

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giac [B]  time = 0.21, size = 414, normalized size = 3.16 \[ -\frac {a^{5} c^{2} \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{2 \, \sqrt {-c}} + \frac {15 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{9} a^{5} c^{2} - 60 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{8} a^{4} \sqrt {-c} c^{2} {\left | a \right |} + 90 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{7} a^{5} c^{3} + 240 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{6} a^{4} \sqrt {-c} c^{3} {\left | a \right |} - 40 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{4} a^{4} \sqrt {-c} c^{4} {\left | a \right |} - 90 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a^{5} c^{5} + 80 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a^{4} \sqrt {-c} c^{5} {\left | a \right |} - 15 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{5} c^{6} - 28 \, a^{4} \sqrt {-c} c^{6} {\left | a \right |}}{30 \, {\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(3/2)/x^6,x, algorithm="giac")

[Out]

-1/2*a^5*c^2*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 1/30*(15*(sqrt(-a^2*c)*x - s
qrt(-a^2*c*x^2 + c))^9*a^5*c^2 - 60*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^8*a^4*sqrt(-c)*c^2*abs(a) + 90*(sq
rt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^7*a^5*c^3 + 240*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^6*a^4*sqrt(-c)*c^
3*abs(a) - 40*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^4*a^4*sqrt(-c)*c^4*abs(a) - 90*(sqrt(-a^2*c)*x - sqrt(-a
^2*c*x^2 + c))^3*a^5*c^5 + 80*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a^4*sqrt(-c)*c^5*abs(a) - 15*(sqrt(-a^
2*c)*x - sqrt(-a^2*c*x^2 + c))*a^5*c^6 - 28*a^4*sqrt(-c)*c^6*abs(a))/((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^
2 - c)^5

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maple [B]  time = 0.07, size = 388, normalized size = 2.96 \[ -\frac {2 a^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{3 c \,x^{3}}-\frac {2 a^{4} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{3 c x}-a^{6} c x \sqrt {-a^{2} c \,x^{2}+c}-\frac {a^{6} c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}+a^{6} c \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}\, x +\frac {a^{6} c^{2} \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2} c}}-\frac {3 a^{3} \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{4 c \,x^{2}}-\frac {2 a^{5} \left (-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 a c \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}-\frac {a^{5} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{12}-\frac {a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{2 c \,x^{4}}-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}{5 c \,x^{5}}+\frac {a^{5} c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{4}-\frac {a^{5} \sqrt {-a^{2} c \,x^{2}+c}\, c}{4}-\frac {2 a^{6} x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a^2/c/x^3*(-a^2*c*x^2+c)^(5/2)-2/3*a^4/c/x*(-a^2*c*x^2+c)^(5/2)-a^6*c*x*(-a^2*c*x^2+c)^(1/2)-a^6*c^2/(a^2
*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+a^6*c*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1/2)*x+a^6*c^2/
(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(1/2))-3/4*a^3/c/x^2*(-a^2*c*x^2+c)^(5/2
)-2/3*a^5*(-(x-1/a)^2*a^2*c-2*a*c*(x-1/a))^(3/2)-1/12*a^5*(-a^2*c*x^2+c)^(3/2)-1/2*a/c/x^4*(-a^2*c*x^2+c)^(5/2
)-1/5/c/x^5*(-a^2*c*x^2+c)^(5/2)+1/4*a^5*c^(3/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-1/4*a^5*(-a^2*c*x^
2+c)^(1/2)*c-2/3*a^6*x*(-a^2*c*x^2+c)^(3/2)

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maxima [B]  time = 0.46, size = 221, normalized size = 1.69 \[ \frac {{\left (a^{2} c^{\frac {3}{2}} x^{2} - c^{\frac {3}{2}}\right )} \sqrt {a x + 1} \sqrt {-a x + 1} a^{2}}{3 \, x^{3}} - \frac {a^{6} c^{\frac {5}{2}} \log \left (\frac {\sqrt {-a^{2} c x^{2} + c} - \sqrt {c}}{\sqrt {-a^{2} c x^{2} + c} + \sqrt {c}}\right ) + \frac {2 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{6} c^{3} + \sqrt {-a^{2} c x^{2} + c} a^{6} c^{4}\right )}}{{\left (a^{2} c x^{2} - c\right )}^{2} + 2 \, {\left (a^{2} c x^{2} - c\right )} c + c^{2}}}{8 \, a c} + \frac {{\left (2 \, a^{4} c^{\frac {3}{2}} x^{4} + a^{2} c^{\frac {3}{2}} x^{2} - 3 \, c^{\frac {3}{2}}\right )} \sqrt {a x + 1} \sqrt {-a x + 1}}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)^(3/2)/x^6,x, algorithm="maxima")

[Out]

1/3*(a^2*c^(3/2)*x^2 - c^(3/2))*sqrt(a*x + 1)*sqrt(-a*x + 1)*a^2/x^3 - 1/8*(a^6*c^(5/2)*log((sqrt(-a^2*c*x^2 +
c) - sqrt(c))/(sqrt(-a^2*c*x^2 + c) + sqrt(c))) + 2*((-a^2*c*x^2 + c)^(3/2)*a^6*c^3 + sqrt(-a^2*c*x^2 + c)*a^
6*c^4)/((a^2*c*x^2 - c)^2 + 2*(a^2*c*x^2 - c)*c + c^2))/(a*c) + 1/15*(2*a^4*c^(3/2)*x^4 + a^2*c^(3/2)*x^2 - 3*
c^(3/2))*sqrt(a*x + 1)*sqrt(-a*x + 1)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\left (c-a^2\,c\,x^2\right )}^{3/2}\,{\left (a\,x+1\right )}^2}{x^6\,\left (a^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 24.93, size = 484, normalized size = 3.69 \[ a^{2} c \left (\begin {cases} \frac {a^{3} \sqrt {c} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{3} - \frac {a \sqrt {c} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{3 x^{2}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{3} \sqrt {c} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{3} - \frac {i a \sqrt {c} \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{3 x^{2}} & \text {otherwise} \end {cases}\right ) + 2 a c \left (\begin {cases} \frac {a^{4} \sqrt {c} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{8} - \frac {a^{3} \sqrt {c}}{8 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} + \frac {3 a \sqrt {c}}{8 x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {\sqrt {c}}{4 a x^{5} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac {i a^{4} \sqrt {c} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{8} + \frac {i a^{3} \sqrt {c}}{8 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} - \frac {3 i a \sqrt {c}}{8 x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i \sqrt {c}}{4 a x^{5} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {2 i a^{4} \sqrt {c} \sqrt {a^{2} x^{2} - 1}}{15 x} + \frac {i a^{2} \sqrt {c} \sqrt {a^{2} x^{2} - 1}}{15 x^{3}} - \frac {i \sqrt {c} \sqrt {a^{2} x^{2} - 1}}{5 x^{5}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {2 a^{4} \sqrt {c} \sqrt {- a^{2} x^{2} + 1}}{15 x} + \frac {a^{2} \sqrt {c} \sqrt {- a^{2} x^{2} + 1}}{15 x^{3}} - \frac {\sqrt {c} \sqrt {- a^{2} x^{2} + 1}}{5 x^{5}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)**(3/2)/x**6,x)

[Out]

a**2*c*Piecewise((a**3*sqrt(c)*sqrt(-1 + 1/(a**2*x**2))/3 - a*sqrt(c)*sqrt(-1 + 1/(a**2*x**2))/(3*x**2), 1/Abs
(a**2*x**2) > 1), (I*a**3*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/3 - I*a*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/(3*x**2), Tr
ue)) + 2*a*c*Piecewise((a**4*sqrt(c)*acosh(1/(a*x))/8 - a**3*sqrt(c)/(8*x*sqrt(-1 + 1/(a**2*x**2))) + 3*a*sqrt
(c)/(8*x**3*sqrt(-1 + 1/(a**2*x**2))) - sqrt(c)/(4*a*x**5*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (-
I*a**4*sqrt(c)*asin(1/(a*x))/8 + I*a**3*sqrt(c)/(8*x*sqrt(1 - 1/(a**2*x**2))) - 3*I*a*sqrt(c)/(8*x**3*sqrt(1 -
1/(a**2*x**2))) + I*sqrt(c)/(4*a*x**5*sqrt(1 - 1/(a**2*x**2))), True)) + c*Piecewise((2*I*a**4*sqrt(c)*sqrt(a
**2*x**2 - 1)/(15*x) + I*a**2*sqrt(c)*sqrt(a**2*x**2 - 1)/(15*x**3) - I*sqrt(c)*sqrt(a**2*x**2 - 1)/(5*x**5),
Abs(a**2*x**2) > 1), (2*a**4*sqrt(c)*sqrt(-a**2*x**2 + 1)/(15*x) + a**2*sqrt(c)*sqrt(-a**2*x**2 + 1)/(15*x**3)
- sqrt(c)*sqrt(-a**2*x**2 + 1)/(5*x**5), True))

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