3.1105 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=155 \[ -\frac{a^2 c^2 (6 a x+1) \sqrt{c-a^2 c x^2}}{2 x}-\frac{1}{2} a^3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+3 a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3} \]

[Out]

-(a^2*c^2*(1 + 6*a*x)*Sqrt[c - a^2*c*x^2])/(2*x) - (a*c*(6 - a*x)*(c - a^2*c*x^2)^(3/2))/(6*x^2) - (c - a^2*c*
x^2)^(5/2)/(3*x^3) - (a^3*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/2 + 3*a^3*c^(5/2)*ArcTanh[Sqrt[c
- a^2*c*x^2]/Sqrt[c]]

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Rubi [A]  time = 0.34214, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6151, 1807, 813, 844, 217, 203, 266, 63, 208} \[ -\frac{a^2 c^2 (6 a x+1) \sqrt{c-a^2 c x^2}}{2 x}-\frac{1}{2} a^3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+3 a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2))/x^4,x]

[Out]

-(a^2*c^2*(1 + 6*a*x)*Sqrt[c - a^2*c*x^2])/(2*x) - (a*c*(6 - a*x)*(c - a^2*c*x^2)^(3/2))/(6*x^2) - (c - a^2*c*
x^2)^(5/2)/(3*x^3) - (a^3*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/2 + 3*a^3*c^(5/2)*ArcTanh[Sqrt[c
- a^2*c*x^2]/Sqrt[c]]

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]

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 217

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

Rule 203

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

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 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]

Rubi steps

\begin{align*} \int \frac{e^{2 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^{5/2}}{x^4} \, dx &=c \int \frac{(1+a x)^2 \left (c-a^2 c x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\frac{1}{3} \int \frac{\left (-6 a c-a^2 c x\right ) \left (c-a^2 c x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}+\frac{1}{8} \int \frac{\left (4 a^2 c^2-24 a^3 c^2 x\right ) \sqrt{c-a^2 c x^2}}{x^2} \, dx\\ &=-\frac{a^2 c^2 (1+6 a x) \sqrt{c-a^2 c x^2}}{2 x}-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\frac{1}{16} \int \frac{48 a^3 c^3+8 a^4 c^3 x}{x \sqrt{c-a^2 c x^2}} \, dx\\ &=-\frac{a^2 c^2 (1+6 a x) \sqrt{c-a^2 c x^2}}{2 x}-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\left (3 a^3 c^3\right ) \int \frac{1}{x \sqrt{c-a^2 c x^2}} \, dx-\frac{1}{2} \left (a^4 c^3\right ) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx\\ &=-\frac{a^2 c^2 (1+6 a x) \sqrt{c-a^2 c x^2}}{2 x}-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\frac{1}{2} \left (3 a^3 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c-a^2 c x}} \, dx,x,x^2\right )-\frac{1}{2} \left (a^4 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )\\ &=-\frac{a^2 c^2 (1+6 a x) \sqrt{c-a^2 c x^2}}{2 x}-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\frac{1}{2} a^3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+\left (3 a c^2\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^2 c^2 (1+6 a x) \sqrt{c-a^2 c x^2}}{2 x}-\frac{a c (6-a x) \left (c-a^2 c x^2\right )^{3/2}}{6 x^2}-\frac{\left (c-a^2 c x^2\right )^{5/2}}{3 x^3}-\frac{1}{2} a^3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )+3 a^3 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-a^2 c x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.263262, size = 149, normalized size = 0.96 \[ -\frac{c^2 \left (3 a^4 x^4+12 a^3 x^3-2 a^2 x^2+6 a x+2\right ) \sqrt{c-a^2 c x^2}}{6 x^3}+3 a^3 c^{5/2} \log \left (\sqrt{c} \sqrt{c-a^2 c x^2}+c\right )+\frac{1}{2} a^3 c^{5/2} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )-3 a^3 c^{5/2} \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^(5/2))/x^4,x]

[Out]

-(c^2*Sqrt[c - a^2*c*x^2]*(2 + 6*a*x - 2*a^2*x^2 + 12*a^3*x^3 + 3*a^4*x^4))/(6*x^3) + (a^3*c^(5/2)*ArcTan[(a*x
*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/2 - 3*a^3*c^(5/2)*Log[x] + 3*a^3*c^(5/2)*Log[c + Sqrt[c]*Sqrt
[c - a^2*c*x^2]]

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Maple [B]  time = 0.05, size = 423, normalized size = 2.7 \begin{align*} -{\frac{2\,{a}^{2}}{3\,cx} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{2\,{a}^{4}x}{3} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{a}^{4}cx}{6} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{a}^{4}{c}^{2}x}{4}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{5\,{a}^{4}{c}^{3}}{4}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{3\,{a}^{3}}{5} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{a}^{3}c \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}+3\,{a}^{3}{c}^{5/2}\ln \left ({\frac{2\,c+2\,\sqrt{c}\sqrt{-{a}^{2}c{x}^{2}+c}}{x}} \right ) -3\,{a}^{3}\sqrt{-{a}^{2}c{x}^{2}+c}{c}^{2}-{\frac{2\,{a}^{3}}{5} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{4}cx}{2} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{4}{c}^{2}x}{4}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}+{\frac{3\,{a}^{4}{c}^{3}}{4}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{a}{c{x}^{2}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{1}{3\,c{x}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}} \end{align*}

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)^(5/2)/x^4,x)

[Out]

-2/3*a^2/c/x*(-a^2*c*x^2+c)^(7/2)-2/3*a^4*x*(-a^2*c*x^2+c)^(5/2)-5/6*a^4*c*x*(-a^2*c*x^2+c)^(3/2)-5/4*a^4*c^2*
x*(-a^2*c*x^2+c)^(1/2)-5/4*a^4*c^3/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-3/5*a^3*(-a^2*c*
x^2+c)^(5/2)-a^3*c*(-a^2*c*x^2+c)^(3/2)+3*a^3*c^(5/2)*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x)-3*a^3*(-a^2*c
*x^2+c)^(1/2)*c^2-2/5*a^3*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(5/2)+1/2*a^4*c*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(3
/2)*x+3/4*a^4*c^2*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)*x+3/4*a^4*c^3/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-
c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2))-a/c/x^2*(-a^2*c*x^2+c)^(7/2)-1/3/c/x^3*(-a^2*c*x^2+c)^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}}{\left (a x + 1\right )}^{2}}{{\left (a^{2} x^{2} - 1\right )} x^{4}}\,{d x} \end{align*}

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)^(5/2)/x^4,x, algorithm="maxima")

[Out]

-integrate((-a^2*c*x^2 + c)^(5/2)*(a*x + 1)^2/((a^2*x^2 - 1)*x^4), x)

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Fricas [A]  time = 2.7942, size = 717, normalized size = 4.63 \begin{align*} \left [\frac{3 \, a^{3} c^{\frac{5}{2}} x^{3} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) + 9 \, a^{3} c^{\frac{5}{2}} x^{3} \log \left (-\frac{a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{c} - 2 \, c}{x^{2}}\right ) -{\left (3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} - 2 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{-a^{2} c x^{2} + c}}{6 \, x^{3}}, \frac{36 \, a^{3} \sqrt{-c} c^{2} x^{3} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + 3 \, a^{3} \sqrt{-c} c^{2} x^{3} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) - 2 \,{\left (3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} - 2 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + 2 \, c^{2}\right )} \sqrt{-a^{2} c x^{2} + c}}{12 \, x^{3}}\right ] \end{align*}

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)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/6*(3*a^3*c^(5/2)*x^3*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^2 - c)) + 9*a^3*c^(5/2)*x^3*log(-(a^2
*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - (3*a^4*c^2*x^4 + 12*a^3*c^2*x^3 - 2*a^2*c^2*x^2 + 6*a*c^
2*x + 2*c^2)*sqrt(-a^2*c*x^2 + c))/x^3, 1/12*(36*a^3*sqrt(-c)*c^2*x^3*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^
2*c*x^2 - c)) + 3*a^3*sqrt(-c)*c^2*x^3*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*(3*a^4*c
^2*x^4 + 12*a^3*c^2*x^3 - 2*a^2*c^2*x^2 + 6*a*c^2*x + 2*c^2)*sqrt(-a^2*c*x^2 + c))/x^3]

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Sympy [C]  time = 14.1235, size = 478, normalized size = 3.08 \begin{align*} - a^{4} c^{2} \left (\begin{cases} \frac{i a^{2} \sqrt{c} x^{3}}{2 \sqrt{a^{2} x^{2} - 1}} - \frac{i \sqrt{c} x}{2 \sqrt{a^{2} x^{2} - 1}} - \frac{i \sqrt{c} \operatorname{acosh}{\left (a x \right )}}{2 a} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac{\sqrt{c} x \sqrt{- a^{2} x^{2} + 1}}{2} + \frac{\sqrt{c} \operatorname{asin}{\left (a x \right )}}{2 a} & \text{otherwise} \end{cases}\right ) - 2 a^{3} c^{2} \left (\begin{cases} i \sqrt{c} \sqrt{a^{2} x^{2} - 1} - \sqrt{c} \log{\left (a x \right )} + \frac{\sqrt{c} \log{\left (a^{2} x^{2} \right )}}{2} + i \sqrt{c} \operatorname{asin}{\left (\frac{1}{a x} \right )} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\\sqrt{c} \sqrt{- a^{2} x^{2} + 1} + \frac{\sqrt{c} \log{\left (a^{2} x^{2} \right )}}{2} - \sqrt{c} \log{\left (\sqrt{- a^{2} x^{2} + 1} + 1 \right )} & \text{otherwise} \end{cases}\right ) + 2 a c^{2} \left (\begin{cases} \frac{a^{2} \sqrt{c} \operatorname{acosh}{\left (\frac{1}{a x} \right )}}{2} + \frac{a \sqrt{c}}{2 x \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} - \frac{\sqrt{c}}{2 a x^{3} \sqrt{-1 + \frac{1}{a^{2} x^{2}}}} & \text{for}\: \frac{1}{\left |{a^{2} x^{2}}\right |} > 1 \\- \frac{i a^{2} \sqrt{c} \operatorname{asin}{\left (\frac{1}{a x} \right )}}{2} - \frac{i a \sqrt{c} \sqrt{1 - \frac{1}{a^{2} x^{2}}}}{2 x} & \text{otherwise} \end{cases}\right ) + c^{2} \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 ) \end{align*}

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)**(5/2)/x**4,x)

[Out]

-a**4*c**2*Piecewise((I*a**2*sqrt(c)*x**3/(2*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*x/(2*sqrt(a**2*x**2 - 1)) - I*sq
rt(c)*acosh(a*x)/(2*a), Abs(a**2*x**2) > 1), (sqrt(c)*x*sqrt(-a**2*x**2 + 1)/2 + sqrt(c)*asin(a*x)/(2*a), True
)) - 2*a**3*c**2*Piecewise((I*sqrt(c)*sqrt(a**2*x**2 - 1) - sqrt(c)*log(a*x) + sqrt(c)*log(a**2*x**2)/2 + I*sq
rt(c)*asin(1/(a*x)), Abs(a**2*x**2) > 1), (sqrt(c)*sqrt(-a**2*x**2 + 1) + sqrt(c)*log(a**2*x**2)/2 - sqrt(c)*l
og(sqrt(-a**2*x**2 + 1) + 1), True)) + 2*a*c**2*Piecewise((a**2*sqrt(c)*acosh(1/(a*x))/2 + a*sqrt(c)/(2*x*sqrt
(-1 + 1/(a**2*x**2))) - sqrt(c)/(2*a*x**3*sqrt(-1 + 1/(a**2*x**2))), 1/Abs(a**2*x**2) > 1), (-I*a**2*sqrt(c)*a
sin(1/(a*x))/2 - I*a*sqrt(c)*sqrt(1 - 1/(a**2*x**2))/(2*x), True)) + c**2*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), True))

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Giac [B]  time = 1.20414, size = 394, normalized size = 2.54 \begin{align*} -\frac{6 \, a^{3} c^{3} \arctan \left (-\frac{\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{a^{4} \sqrt{-c} c^{2} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{2 \,{\left | a \right |}} - \frac{1}{2} \,{\left (a^{4} c^{2} x + 4 \, a^{3} c^{2}\right )} \sqrt{-a^{2} c x^{2} + c} - \frac{2 \,{\left (3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{5} a^{3} c^{3}{\left | a \right |} + 3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{4} a^{4} \sqrt{-c} c^{3} - 3 \,{\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )} a^{3} c^{5}{\left | a \right |} + a^{4} \sqrt{-c} c^{5}\right )}}{3 \,{\left ({\left (\sqrt{-a^{2} c} x - \sqrt{-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{3}{\left | a \right |}} \end{align*}

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)^(5/2)/x^4,x, algorithm="giac")

[Out]

-6*a^3*c^3*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) - 1/2*a^4*sqrt(-c)*c^2*log(abs(-
sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) - 1/2*(a^4*c^2*x + 4*a^3*c^2)*sqrt(-a^2*c*x^2 + c) - 2/3*(3*(sq
rt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^5*a^3*c^3*abs(a) + 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^4*a^4*sqrt(-
c)*c^3 - 3*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))*a^3*c^5*abs(a) + a^4*sqrt(-c)*c^5)/(((sqrt(-a^2*c)*x - sqrt
(-a^2*c*x^2 + c))^2 - c)^3*abs(a))