3.1098 \(\int e^{2 \tanh ^{-1}(a x)} x^3 (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=187 \[ \frac{3 c^2 x \sqrt{c-a^2 c x^2}}{64 a^3}+\frac{3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{64 a^4}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}+\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{(315 a x+208) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4} \]

[Out]

(3*c^2*x*Sqrt[c - a^2*c*x^2])/(64*a^3) + (c*x*(c - a^2*c*x^2)^(3/2))/(32*a^3) - (13*x^2*(c - a^2*c*x^2)^(5/2))
/(63*a^2) - (x^3*(c - a^2*c*x^2)^(5/2))/(4*a) - (x^4*(c - a^2*c*x^2)^(5/2))/9 - ((208 + 315*a*x)*(c - a^2*c*x^
2)^(5/2))/(2520*a^4) + (3*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(64*a^4)

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Rubi [A]  time = 0.37002, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {6151, 1809, 833, 780, 195, 217, 203} \[ \frac{3 c^2 x \sqrt{c-a^2 c x^2}}{64 a^3}+\frac{3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{64 a^4}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}+\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{(315 a x+208) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c^2*x*Sqrt[c - a^2*c*x^2])/(64*a^3) + (c*x*(c - a^2*c*x^2)^(3/2))/(32*a^3) - (13*x^2*(c - a^2*c*x^2)^(5/2))
/(63*a^2) - (x^3*(c - a^2*c*x^2)^(5/2))/(4*a) - (x^4*(c - a^2*c*x^2)^(5/2))/9 - ((208 + 315*a*x)*(c - a^2*c*x^
2)^(5/2))/(2520*a^4) + (3*c^(5/2)*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2]])/(64*a^4)

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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 833

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

Rule 780

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

Rubi steps

\begin{align*} \int e^{2 \tanh ^{-1}(a x)} x^3 \left (c-a^2 c x^2\right )^{5/2} \, dx &=c \int x^3 (1+a x)^2 \left (c-a^2 c x^2\right )^{3/2} \, dx\\ &=-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{\int x^3 \left (-13 a^2 c-18 a^3 c x\right ) \left (c-a^2 c x^2\right )^{3/2} \, dx}{9 a^2}\\ &=-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}+\frac{\int x^2 \left (54 a^3 c^2+104 a^4 c^2 x\right ) \left (c-a^2 c x^2\right )^{3/2} \, dx}{72 a^4 c}\\ &=-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{\int x \left (-208 a^4 c^3-378 a^5 c^3 x\right ) \left (c-a^2 c x^2\right )^{3/2} \, dx}{504 a^6 c^2}\\ &=-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{(208+315 a x) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4}+\frac{c \int \left (c-a^2 c x^2\right )^{3/2} \, dx}{8 a^3}\\ &=\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{(208+315 a x) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4}+\frac{\left (3 c^2\right ) \int \sqrt{c-a^2 c x^2} \, dx}{32 a^3}\\ &=\frac{3 c^2 x \sqrt{c-a^2 c x^2}}{64 a^3}+\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{(208+315 a x) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4}+\frac{\left (3 c^3\right ) \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{64 a^3}\\ &=\frac{3 c^2 x \sqrt{c-a^2 c x^2}}{64 a^3}+\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{(208+315 a x) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4}+\frac{\left (3 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 )}{64 a^3}\\ &=\frac{3 c^2 x \sqrt{c-a^2 c x^2}}{64 a^3}+\frac{c x \left (c-a^2 c x^2\right )^{3/2}}{32 a^3}-\frac{13 x^2 \left (c-a^2 c x^2\right )^{5/2}}{63 a^2}-\frac{x^3 \left (c-a^2 c x^2\right )^{5/2}}{4 a}-\frac{1}{9} x^4 \left (c-a^2 c x^2\right )^{5/2}-\frac{(208+315 a x) \left (c-a^2 c x^2\right )^{5/2}}{2520 a^4}+\frac{3 c^{5/2} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{64 a^4}\\ \end{align*}

Mathematica [A]  time = 0.214483, size = 131, normalized size = 0.7 \[ -\frac{c^2 \left (\left (2240 a^8 x^8+5040 a^7 x^7-320 a^6 x^6-7560 a^5 x^5-4416 a^4 x^4+630 a^3 x^3+832 a^2 x^2+945 a x+1664\right ) \sqrt{c-a^2 c x^2}+945 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )\right )}{20160 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(Sqrt[c - a^2*c*x^2]*(1664 + 945*a*x + 832*a^2*x^2 + 630*a^3*x^3 - 4416*a^4*x^4 - 7560*a^5*x^5 - 320*a^6
*x^6 + 5040*a^7*x^7 + 2240*a^8*x^8) + 945*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))]))
/(20160*a^4)

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Maple [B]  time = 0.046, size = 330, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}}{9\,{a}^{2}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{20}{63\,c{a}^{4}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}}+{\frac{x}{4\,{a}^{3}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{7}{2}}}}-{\frac{3\,x}{8\,{a}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}}-{\frac{15\,cx}{32\,{a}^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{45\,x{c}^{2}}{64\,{a}^{3}}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{45\,{c}^{3}}{64\,{a}^{3}}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-{\frac{2}{5\,{a}^{4}} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{cx}{2\,{a}^{3}} \left ( -c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{3\,x{c}^{2}}{4\,{a}^{3}}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}+{\frac{3\,{c}^{3}}{4\,{a}^{3}}\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^2*(-a^2*c*x^2+c)^(7/2)/a^2/c+20/63/c/a^4*(-a^2*c*x^2+c)^(7/2)+1/4/a^3*x*(-a^2*c*x^2+c)^(7/2)/c-3/8/a^3*x
*(-a^2*c*x^2+c)^(5/2)-15/32*c*x*(-a^2*c*x^2+c)^(3/2)/a^3-45/64*c^2*x*(-a^2*c*x^2+c)^(1/2)/a^3-45/64/a^3*c^3/(a
^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/5/a^4*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(5/2)+1/2/a^
3*c*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(3/2)*x+3/4/a^3*c^2*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)*x+3/4/a^3*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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.88005, size = 726, normalized size = 3.88 \begin{align*} \left [\frac{945 \, \sqrt{-c} c^{2} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) - 2 \,{\left (2240 \, a^{8} c^{2} x^{8} + 5040 \, a^{7} c^{2} x^{7} - 320 \, a^{6} c^{2} x^{6} - 7560 \, a^{5} c^{2} x^{5} - 4416 \, a^{4} c^{2} x^{4} + 630 \, a^{3} c^{2} x^{3} + 832 \, a^{2} c^{2} x^{2} + 945 \, a c^{2} x + 1664 \, c^{2}\right )} \sqrt{-a^{2} c x^{2} + c}}{40320 \, a^{4}}, -\frac{945 \, c^{\frac{5}{2}} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) +{\left (2240 \, a^{8} c^{2} x^{8} + 5040 \, a^{7} c^{2} x^{7} - 320 \, a^{6} c^{2} x^{6} - 7560 \, a^{5} c^{2} x^{5} - 4416 \, a^{4} c^{2} x^{4} + 630 \, a^{3} c^{2} x^{3} + 832 \, a^{2} c^{2} x^{2} + 945 \, a c^{2} x + 1664 \, c^{2}\right )} \sqrt{-a^{2} c x^{2} + c}}{20160 \, a^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40320*(945*sqrt(-c)*c^2*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*(2240*a^8*c^2*x^8 +
5040*a^7*c^2*x^7 - 320*a^6*c^2*x^6 - 7560*a^5*c^2*x^5 - 4416*a^4*c^2*x^4 + 630*a^3*c^2*x^3 + 832*a^2*c^2*x^2 +
945*a*c^2*x + 1664*c^2)*sqrt(-a^2*c*x^2 + c))/a^4, -1/20160*(945*c^(5/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c
)*x/(a^2*c*x^2 - c)) + (2240*a^8*c^2*x^8 + 5040*a^7*c^2*x^7 - 320*a^6*c^2*x^6 - 7560*a^5*c^2*x^5 - 4416*a^4*c^
2*x^4 + 630*a^3*c^2*x^3 + 832*a^2*c^2*x^2 + 945*a*c^2*x + 1664*c^2)*sqrt(-a^2*c*x^2 + c))/a^4]

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Sympy [A]  time = 25.4706, size = 763, normalized size = 4.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**4*c**2*Piecewise((x**8*sqrt(-a**2*c*x**2 + c)/9 - x**6*sqrt(-a**2*c*x**2 + c)/(63*a**2) - 2*x**4*sqrt(-a**
2*c*x**2 + c)/(105*a**4) - 8*x**2*sqrt(-a**2*c*x**2 + c)/(315*a**6) - 16*sqrt(-a**2*c*x**2 + c)/(315*a**8), Ne
(a, 0)), (sqrt(c)*x**8/8, True)) - 2*a**3*c**2*Piecewise((I*a**2*sqrt(c)*x**9/(8*sqrt(a**2*x**2 - 1)) - 7*I*sq
rt(c)*x**7/(48*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*x**5/(192*a**2*sqrt(a**2*x**2 - 1)) - 5*I*sqrt(c)*x**3/(384*a*
*4*sqrt(a**2*x**2 - 1)) + 5*I*sqrt(c)*x/(128*a**6*sqrt(a**2*x**2 - 1)) - 5*I*sqrt(c)*acosh(a*x)/(128*a**7), Ab
s(a**2*x**2) > 1), (-a**2*sqrt(c)*x**9/(8*sqrt(-a**2*x**2 + 1)) + 7*sqrt(c)*x**7/(48*sqrt(-a**2*x**2 + 1)) + s
qrt(c)*x**5/(192*a**2*sqrt(-a**2*x**2 + 1)) + 5*sqrt(c)*x**3/(384*a**4*sqrt(-a**2*x**2 + 1)) - 5*sqrt(c)*x/(12
8*a**6*sqrt(-a**2*x**2 + 1)) + 5*sqrt(c)*asin(a*x)/(128*a**7), True)) + 2*a*c**2*Piecewise((I*a**2*sqrt(c)*x**
7/(6*sqrt(a**2*x**2 - 1)) - 5*I*sqrt(c)*x**5/(24*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*x**3/(48*a**2*sqrt(a**2*x**2
- 1)) + I*sqrt(c)*x/(16*a**4*sqrt(a**2*x**2 - 1)) - I*sqrt(c)*acosh(a*x)/(16*a**5), Abs(a**2*x**2) > 1), (-a*
*2*sqrt(c)*x**7/(6*sqrt(-a**2*x**2 + 1)) + 5*sqrt(c)*x**5/(24*sqrt(-a**2*x**2 + 1)) + sqrt(c)*x**3/(48*a**2*sq
rt(-a**2*x**2 + 1)) - sqrt(c)*x/(16*a**4*sqrt(-a**2*x**2 + 1)) + sqrt(c)*asin(a*x)/(16*a**5), True)) + c**2*Pi
ecewise((x**4*sqrt(-a**2*c*x**2 + c)/5 - x**2*sqrt(-a**2*c*x**2 + c)/(15*a**2) - 2*sqrt(-a**2*c*x**2 + c)/(15*
a**4), Ne(a, 0)), (sqrt(c)*x**4/4, True))

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Giac [A]  time = 1.18195, size = 209, normalized size = 1.12 \begin{align*} \frac{1}{20160} \, \sqrt{-a^{2} c x^{2} + c}{\left ({\left (2 \,{\left ({\left (4 \,{\left (552 \, c^{2} + 5 \,{\left (189 \, a c^{2} + 2 \,{\left (4 \, a^{2} c^{2} - 7 \,{\left (4 \, a^{4} c^{2} x + 9 \, a^{3} c^{2}\right )} x\right )} x\right )} x\right )} x - \frac{315 \, c^{2}}{a}\right )} x - \frac{416 \, c^{2}}{a^{2}}\right )} x - \frac{945 \, c^{2}}{a^{3}}\right )} x - \frac{1664 \, c^{2}}{a^{4}}\right )} - \frac{3 \, c^{3} \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{64 \, a^{3} \sqrt{-c}{\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)*x^3*(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/20160*sqrt(-a^2*c*x^2 + c)*((2*((4*(552*c^2 + 5*(189*a*c^2 + 2*(4*a^2*c^2 - 7*(4*a^4*c^2*x + 9*a^3*c^2)*x)*x
)*x)*x - 315*c^2/a)*x - 416*c^2/a^2)*x - 945*c^2/a^3)*x - 1664*c^2/a^4) - 3/64*c^3*log(abs(-sqrt(-a^2*c)*x + s
qrt(-a^2*c*x^2 + c)))/(a^3*sqrt(-c)*abs(a))