3.25 \(\int x^2 (a+b \sec ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=236 \[ -\frac{i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac{i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac{b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b x^2 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c^3}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3} \]

[Out]

(b^2*x*(a + b*ArcSec[c*x]))/c^2 - (b*Sqrt[1 - 1/(c^2*x^2)]*x^2*(a + b*ArcSec[c*x])^2)/(2*c) + (x^3*(a + b*ArcS
ec[c*x])^3)/3 + (I*b*(a + b*ArcSec[c*x])^2*ArcTan[E^(I*ArcSec[c*x])])/c^3 - (b^3*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]
])/c^3 - (I*b^2*(a + b*ArcSec[c*x])*PolyLog[2, (-I)*E^(I*ArcSec[c*x])])/c^3 + (I*b^2*(a + b*ArcSec[c*x])*PolyL
og[2, I*E^(I*ArcSec[c*x])])/c^3 + (b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])])/c^3 - (b^3*PolyLog[3, I*E^(I*ArcSec
[c*x])])/c^3

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Rubi [A]  time = 0.192231, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5222, 4409, 4186, 3770, 4181, 2531, 2282, 6589} \[ -\frac{i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac{i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^3}+\frac{b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b x^2 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{i b \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )^2}{c^3}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSec[c*x])^3,x]

[Out]

(b^2*x*(a + b*ArcSec[c*x]))/c^2 - (b*Sqrt[1 - 1/(c^2*x^2)]*x^2*(a + b*ArcSec[c*x])^2)/(2*c) + (x^3*(a + b*ArcS
ec[c*x])^3)/3 + (I*b*(a + b*ArcSec[c*x])^2*ArcTan[E^(I*ArcSec[c*x])])/c^3 - (b^3*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]
])/c^3 - (I*b^2*(a + b*ArcSec[c*x])*PolyLog[2, (-I)*E^(I*ArcSec[c*x])])/c^3 + (I*b^2*(a + b*ArcSec[c*x])*PolyL
og[2, I*E^(I*ArcSec[c*x])])/c^3 + (b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])])/c^3 - (b^3*PolyLog[3, I*E^(I*ArcSec
[c*x])])/c^3

Rule 5222

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x^2 \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^3 \sec ^3(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b \operatorname{Subst}\left (\int (a+b x)^2 \sec ^3(x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b \operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{2 c^3}-\frac{b^3 \operatorname{Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3}+\frac{b^2 \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}-\frac{b^2 \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3}-\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{\left (i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}-\frac{\left (i b^3\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^3}\\ &=\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3}-\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{b^3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sec ^{-1}(c x)}\right )}{c^3}\\ &=\frac{b^2 x \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^2 \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^3}-\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{i b^2 \left (a+b \sec ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}+\frac{b^3 \text{Li}_3\left (-i e^{i \sec ^{-1}(c x)}\right )}{c^3}-\frac{b^3 \text{Li}_3\left (i e^{i \sec ^{-1}(c x)}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 1.32206, size = 403, normalized size = 1.71 \[ \frac{-6 i b^2 \text{PolyLog}\left (2,-i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 i b^2 \text{PolyLog}\left (2,i e^{i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )+6 b^3 \text{PolyLog}\left (3,-i e^{i \sec ^{-1}(c x)}\right )-6 b^3 \text{PolyLog}\left (3,i e^{i \sec ^{-1}(c x)}\right )-3 a^2 b c^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}}-3 a^2 b \log \left (x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )+6 a^2 b c^3 x^3 \sec ^{-1}(c x)+2 a^3 c^3 x^3+6 a b^2 c^3 x^3 \sec ^{-1}(c x)^2-6 a b^2 c^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}} \sec ^{-1}(c x)+6 a b^2 c x-6 a b^2 \sec ^{-1}(c x) \log \left (1-i e^{i \sec ^{-1}(c x)}\right )+6 a b^2 \sec ^{-1}(c x) \log \left (1+i e^{i \sec ^{-1}(c x)}\right )+2 b^3 c^3 x^3 \sec ^{-1}(c x)^3-3 b^3 c^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}} \sec ^{-1}(c x)^2-6 b^3 \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )+6 b^3 c x \sec ^{-1}(c x)+6 i b^3 \sec ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sec ^{-1}(c x)}\right )}{6 c^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*(a + b*ArcSec[c*x])^3,x]

[Out]

(6*a*b^2*c*x - 3*a^2*b*c^2*Sqrt[1 - 1/(c^2*x^2)]*x^2 + 2*a^3*c^3*x^3 + 6*b^3*c*x*ArcSec[c*x] - 6*a*b^2*c^2*Sqr
t[1 - 1/(c^2*x^2)]*x^2*ArcSec[c*x] + 6*a^2*b*c^3*x^3*ArcSec[c*x] - 3*b^3*c^2*Sqrt[1 - 1/(c^2*x^2)]*x^2*ArcSec[
c*x]^2 + 6*a*b^2*c^3*x^3*ArcSec[c*x]^2 + 2*b^3*c^3*x^3*ArcSec[c*x]^3 + (6*I)*b^3*ArcSec[c*x]^2*ArcTan[E^(I*Arc
Sec[c*x])] - 6*b^3*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]] - 6*a*b^2*ArcSec[c*x]*Log[1 - I*E^(I*ArcSec[c*x])] + 6*a*b^2
*ArcSec[c*x]*Log[1 + I*E^(I*ArcSec[c*x])] - 3*a^2*b*Log[(1 + Sqrt[1 - 1/(c^2*x^2)])*x] - (6*I)*b^2*(a + b*ArcS
ec[c*x])*PolyLog[2, (-I)*E^(I*ArcSec[c*x])] + (6*I)*b^2*(a + b*ArcSec[c*x])*PolyLog[2, I*E^(I*ArcSec[c*x])] +
6*b^3*PolyLog[3, (-I)*E^(I*ArcSec[c*x])] - 6*b^3*PolyLog[3, I*E^(I*ArcSec[c*x])])/(6*c^3)

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Maple [B]  time = 0.52, size = 687, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsec(c*x))^3,x)

[Out]

1/3*a^3*x^3+1/3*x^3*b^3*arcsec(c*x)^3-1/2/c*b^3*((c^2*x^2-1)/c^2/x^2)^(1/2)*arcsec(c*x)^2*x^2+1/c^2*b^3*arcsec
(c*x)*x+1/2/c^3*b^3*arcsec(c*x)^2*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-I/c^3*a*b^2*dilog(1+I*(1/c/x+I*(1-1/c^
2/x^2)^(1/2)))+b^3*polylog(3,-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))/c^3-1/2/c^3*b^3*arcsec(c*x)^2*ln(1-I*(1/c/x+I*(
1-1/c^2/x^2)^(1/2)))+I/c^3*b^3*arcsec(c*x)*polylog(2,I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))-b^3*polylog(3,I*(1/c/x+I
*(1-1/c^2/x^2)^(1/2)))/c^3+2*I/c^3*b^3*arctan(1/c/x+I*(1-1/c^2/x^2)^(1/2))+b^2*x^3*a*arcsec(c*x)^2-1/c*a*b^2*(
(c^2*x^2-1)/c^2/x^2)^(1/2)*arcsec(c*x)*x^2-I/c^3*b^3*arcsec(c*x)*polylog(2,-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+I
/c^3*a*b^2*dilog(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+1/c^3*a*b^2*arcsec(c*x)*ln(1+I*(1/c/x+I*(1-1/c^2/x^2)^(1/2
)))-1/c^3*a*b^2*arcsec(c*x)*ln(1-I*(1/c/x+I*(1-1/c^2/x^2)^(1/2)))+1/c^2*x*a*b^2+x^3*a^2*b*arcsec(c*x)-1/2/c*a^
2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x^2+1/2/c^3*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)-1/2/c^4*a^2*b*(c^2*x^2-1)^(1/2)/
((c^2*x^2-1)/c^2/x^2)^(1/2)/x*ln(c*x+(c^2*x^2-1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsec(c*x))^3,x, algorithm="maxima")

[Out]

1/3*b^3*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 1/4*b^3*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^
2)^2 - 1/2*a*b^2*c^2*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c^5)*log(c)^2 - 12*b^3*c^2*i
ntegrate(1/4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2 - 1), x)*log(c)^2 + 12*b^3*c^2*integrate(1/4*x^4
*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) - 24*b^3*c^2*integrate(1/4*x^4*arct
an(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^2 - 1), x)*log(c) + 12*a*b^2*c^2*integrate(1/4*x^4*log(c^2*x^2)/
(c^2*x^2 - 1), x)*log(c) - 24*a*b^2*c^2*integrate(1/4*x^4*log(x)/(c^2*x^2 - 1), x)*log(c) + 1/3*a^3*x^3 + 12*b
^3*c^2*integrate(1/4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 12*b^3*c^
2*integrate(1/4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^2 - 1), x) + 12*a*b^2*c^2*integrate(1/
4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^2 - 1), x) + 4*b^3*c^2*integrate(1/4*x^4*arctan(sqrt(c*x +
1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x) - 3*a*b^2*c^2*integrate(1/4*x^4*log(c^2*x^2)^2/(c^2*x^2 - 1),
x) + 12*a*b^2*c^2*integrate(1/4*x^4*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) - 12*a*b^2*c^2*integrate(1/4*x^4*lo
g(x)^2/(c^2*x^2 - 1), x) + 3/2*a*b^2*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3)*log(c)^2 + 12*b^3*integra
te(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^2 - 1), x)*log(c)^2 - 12*b^3*integrate(1/4*x^2*arctan(sq
rt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^2 - 1), x)*log(c) + 24*b^3*integrate(1/4*x^2*arctan(sqrt(c*x +
1)*sqrt(c*x - 1))*log(x)/(c^2*x^2 - 1), x)*log(c) - 12*a*b^2*integrate(1/4*x^2*log(c^2*x^2)/(c^2*x^2 - 1), x)*
log(c) + 24*a*b^2*integrate(1/4*x^2*log(x)/(c^2*x^2 - 1), x)*log(c) + 1/4*(4*x^3*arcsec(c*x) - (2*sqrt(-1/(c^2
*x^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) -
1)/c^2)/c)*a^2*b - 4*b^3*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(
c^2*x^2 - 1), x) + b^3*integrate(1/4*sqrt(c*x + 1)*sqrt(c*x - 1)*x^2*log(c^2*x^2)^2/(c^2*x^2 - 1), x) - 12*b^3
*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) + 12*b^3*integrat
e(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^2 - 1), x) - 12*a*b^2*integrate(1/4*x^2*arctan(s
qrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^2 - 1), x) - 4*b^3*integrate(1/4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*
log(c^2*x^2)/(c^2*x^2 - 1), x) + 3*a*b^2*integrate(1/4*x^2*log(c^2*x^2)^2/(c^2*x^2 - 1), x) - 12*a*b^2*integra
te(1/4*x^2*log(c^2*x^2)*log(x)/(c^2*x^2 - 1), x) + 12*a*b^2*integrate(1/4*x^2*log(x)^2/(c^2*x^2 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x^{2} \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} x^{2} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b x^{2} \operatorname{arcsec}\left (c x\right ) + a^{3} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsec(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^2*arcsec(c*x)^3 + 3*a*b^2*x^2*arcsec(c*x)^2 + 3*a^2*b*x^2*arcsec(c*x) + a^3*x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \operatorname{asec}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asec(c*x))**3,x)

[Out]

Integral(x**2*(a + b*asec(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{3} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsec(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)^3*x^2, x)