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

Optimal. Leaf size=207 \[ \frac{i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^4}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}-\frac{b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^4}-\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^3 x \sqrt{1-\frac{1}{c^2 x^2}}}{4 c^3} \]

[Out]

-(b^3*Sqrt[1 - 1/(c^2*x^2)]*x)/(4*c^3) + (b^2*x^2*(a + b*ArcSec[c*x]))/(4*c^2) + ((I/2)*b*(a + b*ArcSec[c*x])^
2)/c^4 - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(a + b*ArcSec[c*x])^2)/(2*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3*(a + b*ArcSe
c[c*x])^2)/(4*c) + (x^4*(a + b*ArcSec[c*x])^3)/4 - (b^2*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])])/c^
4 + ((I/2)*b^3*PolyLog[2, -E^((2*I)*ArcSec[c*x])])/c^4

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Rubi [A]  time = 0.211203, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {5222, 4409, 4186, 3767, 8, 4184, 3719, 2190, 2279, 2391} \[ \frac{i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^4}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}-\frac{b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^4}-\frac{b x^3 \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^3 x \sqrt{1-\frac{1}{c^2 x^2}}}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSec[c*x])^3,x]

[Out]

-(b^3*Sqrt[1 - 1/(c^2*x^2)]*x)/(4*c^3) + (b^2*x^2*(a + b*ArcSec[c*x]))/(4*c^2) + ((I/2)*b*(a + b*ArcSec[c*x])^
2)/c^4 - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(a + b*ArcSec[c*x])^2)/(2*c^3) - (b*Sqrt[1 - 1/(c^2*x^2)]*x^3*(a + b*ArcSe
c[c*x])^2)/(4*c) + (x^4*(a + b*ArcSec[c*x])^3)/4 - (b^2*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])])/c^
4 + ((I/2)*b^3*PolyLog[2, -E^((2*I)*ArcSec[c*x])])/c^4

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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3719

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

Rule 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x^3 \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^3 \sec ^4(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^4}\\ &=\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \sec ^4(x) \, dx,x,\sec ^{-1}(c x)\right )}{4 c^4}\\ &=\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b \operatorname{Subst}\left (\int (a+b x)^2 \sec ^2(x) \, dx,x,\sec ^{-1}(c x)\right )}{2 c^4}-\frac{b^3 \operatorname{Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(c x)\right )}{4 c^4}\\ &=\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{b^2 \operatorname{Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^4}+\frac{b^3 \operatorname{Subst}\left (\int 1 \, dx,x,-c \sqrt{1-\frac{1}{c^2 x^2}} x\right )}{4 c^4}\\ &=-\frac{b^3 \sqrt{1-\frac{1}{c^2 x^2}} x}{4 c^3}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )}{c^4}\\ &=-\frac{b^3 \sqrt{1-\frac{1}{c^2 x^2}} x}{4 c^3}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^4}+\frac{b^3 \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^4}\\ &=-\frac{b^3 \sqrt{1-\frac{1}{c^2 x^2}} x}{4 c^3}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^4}-\frac{\left (i b^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )}{2 c^4}\\ &=-\frac{b^3 \sqrt{1-\frac{1}{c^2 x^2}} x}{4 c^3}+\frac{b^2 x^2 \left (a+b \sec ^{-1}(c x)\right )}{4 c^2}+\frac{i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^4}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^3}-\frac{b \sqrt{1-\frac{1}{c^2 x^2}} x^3 \left (a+b \sec ^{-1}(c x)\right )^2}{4 c}+\frac{1}{4} x^4 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^4}+\frac{i b^3 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.835647, size = 288, normalized size = 1.39 \[ \frac{2 i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+b \sec ^{-1}(c x) \left (c x \left (3 a^2 c^3 x^3-2 a b \sqrt{1-\frac{1}{c^2 x^2}} \left (c^2 x^2+2\right )+b^2 c x\right )-4 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )-a^2 b c^3 x^3 \sqrt{1-\frac{1}{c^2 x^2}}-2 a^2 b c x \sqrt{1-\frac{1}{c^2 x^2}}+a^3 c^4 x^4+a b^2 c^2 x^2-b^2 \sec ^{-1}(c x)^2 \left (-3 a c^4 x^4+b \left (c^3 x^3 \sqrt{1-\frac{1}{c^2 x^2}}+2 c x \sqrt{1-\frac{1}{c^2 x^2}}-2 i\right )\right )-4 a b^2 \log \left (\frac{1}{c x}\right )-b^3 c x \sqrt{1-\frac{1}{c^2 x^2}}+b^3 c^4 x^4 \sec ^{-1}(c x)^3}{4 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*a^2*b*c*Sqrt[1 - 1/(c^2*x^2)]*x - b^3*c*Sqrt[1 - 1/(c^2*x^2)]*x + a*b^2*c^2*x^2 - a^2*b*c^3*Sqrt[1 - 1/(c^
2*x^2)]*x^3 + a^3*c^4*x^4 - b^2*(-3*a*c^4*x^4 + b*(-2*I + 2*c*Sqrt[1 - 1/(c^2*x^2)]*x + c^3*Sqrt[1 - 1/(c^2*x^
2)]*x^3))*ArcSec[c*x]^2 + b^3*c^4*x^4*ArcSec[c*x]^3 + b*ArcSec[c*x]*(c*x*(b^2*c*x + 3*a^2*c^3*x^3 - 2*a*b*Sqrt
[1 - 1/(c^2*x^2)]*(2 + c^2*x^2)) - 4*b^2*Log[1 + E^((2*I)*ArcSec[c*x])]) - 4*a*b^2*Log[1/(c*x)] + (2*I)*b^3*Po
lyLog[2, -E^((2*I)*ArcSec[c*x])])/(4*c^4)

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Maple [B]  time = 0.472, size = 447, normalized size = 2.2 \begin{align*}{\frac{{x}^{4}{a}^{3}}{4}}+{\frac{{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}{x}^{4}}{4}}-{\frac{{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}{x}^{3}}{4\,c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}x}{2\,{c}^{3}}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{\frac{i}{2}}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{{c}^{4}}}+{\frac{{b}^{3}{\rm arcsec} \left (cx\right ){x}^{2}}{4\,{c}^{2}}}-{\frac{{b}^{3}x}{4\,{c}^{3}}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{\frac{i}{2}}{b}^{3}}{{c}^{4}}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }-{\frac{{b}^{3}{\rm arcsec} \left (cx\right )}{{c}^{4}}\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }-{\frac{{\frac{i}{4}}{b}^{3}}{{c}^{4}}}+{\frac{3\,{a}^{2}b{x}^{4}{\rm arcsec} \left (cx\right )}{4}}-{\frac{{a}^{2}b{x}^{3}}{4\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{{a}^{2}bx}{4\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{{a}^{2}b}{2\,{c}^{5}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,a{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}{x}^{4}}{4}}-{\frac{a{b}^{2}{\rm arcsec} \left (cx\right ){x}^{3}}{2\,c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{a{x}^{2}{b}^{2}}{4\,{c}^{2}}}-{\frac{a{b}^{2}{\rm arcsec} \left (cx\right )x}{{c}^{3}}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{a{b}^{2}}{{c}^{4}}\ln \left ({\frac{1}{cx}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*a^3+1/4*b^3*arcsec(c*x)^3*x^4-1/4/c*b^3*((c^2*x^2-1)/c^2/x^2)^(1/2)*arcsec(c*x)^2*x^3-1/2/c^3*b^3*arcs
ec(c*x)^2*((c^2*x^2-1)/c^2/x^2)^(1/2)*x+1/2*I/c^4*b^3*arcsec(c*x)^2+1/4/c^2*b^3*arcsec(c*x)*x^2-1/4/c^3*b^3*((
c^2*x^2-1)/c^2/x^2)^(1/2)*x+1/2*I*b^3*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)/c^4-1/c^4*b^3*arcsec(c*x)*ln
(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)-1/4*I/c^4*b^3+3/4*a^2*b*x^4*arcsec(c*x)-1/4/c*a^2*b/((c^2*x^2-1)/c^2/x^2)^
(1/2)*x^3-1/4/c^3*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x+1/2/c^5*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x+3/4*a*b^2*ar
csec(c*x)^2*x^4-1/2/c*a*b^2*((c^2*x^2-1)/c^2/x^2)^(1/2)*arcsec(c*x)*x^3+1/4/c^2*x^2*a*b^2-1/c^3*a*b^2*((c^2*x^
2-1)/c^2/x^2)^(1/2)*arcsec(c*x)*x-1/c^4*a*b^2*ln(1/c/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{4} \, a b^{2} x^{4} \operatorname{arcsec}\left (c x\right )^{2} + \frac{1}{4} \, a^{3} x^{4} + \frac{1}{4} \,{\left (3 \, x^{4} \operatorname{arcsec}\left (c x\right ) - \frac{c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 3 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} a^{2} b + \frac{1}{16} \,{\left (4 \, x^{4} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{3} - 3 \, x^{4} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (c^{2} x^{2}\right )^{2} - 16 \, \int \frac{3 \,{\left ({\left (4 \, x^{3} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} - x^{3} \log \left (c^{2} x^{2}\right )^{2}\right )} \sqrt{c x + 1} \sqrt{c x - 1} + 4 \,{\left (4 \, c^{2} x^{5} \log \left (c\right )^{2} - 4 \, x^{3} \log \left (c\right )^{2} + 4 \,{\left (c^{2} x^{5} - x^{3}\right )} \log \left (x\right )^{2} -{\left ({\left (4 \, c^{2} \log \left (c\right ) + c^{2}\right )} x^{5} - x^{3}{\left (4 \, \log \left (c\right ) + 1\right )} + 4 \,{\left (c^{2} x^{5} - x^{3}\right )} \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right ) + 8 \,{\left (c^{2} x^{5} \log \left (c\right ) - x^{3} \log \left (c\right )\right )} \log \left (x\right )\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )}}{16 \,{\left (c^{2} x^{2} - 1\right )}}\,{d x}\right )} b^{3} + \frac{{\left ({\left (c^{2} x^{2} + 2 \, \log \left (x^{2}\right )\right )} \sqrt{c x + 1} \sqrt{c x - 1} - 2 \,{\left (c^{4} x^{4} + c^{2} x^{2} - 2\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )} a b^{2}}{4 \, \sqrt{c x + 1} \sqrt{c x - 1} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*a*b^2*x^4*arcsec(c*x)^2 + 1/4*a^3*x^4 + 1/4*(3*x^4*arcsec(c*x) - (c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*s
qrt(-1/(c^2*x^2) + 1))/c^3)*a^2*b + 1/16*(4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3*x^4*arctan(sqrt(c*x
+ 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 - 16*integrate(3/16*((4*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 - x^3*log
(c^2*x^2)^2)*sqrt(c*x + 1)*sqrt(c*x - 1) + 4*(4*c^2*x^5*log(c)^2 - 4*x^3*log(c)^2 + 4*(c^2*x^5 - x^3)*log(x)^2
- ((4*c^2*log(c) + c^2)*x^5 - x^3*(4*log(c) + 1) + 4*(c^2*x^5 - x^3)*log(x))*log(c^2*x^2) + 8*(c^2*x^5*log(c)
- x^3*log(c))*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))/(c^2*x^2 - 1), x))*b^3 + 1/4*((c^2*x^2 + 2*log(x^2
))*sqrt(c*x + 1)*sqrt(c*x - 1) - 2*(c^4*x^4 + c^2*x^2 - 2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*a*b^2/(sqrt(c*
x + 1)*sqrt(c*x - 1)*c^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral(x**3*(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^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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