\(\int x^2 (a+b \sec ^{-1}(c x)) \, dx\) [5]

Optimal result

Integrand size = 12, antiderivative size = 64 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c^3} \]

[Out]

1/3*x^3*(a+b*arcsec(c*x))-1/6*b*arctanh((1-1/c^2/x^2)^(1/2))/c^3-1/6*b*x^2*(1-1/c^2/x^2)^(1/2)/c

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5328, 272, 44, 65, 214} \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c^3}-\frac {b x^2 \sqrt {1-\frac {1}{c^2 x^2}}}{6 c} \]

[In]

Int[x^2*(a + b*ArcSec[c*x]),x]

[Out]

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

Rule 44

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

Rule 65

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 214

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

Rule 272

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 5328

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSec[c*x]
)/(d*(m + 1))), x] - Dist[b*(d/(c*(m + 1))), Int[(d*x)^(m - 1)/Sqrt[1 - 1/(c^2*x^2)], x], x] /; FreeQ[{a, b, c
, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \int \frac {x}{\sqrt {1-\frac {1}{c^2 x^2}}} \, dx}{3 c} \\ & = \frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac {b \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{6 c} \\ & = -\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{12 c^3} \\ & = -\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \text {Subst}\left (\int \frac {1}{c^2-c^2 x^2} \, dx,x,\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c} \\ & = -\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.33 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {a x^3}{3}-\frac {b x^2 \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}}{6 c}+\frac {1}{3} b x^3 \sec ^{-1}(c x)-\frac {b \log \left (x \left (1+\sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}\right )\right )}{6 c^3} \]

[In]

Integrate[x^2*(a + b*ArcSec[c*x]),x]

[Out]

(a*x^3)/3 - (b*x^2*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])/(6*c) + (b*x^3*ArcSec[c*x])/3 - (b*Log[x*(1 + Sqrt[(-1 + c^
2*x^2)/(c^2*x^2)])])/(6*c^3)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.47

method	result	size
parts	\(\frac {x^{3} a}{3}+\frac {b \left (\frac {c^{3} x^{3} \operatorname {arcsec}\left (c x \right )}{3}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (c x \sqrt {c^{2} x^{2}-1}+\ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )\right )}{6 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{3}}\)	\(94\)
derivativedivides	\(\frac {\frac {a \,c^{3} x^{3}}{3}+b \left (\frac {c^{3} x^{3} \operatorname {arcsec}\left (c x \right )}{3}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (c x \sqrt {c^{2} x^{2}-1}+\ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )\right )}{6 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{3}}\)	\(98\)
default	\(\frac {\frac {a \,c^{3} x^{3}}{3}+b \left (\frac {c^{3} x^{3} \operatorname {arcsec}\left (c x \right )}{3}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (c x \sqrt {c^{2} x^{2}-1}+\ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )\right )}{6 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{3}}\)	\(98\)

[In]

int(x^2*(a+b*arcsec(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3*a+b/c^3*(1/3*c^3*x^3*arcsec(c*x)-1/6*(c^2*x^2-1)^(1/2)*(c*x*(c^2*x^2-1)^(1/2)+ln(c*x+(c^2*x^2-1)^(1/2)
))/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.47 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {2 \, a c^{3} x^{3} + 4 \, b c^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} b c x + 2 \, {\left (b c^{3} x^{3} - b c^{3}\right )} \operatorname {arcsec}\left (c x\right ) + b \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, c^{3}} \]

[In]

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

[Out]

1/6*(2*a*c^3*x^3 + 4*b*c^3*arctan(-c*x + sqrt(c^2*x^2 - 1)) - sqrt(c^2*x^2 - 1)*b*c*x + 2*(b*c^3*x^3 - b*c^3)*
arcsec(c*x) + b*log(-c*x + sqrt(c^2*x^2 - 1)))/c^3

Sympy [A] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.67 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {asec}{\left (c x \right )}}{3} - \frac {b \left (\begin {cases} \frac {x \sqrt {c^{2} x^{2} - 1}}{2 c} + \frac {\operatorname {acosh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- \frac {i c x^{3}}{2 \sqrt {- c^{2} x^{2} + 1}} + \frac {i x}{2 c \sqrt {- c^{2} x^{2} + 1}} - \frac {i \operatorname {asin}{\left (c x \right )}}{2 c^{2}} & \text {otherwise} \end {cases}\right )}{3 c} \]

[In]

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

[Out]

a*x**3/3 + b*x**3*asec(c*x)/3 - b*Piecewise((x*sqrt(c**2*x**2 - 1)/(2*c) + acosh(c*x)/(2*c**2), Abs(c**2*x**2)
> 1), (-I*c*x**3/(2*sqrt(-c**2*x**2 + 1)) + I*x/(2*c*sqrt(-c**2*x**2 + 1)) - I*asin(c*x)/(2*c**2), True))/(3*
c)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsec}\left (c x\right ) - \frac {\frac {2 \, \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b \]

[In]

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

[Out]

1/3*a*x^3 + 1/12*(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)*b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2101 vs. \(2 (54) = 108\).

Time = 0.59 (sec) , antiderivative size = 2101, normalized size of antiderivative = 32.83 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/6*c*(2*b*arccos(1/(c*x))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x)
+ 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) - b*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/(c^4 + 3
*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(
1/(c*x) + 1)^6) + b*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/(c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1
)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6) + 2*a/(c^4 + 3*c^4*
(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*
x) + 1)^6) - 6*b*(1/(c^2*x^2) - 1)*arccos(1/(c*x))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/
(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) - 3*b*(1/(c^2*x^2
) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*
(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) + 3*b*(1/(c^2*
x^2) - 1)*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c
^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^2) - 2*b*sqrt(
-1/(c^2*x^2) + 1)/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4
+ c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)) - 6*a*(1/(c^2*x^2) - 1)/((c^4 + 3*c^4*(1/(c^2*x^2) -
1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/
(c*x) + 1)^2) + 6*b*(1/(c^2*x^2) - 1)^2*arccos(1/(c*x))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^
4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) - 3*b*(1/(c^
2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 +
3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) + 3*b*(
1/(c^2*x^2) - 1)^2*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) - 1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1
)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) +
6*a*(1/(c^2*x^2) - 1)^2/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) +
1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^4) - 2*b*(1/(c^2*x^2) - 1)^3*arccos(1/(c*x))/((
c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) -
1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) - b*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-1/(c^2*x^2) + 1) + 1/(c*x) + 1))
/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2
) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) + b*(1/(c^2*x^2) - 1)^3*log(abs(sqrt(-1/(c^2*x^2) + 1) - 1/(c*x) -
1))/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*
x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6) + 2*b*(1/(c^2*x^2) - 1)^2*sqrt(-1/(c^2*x^2) + 1)/((c^4 + 3*c^4*(
1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x
) + 1)^6)*(1/(c*x) + 1)^5) - 2*a*(1/(c^2*x^2) - 1)^3/((c^4 + 3*c^4*(1/(c^2*x^2) - 1)/(1/(c*x) + 1)^2 + 3*c^4*(
1/(c^2*x^2) - 1)^2/(1/(c*x) + 1)^4 + c^4*(1/(c^2*x^2) - 1)^3/(1/(c*x) + 1)^6)*(1/(c*x) + 1)^6))

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

[In]

int(x^2*(a + b*acos(1/(c*x))),x)

[Out]

int(x^2*(a + b*acos(1/(c*x))), x)