Integrand size = 12, antiderivative size = 64 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c^3} \]
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
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.33 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {a x^3}{3}-\frac {b x^2 \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}}{6 c}+\frac {1}{3} b x^3 \sec ^{-1}(c x)-\frac {b \log \left (x \left (1+\sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}\right )\right )}{6 c^3} \]
(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)
Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5743, 798, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx\) |
\(\Big \downarrow \) 5743 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac {b \int \frac {x}{\sqrt {1-\frac {1}{c^2 x^2}}}dx}{3 c}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {b \int \frac {x^4}{\sqrt {1-\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {b \left (\frac {\int \frac {x^2}{\sqrt {1-\frac {1}{c^2 x^2}}}d\frac {1}{x^2}}{2 c^2}-x^2 \sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {b \left (x^2 \left (-\sqrt {1-\frac {1}{c^2 x^2}}\right )-\int \frac {1}{c^2-\frac {c^2}{x^4}}d\sqrt {1-\frac {1}{c^2 x^2}}\right )}{6 c}+\frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \sec ^{-1}(c x)\right )+\frac {b \left (x^2 \left (-\sqrt {1-\frac {1}{c^2 x^2}}\right )-\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{c^2 x^2}}\right )}{c^2}\right )}{6 c}\) |
(x^3*(a + b*ArcSec[c*x]))/3 + (b*(-(Sqrt[1 - 1/(c^2*x^2)]*x^2) - ArcTanh[S qrt[1 - 1/(c^2*x^2)]]/c^2))/(6*c)
3.1.5.3.1 Defintions of rubi rules used
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(d*x)^(m + 1)*((a + b*ArcSec[c*x])/(d*(m + 1))), x] - Simp[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]
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 | \(\frac {\frac {a \,c^{3} x^{3}}{3}+b \left (\frac {c^{3} x^{3} \operatorname {arcsec}\left (c x \right )}{3}-\frac {\sqrt {c^{2} x^{2}-1}\, \left (c x \sqrt {c^{2} x^{2}-1}+\ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )\right )}{6 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{3}}\) | \(98\) |
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)
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.47 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {2 \, a c^{3} x^{3} + 4 \, b c^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} b c x + 2 \, {\left (b c^{3} x^{3} - b c^{3}\right )} \operatorname {arcsec}\left (c x\right ) + b \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, c^{3}} \]
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
Time = 1.80 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.67 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {asec}{\left (c x \right )}}{3} - \frac {b \left (\begin {cases} \frac {x \sqrt {c^{2} x^{2} - 1}}{2 c} + \frac {\operatorname {acosh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\- \frac {i c x^{3}}{2 \sqrt {- c^{2} x^{2} + 1}} + \frac {i x}{2 c \sqrt {- c^{2} x^{2} + 1}} - \frac {i \operatorname {asin}{\left (c x \right )}}{2 c^{2}} & \text {otherwise} \end {cases}\right )}{3 c} \]
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)
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.53 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsec}\left (c x\right ) - \frac {\frac {2 \, \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac {\log \left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b \]
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
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 \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Too large to display} \]
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)*l og(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/(...
Timed out. \[ \int x^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x^2\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]