Integrand size = 10, antiderivative size = 98 \[ \int x^3 \arccos (a x)^2 \, dx=-\frac {3 x^2}{32 a^2}-\frac {x^4}{32}-\frac {3 x \sqrt {1-a^2 x^2} \arccos (a x)}{16 a^3}-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{8 a}-\frac {3 \arccos (a x)^2}{32 a^4}+\frac {1}{4} x^4 \arccos (a x)^2 \]
-3/32*x^2/a^2-1/32*x^4-3/32*arccos(a*x)^2/a^4+1/4*x^4*arccos(a*x)^2-3/16*x *arccos(a*x)*(-a^2*x^2+1)^(1/2)/a^3-1/8*x^3*arccos(a*x)*(-a^2*x^2+1)^(1/2) /a
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int x^3 \arccos (a x)^2 \, dx=\frac {-a^2 x^2 \left (3+a^2 x^2\right )-2 a x \sqrt {1-a^2 x^2} \left (3+2 a^2 x^2\right ) \arccos (a x)+\left (-3+8 a^4 x^4\right ) \arccos (a x)^2}{32 a^4} \]
(-(a^2*x^2*(3 + a^2*x^2)) - 2*a*x*Sqrt[1 - a^2*x^2]*(3 + 2*a^2*x^2)*ArcCos [a*x] + (-3 + 8*a^4*x^4)*ArcCos[a*x]^2)/(32*a^4)
Time = 0.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5139, 5211, 15, 5211, 15, 5153}
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.
\(\displaystyle \int x^3 \arccos (a x)^2 \, dx\) |
\(\Big \downarrow \) 5139 |
\(\displaystyle \frac {1}{2} a \int \frac {x^4 \arccos (a x)}{\sqrt {1-a^2 x^2}}dx+\frac {1}{4} x^4 \arccos (a x)^2\) |
\(\Big \downarrow \) 5211 |
\(\displaystyle \frac {1}{2} a \left (\frac {3 \int \frac {x^2 \arccos (a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {\int x^3dx}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{4 a^2}\right )+\frac {1}{4} x^4 \arccos (a x)^2\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{2} a \left (\frac {3 \int \frac {x^2 \arccos (a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{4 a^2}-\frac {x^4}{16 a}\right )+\frac {1}{4} x^4 \arccos (a x)^2\) |
\(\Big \downarrow \) 5211 |
\(\displaystyle \frac {1}{2} a \left (\frac {3 \left (\frac {\int \frac {\arccos (a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {\int xdx}{2 a}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)}{2 a^2}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{4 a^2}-\frac {x^4}{16 a}\right )+\frac {1}{4} x^4 \arccos (a x)^2\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {1}{2} a \left (\frac {3 \left (\frac {\int \frac {\arccos (a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)}{2 a^2}-\frac {x^2}{4 a}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{4 a^2}-\frac {x^4}{16 a}\right )+\frac {1}{4} x^4 \arccos (a x)^2\) |
\(\Big \downarrow \) 5153 |
\(\displaystyle \frac {1}{2} a \left (-\frac {x^3 \sqrt {1-a^2 x^2} \arccos (a x)}{4 a^2}+\frac {3 \left (-\frac {\arccos (a x)^2}{4 a^3}-\frac {x \sqrt {1-a^2 x^2} \arccos (a x)}{2 a^2}-\frac {x^2}{4 a}\right )}{4 a^2}-\frac {x^4}{16 a}\right )+\frac {1}{4} x^4 \arccos (a x)^2\) |
(x^4*ArcCos[a*x]^2)/4 + (a*(-1/16*x^4/a - (x^3*Sqrt[1 - a^2*x^2]*ArcCos[a* x])/(4*a^2) + (3*(-1/4*x^2/a - (x*Sqrt[1 - a^2*x^2]*ArcCos[a*x])/(2*a^2) - ArcCos[a*x]^2/(4*a^3)))/(4*a^2)))/2
3.1.13.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcCos[c*x])^n/(d*(m + 1))), x] + Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(-(b*c*(n + 1))^(-1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2] ]*(a + b*ArcCos[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^ 2*d + e, 0] && NeQ[n, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcCos[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] - S imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f* x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m , 1] && NeQ[m + 2*p + 1, 0]
Time = 0.91 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(\frac {\frac {a^{4} x^{4} \arccos \left (a x \right )^{2}}{4}-\frac {\arccos \left (a x \right ) \left (2 a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}+3 a x \sqrt {-a^{2} x^{2}+1}+3 \arccos \left (a x \right )\right )}{16}+\frac {3 \arccos \left (a x \right )^{2}}{32}-\frac {\left (2 a^{2} x^{2}+3\right )^{2}}{128}}{a^{4}}\) | \(91\) |
default | \(\frac {\frac {a^{4} x^{4} \arccos \left (a x \right )^{2}}{4}-\frac {\arccos \left (a x \right ) \left (2 a^{3} x^{3} \sqrt {-a^{2} x^{2}+1}+3 a x \sqrt {-a^{2} x^{2}+1}+3 \arccos \left (a x \right )\right )}{16}+\frac {3 \arccos \left (a x \right )^{2}}{32}-\frac {\left (2 a^{2} x^{2}+3\right )^{2}}{128}}{a^{4}}\) | \(91\) |
1/a^4*(1/4*a^4*x^4*arccos(a*x)^2-1/16*arccos(a*x)*(2*a^3*x^3*(-a^2*x^2+1)^ (1/2)+3*a*x*(-a^2*x^2+1)^(1/2)+3*arccos(a*x))+3/32*arccos(a*x)^2-1/128*(2* a^2*x^2+3)^2)
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.71 \[ \int x^3 \arccos (a x)^2 \, dx=-\frac {a^{4} x^{4} + 3 \, a^{2} x^{2} - {\left (8 \, a^{4} x^{4} - 3\right )} \arccos \left (a x\right )^{2} + 2 \, {\left (2 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \arccos \left (a x\right )}{32 \, a^{4}} \]
-1/32*(a^4*x^4 + 3*a^2*x^2 - (8*a^4*x^4 - 3)*arccos(a*x)^2 + 2*(2*a^3*x^3 + 3*a*x)*sqrt(-a^2*x^2 + 1)*arccos(a*x))/a^4
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99 \[ \int x^3 \arccos (a x)^2 \, dx=\begin {cases} \frac {x^{4} \operatorname {acos}^{2}{\left (a x \right )}}{4} - \frac {x^{4}}{32} - \frac {x^{3} \sqrt {- a^{2} x^{2} + 1} \operatorname {acos}{\left (a x \right )}}{8 a} - \frac {3 x^{2}}{32 a^{2}} - \frac {3 x \sqrt {- a^{2} x^{2} + 1} \operatorname {acos}{\left (a x \right )}}{16 a^{3}} - \frac {3 \operatorname {acos}^{2}{\left (a x \right )}}{32 a^{4}} & \text {for}\: a \neq 0 \\\frac {\pi ^{2} x^{4}}{16} & \text {otherwise} \end {cases} \]
Piecewise((x**4*acos(a*x)**2/4 - x**4/32 - x**3*sqrt(-a**2*x**2 + 1)*acos( a*x)/(8*a) - 3*x**2/(32*a**2) - 3*x*sqrt(-a**2*x**2 + 1)*acos(a*x)/(16*a** 3) - 3*acos(a*x)**2/(32*a**4), Ne(a, 0)), (pi**2*x**4/16, True))
\[ \int x^3 \arccos (a x)^2 \, dx=\int { x^{3} \arccos \left (a x\right )^{2} \,d x } \]
1/4*x^4*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2 - a*integrate(1/2*sqr t(a*x + 1)*sqrt(-a*x + 1)*x^4*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)/( a^2*x^2 - 1), x)
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89 \[ \int x^3 \arccos (a x)^2 \, dx=\frac {1}{4} \, x^{4} \arccos \left (a x\right )^{2} - \frac {1}{32} \, x^{4} - \frac {\sqrt {-a^{2} x^{2} + 1} x^{3} \arccos \left (a x\right )}{8 \, a} - \frac {3 \, x^{2}}{32 \, a^{2}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1} x \arccos \left (a x\right )}{16 \, a^{3}} - \frac {3 \, \arccos \left (a x\right )^{2}}{32 \, a^{4}} + \frac {15}{256 \, a^{4}} \]
1/4*x^4*arccos(a*x)^2 - 1/32*x^4 - 1/8*sqrt(-a^2*x^2 + 1)*x^3*arccos(a*x)/ a - 3/32*x^2/a^2 - 3/16*sqrt(-a^2*x^2 + 1)*x*arccos(a*x)/a^3 - 3/32*arccos (a*x)^2/a^4 + 15/256/a^4
Timed out. \[ \int x^3 \arccos (a x)^2 \, dx=\int x^3\,{\mathrm {acos}\left (a\,x\right )}^2 \,d x \]