Integrand size = 10, antiderivative size = 89 \[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}+\frac {8 x^2}{3 \sqrt {\arccos (a x)}}+\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{3 a^2} \]
8/3*FresnelS(2*arccos(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2+2/3*x*(-a^2*x^2+1) ^(1/2)/a/arccos(a*x)^(3/2)-4/3/a^2/arccos(a*x)^(1/2)+8/3*x^2/arccos(a*x)^( 1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.69 \[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\frac {8 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )+\frac {4 \arccos (a x) \cos (2 \arccos (a x))+\sin (2 \arccos (a x))}{\arccos (a x)^{3/2}}}{3 a^2} \]
(8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]] + (4*ArcCos[a*x]*Cos[ 2*ArcCos[a*x]] + Sin[2*ArcCos[a*x]])/ArcCos[a*x]^(3/2))/(3*a^2)
Time = 0.69 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {5145, 5153, 5223, 5147, 4906, 27, 3042, 3786, 3832}
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.
\(\displaystyle \int \frac {x}{\arccos (a x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 5145 |
\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}dx}{3 a}+\frac {4}{3} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 5153 |
\(\displaystyle \frac {4}{3} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{3/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 5223 |
\(\displaystyle \frac {4}{3} a \left (\frac {2 x^2}{a \sqrt {\arccos (a x)}}-\frac {4 \int \frac {x}{\sqrt {\arccos (a x)}}dx}{a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 5147 |
\(\displaystyle \frac {4}{3} a \left (\frac {4 \int \frac {a x \sqrt {1-a^2 x^2}}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \frac {4}{3} a \left (\frac {4 \int \frac {\sin (2 \arccos (a x))}{2 \sqrt {\arccos (a x)}}d\arccos (a x)}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4}{3} a \left (\frac {2 \int \frac {\sin (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{3} a \left (\frac {2 \int \frac {\sin (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle \frac {4}{3} a \left (\frac {4 \int \sin (2 \arccos (a x))d\sqrt {\arccos (a x)}}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle \frac {4}{3} a \left (\frac {2 \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{a^3}+\frac {2 x^2}{a \sqrt {\arccos (a x)}}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}-\frac {4}{3 a^2 \sqrt {\arccos (a x)}}\) |
(2*x*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^(3/2)) - 4/(3*a^2*Sqrt[ArcCos[a*x ]]) + (4*a*((2*x^2)/(a*Sqrt[ArcCos[a*x]]) + (2*Sqrt[Pi]*FresnelS[(2*Sqrt[A rcCos[a*x]])/Sqrt[Pi]])/a^3))/3
3.2.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f }, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[( -x^m)*Sqrt[1 - c^2*x^2]*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] + ( -Simp[c*((m + 1)/(b*(n + 1))) Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n + 1)/ Sqrt[1 - c^2*x^2]), x], x] + Simp[m/(b*c*(n + 1)) Int[x^(m - 1)*((a + b*A rcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && I GtQ[m, 0] && LtQ[n, -2]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[- (b*c^(m + 1))^(-1) Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b], x], x , a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]
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_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-(f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c ^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] + Simp[f*(m/(b*c*( n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] Int[(f*x)^(m - 1)*(a + b *ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2 *d + e, 0] && LtQ[n, -1]
Time = 0.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63
method | result | size |
default | \(\frac {8 \sqrt {\pi }\, \operatorname {FresnelS}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {3}{2}}+4 \arccos \left (a x \right ) \cos \left (2 \arccos \left (a x \right )\right )+\sin \left (2 \arccos \left (a x \right )\right )}{3 a^{2} \arccos \left (a x \right )^{\frac {3}{2}}}\) | \(56\) |
1/3/a^2*(8*Pi^(1/2)*FresnelS(2*arccos(a*x)^(1/2)/Pi^(1/2))*arccos(a*x)^(3/ 2)+4*arccos(a*x)*cos(2*arccos(a*x))+sin(2*arccos(a*x)))/arccos(a*x)^(3/2)
Exception generated. \[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\int \frac {x}{\operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]
Exception generated. \[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\int { \frac {x}{\arccos \left (a x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x}{\arccos (a x)^{5/2}} \, dx=\int \frac {x}{{\mathrm {acos}\left (a\,x\right )}^{5/2}} \,d x \]