Integrand size = 10, antiderivative size = 119 \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}+\frac {8 x^2}{15 \arccos (a x)^{3/2}}-\frac {32 x \sqrt {1-a^2 x^2}}{15 a \sqrt {\arccos (a x)}}+\frac {32 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{15 a^2} \]
-4/15/a^2/arccos(a*x)^(3/2)+8/15*x^2/arccos(a*x)^(3/2)+32/15*FresnelC(2*ar ccos(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2+2/5*x*(-a^2*x^2+1)^(1/2)/a/arccos(a *x)^(5/2)-32/15*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(1/2)
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\frac {\frac {4 \cos (2 \arccos (a x))}{\arccos (a x)^{3/2}}+32 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )-\frac {\left (-3+16 \arccos (a x)^2\right ) \sin (2 \arccos (a x))}{\arccos (a x)^{5/2}}}{15 a^2} \]
((4*Cos[2*ArcCos[a*x]])/ArcCos[a*x]^(3/2) + 32*Sqrt[Pi]*FresnelC[(2*Sqrt[A rcCos[a*x]])/Sqrt[Pi]] - ((-3 + 16*ArcCos[a*x]^2)*Sin[2*ArcCos[a*x]])/ArcC os[a*x]^(5/2))/(15*a^2)
Time = 0.67 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5145, 5153, 5223, 5143, 25, 3042, 3785, 3833}
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.
\(\displaystyle \int \frac {x}{\arccos (a x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 5145 |
\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx}{5 a}+\frac {4}{5} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}\) |
\(\Big \downarrow \) 5153 |
\(\displaystyle \frac {4}{5} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 5223 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \int \frac {x}{\arccos (a x)^{3/2}}dx}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 5143 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 \int -\frac {\cos (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}+\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \int \frac {\cos (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \int \frac {\sin \left (2 \arccos (a x)+\frac {\pi }{2}\right )}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 3785 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {4 \int \cos (2 \arccos (a x))d\sqrt {\arccos (a x)}}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\) |
(2*x*Sqrt[1 - a^2*x^2])/(5*a*ArcCos[a*x]^(5/2)) - 4/(15*a^2*ArcCos[a*x]^(3 /2)) + (4*a*((2*x^2)/(3*a*ArcCos[a*x]^(3/2)) - (4*((2*x*Sqrt[1 - a^2*x^2]) /(a*Sqrt[ArcCos[a*x]]) - (2*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[P i]])/a^2))/(3*a)))/5
3.2.16.3.1 Defintions of rubi rules used
Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S imp[2/d Subst[Int[Cos[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[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
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] - S imp[1/(b^2*c^(m + 1)*(n + 1)) Subst[Int[ExpandTrigReduce[x^(n + 1), Cos[- a/b + x/b]^(m - 1)*(m - (m + 1)*Cos[-a/b + x/b]^2), x], x], x, a + b*ArcCos [c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
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_.)/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 = 0.83 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61
method | result | size |
default | \(-\frac {-32 \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {5}{2}}+16 \sin \left (2 \arccos \left (a x \right )\right ) \arccos \left (a x \right )^{2}-4 \arccos \left (a x \right ) \cos \left (2 \arccos \left (a x \right )\right )-3 \sin \left (2 \arccos \left (a x \right )\right )}{15 a^{2} \arccos \left (a x \right )^{\frac {5}{2}}}\) | \(73\) |
-1/15/a^2*(-32*Pi^(1/2)*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))*arccos(a*x) ^(5/2)+16*sin(2*arccos(a*x))*arccos(a*x)^2-4*arccos(a*x)*cos(2*arccos(a*x) )-3*sin(2*arccos(a*x)))/arccos(a*x)^(5/2)
Exception generated. \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\int \frac {x}{\operatorname {acos}^{\frac {7}{2}}{\left (a x \right )}}\, dx \]
Exception generated. \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\int { \frac {x}{\arccos \left (a x\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\int \frac {x}{{\mathrm {acos}\left (a\,x\right )}^{7/2}} \,d x \]