Integrand size = 10, antiderivative size = 90 \[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\frac {2 \sqrt {1-(a+b x)^2}}{3 b \arccos (a+b x)^{3/2}}+\frac {4 (a+b x)}{3 b \sqrt {\arccos (a+b x)}}+\frac {4 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a+b x)}\right )}{3 b} \] Output:
2/3*(1-(b*x+a)^2)^(1/2)/b/arccos(b*x+a)^(3/2)+4/3*(b*x+a)/b/arccos(b*x+a)^ (1/2)+4/3*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2))/ b
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.54 \[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=-\frac {2 \left (-\sqrt {1-(a+b x)^2}-e^{-i \arccos (a+b x)} \arccos (a+b x)-e^{i \arccos (a+b x)} \arccos (a+b x)+i (-i \arccos (a+b x))^{3/2} \Gamma \left (\frac {1}{2},-i \arccos (a+b x)\right )-i (i \arccos (a+b x))^{3/2} \Gamma \left (\frac {1}{2},i \arccos (a+b x)\right )\right )}{3 b \arccos (a+b x)^{3/2}} \] Input:
Integrate[ArcCos[a + b*x]^(-5/2),x]
Output:
(-2*(-Sqrt[1 - (a + b*x)^2] - ArcCos[a + b*x]/E^(I*ArcCos[a + b*x]) - E^(I *ArcCos[a + b*x])*ArcCos[a + b*x] + I*((-I)*ArcCos[a + b*x])^(3/2)*Gamma[1 /2, (-I)*ArcCos[a + b*x]] - I*(I*ArcCos[a + b*x])^(3/2)*Gamma[1/2, I*ArcCo s[a + b*x]]))/(3*b*ArcCos[a + b*x]^(3/2))
Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5303, 5133, 5223, 5135, 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.
\(\displaystyle \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 5303 |
\(\displaystyle \frac {\int \frac {1}{\arccos (a+b x)^{5/2}}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5133 |
\(\displaystyle \frac {\frac {2}{3} \int \frac {a+b x}{\sqrt {1-(a+b x)^2} \arccos (a+b x)^{3/2}}d(a+b x)+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
\(\Big \downarrow \) 5223 |
\(\displaystyle \frac {\frac {2}{3} \left (\frac {2 (a+b x)}{\sqrt {\arccos (a+b x)}}-2 \int \frac {1}{\sqrt {\arccos (a+b x)}}d(a+b x)\right )+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
\(\Big \downarrow \) 5135 |
\(\displaystyle \frac {\frac {2}{3} \left (2 \int \frac {\sqrt {1-(a+b x)^2}}{\sqrt {\arccos (a+b x)}}d\arccos (a+b x)+\frac {2 (a+b x)}{\sqrt {\arccos (a+b x)}}\right )+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2}{3} \left (2 \int \frac {\sin (\arccos (a+b x))}{\sqrt {\arccos (a+b x)}}d\arccos (a+b x)+\frac {2 (a+b x)}{\sqrt {\arccos (a+b x)}}\right )+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
\(\Big \downarrow \) 3786 |
\(\displaystyle \frac {\frac {2}{3} \left (4 \int \sqrt {1-(a+b x)^2}d\sqrt {\arccos (a+b x)}+\frac {2 (a+b x)}{\sqrt {\arccos (a+b x)}}\right )+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle \frac {\frac {2}{3} \left (2 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a+b x)}\right )+\frac {2 (a+b x)}{\sqrt {\arccos (a+b x)}}\right )+\frac {2 \sqrt {1-(a+b x)^2}}{3 \arccos (a+b x)^{3/2}}}{b}\) |
Input:
Int[ArcCos[a + b*x]^(-5/2),x]
Output:
((2*Sqrt[1 - (a + b*x)^2])/(3*ArcCos[a + b*x]^(3/2)) + (2*((2*(a + b*x))/S qrt[ArcCos[a + b*x]] + 2*Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b* x]]]))/3)/b
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[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-Sqrt[1 - c ^2*x^2])*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp[c/(b*(n + 1 )) Int[x*((a + b*ArcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ [{a, b, c}, x] && LtQ[n, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[-(b*c)^(-1) Subst[Int[x^n*Sin[-a/b + x/b], x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x]
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]
Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d Subst[Int[(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]
Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.33
method | result | size |
default | \(\frac {\sqrt {2}\, \left (4 \pi \arccos \left (b x +a \right )^{2} \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )+2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, b x +2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, a +\sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\right )}{3 b \sqrt {\pi }\, \arccos \left (b x +a \right )^{2}}\) | \(120\) |
Input:
int(1/arccos(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/b*2^(1/2)/Pi^(1/2)*(4*Pi*arccos(b*x+a)^2*FresnelS(2^(1/2)/Pi^(1/2)*arc cos(b*x+a)^(1/2))+2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*b*x+2*arccos(b*x+ a)^(3/2)*2^(1/2)*Pi^(1/2)*a+2^(1/2)*Pi^(1/2)*arccos(b*x+a)^(1/2)*(-b^2*x^2 -2*a*b*x-a^2+1)^(1/2))/arccos(b*x+a)^2
Exception generated. \[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/arccos(b*x+a)^(5/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
\[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\int \frac {1}{\operatorname {acos}^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/acos(b*x+a)**(5/2),x)
Output:
Integral(acos(a + b*x)**(-5/2), x)
Exception generated. \[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/arccos(b*x+a)^(5/2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
\[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\int { \frac {1}{\arccos \left (b x + a\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/arccos(b*x+a)^(5/2),x, algorithm="giac")
Output:
integrate(arccos(b*x + a)^(-5/2), x)
Timed out. \[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^{5/2}} \,d x \] Input:
int(1/acos(a + b*x)^(5/2),x)
Output:
int(1/acos(a + b*x)^(5/2), x)
\[ \int \frac {1}{\arccos (a+b x)^{5/2}} \, dx=\frac {-\frac {2 \mathit {acos} \left (b x +a \right )^{2} \left (\int \frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \sqrt {\mathit {acos} \left (b x +a \right )}\, x}{\mathit {acos} \left (b x +a \right )^{2} a^{2}+2 \mathit {acos} \left (b x +a \right )^{2} a b x +\mathit {acos} \left (b x +a \right )^{2} b^{2} x^{2}-\mathit {acos} \left (b x +a \right )^{2}}d x \right ) b^{2}}{3}+\frac {4 \sqrt {\mathit {acos} \left (b x +a \right )}\, \mathit {acos} \left (b x +a \right ) a}{3}+\frac {2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \sqrt {\mathit {acos} \left (b x +a \right )}}{3}}{\mathit {acos} \left (b x +a \right )^{2} b} \] Input:
int(1/acos(b*x+a)^(5/2),x)
Output:
(2*( - acos(a + b*x)**2*int((sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*sqrt( acos(a + b*x))*x)/(acos(a + b*x)**2*a**2 + 2*acos(a + b*x)**2*a*b*x + acos (a + b*x)**2*b**2*x**2 - acos(a + b*x)**2),x)*b**2 + 2*sqrt(acos(a + b*x)) *acos(a + b*x)*a + sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*sqrt(acos(a + b *x))))/(3*acos(a + b*x)**2*b)