\(\int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx\) [115]

Optimal result

Integrand size = 19, antiderivative size = 26 \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=-\sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (x)}\right ) \]

[Out]

-FresnelC(2^(1/2)/Pi^(1/2)*arccos(x)^(1/2))*2^(1/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4810, 3385, 3433} \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=-\sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (x)}\right ) \]

[In]

Int[x/(Sqrt[1 - x^2]*Sqrt[ArcCos[x]]),x]

[Out]

-(Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[x]]])

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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]

Rule 3433

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]

Rule 4810

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(-(b*c^
(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGt
Q[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\arccos (x)\right ) \\ & = -\left (2 \text {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\arccos (x)}\right )\right ) \\ & = -\sqrt {2 \pi } \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (x)}\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\frac {i \left (\sqrt {-i \arccos (x)} \Gamma \left (\frac {1}{2},-i \arccos (x)\right )-\sqrt {i \arccos (x)} \Gamma \left (\frac {1}{2},i \arccos (x)\right )\right )}{2 \sqrt {\arccos (x)}} \]

[In]

Integrate[x/(Sqrt[1 - x^2]*Sqrt[ArcCos[x]]),x]

[Out]

((I/2)*(Sqrt[(-I)*ArcCos[x]]*Gamma[1/2, (-I)*ArcCos[x]] - Sqrt[I*ArcCos[x]]*Gamma[1/2, I*ArcCos[x]]))/Sqrt[Arc
Cos[x]]

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
default	\(-\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (x \right )}}{\sqrt {\pi }}\right ) \sqrt {2}\, \sqrt {\pi }\)	\(21\)

[In]

int(x/(-x^2+1)^(1/2)/arccos(x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-FresnelC(2^(1/2)/Pi^(1/2)*arccos(x)^(1/2))*2^(1/2)*Pi^(1/2)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x/(-x^2+1)^(1/2)/arccos(x)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\int \frac {x}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \sqrt {\operatorname {acos}{\left (x \right )}}}\, dx \]

[In]

integrate(x/(-x**2+1)**(1/2)/acos(x)**(1/2),x)

[Out]

Integral(x/(sqrt(-(x - 1)*(x + 1))*sqrt(acos(x))), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x/(-x^2+1)^(1/2)/arccos(x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42 \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (x\right )}\right ) - \left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (x\right )}\right ) \]

[In]

integrate(x/(-x^2+1)^(1/2)/arccos(x)^(1/2),x, algorithm="giac")

[Out]

(1/4*I + 1/4)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(arccos(x))) - (1/4*I - 1/4)*sqrt(2)*sqrt(pi)*erf
(-(1/2*I + 1/2)*sqrt(2)*sqrt(arccos(x)))

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {1-x^2} \sqrt {\arccos (x)}} \, dx=\int \frac {x}{\sqrt {\mathrm {acos}\left (x\right )}\,\sqrt {1-x^2}} \,d x \]

[In]

int(x/(acos(x)^(1/2)*(1 - x^2)^(1/2)),x)

[Out]

int(x/(acos(x)^(1/2)*(1 - x^2)^(1/2)), x)