3.1.77 \(\int x \sqrt {\text {ArcCos}(a x)} \, dx\) [77]

Optimal. Leaf size=59 \[ -\frac {\sqrt {\text {ArcCos}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\text {ArcCos}(a x)}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{8 a^2} \]

[Out]

-1/8*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2-1/4*arccos(a*x)^(1/2)/a^2+1/2*x^2*arccos(a*x)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4726, 4810, 3393, 3385, 3433} \begin {gather*} -\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{8 a^2}-\frac {\sqrt {\text {ArcCos}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\text {ArcCos}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[ArcCos[a*x]],x]

[Out]

-1/4*Sqrt[ArcCos[a*x]]/a^2 + (x^2*Sqrt[ArcCos[a*x]])/2 - (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]])/(
8*a^2)

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 4726

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCos[c*x])^n/(m
+ 1)), x] + Dist[b*c*(n/(m + 1)), Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; Fre
eQ[{a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]

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*} \int x \sqrt {\cos ^{-1}(a x)} \, dx &=\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}+\frac {1}{4} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \sqrt {\cos ^{-1}(a x)}} \, dx\\ &=\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{4 a^2}\\ &=\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{4 a^2}\\ &=-\frac {\sqrt {\cos ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^2}\\ &=-\frac {\sqrt {\cos ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{4 a^2}\\ &=-\frac {\sqrt {\cos ^{-1}(a x)}}{4 a^2}+\frac {1}{2} x^2 \sqrt {\cos ^{-1}(a x)}-\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{8 a^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 49, normalized size = 0.83 \begin {gather*} \frac {\frac {1}{4} \sqrt {\text {ArcCos}(a x)} \cos (2 \text {ArcCos}(a x))-\frac {1}{8} \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[ArcCos[a*x]],x]

[Out]

((Sqrt[ArcCos[a*x]]*Cos[2*ArcCos[a*x]])/4 - (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]])/8)/a^2

________________________________________________________________________________________

Maple [A]
time = 0.15, size = 42, normalized size = 0.71

method	result	size
default	\(-\frac {-2 \cos \left (2 \arccos \left (a x \right )\right ) \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }+\pi \FresnelC \left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )}{8 a^{2} \sqrt {\pi }}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccos(a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/a^2/Pi^(1/2)*(-2*cos(2*arccos(a*x))*arccos(a*x)^(1/2)*Pi^(1/2)+Pi*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2)))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {\operatorname {acos}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acos(a*x)**(1/2),x)

[Out]

Integral(x*sqrt(acos(a*x)), x)

________________________________________________________________________________________

Giac [C] Result contains complex when optimal does not.
time = 0.46, size = 71, normalized size = 1.20 \begin {gather*} \frac {\left (i + 1\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{32 \, a^{2}} - \frac {\left (i - 1\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{32 \, a^{2}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{8 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccos(a*x)^(1/2),x, algorithm="giac")

[Out]

(1/32*I + 1/32)*sqrt(pi)*erf((I - 1)*sqrt(arccos(a*x)))/a^2 - (1/32*I - 1/32)*sqrt(pi)*erf(-(I + 1)*sqrt(arcco
s(a*x)))/a^2 + 1/8*sqrt(arccos(a*x))*e^(2*I*arccos(a*x))/a^2 + 1/8*sqrt(arccos(a*x))*e^(-2*I*arccos(a*x))/a^2

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,\sqrt {\mathrm {acos}\left (a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*acos(a*x)^(1/2), x)

________________________________________________________________________________________