3.2.74 \(\int x \sqrt {a+b \text {ArcCos}(c x)} \, dx\) [174]

Optimal. Leaf size=137 \[ -\frac {\sqrt {a+b \text {ArcCos}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \text {ArcCos}(c x)}-\frac {\sqrt {b} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {b} \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{8 c^2} \]

[Out]

-1/8*cos(2*a/b)*FresnelC(2*(a+b*arccos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*b^(1/2)*Pi^(1/2)/c^2-1/8*FresnelS(2*(a+b*
arccos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*b^(1/2)*Pi^(1/2)/c^2-1/4*(a+b*arccos(c*x))^(1/2)/c^2+1/2*x^2*(
a+b*arccos(c*x))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4726, 4810, 3393, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {\sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{8 c^2}-\frac {\sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {a+b \text {ArcCos}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \text {ArcCos}(c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*ArcCos[c*x]],x]

[Out]

-1/4*Sqrt[a + b*ArcCos[c*x]]/c^2 + (x^2*Sqrt[a + b*ArcCos[c*x]])/2 - (Sqrt[b]*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(
2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi])])/(8*c^2) - (Sqrt[b]*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]
])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(8*c^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 3386

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

Rule 3387

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[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 3432

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]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 123, normalized size = 0.90 \begin {gather*} -\frac {-2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)} \cos (2 \text {ArcCos}(c x))+\sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right )+\sqrt {\pi } S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{8 \sqrt {\frac {1}{b}} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a + b*ArcCos[c*x]],x]

[Out]

-1/8*(-2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]]*Cos[2*ArcCos[c*x]] + Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[b^(-
1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]] + Sqrt[Pi]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]]*
Sin[(2*a)/b])/(Sqrt[b^(-1)]*c^2)

________________________________________________________________________________________

Maple [A]
time = 0.31, size = 192, normalized size = 1.40

method	result	size
default	\(\frac {-\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b +\sqrt {2}\, \sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) b +4 \arccos \left (c x \right ) \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) b +4 \cos \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right ) a}{16 c^{2} \sqrt {a +b \arccos \left (c x \right )}}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/c^2/(a+b*arccos(c*x))^(1/2)*(-2^(1/2)*Pi^(1/2)*(-2/b)^(1/2)*(a+b*arccos(c*x))^(1/2)*cos(2*a/b)*FresnelC(2
*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b+2^(1/2)*Pi^(1/2)*(-2/b)^(1/2)*(a+b*arccos(c*x))^(1
/2)*sin(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arccos(c*x))^(1/2)/b)*b+4*arccos(c*x)*cos(-2*(a+b
*arccos(c*x))/b+2*a/b)*b+4*cos(-2*(a+b*arccos(c*x))/b+2*a/b)*a)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*arccos(c*x) + a)*x, x)

________________________________________________________________________________________

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*(a+b*arccos(c*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 {a + b \operatorname {acos}{\left (c x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acos(c*x))**(1/2),x)

[Out]

Integral(x*sqrt(a + b*acos(c*x)), x)

________________________________________________________________________________________

Giac [C] Result contains complex when optimal does not.
time = 0.74, size = 448, normalized size = 3.27 \begin {gather*} -\frac {i \, \sqrt {\pi } a \sqrt {b} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{4 \, {\left (b + \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{16 \, {\left (b + \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {i \, \sqrt {\pi } a \sqrt {b} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{4 \, {\left (b - \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} + \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{16 \, {\left (b - \frac {i \, b^{2}}{{\left | b \right |}}\right )} c^{2}} - \frac {i \, \sqrt {\pi } a \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{4 \, c^{2} {\left (\sqrt {b} - \frac {i \, b^{\frac {3}{2}}}{{\left | b \right |}}\right )}} + \frac {i \, \sqrt {\pi } a \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arccos \left (c x\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{4 \, \sqrt {b} c^{2} {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} + \frac {\sqrt {b \arccos \left (c x\right ) + a} e^{\left (2 i \, \arccos \left (c x\right )\right )}}{8 \, c^{2}} + \frac {\sqrt {b \arccos \left (c x\right ) + a} e^{\left (-2 i \, \arccos \left (c x\right )\right )}}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*sqrt(pi)*a*sqrt(b)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(
2*I*a/b)/((b + I*b^2/abs(b))*c^2) + 1/16*sqrt(pi)*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) - I*sqrt(b*arcc
os(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/((b + I*b^2/abs(b))*c^2) + 1/4*I*sqrt(pi)*a*sqrt(b)*erf(-sqrt(b*arcco
s(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/((b - I*b^2/abs(b))*c^2) + 1/16*s
qrt(pi)*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/
((b - I*b^2/abs(b))*c^2) - 1/4*I*sqrt(pi)*a*erf(-sqrt(b*arccos(c*x) + a)/sqrt(b) + I*sqrt(b*arccos(c*x) + a)*s
qrt(b)/abs(b))*e^(-2*I*a/b)/(c^2*(sqrt(b) - I*b^(3/2)/abs(b))) + 1/4*I*sqrt(pi)*a*erf(-sqrt(b*arccos(c*x) + a)
/sqrt(b) - I*sqrt(b*arccos(c*x) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/(sqrt(b)*c^2*(I*b/abs(b) + 1)) + 1/8*sqrt(b*a
rccos(c*x) + a)*e^(2*I*arccos(c*x))/c^2 + 1/8*sqrt(b*arccos(c*x) + a)*e^(-2*I*arccos(c*x))/c^2

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\sqrt {a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*acos(c*x))^(1/2),x)

[Out]

int(x*(a + b*acos(c*x))^(1/2), x)

________________________________________________________________________________________