3.174 \(\int x \sqrt {a+b \cos ^{-1}(c x)} \, dx\)

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

[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 = 0.40, 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 = {4630, 4724, 3312, 3306, 3305, 3351, 3304, 3352} \[ -\frac {\sqrt {\pi } \sqrt {b} \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi } \sqrt {b}}\right )}{8 c^2}-\frac {\sqrt {\pi } \sqrt {b} \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {b} \sqrt {\pi }}\right )}{8 c^2}-\frac {\sqrt {a+b \cos ^{-1}(c x)}}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \cos ^{-1}(c x)} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a + b*ArcCos[c*x]],x]

[Out]

-Sqrt[a + b*ArcCos[c*x]]/(4*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 3304

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 3305

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 3306

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 3312

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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 4630

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 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 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 \operatorname {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 \operatorname {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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.24, size = 123, normalized size = 0.90 \[ -\frac {\sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi }}\right )+\sqrt {\pi } \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \cos ^{-1}(c x)}}{\sqrt {\pi }}\right )-2 \sqrt {\frac {1}{b}} \cos \left (2 \cos ^{-1}(c x)\right ) \sqrt {a+b \cos ^{-1}(c x)}}{8 \sqrt {\frac {1}{b}} c^2} \]

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)

________________________________________________________________________________________

fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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)

________________________________________________________________________________________

giac [B]  time = 1.15, size = 469, normalized size = 3.42 \[ -\frac {\sqrt {\pi } a \sqrt {b} i \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (\frac {2 \, a i}{b}\right )}}{4 \, {\left (\frac {b^{2} i}{{\left | b \right |}} + b\right )} c^{2}} - \frac {\sqrt {\pi } a \sqrt {b} i \operatorname {erf}\left (\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (-\frac {2 \, a i}{b}\right )}}{4 \, {\left (\frac {b^{2} i}{{\left | b \right |}} - b\right )} c^{2}} + \frac {\sqrt {\pi } a i \operatorname {erf}\left (\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (-\frac {2 \, a i}{b}\right )}}{4 \, {\left (\frac {b^{\frac {3}{2}} i}{{\left | b \right |}} - \sqrt {b}\right )} c^{2}} + \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (\frac {2 \, a i}{b}\right )}}{16 \, {\left (\frac {b^{2} i}{{\left | b \right |}} + b\right )} c^{2}} + \frac {\sqrt {\pi } a i \operatorname {erf}\left (-\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (\frac {2 \, a i}{b}\right )}}{4 \, \sqrt {b} c^{2} {\left (\frac {b i}{{\left | b \right |}} + 1\right )}} - \frac {\sqrt {\pi } b^{\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {b \arccos \left (c x\right ) + a} \sqrt {b} i}{{\left | b \right |}} - \frac {\sqrt {b \arccos \left (c x\right ) + a}}{\sqrt {b}}\right ) e^{\left (-\frac {2 \, a i}{b}\right )}}{16 \, {\left (\frac {b^{2} i}{{\left | b \right |}} - b\right )} c^{2}} + \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}} \]

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*sqrt(pi)*a*sqrt(b)*i*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(
2*a*i/b)/((b^2*i/abs(b) + b)*c^2) - 1/4*sqrt(pi)*a*sqrt(b)*i*erf(sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sq
rt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^2*i/abs(b) - b)*c^2) + 1/4*sqrt(pi)*a*i*erf(sqrt(b*arccos(c*x)
+ a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^(3/2)*i/abs(b) - sqrt(b))*c^2) + 1/
16*sqrt(pi)*b^(3/2)*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(2*a*i/
b)/((b^2*i/abs(b) + b)*c^2) + 1/4*sqrt(pi)*a*i*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c
*x) + a)/sqrt(b))*e^(2*a*i/b)/(sqrt(b)*c^2*(b*i/abs(b) + 1)) - 1/16*sqrt(pi)*b^(3/2)*erf(sqrt(b*arccos(c*x) +
a)*sqrt(b)*i/abs(b) - sqrt(b*arccos(c*x) + a)/sqrt(b))*e^(-2*a*i/b)/((b^2*i/abs(b) - b)*c^2) + 1/8*sqrt(b*arcc
os(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

________________________________________________________________________________________

maple [A]  time = 0.27, size = 173, normalized size = 1.26 \[ \frac {-\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b -\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b +2 \arccos \left (c x \right ) \cos \left (\frac {2 a +2 b \arccos \left (c x \right )}{b}-\frac {2 a}{b}\right ) b +2 \cos \left (\frac {2 a +2 b \arccos \left (c x \right )}{b}-\frac {2 a}{b}\right ) a}{8 c^{2} \sqrt {a +b \arccos \left (c x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \arccos \left (c x\right ) + a} x\,{d x} \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sqrt {a + b \operatorname {acos}{\left (c x \right )}}\, dx \]

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)

________________________________________________________________________________________