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

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

[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)

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Rubi [A]  time = 0.39969, antiderivative size = 137, normalized size of antiderivative = 1., 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 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])

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 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 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 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 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 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]

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 = 0.200957, size = 123, normalized size = 0.9 \[ -\frac{\sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )+\sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )-2 \sqrt{\frac{1}{b}} \cos \left (2 \cos ^{-1}(c x)\right ) \sqrt{a+b \cos ^{-1}(c x)}}{8 \sqrt{\frac{1}{b}} c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-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])/(8*Sqrt[b^(-1)]*c^2)

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Maple [A]  time = 0.102, size = 173, normalized size = 1.3 \begin{align*}{\frac{1}{8\,{c}^{2}} \left ( -{\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}\sqrt{a+b\arccos \left ( cx \right ) }b-{\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) \sqrt{\pi }\sqrt{{b}^{-1}}\sqrt{a+b\arccos \left ( cx \right ) }b+2\,\arccos \left ( cx \right ) \cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) b+2\,\cos \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) a \right ){\frac{1}{\sqrt{a+b\arccos \left ( cx \right ) }}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \arccos \left (c x\right ) + a} x\,{d x} \end{align*}

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)

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

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: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{a + b \operatorname{acos}{\left (c x \right )}}\, dx \end{align*}

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)

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Giac [A]  time = 1.57599, size = 243, normalized size = 1.77 \begin{align*} \frac{\sqrt{\pi } \sqrt{b} \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 \, c^{2}{\left (\frac{b i}{{\left | b \right |}} + 1\right )}} - \frac{\sqrt{\pi } \sqrt{b} \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 \, c^{2}{\left (\frac{b i}{{\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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