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

Optimal. Leaf size=99 \[ \frac{\sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{2 \sqrt{b} c^2}-\frac{\sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{2 \sqrt{b} c^2} \]

[Out]

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

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Rubi [A]  time = 0.17077, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {4636, 4406, 12, 3306, 3305, 3351, 3304, 3352} \[ \frac{\sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{2 \sqrt{b} c^2}-\frac{\sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{2 \sqrt{b} c^2} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*ArcCos[c*x]],x]

[Out]

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

Rule 4636

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Cos[x]^m*Sin[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.159688, size = 91, normalized size = 0.92 \[ -\frac{\sqrt{\pi } \sqrt{\frac{1}{b}} \left (\cos \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )-\sin \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 80, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{\pi }}{2\,{c}^{2}}\sqrt{{b}^{-1}} \left ( \cos \left ( 2\,{\frac{a}{b}} \right ){\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 ){\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{b \arccos \left (c x\right ) + a}}\,{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(x/sqrt(b*arccos(c*x) + a), 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 \frac{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 = 1.82085, size = 188, normalized size = 1.9 \begin{align*} \frac{\sqrt{\pi } 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 } 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 )}} \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]

1/4*sqrt(pi)*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) + 1/4*sqrt(pi)*i*erf(-sqrt(b*arccos(c*x) + a)*sqrt(b)*i/abs(b) - sqrt(b*arcco
s(c*x) + a)/sqrt(b))*e^(2*a*i/b)/(sqrt(b)*c^2*(b*i/abs(b) + 1))