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

Optimal. Leaf size=63 \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{2 b c^2}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{2 b c^2} \]

[Out]

(CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2*a)/b])/(2*b*c^2) - (Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c
*x]))/b])/(2*b*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.120868, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4636, 4406, 12, 3303, 3299, 3302} \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcCos[c*x]),x]

[Out]

(CosIntegral[(2*a)/b + 2*ArcCos[c*x]]*Sin[(2*a)/b])/(2*b*c^2) - (Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcCos[c
*x]])/(2*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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{x}{a+b \cos ^{-1}(c x)} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{c^2}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 (a+b x)} \, dx,x,\cos ^{-1}(c x)\right )}{c^2}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sin (2 x)}{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 )}{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 )}{a+b x} \, dx,x,\cos ^{-1}(c x)\right )}{2 c^2}\\ &=\frac{\text{Ci}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right ) \sin \left (\frac{2 a}{b}\right )}{2 b c^2}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2}\\ \end{align*}

Mathematica [A]  time = 0.067942, size = 56, normalized size = 0.89 \[ -\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )-\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{2 b c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*ArcCos[c*x]),x]

[Out]

-(-(CosIntegral[(2*a)/b + 2*ArcCos[c*x]]*Sin[(2*a)/b]) + Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcCos[c*x]])/(2
*b*c^2)

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 58, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{1}{2\,b}{\it Si} \left ( 2\,\arccos \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) }+{\frac{1}{2\,b}{\it Ci} \left ( 2\,\arccos \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/2*Si(2*arccos(c*x)+2*a/b)*cos(2*a/b)/b+1/2*Ci(2*arccos(c*x)+2*a/b)*sin(2*a/b)/b)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{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)),x, algorithm="maxima")

[Out]

integrate(x/(b*arccos(c*x) + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{b \arccos \left (c x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arccos(c*x)),x, algorithm="fricas")

[Out]

integral(x/(b*arccos(c*x) + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{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)),x)

[Out]

Integral(x/(a + b*acos(c*x)), x)

________________________________________________________________________________________

Giac [A]  time = 1.17329, size = 116, normalized size = 1.84 \begin{align*} \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right ) \sin \left (\frac{a}{b}\right )}{b c^{2}} - \frac{\cos \left (\frac{a}{b}\right )^{2} \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac{\operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{2 \, b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(a/b)*cos_integral(2*a/b + 2*arccos(c*x))*sin(a/b)/(b*c^2) - cos(a/b)^2*sin_integral(2*a/b + 2*arccos(c*x))
/(b*c^2) + 1/2*sin_integral(2*a/b + 2*arccos(c*x))/(b*c^2)