3.164 \(\int \frac{x}{(a+b \cos ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=91 \[ -\frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}-\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \cos ^{-1}(c x)\right )}{b}\right )}{b^2 c^2}+\frac{x \sqrt{1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )} \]

[Out]

(x*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcCos[c*x])) - (Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcCos[c*x]))/b])/(b^2*c
^2) - (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcCos[c*x]))/b])/(b^2*c^2)

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Rubi [A]  time = 0.0968336, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4632, 3303, 3299, 3302} \[ -\frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{b^2 c^2}-\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \cos ^{-1}(c x)\right )}{b^2 c^2}+\frac{x \sqrt{1-c^2 x^2}}{b c \left (a+b \cos ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcCos[c*x])^2,x]

[Out]

(x*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcCos[c*x])) - (Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcCos[c*x]])/(b^2*c^2
) - (Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcCos[c*x]])/(b^2*c^2)

Rule 4632

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1
), Cos[x]^(m - 1)*(m - (m + 1)*Cos[x]^2), x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] &&
GeQ[n, -2] && LtQ[n, -1]

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

Mathematica [A]  time = 0.257513, size = 80, normalized size = 0.88 \[ \frac{\frac{b c x \sqrt{1-c^2 x^2}}{a+b \cos ^{-1}(c x)}-\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )-\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\cos ^{-1}(c x)\right )\right )}{b^2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*ArcCos[c*x])^2,x]

[Out]

((b*c*x*Sqrt[1 - c^2*x^2])/(a + b*ArcCos[c*x]) - Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcCos[c*x])] - Sin[(2*a)/b
]*SinIntegral[2*(a/b + ArcCos[c*x])])/(b^2*c^2)

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Maple [A]  time = 0.052, size = 78, normalized size = 0.9 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{\sin \left ( 2\,\arccos \left ( cx \right ) \right ) }{ \left ( 2\,a+2\,b\arccos \left ( cx \right ) \right ) b}}-{\frac{1}{{b}^{2}} \left ({\it Si} \left ( 2\,\arccos \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) +{\it Ci} \left ( 2\,\arccos \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x}{b^{2} \arccos \left (c x\right )^{2} + 2 \, a b \arccos \left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a + b \operatorname{acos}{\left (c x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*acos(c*x))**2, x)

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Giac [B]  time = 1.21624, size = 436, normalized size = 4.79 \begin{align*} -\frac{2 \, b \arccos \left (c x\right ) \cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac{2 \, b \arccos \left (c x\right ) \cos \left (\frac{a}{b}\right ) \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac{2 \, a \cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} - \frac{2 \, a \cos \left (\frac{a}{b}\right ) \sin \left (\frac{a}{b}\right ) \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac{\sqrt{-c^{2} x^{2} + 1} b c x}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac{b \arccos \left (c x\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} + \frac{a \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b^{3} c^{2} \arccos \left (c x\right ) + a b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*arccos(c*x)*cos(a/b)^2*cos_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 2*b*arccos
(c*x)*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 2*a*cos(a/b)^2
*cos_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) - 2*a*cos(a/b)*sin(a/b)*sin_integral(2*
a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + sqrt(-c^2*x^2 + 1)*b*c*x/(b^3*c^2*arccos(c*x) + a*b^2
*c^2) + b*arccos(c*x)*cos_integral(2*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2) + a*cos_integral(2
*a/b + 2*arccos(c*x))/(b^3*c^2*arccos(c*x) + a*b^2*c^2)