3.194 \(\int \frac{x}{(a+b \cos ^{-1}(c x))^{3/2}} \, dx\)

Optimal. Leaf size=130 \[ -\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} c^2}-\frac{2 \sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} c^2}+\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}} \]

[Out]

(2*x*Sqrt[1 - c^2*x^2])/(b*c*Sqrt[a + b*ArcCos[c*x]]) - (2*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcCos
[c*x]])/(Sqrt[b]*Sqrt[Pi])])/(b^(3/2)*c^2) - (2*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi
])]*Sin[(2*a)/b])/(b^(3/2)*c^2)

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Rubi [A]  time = 0.153552, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4632, 3306, 3305, 3351, 3304, 3352} \[ -\frac{2 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi } \sqrt{b}}\right )}{b^{3/2} c^2}-\frac{2 \sqrt{\pi } \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b} \sqrt{\pi }}\right )}{b^{3/2} c^2}+\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcCos[c*x])^(3/2),x]

[Out]

(2*x*Sqrt[1 - c^2*x^2])/(b*c*Sqrt[a + b*ArcCos[c*x]]) - (2*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[a + b*ArcCos
[c*x]])/(Sqrt[b]*Sqrt[Pi])])/(b^(3/2)*c^2) - (2*Sqrt[Pi]*FresnelS[(2*Sqrt[a + b*ArcCos[c*x]])/(Sqrt[b]*Sqrt[Pi
])]*Sin[(2*a)/b])/(b^(3/2)*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 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}{\left (a+b \cos ^{-1}(c x)\right )^{3/2}} \, dx &=\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{b c^2}\\ &=\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{\left (2 \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 )}{b c^2}-\frac{\left (2 \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 )}{b c^2}\\ &=\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{\left (4 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{b^2 c^2}-\frac{\left (4 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{b^2 c^2}\\ &=\frac{2 x \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{2 \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 )}{b^{3/2} c^2}-\frac{2 \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 )}{b^{3/2} c^2}\\ \end{align*}

Mathematica [A]  time = 0.329857, size = 124, normalized size = 0.95 \[ \frac{-2 \sqrt{\pi } \left (\frac{1}{b}\right )^{3/2} \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )-2 \sqrt{\pi } \left (\frac{1}{b}\right )^{3/2} \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{\pi }}\right )+\frac{\sin \left (2 \cos ^{-1}(c x)\right )}{b \sqrt{a+b \cos ^{-1}(c x)}}}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*ArcCos[c*x])^(3/2),x]

[Out]

(-2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[(2*a)/b]*FresnelC[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]] - 2*(b^(-
1))^(3/2)*Sqrt[Pi]*FresnelS[(2*Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[c*x]])/Sqrt[Pi]]*Sin[(2*a)/b] + Sin[2*ArcCos[c*x
]]/(b*Sqrt[a + b*ArcCos[c*x]]))/c^2

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Maple [A]  time = 0.102, size = 142, normalized size = 1.1 \begin{align*}{\frac{1}{b{c}^{2}} \left ( -2\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ( 2\,{\frac{a}{b}} \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ) -2\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{a+b\arccos \left ( cx \right ) }\sin \left ( 2\,{\frac{a}{b}} \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ) +\sin \left ( 2\,{\frac{a+b\arccos \left ( cx \right ) }{b}}-2\,{\frac{a}{b}} \right ) \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))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arccos(c*x))^(3/2),x, algorithm="maxima")

[Out]

integrate(x/(b*arccos(c*x) + a)^(3/2), 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))^(3/2),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}{\left (a + b \operatorname{acos}{\left (c x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*acos(c*x))**(3/2),x)

[Out]

Integral(x/(a + b*acos(c*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (b \arccos \left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*arccos(c*x))^(3/2),x, algorithm="giac")

[Out]

integrate(x/(b*arccos(c*x) + a)^(3/2), x)