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

Optimal. Leaf size=137 \[ -\frac{2 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c}-\frac{2 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c}+\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.265989, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {4622, 4724, 3306, 3305, 3351, 3304, 3352} \[ -\frac{2 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c}-\frac{2 \sqrt{2 \pi } \sin \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c}+\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCos[c*x])^(-3/2),x]

[Out]

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

Rule 4622

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])^(n + 1)
)/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*x^2], x], x] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -1]

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 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{1}{\left (a+b \cos ^{-1}(c x)\right )^{3/2}} \, dx &=\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}+\frac{(2 c) \int \frac{x}{\sqrt{1-c^2 x^2} \sqrt{a+b \cos ^{-1}(c x)}} \, dx}{b}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{b c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{\left (2 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{b c}-\frac{\left (2 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\cos ^{-1}(c x)\right )}{b c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{\left (4 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{b^2 c}-\frac{\left (4 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \cos ^{-1}(c x)}\right )}{b^2 c}\\ &=\frac{2 \sqrt{1-c^2 x^2}}{b c \sqrt{a+b \cos ^{-1}(c x)}}-\frac{2 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right )}{b^{3/2} c}-\frac{2 \sqrt{2 \pi } S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \cos ^{-1}(c x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{b^{3/2} c}\\ \end{align*}

Mathematica [C]  time = 0.166727, size = 150, normalized size = 1.09 \[ -\frac{i e^{-\frac{i a}{b}} \left (-\sqrt{-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+e^{\frac{2 i a}{b}} \sqrt{\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \cos ^{-1}(c x)\right )}{b}\right )+2 i e^{\frac{i a}{b}} \sqrt{1-c^2 x^2}\right )}{b c \sqrt{a+b \cos ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*((2*I)*E^((I*a)/b)*Sqrt[1 - c^2*x^2] - Sqrt[((-I)*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, ((-I)*(a + b*ArcCos
[c*x]))/b] + E^(((2*I)*a)/b)*Sqrt[(I*(a + b*ArcCos[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcCos[c*x]))/b]))/(b*c*E^((
I*a)/b)*Sqrt[a + b*ArcCos[c*x]])

________________________________________________________________________________________

Maple [A]  time = 0.092, size = 150, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{bc\sqrt{a+b\arccos \left ( cx \right ) }} \left ( -\sqrt{\pi }\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }\cos \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{{b}^{-1}}-\sqrt{\pi }\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }\sin \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arccos \left ( cx \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \sqrt{{b}^{-1}}+\sin \left ({\frac{a+b\arccos \left ( cx \right ) }{b}}-{\frac{a}{b}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arccos(c*x))^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a+b*arccos(c*x))^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(1/(a+b*acos(c*x))**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a+b*arccos(c*x))^(3/2),x, algorithm="giac")

[Out]

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