3.1.41 \(\int \frac {1}{\text {ArcCos}(a+b x)^{3/2}} \, dx\) [41]

Optimal. Leaf size=64 \[ \frac {2 \sqrt {1-(a+b x)^2}}{b \sqrt {\text {ArcCos}(a+b x)}}-\frac {2 \sqrt {2 \pi } \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4888, 4718, 4810, 3385, 3433} \begin {gather*} \frac {2 \sqrt {1-(a+b x)^2}}{b \sqrt {\text {ArcCos}(a+b x)}}-\frac {2 \sqrt {2 \pi } \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(-3/2),x]

[Out]

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

Rule 3385

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 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4718

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] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4810

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(-(b*c^
(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGt
Q[m, 0]

Rule 4888

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCos[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.04, size = 97, normalized size = 1.52 \begin {gather*} -\frac {-2 \sqrt {1-(a+b x)^2}-i \sqrt {-i \text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {1}{2},-i \text {ArcCos}(a+b x)\right )+i \sqrt {i \text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {1}{2},i \text {ArcCos}(a+b x)\right )}{b \sqrt {\text {ArcCos}(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^(-3/2),x]

[Out]

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

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Maple [A]
time = 0.17, size = 84, normalized size = 1.31

method	result	size
default	\(-\frac {\sqrt {2}\, \left (2 \arccos \left (b x +a \right ) \pi \FresnelC \left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )-\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\right )}{b \sqrt {\pi }\, \arccos \left (b x +a \right )}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccos(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\operatorname {acos}^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/acos(b*x+a)**(3/2),x)

[Out]

Integral(acos(a + b*x)**(-3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(b*x + a)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\mathrm {acos}\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/acos(a + b*x)^(3/2),x)

[Out]

int(1/acos(a + b*x)^(3/2), x)

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