3.1.38 \(\int \text {ArcCos}(a+b x)^{3/2} \, dx\) [38]

Optimal. Leaf size=89 \[ -\frac {3 \sqrt {1-(a+b x)^2} \sqrt {\text {ArcCos}(a+b x)}}{2 b}+\frac {(a+b x) \text {ArcCos}(a+b x)^{3/2}}{b}+\frac {3 \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{2 b} \]

[Out]

(b*x+a)*arccos(b*x+a)^(3/2)/b+3/4*FresnelS(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2))*2^(1/2)*Pi^(1/2)/b-3/2*(1-(b*
x+a)^2)^(1/2)*arccos(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4888, 4716, 4768, 4720, 3386, 3432} \begin {gather*} \frac {3 \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a+b x)}\right )}{2 b}+\frac {(a+b x) \text {ArcCos}(a+b x)^{3/2}}{b}-\frac {3 \sqrt {1-(a+b x)^2} \sqrt {\text {ArcCos}(a+b x)}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a + b*x]^(3/2),x]

[Out]

(-3*Sqrt[1 - (a + b*x)^2]*Sqrt[ArcCos[a + b*x]])/(2*b) + ((a + b*x)*ArcCos[a + b*x]^(3/2))/b + (3*Sqrt[Pi/2]*F
resnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a + b*x]]])/(2*b)

Rule 3386

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 3432

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

Rule 4716

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcCos[c*x])^n, x] + Dist[b*c*n, Int[
x*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 4720

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

Rule 4768

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

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.03, size = 76, normalized size = 0.85 \begin {gather*} -\frac {\sqrt {-i \text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {5}{2},-i \text {ArcCos}(a+b x)\right )+\sqrt {i \text {ArcCos}(a+b x)} \text {Gamma}\left (\frac {5}{2},i \text {ArcCos}(a+b x)\right )}{2 b \sqrt {\text {ArcCos}(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a + b*x]^(3/2),x]

[Out]

-1/2*(Sqrt[(-I)*ArcCos[a + b*x]]*Gamma[5/2, (-I)*ArcCos[a + b*x]] + Sqrt[I*ArcCos[a + b*x]]*Gamma[5/2, I*ArcCo
s[a + b*x]])/(b*Sqrt[ArcCos[a + b*x]])

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Maple [A]
time = 0.17, size = 105, normalized size = 1.18

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, b x -2 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, a +3 \sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-3 \pi \,\mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )\right )}{4 b \sqrt {\pi }}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b*2^(1/2)*(-2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*b*x-2*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*a+3*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)-3*Pi*FresnelS(2^(1/2)/Pi^(1/2)*arccos(b*x+a)^(1/2
)))/Pi^(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(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(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 \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(acos(b*x+a)**(3/2),x)

[Out]

Integral(acos(a + b*x)**(3/2), x)

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Giac [C] Result contains complex when optimal does not.
time = 0.52, size = 139, normalized size = 1.56 \begin {gather*} \frac {\arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {\arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {\left (3 i - 3\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (b x + a\right )}\right )}{16 \, b} - \frac {\left (3 i + 3\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (b x + a\right )}\right )}{16 \, b} + \frac {3 i \, \sqrt {\arccos \left (b x + a\right )} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{4 \, b} - \frac {3 i \, \sqrt {\arccos \left (b x + a\right )} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arccos(b*x + a)^(3/2)*e^(I*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(3/2)*e^(-I*arccos(b*x + a))/b + (3/16
*I - 3/16)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(arccos(b*x + a)))/b - (3/16*I + 3/16)*sqrt(2)*sqrt(
pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(arccos(b*x + a)))/b + 3/4*I*sqrt(arccos(b*x + a))*e^(I*arccos(b*x + a))/b
- 3/4*I*sqrt(arccos(b*x + a))*e^(-I*arccos(b*x + a))/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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(acos(a + b*x)^(3/2),x)

[Out]

int(acos(a + b*x)^(3/2), x)

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