3.2.94 \(\int \frac {x}{(a+b \text {ArcCos}(c x))^{3/2}} \, dx\) [194]

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

[Out]

-2*cos(2*a/b)*FresnelC(2*(a+b*arccos(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(3/2)/c^2-2*FresnelS(2*(a+b*arcc
os(c*x))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/b^(3/2)/c^2+2*x*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arccos(c*x))^
(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 130, normalized size of antiderivative = 1.00, 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 = {4728, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {2 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcCos}(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 \text {ArcCos}(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 \text {ArcCos}(c x)}} \end {gather*}

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 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 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 3387

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 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 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 4728

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

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 \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.24, size = 124, normalized size = 0.95 \begin {gather*} \frac {-2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right )-2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } S\left (\frac {2 \sqrt {\frac {1}{b}} \sqrt {a+b \text {ArcCos}(c x)}}{\sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )+\frac {\sin (2 \text {ArcCos}(c x))}{b \sqrt {a+b \text {ArcCos}(c x)}}}{c^2} \end {gather*}

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 = 0.22, size = 162, normalized size = 1.25

method	result	size
default	\(-\frac {\sqrt {-\frac {2}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right )-\sqrt {-\frac {2}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arccos \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right )+\sin \left (-\frac {2 \left (a +b \arccos \left (c x \right )\right )}{b}+\frac {2 a}{b}\right )}{c^{2} b \sqrt {a +b \arccos \left (c x \right )}}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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(x/(a+b*arccos(c*x))^(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 {x}{\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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 = 0.00, 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(x/(a+b*arccos(c*x))^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*acos(c*x))^(3/2),x)

[Out]

int(x/(a + b*acos(c*x))^(3/2), x)

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