3.8.1 \(\int \frac {1}{(a+b \text {ArcSin}(c x))^{3/2}} \, dx\) [701]

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

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {4717, 4809, 3387, 3386, 3432, 3385, 3433} \begin {gather*} \frac {2 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c x)}}{\sqrt {b}}\right )}{b^{3/2} c}-\frac {2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \text {ArcSin}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^(-3/2),x]

[Out]

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

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 4717

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

Rule 4809

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.21, size = 167, normalized size = 1.22 \begin {gather*} \frac {e^{-\frac {i (a+b \text {ArcSin}(c x))}{b}} \left (e^{i \text {ArcSin}(c x)} \sqrt {-\frac {i (a+b \text {ArcSin}(c x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {i (a+b \text {ArcSin}(c x))}{b}\right )+e^{\frac {i a}{b}} \left (-1-e^{2 i \text {ArcSin}(c x)}+e^{\frac {i (a+b \text {ArcSin}(c x))}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {i (a+b \text {ArcSin}(c x))}{b}\right )\right )\right )}{b c \sqrt {a+b \text {ArcSin}(c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 158, normalized size = 1.15

method	result	size
default	\(-\frac {2 \left (-\sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}-\sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}+\cos \left (-\frac {a +b \arcsin \left (c x \right )}{b}+\frac {a}{b}\right )\right )}{c b \sqrt {a +b \arcsin \left (c x \right )}}\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((b*arcsin(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(1/(a+b*arcsin(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 {1}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(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(1/(a+b*arcsin(c*x))^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(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 {1}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c*x))^(3/2),x)

[Out]

int(1/(a + b*asin(c*x))^(3/2), x)

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