3.2.67 \(\int \frac {1}{(a+b \text {ArcSin}(c+d x))^{3/2}} \, dx\) [167]

Optimal. Leaf size=144 \[ -\frac {2 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \text {ArcSin}(c+d x)}}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {2 \sqrt {2 \pi } \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{b^{3/2} d} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {4887, 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+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-3/2),x]

[Out]

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

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]

Rule 4887

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.19, size = 185, normalized size = 1.28 \begin {gather*} \frac {e^{-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \left (e^{i \text {ArcSin}(c+d x)} \sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )+e^{\frac {i a}{b}} \left (-1-e^{2 i \text {ArcSin}(c+d x)}+e^{\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {i (a+b \text {ArcSin}(c+d x))}{b}\right )\right )\right )}{b d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.25, size = 170, normalized size = 1.18

method	result	size
default	\(-\frac {2 \left (-\sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}-\sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}+\cos \left (-\frac {a +b \arcsin \left (d x +c \right )}{b}+\frac {a}{b}\right )\right )}{d b \sqrt {a +b \arcsin \left (d x +c \right )}}\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b/(a+b*arcsin(d*x+c))^(1/2)*(-(a+b*arcsin(d*x+c))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(
a+b*arcsin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)-(a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/
Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)+cos(-(a+b*arcsin(d*x+c))/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(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*arcsin(d*x + c) + 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(d*x+c))^(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 + d 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(d*x+c))**(3/2),x)

[Out]

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

[Out]

integrate((b*arcsin(d*x + c) + 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+d\,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 + d*x))^(3/2),x)

[Out]

int(1/(a + b*asin(c + d*x))^(3/2), x)

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