3.231 \(\int \frac {1}{(a+b \sin ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

[Out]

1/2*(d*x+c)/b^2/d/(a+b*arcsin(d*x+c))-1/2*Ci((a+b*arcsin(d*x+c))/b)*cos(a/b)/b^3/d-1/2*Si((a+b*arcsin(d*x+c))/
b)*sin(a/b)/b^3/d-1/2*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))^2

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Rubi [A]  time = 0.17, antiderivative size = 127, 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 = {4803, 4621, 4719, 4623, 3303, 3299, 3302} \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-3),x]

[Out]

-Sqrt[1 - (c + d*x)^2]/(2*b*d*(a + b*ArcSin[c + d*x])^2) + (c + d*x)/(2*b^2*d*(a + b*ArcSin[c + d*x])) - (Cos[
a/b]*CosIntegral[(a + b*ArcSin[c + d*x])/b])/(2*b^3*d) - (Sin[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(2*
b^3*d)

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4621

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 4623

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

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4803

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} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{2 b^3 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{2 b^3 d}-\frac {\sin \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {x}{b}\right )}{x} \, dx,x,a+b \sin ^{-1}(c+d x)\right )}{2 b^3 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {c+d x}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{2 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 100, normalized size = 0.79 \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+\frac {b \left (b \sqrt {1-(c+d x)^2}-(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )\right )}{\left (a+b \sin ^{-1}(c+d x)\right )^2}}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((b*(b*Sqrt[1 - (c + d*x)^2] - (c + d*x)*(a + b*ArcSin[c + d*x])))/(a + b*ArcSin[c + d*x])^2 + Cos[a/b]*C
osIntegral[a/b + ArcSin[c + d*x]] + Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(b^3*d)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3), x)

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giac [B]  time = 0.22, size = 547, normalized size = 4.31 \[ -\frac {b^{2} \arcsin \left (d x + c\right )^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} - \frac {b^{2} \arcsin \left (d x + c\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} - \frac {a b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d} - \frac {a b \arcsin \left (d x + c\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d} + \frac {{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} - \frac {a^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} - \frac {a^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} + \frac {{\left (d x + c\right )} a b}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} - \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} b^{2}}{2 \, {\left (b^{5} d \arcsin \left (d x + c\right )^{2} + 2 \, a b^{4} d \arcsin \left (d x + c\right ) + a^{2} b^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*a
rcsin(d*x + c) + a^2*b^3*d) - 1/2*b^2*arcsin(d*x + c)^2*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*ar
csin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - a*b*arcsin(d*x + c)*cos(a/b)*cos_integral(a/b + arc
sin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - a*b*arcsin(d*x + c)*sin(a/b)
*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*(
d*x + c)*b^2*arcsin(d*x + c)/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/2*a^2*cos(a
/b)*cos_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/
2*a^2*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*
b^3*d) + 1/2*(d*x + c)*a*b/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/2*sqrt(-(d*x
+ c)^2 + 1)*b^2/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d)

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maple [A]  time = 0.07, size = 158, normalized size = 1.24 \[ \frac {-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (a +b \arcsin \left (d x +c \right )\right )^{2} b}-\frac {\arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b +\Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +\Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -b \left (d x +c \right )}{2 \left (a +b \arcsin \left (d x +c \right )\right ) b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/2*(1-(d*x+c)^2)^(1/2)/(a+b*arcsin(d*x+c))^2/b-1/2*(arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*sin(a/b)*b+arcs
in(d*x+c)*Ci(arcsin(d*x+c)+a/b)*cos(a/b)*b+Si(arcsin(d*x+c)+a/b)*sin(a/b)*a+Ci(arcsin(d*x+c)+a/b)*cos(a/b)*a-b
*(d*x+c))/(a+b*arcsin(d*x+c))/b^3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a d x - \sqrt {d x + c + 1} \sqrt {-d x - c + 1} b + a c + {\left (b d x + b c\right )} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) - {\left (b^{4} d \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, a b^{3} d \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + a^{2} b^{2} d\right )} \int \frac {1}{b^{3} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + a b^{2}}\,{d x}}{2 \, {\left (b^{4} d \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, a b^{3} d \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right ) + a^{2} b^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*d*x - sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*b + a*c + (b*d*x + b*c)*arctan2(d*x + c, sqrt(d*x + c + 1)*s
qrt(-d*x - c + 1)) - 2*(b^4*d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 2*a*b^3*d*arctan2(d*x
+ c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a^2*b^2*d)*integrate(1/2/(b^3*arctan2(d*x + c, sqrt(d*x + c + 1)
*sqrt(-d*x - c + 1)) + a*b^2), x))/(b^4*d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 2*a*b^3*d
*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a^2*b^2*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c + d*x))^3,x)

[Out]

int(1/(a + b*asin(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c + d*x))**(-3), x)

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