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

Optimal. Leaf size=111 \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{2 b^3 c}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{2 b^3 c}+\frac {x}{2 b^2 \left (a+b \sin ^{-1}(c x)\right )}-\frac {\sqrt {1-c^2 x^2}}{2 b c \left (a+b \sin ^{-1}(c x)\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 111, 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 = {4621, 4719, 4623, 3303, 3299, 3302} \[ -\frac {\cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{2 b^3 c}-\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{2 b^3 c}+\frac {x}{2 b^2 \left (a+b \sin ^{-1}(c x)\right )}-\frac {\sqrt {1-c^2 x^2}}{2 b c \left (a+b \sin ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^(-3),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 93, normalized size = 0.84 \[ -\frac {\frac {b \left (\frac {b \sqrt {1-c^2 x^2}}{c}-x \left (a+b \sin ^{-1}(c x)\right )\right )}{\left (a+b \sin ^{-1}(c x)\right )^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{c}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{c}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])^(-3),x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} \arcsin \left (c x\right )^{3} + 3 \, a b^{2} \arcsin \left (c x\right )^{2} + 3 \, a^{2} b \arcsin \left (c x\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.07, size = 482, normalized size = 4.34 \[ -\frac {b^{2} \arcsin \left (c x\right )^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {b^{2} \arcsin \left (c x\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} + \frac {b^{2} c x \arcsin \left (c x\right )}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {a b \arcsin \left (c x\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c} - \frac {a b \arcsin \left (c x\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c} + \frac {a b c x}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {a^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {a^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} - \frac {\sqrt {-c^{2} x^{2} + 1} b^{2}}{2 \, {\left (b^{5} c \arcsin \left (c x\right )^{2} + 2 \, a b^{4} c \arcsin \left (c x\right ) + a^{2} b^{3} c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2*arcsin(c*x)^2*cos(a/b)*cos_integral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) +
a^2*b^3*c) - 1/2*b^2*arcsin(c*x)^2*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*
arcsin(c*x) + a^2*b^3*c) + 1/2*b^2*c*x*arcsin(c*x)/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b^3*c) -
a*b*arcsin(c*x)*cos(a/b)*cos_integral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b
^3*c) - a*b*arcsin(c*x)*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x)
+ a^2*b^3*c) + 1/2*a*b*c*x/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b^3*c) - 1/2*a^2*cos(a/b)*cos_in
tegral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b^3*c) - 1/2*a^2*sin(a/b)*sin_int
egral(a/b + arcsin(c*x))/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b^3*c) - 1/2*sqrt(-c^2*x^2 + 1)*b^
2/(b^5*c*arcsin(c*x)^2 + 2*a*b^4*c*arcsin(c*x) + a^2*b^3*c)

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maple [A]  time = 0.04, size = 138, normalized size = 1.24 \[ \frac {-\frac {\sqrt {-c^{2} x^{2}+1}}{2 \left (a +b \arcsin \left (c x \right )\right )^{2} b}-\frac {\arcsin \left (c x \right ) \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +\arcsin \left (c x \right ) \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b +\Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +\Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -x b c}{2 \left (a +b \arcsin \left (c x \right )\right ) b^{3}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*c*x - sqrt(c*x + 1)*sqrt(-c*x + 1)*b - 2*(b^4*c*arct
an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b^3*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a^2*b^2*c)*i
ntegrate(1/2/(b^3*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b^2), x))/(b^4*c*arctan2(c*x, sqrt(c*x + 1)*s
qrt(-c*x + 1))^2 + 2*a*b^3*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a^2*b^2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c*x))^3,x)

[Out]

int(1/(a + b*asin(c*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**(-3), x)

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