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

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

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {4633, 4719, 4635, 4406, 12, 3303, 3299, 3302, 4641} \[ \frac {\sin \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^3 c^2}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^3 c^2}-\frac {1}{2 b^2 c^2 \left (a+b \sin ^{-1}(c x)\right )}+\frac {x^2}{b^2 \left (a+b \sin ^{-1}(c x)\right )}-\frac {x \sqrt {1-c^2 x^2}}{2 b c \left (a+b \sin ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*ArcSin[c*x])^3,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4633

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

Rule 4635

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
in[x]^m*Cos[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4641

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

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

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Mathematica [A]  time = 0.39, size = 108, normalized size = 0.83 \[ \frac {-\frac {b^2 c x \sqrt {1-c^2 x^2}}{\left (a+b \sin ^{-1}(c x)\right )^2}+\frac {b \left (2 c^2 x^2-1\right )}{a+b \sin ^{-1}(c x)}+2 \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )-2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{2 b^3 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*ArcSin[c*x])^3,x]

[Out]

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

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b^{3} \arcsin \left (c x\right )^{3} + 3 \, a b^{2} \arcsin \left (c x\right )^{2} + 3 \, a^{2} b \arcsin \left (c x\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/(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 = 1.52, size = 864, normalized size = 6.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*arcsin(c*x)^2*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2
*arcsin(c*x) + a^2*b^3*c^2) - 2*b^2*arcsin(c*x)^2*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^5*c^2*arcs
in(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) + 4*a*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arcsin(
c*x))*sin(a/b)/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) - 4*a*b*arcsin(c*x)*cos(a/b)^2*
sin_integral(2*a/b + 2*arcsin(c*x))/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) + 2*a^2*co
s(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3
*c^2) + b^2*arcsin(c*x)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x)
+ a^2*b^3*c^2) - 2*a^2*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*ar
csin(c*x) + a^2*b^3*c^2) - 1/2*sqrt(-c^2*x^2 + 1)*b^2*c*x/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a
^2*b^3*c^2) + (c^2*x^2 - 1)*b^2*arcsin(c*x)/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) +
2*a*b*arcsin(c*x)*sin_integral(2*a/b + 2*arcsin(c*x))/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b
^3*c^2) + (c^2*x^2 - 1)*a*b/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) + 1/2*b^2*arcsin(c
*x)/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) + a^2*sin_integral(2*a/b + 2*arcsin(c*x))/
(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2*arcsin(c*x) + a^2*b^3*c^2) + 1/2*a*b/(b^5*c^2*arcsin(c*x)^2 + 2*a*b^4*c^2
*arcsin(c*x) + a^2*b^3*c^2)

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maple [A]  time = 0.03, size = 157, normalized size = 1.21 \[ \frac {-\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{4 \left (a +b \arcsin \left (c x \right )\right )^{2} b}-\frac {2 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b -2 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) \arcsin \left (c x \right ) b +2 \Si \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -2 \Ci \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (2 \arcsin \left (c x \right )\right ) b}{2 \left (a +b \arcsin \left (c x \right )\right ) b^{3}}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/4*sin(2*arcsin(c*x))/(a+b*arcsin(c*x))^2/b-1/2*(2*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*arcsin(c*x)*b-2
*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*arcsin(c*x)*b+2*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a-2*Ci(2*arcsin(c*x)+2*
a/b)*sin(2*a/b)*a+cos(2*arcsin(c*x))*b)/(a+b*arcsin(c*x))/b^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*asin(c*x))^3,x)

[Out]

int(x/(a + b*asin(c*x))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(a + b*asin(c*x))**3, x)

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