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

Optimal. Leaf size=108 \[ \frac {a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}-\frac {1-2 (a+b x)^2}{2 b^2 \sin ^{-1}(a+b x)}-\frac {x \sqrt {1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]

[Out]

-1/2*a*(b*x+a)/b^2/arcsin(b*x+a)+1/2*(-1+2*(b*x+a)^2)/b^2/arcsin(b*x+a)+1/2*a*Ci(arcsin(b*x+a))/b^2-Si(2*arcsi
n(b*x+a))/b^2-1/2*x*(1-(b*x+a)^2)^(1/2)/b/arcsin(b*x+a)^2

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Rubi [A]  time = 0.27, antiderivative size = 151, normalized size of antiderivative = 1.40, number of steps used = 14, number of rules used = 12, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.200, Rules used = {4805, 4745, 4621, 4719, 4623, 3302, 4633, 4635, 4406, 12, 3299, 4641} \[ \frac {a \text {CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{b^2}+\frac {(a+b x)^2}{b^2 \sin ^{-1}(a+b x)}-\frac {a (a+b x)}{2 b^2 \sin ^{-1}(a+b x)}-\frac {\sqrt {1-(a+b x)^2} (a+b x)}{2 b^2 \sin ^{-1}(a+b x)^2}-\frac {1}{2 b^2 \sin ^{-1}(a+b x)}+\frac {a \sqrt {1-(a+b x)^2}}{2 b^2 \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSin[a + b*x]^3,x]

[Out]

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

Rule 4745

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 121, normalized size = 1.12 \[ -\frac {x \sqrt {-a^2-2 a b x-b^2 x^2+1}}{2 b \sin ^{-1}(a+b x)^2}+\frac {a^2+3 a b x+2 b^2 x^2-1}{2 b^2 \sin ^{-1}(a+b x)}-2 \left (\frac {\text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{b^2}\right )-\frac {3 a \text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSin[a + b*x]^3,x]

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\arcsin \left (b x + a\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsin(b*x + a)^3, x)

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giac [A]  time = 0.49, size = 139, normalized size = 1.29 \[ \frac {a \operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b^{2}} - \frac {{\left (b x + a\right )} a}{2 \, b^{2} \arcsin \left (b x + a\right )} - \frac {\operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{2} - 1}{b^{2} \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a}{2 \, b^{2} \arcsin \left (b x + a\right )^{2}} + \frac {1}{2 \, b^{2} \arcsin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*cos_integral(arcsin(b*x + a))/b^2 - 1/2*(b*x + a)*a/(b^2*arcsin(b*x + a)) - sin_integral(2*arcsin(b*x +
a))/b^2 + ((b*x + a)^2 - 1)/(b^2*arcsin(b*x + a)) - 1/2*sqrt(-(b*x + a)^2 + 1)*(b*x + a)/(b^2*arcsin(b*x + a)^
2) + 1/2*sqrt(-(b*x + a)^2 + 1)*a/(b^2*arcsin(b*x + a)^2) + 1/2/(b^2*arcsin(b*x + a))

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maple [A]  time = 0.11, size = 109, normalized size = 1.01 \[ \frac {-\frac {\sin \left (2 \arcsin \left (b x +a \right )\right )}{4 \arcsin \left (b x +a \right )^{2}}-\frac {\cos \left (2 \arcsin \left (b x +a \right )\right )}{2 \arcsin \left (b x +a \right )}-\Si \left (2 \arcsin \left (b x +a \right )\right )+\frac {a \left (\Ci \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}-\left (b x +a \right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{2 \arcsin \left (b x +a \right )^{2}}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/asin(a + b*x)^3,x)

[Out]

int(x/asin(a + b*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asin(a + b*x)**3, x)

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