3.2.48 \(\int \frac {x}{\text {ArcSin}(a+b x)^2} \, dx\) [148]

Optimal. Leaf size=55 \[ -\frac {x \sqrt {1-(a+b x)^2}}{b \text {ArcSin}(a+b x)}+\frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b^2}+\frac {a \text {Si}(\text {ArcSin}(a+b x))}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.58, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4889, 4829, 4717, 4809, 3380, 4727, 3383} \begin {gather*} \frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{b^2}+\frac {a \text {Si}(\text {ArcSin}(a+b x))}{b^2}+\frac {a \sqrt {1-(a+b x)^2}}{b^2 \text {ArcSin}(a+b x)}-\frac {(a+b x) \sqrt {1-(a+b x)^2}}{b^2 \text {ArcSin}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/ArcSin[a + b*x]^2,x]

[Out]

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

Rule 3380

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 3383

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 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 4727

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[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[
-a/b + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x]
&& IGtQ[m, 0] && GeQ[n, -2] && 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 4829

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 4889

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

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Mathematica [A]
time = 0.10, size = 63, normalized size = 1.15 \begin {gather*} \frac {-b x \sqrt {1-(a+b x)^2}+\text {ArcSin}(a+b x) \text {CosIntegral}(2 \text {ArcSin}(a+b x))+a \text {ArcSin}(a+b x) \text {Si}(\text {ArcSin}(a+b x))}{b^2 \text {ArcSin}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSin[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 0.10, size = 72, normalized size = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (2 \arcsin \left (b x +a \right )\right )}{2 \arcsin \left (b x +a \right )}+\cosineIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {a \left (\sinIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{\arcsin \left (b x +a \right )}}{b^{2}}\)	\(72\)
default	\(\frac {-\frac {\sin \left (2 \arcsin \left (b x +a \right )\right )}{2 \arcsin \left (b x +a \right )}+\cosineIntegral \left (2 \arcsin \left (b x +a \right )\right )+\frac {a \left (\sinIntegral \left (\arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )}{\arcsin \left (b x +a \right )}}{b^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x/arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [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(x/arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x/arcsin(b*x + a)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/asin(b*x+a)**2,x)

[Out]

Integral(x/asin(a + b*x)**2, x)

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Giac [A]
time = 0.42, size = 83, normalized size = 1.51 \begin {gather*} \frac {a \operatorname {Si}\left (\arcsin \left (b x + a\right )\right )}{b^{2}} + \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b^{2}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )}}{b^{2} \arcsin \left (b x + a\right )} + \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} a}{b^{2} \arcsin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x}{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/asin(a + b*x)^2,x)

[Out]

int(x/asin(a + b*x)^2, x)

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