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

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

[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)

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Rubi [A]  time = 0.172113, antiderivative size = 127, normalized size of antiderivative = 1., 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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.621288, size = 100, normalized size = 0.79 \[ -\frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\frac{b \left (b \sqrt{1-(c+d x)^2}-(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )\right )}{\left (a+b \sin ^{-1}(c+d x)\right )^2}}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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]*CosIn
tegral[a/b + ArcSin[c + d*x]] + Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(2*b^3*d)

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Maple [A]  time = 0.036, size = 158, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{2\, \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}b}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{1}{ \left ( 2\,a+2\,b\arcsin \left ( dx+c \right ) \right ){b}^{3}} \left ( \arcsin \left ( dx+c \right ){\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b+\arcsin \left ( dx+c \right ){\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b+{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a+{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-b \left ( dx+c \right ) \right ) } \right ) } \end{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \operatorname{asin}{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

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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Giac [B]  time = 1.2306, size = 738, normalized size = 5.81 \begin{align*} \text{result too large to display} \end{align*}

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)