3.3.25 \(\int \frac {1}{(a+b \text {ArcSin}(c+d x))^2} \, dx\) [225]

Optimal. Leaf size=93 \[ -\frac {\sqrt {1-(c+d x)^2}}{b d (a+b \text {ArcSin}(c+d x))}+\frac {\text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c+d x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b^2 d}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c+d x)}{b}\right )}{b^2 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4887, 4717, 4809, 3384, 3380, 3383} \begin {gather*} \frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a+b \text {ArcSin}(c+d x)}{b}\right )}{b^2 d}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \text {ArcSin}(c+d x)}{b}\right )}{b^2 d}-\frac {\sqrt {1-(c+d x)^2}}{b d (a+b \text {ArcSin}(c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-2),x]

[Out]

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

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 3384

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 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 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 4887

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]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 79, normalized size = 0.85 \begin {gather*} \frac {-\frac {b \sqrt {1-(c+d x)^2}}{a+b \text {ArcSin}(c+d x)}+\text {CosIntegral}\left (\frac {a}{b}+\text {ArcSin}(c+d x)\right ) \sin \left (\frac {a}{b}\right )-\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\text {ArcSin}(c+d x)\right )}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])^(-2),x]

[Out]

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

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Maple [A]
time = 0.08, size = 83, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{\left (a +b \arcsin \left (d x +c \right )\right ) b}-\frac {\sinIntegral \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\cosineIntegral \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{b^{2}}}{d}\)	\(83\)
default	\(\frac {-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{\left (a +b \arcsin \left (d x +c \right )\right ) b}-\frac {\sinIntegral \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\cosineIntegral \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{b^{2}}}{d}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

((b^2*d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a*b*d)*integrate(sqrt(d*x + c + 1)*(d*x + c)*
sqrt(-d*x - c + 1)/(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - a*b + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - b^2)*ar
ctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))), x) - sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))/(b^2*d*arcta
n2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a*b*d)

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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(1/(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x+c))**2,x)

[Out]

Integral((a + b*asin(c + d*x))**(-2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (91) = 182\).
time = 0.43, size = 215, normalized size = 2.31 \begin {gather*} \frac {b \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {b \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {a \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right ) \sin \left (\frac {a}{b}\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {a \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} b}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

b*arcsin(d*x + c)*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - b*arcsin(d*
x + c)*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + a*cos_integral(a/b + a
rcsin(d*x + c))*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - a*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b
^3*d*arcsin(d*x + c) + a*b^2*d) - sqrt(-(d*x + c)^2 + 1)*b/(b^3*d*arcsin(d*x + c) + a*b^2*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c + d*x))^2,x)

[Out]

int(1/(a + b*asin(c + d*x))^2, x)

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