3.237 \(\int \frac {1}{(a+b \sin ^{-1}(c+d x))^4} \, dx\)

Optimal. Leaf size=164 \[ -\frac {\sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{6 b^4 d}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{6 b^4 d}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3} \]

[Out]

1/6*(d*x+c)/b^2/d/(a+b*arcsin(d*x+c))^2+1/6*cos(a/b)*Si((a+b*arcsin(d*x+c))/b)/b^4/d-1/6*Ci((a+b*arcsin(d*x+c)
)/b)*sin(a/b)/b^4/d-1/3*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))^3+1/6*(1-(d*x+c)^2)^(1/2)/b^3/d/(a+b*arcsi
n(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 160, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {4803, 4621, 4719, 4723, 3303, 3299, 3302} \[ -\frac {\sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{6 b^4 d}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{6 b^4 d}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^(-4),x]

[Out]

-Sqrt[1 - (c + d*x)^2]/(3*b*d*(a + b*ArcSin[c + d*x])^3) + (c + d*x)/(6*b^2*d*(a + b*ArcSin[c + d*x])^2) + Sqr
t[1 - (c + d*x)^2]/(6*b^3*d*(a + b*ArcSin[c + d*x])) - (CosIntegral[a/b + ArcSin[c + d*x]]*Sin[a/b])/(6*b^4*d)
+ (Cos[a/b]*SinIntegral[a/b + ArcSin[c + d*x]])/(6*b^4*d)

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 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 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \sin ^{-1}(c+d x)\right )^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sin ^{-1}(x)\right )^4} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}-\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{6 b^2 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )} \, dx,x,c+d x\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{6 b^3 d}-\frac {\sin \left (\frac {a}{b}\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{6 b^3 d}\\ &=-\frac {\sqrt {1-(c+d x)^2}}{3 b d \left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {c+d x}{6 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\sqrt {1-(c+d x)^2}}{6 b^3 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 )}{6 b^4 d}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{6 b^4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 134, normalized size = 0.82 \[ \frac {-\frac {2 b^3 \sqrt {1-(c+d x)^2}}{\left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac {b^2 (c+d x)}{\left (a+b \sin ^{-1}(c+d x)\right )^2}-\sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+\frac {b \sqrt {1-(c+d x)^2}}{a+b \sin ^{-1}(c+d x)}}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])^(-4),x]

[Out]

((-2*b^3*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x])^3 + (b^2*(c + d*x))/(a + b*ArcSin[c + d*x])^2 + (b*Sqr
t[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x]) - CosIntegral[a/b + ArcSin[c + d*x]]*Sin[a/b] + Cos[a/b]*SinIntegr
al[a/b + ArcSin[c + d*x]])/(6*b^4*d)

________________________________________________________________________________________

fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(1/(b^4*arcsin(d*x + c)^4 + 4*a*b^3*arcsin(d*x + c)^3 + 6*a^2*b^2*arcsin(d*x + c)^2 + 4*a^3*b*arcsin(d
*x + c) + a^4), x)

________________________________________________________________________________________

giac [B]  time = 0.25, size = 1112, normalized size = 6.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/6*b^3*arcsin(d*x + c)^3*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*a
rcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/6*b^3*arcsin(d*x + c)^3*cos(a/b)*sin_integral(
a/b + arcsin(d*x + c))/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) +
a^3*b^4*d) - 1/2*a*b^2*arcsin(d*x + c)^2*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^7*d*arcsin(d*x + c)^3
+ 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/2*a*b^2*arcsin(d*x + c)^2*cos(a/
b)*sin_integral(a/b + arcsin(d*x + c))/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*ar
csin(d*x + c) + a^3*b^4*d) - 1/2*a^2*b*arcsin(d*x + c)*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^7*d*arc
sin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/2*a^2*b*arcsin(d*x
+ c)*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*
a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/6*sqrt(-(d*x + c)^2 + 1)*b^3*arcsin(d*x + c)^2/(b^7*d*arcsin(d*x +
c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/6*(d*x + c)*b^3*arcsin(d*x +
c)/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) - 1/6*a^
3*cos_integral(a/b + arcsin(d*x + c))*sin(a/b)/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*
b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/6*a^3*cos(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^7*d*arcsin(d*x +
c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) + 1/3*sqrt(-(d*x + c)^2 + 1)*a*b
^2*arcsin(d*x + c)/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*
b^4*d) + 1/6*(d*x + c)*a*b^2/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x +
c) + a^3*b^4*d) + 1/6*sqrt(-(d*x + c)^2 + 1)*a^2*b/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*arcsin(d*x + c)^2 + 3
*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d) - 1/3*sqrt(-(d*x + c)^2 + 1)*b^3/(b^7*d*arcsin(d*x + c)^3 + 3*a*b^6*d*
arcsin(d*x + c)^2 + 3*a^2*b^5*d*arcsin(d*x + c) + a^3*b^4*d)

________________________________________________________________________________________

maple [A]  time = 0.17, size = 270, normalized size = 1.65 \[ \frac {-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{3 \left (a +b \arcsin \left (d x +c \right )\right )^{3} b}+\frac {\arcsin \left (d x +c \right )^{2} \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b^{2}-\arcsin \left (d x +c \right )^{2} \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b^{2}+2 \arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a b -2 \arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a b +\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}\, b^{2}+\Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a^{2}-\Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a^{2}+\sqrt {1-\left (d x +c \right )^{2}}\, a b +\left (d x +c \right ) b^{2}}{6 \left (a +b \arcsin \left (d x +c \right )\right )^{2} b^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*arcsin(d*x+c))^4,x)

[Out]

1/d*(-1/3*(1-(d*x+c)^2)^(1/2)/(a+b*arcsin(d*x+c))^3/b+1/6*(arcsin(d*x+c)^2*Si(arcsin(d*x+c)+a/b)*cos(a/b)*b^2-
arcsin(d*x+c)^2*Ci(arcsin(d*x+c)+a/b)*sin(a/b)*b^2+2*arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*cos(a/b)*a*b-2*arcsin
(d*x+c)*Ci(arcsin(d*x+c)+a/b)*sin(a/b)*a*b+arcsin(d*x+c)*(1-(d*x+c)^2)^(1/2)*b^2+Si(arcsin(d*x+c)+a/b)*cos(a/b
)*a^2-Ci(arcsin(d*x+c)+a/b)*sin(a/b)*a^2+(1-(d*x+c)^2)^(1/2)*a*b+(d*x+c)*b^2)/(a+b*arcsin(d*x+c))^2/b^4)

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*arcsin(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asin(c + d*x))^4,x)

[Out]

int(1/(a + b*asin(c + d*x))^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*asin(d*x+c))**4,x)

[Out]

Integral((a + b*asin(c + d*x))**(-4), x)

________________________________________________________________________________________