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

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

[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])/b]*Sin[a/b])/(6*b^
4*d) + (Cos[a/b]*SinIntegral[(a + b*ArcSin[c + d*x])/b])/(6*b^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.268409, 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 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 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 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 )^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 = 0.308903, size = 134, normalized size = 0.82 \[ \frac{-\frac{2 b^3 \sqrt{1-(c+d x)^2}}{\left (a+b \sin ^{-1}(c+d x)\right )^3}+\frac{b^2 (c+d x)}{\left (a+b \sin ^{-1}(c+d x)\right )^2}-\sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )+\frac{b \sqrt{1-(c+d x)^2}}{a+b \sin ^{-1}(c+d x)}}{6 b^4 d} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.065, size = 270, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{3\, \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{3}b}\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{1}{6\, \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}{b}^{4}} \left ( \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ){b}^{2}- \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ){b}^{2}+2\,\arcsin \left ( dx+c \right ){\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) ab-2\,\arcsin \left ( dx+c \right ){\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) ab+\arcsin \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}{b}^{2}+{\it Si} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ){a}^{2}-{\it Ci} \left ( \arcsin \left ( dx+c \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ){a}^{2}+\sqrt{1- \left ( dx+c \right ) ^{2}}ab+ \left ( dx+c \right ){b}^{2} \right ) } \right ) } \end{align*}

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., 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))^4,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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 )^{4}}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [B]  time = 1.29513, size = 1501, normalized size = 9.15 \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))^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)