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

Optimal. Leaf size=144 \[ -\frac{a b^2 \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac{\left (2 a^2+1\right ) b^3 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}}-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{\sin ^{-1}(a+b x)}{3 x^3} \]

[Out]

-(b*Sqrt[1 - (a + b*x)^2])/(6*(1 - a^2)*x^2) - (a*b^2*Sqrt[1 - (a + b*x)^2])/(2*(1 - a^2)^2*x) - ArcSin[a + b*
x]/(3*x^3) - ((1 + 2*a^2)*b^3*ArcTanh[(1 - a*(a + b*x))/(Sqrt[1 - a^2]*Sqrt[1 - (a + b*x)^2])])/(6*(1 - a^2)^(
5/2))

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Rubi [A]  time = 0.18048, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4805, 4743, 745, 807, 725, 206} \[ -\frac{a b^2 \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac{\left (2 a^2+1\right ) b^3 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}}-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{\sin ^{-1}(a+b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]/x^4,x]

[Out]

-(b*Sqrt[1 - (a + b*x)^2])/(6*(1 - a^2)*x^2) - (a*b^2*Sqrt[1 - (a + b*x)^2])/(2*(1 - a^2)^2*x) - ArcSin[a + b*
x]/(3*x^3) - ((1 + 2*a^2)*b^3*ArcTanh[(1 - a*(a + b*x))/(Sqrt[1 - a^2]*Sqrt[1 - (a + b*x)^2])])/(6*(1 - a^2)^(
5/2))

Rule 4805

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]

Rule 4743

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcSin[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSin[c*x])^(
n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^{-1}(a+b x)}{x^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sin ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sqrt{1-x^2}} \, dx,x,a+b x\right )\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{\sin ^{-1}(a+b x)}{3 x^3}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\frac{2 a}{b}+\frac{x}{b}}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1-x^2}} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )}\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{a b^2 \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac{\sin ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (1+2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right ) \sqrt{1-x^2}} \, dx,x,a+b x\right )}{6 \left (1-a^2\right )^2}\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{a b^2 \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac{\sin ^{-1}(a+b x)}{3 x^3}-\frac{\left (\left (1+2 a^2\right ) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{b^2}-\frac{a^2}{b^2}-x^2} \, dx,x,\frac{\frac{1}{b}-\frac{a (a+b x)}{b}}{\sqrt{1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^2}\\ &=-\frac{b \sqrt{1-(a+b x)^2}}{6 \left (1-a^2\right ) x^2}-\frac{a b^2 \sqrt{1-(a+b x)^2}}{2 \left (1-a^2\right )^2 x}-\frac{\sin ^{-1}(a+b x)}{3 x^3}-\frac{\left (1+2 a^2\right ) b^3 \tanh ^{-1}\left (\frac{1-a (a+b x)}{\sqrt{1-a^2} \sqrt{1-(a+b x)^2}}\right )}{6 \left (1-a^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.20994, size = 166, normalized size = 1.15 \[ \frac{\sqrt{1-a^2} b x \left (a^2-3 a b x-1\right ) \sqrt{-a^2-2 a b x-b^2 x^2+1}+\left (2 a^2+1\right ) b^3 x^3 \log (x)-\left (2 a^2+1\right ) b^3 x^3 \log \left (\sqrt{1-a^2} \sqrt{-a^2-2 a b x-b^2 x^2+1}-a^2-a b x+1\right )-2 \left (1-a^2\right )^{5/2} \sin ^{-1}(a+b x)}{6 \left (1-a^2\right )^{5/2} x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]/x^4,x]

[Out]

(Sqrt[1 - a^2]*b*x*(-1 + a^2 - 3*a*b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] - 2*(1 - a^2)^(5/2)*ArcSin[a + b*x]
+ (1 + 2*a^2)*b^3*x^3*Log[x] - (1 + 2*a^2)*b^3*x^3*Log[1 - a^2 - a*b*x + Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x
- b^2*x^2]])/(6*(1 - a^2)^(5/2)*x^3)

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Maple [A]  time = 0.008, size = 227, normalized size = 1.6 \begin{align*} -{\frac{\arcsin \left ( bx+a \right ) }{3\,{x}^{3}}}-{\frac{b}{ \left ( -6\,{a}^{2}+6 \right ){x}^{2}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{b}^{2}a}{2\, \left ( -{a}^{2}+1 \right ) ^{2}x}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{b}^{3}{a}^{2}}{2}\ln \left ({\frac{1}{bx} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ) \left ( -{a}^{2}+1 \right ) ^{-{\frac{5}{2}}}}-{\frac{{b}^{3}}{6}\ln \left ({\frac{1}{bx} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ) \left ( -{a}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsin(b*x+a)/x^3-1/6*b/(-a^2+1)/x^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/2*b^2*a/(-a^2+1)^2/x*(-b^2*x^2-2*a*
b*x-a^2+1)^(1/2)-1/2*b^3*a^2/(-a^2+1)^(5/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/
2))/b/x)-1/6*b^3/(-a^2+1)^(3/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/b/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.60826, size = 925, normalized size = 6.42 \begin{align*} \left [-\frac{{\left (2 \, a^{2} + 1\right )} \sqrt{-a^{2} + 1} b^{3} x^{3} \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x + 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 4 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) + 2 \,{\left (3 \,{\left (a^{3} - a\right )} b^{2} x^{2} -{\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{12 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}, \frac{{\left (2 \, a^{2} + 1\right )} \sqrt{a^{2} - 1} b^{3} x^{3} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - 2 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} \arcsin \left (b x + a\right ) -{\left (3 \,{\left (a^{3} - a\right )} b^{2} x^{2} -{\left (a^{4} - 2 \, a^{2} + 1\right )} b x\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*((2*a^2 + 1)*sqrt(-a^2 + 1)*b^3*x^3*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x + 2*sqrt(-b^2*x^
2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) + 4*(a^6 - 3*a^4 + 3*a^2 - 1)*arcsin
(b*x + a) + 2*(3*(a^3 - a)*b^2*x^2 - (a^4 - 2*a^2 + 1)*b*x)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^6 - 3*a^4
+ 3*a^2 - 1)*x^3), 1/6*((2*a^2 + 1)*sqrt(a^2 - 1)*b^3*x^3*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a
^2 - 1)*sqrt(a^2 - 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) - 2*(a^6 - 3*a^4 + 3*a^2 - 1)*a
rcsin(b*x + a) - (3*(a^3 - a)*b^2*x^2 - (a^4 - 2*a^2 + 1)*b*x)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^6 - 3*a
^4 + 3*a^2 - 1)*x^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a + b*x)/x**4, x)

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Giac [B]  time = 1.21661, size = 752, normalized size = 5.22 \begin{align*} \frac{1}{3} \, b{\left (\frac{{\left (2 \, a^{2} b^{3} + b^{3}\right )} \arctan \left (\frac{\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt{a^{2} - 1}}\right )}{{\left (a^{4}{\left | b \right |} - 2 \, a^{2}{\left | b \right |} +{\left | b \right |}\right )} \sqrt{a^{2} - 1}} - \frac{\frac{4 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} a^{4} b^{3}}{{\left (b^{2} x + a b\right )}^{2}} + 4 \, a^{4} b^{3} - \frac{11 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} a^{3} b^{3}}{b^{2} x + a b} - \frac{5 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{3} a^{3} b^{3}}{{\left (b^{2} x + a b\right )}^{3}} + \frac{7 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} a^{2} b^{3}}{{\left (b^{2} x + a b\right )}^{2}} - a^{2} b^{3} + \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} a b^{3}}{b^{2} x + a b} + \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{3} a b^{3}}{{\left (b^{2} x + a b\right )}^{3}} - \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} b^{3}}{{\left (b^{2} x + a b\right )}^{2}}}{{\left (a^{6}{\left | b \right |} - 2 \, a^{4}{\left | b \right |} + a^{2}{\left | b \right |}\right )}{\left (\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}^{2}}\right )} - \frac{\arcsin \left (b x + a\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*((2*a^2*b^3 + b^3)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - 1)/sqrt(a^2
- 1))/((a^4*abs(b) - 2*a^2*abs(b) + abs(b))*sqrt(a^2 - 1)) - (4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) +
b)^2*a^4*b^3/(b^2*x + a*b)^2 + 4*a^4*b^3 - 11*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^3*b^3/(b^2*x +
a*b) - 5*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^3*b^3/(b^2*x + a*b)^3 + 7*(sqrt(-b^2*x^2 - 2*a*b
*x - a^2 + 1)*abs(b) + b)^2*a^2*b^3/(b^2*x + a*b)^2 - a^2*b^3 + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) +
b)*a*b^3/(b^2*x + a*b) + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a*b^3/(b^2*x + a*b)^3 - 2*(sqrt(
-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*b^3/(b^2*x + a*b)^2)/((a^6*abs(b) - 2*a^4*abs(b) + a^2*abs(b))*((s
qrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + a - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*
abs(b) + b)/(b^2*x + a*b))^2)) - 1/3*arcsin(b*x + a)/x^3