Integrand size = 8, antiderivative size = 65 \[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=-\frac {\sqrt {1-(a+b x)^2}}{2 b \arcsin (a+b x)^2}+\frac {a+b x}{2 b \arcsin (a+b x)}-\frac {\operatorname {CosIntegral}(\arcsin (a+b x))}{2 b} \] Output:
-1/2*(1-(b*x+a)^2)^(1/2)/b/arcsin(b*x+a)^2+1/2*(b*x+a)/b/arcsin(b*x+a)-1/2 *Ci(arcsin(b*x+a))/b
Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=-\frac {\sqrt {1-(a+b x)^2}}{2 b \arcsin (a+b x)^2}+\frac {a+b x}{2 b \arcsin (a+b x)}-\frac {\operatorname {CosIntegral}(\arcsin (a+b x))}{2 b} \] Input:
Integrate[ArcSin[a + b*x]^(-3),x]
Output:
-1/2*Sqrt[1 - (a + b*x)^2]/(b*ArcSin[a + b*x]^2) + (a + b*x)/(2*b*ArcSin[a + b*x]) - CosIntegral[ArcSin[a + b*x]]/(2*b)
Time = 0.41 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5302, 5132, 5222, 5134, 3042, 3783}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\arcsin (a+b x)^3} \, dx\) |
\(\Big \downarrow \) 5302 |
\(\displaystyle \frac {\int \frac {1}{\arcsin (a+b x)^3}d(a+b x)}{b}\) |
\(\Big \downarrow \) 5132 |
\(\displaystyle \frac {-\frac {1}{2} \int \frac {a+b x}{\sqrt {1-(a+b x)^2} \arcsin (a+b x)^2}d(a+b x)-\frac {\sqrt {1-(a+b x)^2}}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 5222 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {a+b x}{\arcsin (a+b x)}-\int \frac {1}{\arcsin (a+b x)}d(a+b x)\right )-\frac {\sqrt {1-(a+b x)^2}}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 5134 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {a+b x}{\arcsin (a+b x)}-\int \frac {\sqrt {1-(a+b x)^2}}{\arcsin (a+b x)}d\arcsin (a+b x)\right )-\frac {\sqrt {1-(a+b x)^2}}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {a+b x}{\arcsin (a+b x)}-\int \frac {\sin \left (\arcsin (a+b x)+\frac {\pi }{2}\right )}{\arcsin (a+b x)}d\arcsin (a+b x)\right )-\frac {\sqrt {1-(a+b x)^2}}{2 \arcsin (a+b x)^2}}{b}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {a+b x}{\arcsin (a+b x)}-\operatorname {CosIntegral}(\arcsin (a+b x))\right )-\frac {\sqrt {1-(a+b x)^2}}{2 \arcsin (a+b x)^2}}{b}\) |
Input:
Int[ArcSin[a + b*x]^(-3),x]
Output:
(-1/2*Sqrt[1 - (a + b*x)^2]/ArcSin[a + b*x]^2 + ((a + b*x)/ArcSin[a + b*x] - CosIntegral[ArcSin[a + b*x]])/2)/b
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
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] + Simp[c/(b*(n + 1)) Int[x*((a + b*ArcSin[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; FreeQ[{a , b, c}, x] && LtQ[n, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[1/(b*c) Su bst[Int[x^n*Cos[-a/b + x/b], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^ 2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] - Simp[f*(m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] 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]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d Subst[Int[(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arcsin \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arcsin \left (b x +a \right )}-\frac {\operatorname {Ci}\left (\arcsin \left (b x +a \right )\right )}{2}}{b}\) | \(53\) |
default | \(\frac {-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arcsin \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arcsin \left (b x +a \right )}-\frac {\operatorname {Ci}\left (\arcsin \left (b x +a \right )\right )}{2}}{b}\) | \(53\) |
Input:
int(1/arcsin(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/2/arcsin(b*x+a)^2*(1-(b*x+a)^2)^(1/2)+1/2*(b*x+a)/arcsin(b*x+a)-1/ 2*Ci(arcsin(b*x+a)))
\[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=\int { \frac {1}{\arcsin \left (b x + a\right )^{3}} \,d x } \] Input:
integrate(1/arcsin(b*x+a)^3,x, algorithm="fricas")
Output:
integral(arcsin(b*x + a)^(-3), x)
\[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=\int \frac {1}{\operatorname {asin}^{3}{\left (a + b x \right )}}\, dx \] Input:
integrate(1/asin(b*x+a)**3,x)
Output:
Integral(asin(a + b*x)**(-3), x)
\[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=\int { \frac {1}{\arcsin \left (b x + a\right )^{3}} \,d x } \] Input:
integrate(1/arcsin(b*x+a)^3,x, algorithm="maxima")
Output:
-1/2*(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2*integrate (1/arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)), x) - (b*x + a)* arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)) + sqrt(b*x + a + 1) *sqrt(-b*x - a + 1))/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=-\frac {\operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b} + \frac {b x + a}{2 \, b \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{2 \, b \arcsin \left (b x + a\right )^{2}} \] Input:
integrate(1/arcsin(b*x+a)^3,x, algorithm="giac")
Output:
-1/2*cos_integral(arcsin(b*x + a))/b + 1/2*(b*x + a)/(b*arcsin(b*x + a)) - 1/2*sqrt(-(b*x + a)^2 + 1)/(b*arcsin(b*x + a)^2)
Timed out. \[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=\int \frac {1}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \] Input:
int(1/asin(a + b*x)^3,x)
Output:
int(1/asin(a + b*x)^3, x)
\[ \int \frac {1}{\arcsin (a+b x)^3} \, dx=\int \frac {1}{\mathit {asin} \left (b x +a \right )^{3}}d x \] Input:
int(1/asin(b*x+a)^3,x)
Output:
int(1/asin(a + b*x)**3,x)