Integrand size = 14, antiderivative size = 227 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=-\frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )}+\frac {x \operatorname {CosIntegral}\left (\frac {a+b \arcsin \left (1+d x^2\right )}{2 b}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )}+\frac {x \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \arcsin \left (1+d x^2\right )}{2 b}\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
1/8*x/b^2/(a+b*arcsin(d*x^2+1))+1/16*x*Ci(1/2*(a+b*arcsin(d*x^2+1))/b)*(co s(1/2*a/b)-sin(1/2*a/b))/b^3/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^ 2+1)))+1/16*x*Si(1/2*(a+b*arcsin(d*x^2+1))/b)*(cos(1/2*a/b)+sin(1/2*a/b))/ b^3/(cos(1/2*arcsin(d*x^2+1))-sin(1/2*arcsin(d*x^2+1)))-1/4*(-d^2*x^4-2*d* x^2)^(1/2)/b/d/x/(a+b*arcsin(d*x^2+1))^2
Time = 0.34 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=-\frac {\sqrt {-d x^2 \left (2+d x^2\right )}}{4 b d x \left (a+b \arcsin \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \arcsin \left (1+d x^2\right )\right )}+\frac {x \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (\frac {a}{b}+\arcsin \left (1+d x^2\right )\right )\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )+\left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {1}{2} \left (\frac {a}{b}+\arcsin \left (1+d x^2\right )\right )\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1+d x^2\right )\right )\right )} \]
-1/4*Sqrt[-(d*x^2*(2 + d*x^2))]/(b*d*x*(a + b*ArcSin[1 + d*x^2])^2) + x/(8 *b^2*(a + b*ArcSin[1 + d*x^2])) + (x*(CosIntegral[(a/b + ArcSin[1 + d*x^2] )/2]*(Cos[a/(2*b)] - Sin[a/(2*b)]) + (Cos[a/(2*b)] + Sin[a/(2*b)])*SinInte gral[(a/b + ArcSin[1 + d*x^2])/2]))/(16*b^3*(Cos[ArcSin[1 + d*x^2]/2] - Si n[ArcSin[1 + d*x^2]/2]))
Time = 0.34 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5327, 5315}
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.
\(\displaystyle \int \frac {1}{\left (a+b \arcsin \left (d x^2+1\right )\right )^3} \, dx\) |
\(\Big \downarrow \) 5327 |
\(\displaystyle -\frac {\int \frac {1}{a+b \arcsin \left (d x^2+1\right )}dx}{8 b^2}+\frac {x}{8 b^2 \left (a+b \arcsin \left (d x^2+1\right )\right )}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^2}\) |
\(\Big \downarrow \) 5315 |
\(\displaystyle -\frac {-\frac {x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin \left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}-\frac {x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {a+b \arcsin \left (d x^2+1\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (d x^2+1\right )\right )\right )}}{8 b^2}+\frac {x}{8 b^2 \left (a+b \arcsin \left (d x^2+1\right )\right )}-\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \arcsin \left (d x^2+1\right )\right )^2}\) |
-1/4*Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*(a + b*ArcSin[1 + d*x^2])^2) + x/(8*b ^2*(a + b*ArcSin[1 + d*x^2])) - (-1/2*(x*CosIntegral[(a + b*ArcSin[1 + d*x ^2])/(2*b)]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(b*(Cos[ArcSin[1 + d*x^2]/2] - Sin[ArcSin[1 + d*x^2]/2])) - (x*(Cos[a/(2*b)] + Sin[a/(2*b)])*SinIntegral[ (a + b*ArcSin[1 + d*x^2])/(2*b)])/(2*b*(Cos[ArcSin[1 + d*x^2]/2] - Sin[Arc Sin[1 + d*x^2]/2])))/(8*b^2)
3.5.7.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(-1), x_Symbol] :> Simp[(-x )*(c*Cos[a/(2*b)] - Sin[a/(2*b)])*(CosIntegral[(c/(2*b))*(a + b*ArcSin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] - Simp[x*(c*Cos[a/(2*b)] + Sin[a/(2*b)])*(SinIntegral[(c/(2*b))*(a + b*Arc Sin[c + d*x^2])]/(2*b*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2 ]))), x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*( (a + b*ArcSin[c + d*x^2])^(n + 2)/(4*b^2*(n + 1)*(n + 2))), x] + (Simp[Sqrt [-2*c*d*x^2 - d^2*x^4]*((a + b*ArcSin[c + d*x^2])^(n + 1)/(2*b*d*(n + 1)*x) ), x] - Simp[1/(4*b^2*(n + 1)*(n + 2)) Int[(a + b*ArcSin[c + d*x^2])^(n + 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[ n, -2]
\[\int \frac {1}{{\left (a +b \arcsin \left (d \,x^{2}+1\right )\right )}^{3}}d x\]
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{3}} \,d x } \]
integral(1/(b^3*arcsin(d*x^2 + 1)^3 + 3*a*b^2*arcsin(d*x^2 + 1)^2 + 3*a^2* b*arcsin(d*x^2 + 1) + a^3), x)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {asin}{\left (d x^{2} + 1 \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
\[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \arcsin \left (d x^{2} + 1\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (a+b \arcsin \left (1+d x^2\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asin}\left (d\,x^2+1\right )\right )}^3} \,d x \]