Integrand size = 20, antiderivative size = 275 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{4 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^2}-\frac {x}{8 b^2 \left (a+i b \arcsin \left (1-i d x^2\right )\right )}+\frac {x \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {x \left (i \cosh \left (\frac {a}{2 b}\right )+\sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{16 b^3 \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )} \]
-1/8*x/b^2/(a-I*b*arcsin(-1+I*d*x^2))+1/16*x*Ci(-1/2*I*(a-I*b*arcsin(-1+I* d*x^2))/b)*(I*cosh(1/2*a/b)-sinh(1/2*a/b))/b^3/(cos(1/2*arcsin(-1+I*d*x^2) )+sin(1/2*arcsin(-1+I*d*x^2)))-1/16*x*Si(1/2*I*a/b+1/2*arcsin(-1+I*d*x^2)) *(I*cosh(1/2*a/b)+sinh(1/2*a/b))/b^3/(cos(1/2*arcsin(-1+I*d*x^2))+sin(1/2* arcsin(-1+I*d*x^2)))-1/4*(2*I*d*x^2+d^2*x^4)^(1/2)/b/d/x/(a-I*b*arcsin(-1+ I*d*x^2))^2
Time = 0.40 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\frac {-\frac {8 b^2 \sqrt {d x^2 \left (2 i+d x^2\right )}}{d \left (a+i b \arcsin \left (1-i d x^2\right )\right )^2}-\frac {4 b x^2}{a+i b \arcsin \left (1-i d x^2\right )}+\frac {2 i x^2 \left (\operatorname {CosIntegral}\left (\frac {1}{2} \left (-\frac {i a}{b}+\arcsin \left (1-i d x^2\right )\right )\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )-\left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}{\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}}{32 b^3 x} \]
((-8*b^2*Sqrt[d*x^2*(2*I + d*x^2)])/(d*(a + I*b*ArcSin[1 - I*d*x^2])^2) - (4*b*x^2)/(a + I*b*ArcSin[1 - I*d*x^2]) + ((2*I)*x^2*(CosIntegral[(((-I)*a )/b + ArcSin[1 - I*d*x^2])/2]*(Cosh[a/(2*b)] + I*Sinh[a/(2*b)]) - (Cosh[a/ (2*b)] - I*Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2] ))/(Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2]))/(32*b^3*x)
Time = 0.37 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx\) |
| \(\Big \downarrow \) 5327 |
| \(\displaystyle \frac {\int \frac {1}{a+i b \arcsin \left (1-i d x^2\right )}dx}{8 b^2}-\frac {x}{8 b^2 \left (a+i b \arcsin \left (1-i d x^2\right )\right )}-\frac {\sqrt {d^2 x^4+2 i d x^2}}{4 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^2}\) |
| \(\Big \downarrow \) 5315 |
| \(\displaystyle \frac {\frac {x \left (-\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \operatorname {CosIntegral}\left (-\frac {i \left (a+i b \arcsin \left (1-i d x^2\right )\right )}{2 b}\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}-\frac {x \left (\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {Si}\left (\frac {i a}{2 b}-\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )}{2 b \left (\cos \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-i d x^2\right )\right )\right )}}{8 b^2}-\frac {x}{8 b^2 \left (a+i b \arcsin \left (1-i d x^2\right )\right )}-\frac {\sqrt {d^2 x^4+2 i d x^2}}{4 b d x \left (a+i b \arcsin \left (1-i d x^2\right )\right )^2}\) |
-1/4*Sqrt[(2*I)*d*x^2 + d^2*x^4]/(b*d*x*(a + I*b*ArcSin[1 - I*d*x^2])^2) - x/(8*b^2*(a + I*b*ArcSin[1 - I*d*x^2])) + ((x*CosIntegral[((-1/2*I)*(a + I*b*ArcSin[1 - I*d*x^2]))/b]*(I*Cosh[a/(2*b)] - Sinh[a/(2*b)]))/(2*b*(Cos[ ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])) - (x*(I*Cosh[a/(2*b) ] + Sinh[a/(2*b)])*SinIntegral[((I/2)*a)/b - ArcSin[1 - I*d*x^2]/2])/(2*b* (Cos[ArcSin[1 - I*d*x^2]/2] - Sin[ArcSin[1 - I*d*x^2]/2])))/(8*b^2)
3.4.20.3.1 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]
\[\int \frac {1}{{\left (a +b \,\operatorname {arcsinh}\left (d \,x^{2}+i\right )\right )}^{3}}d x\]
\[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{3}} \,d x } \]
-1/8*(b*d*x*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a*d*x - 8*(b^4*d*lo g(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I)^2 + 2*a*b^3*d*log(d*x^2 + sqrt(d^2* x^2 + 2*I*d)*x + I) + a^2*b^2*d)*integral(1/8/(b^3*log(d*x^2 + sqrt(d^2*x^ 2 + 2*I*d)*x + I) + a*b^2), x) + 2*sqrt(d^2*x^2 + 2*I*d)*b)/(b^4*d*log(d*x ^2 + sqrt(d^2*x^2 + 2*I*d)*x + I)^2 + 2*a*b^3*d*log(d*x^2 + sqrt(d^2*x^2 + 2*I*d)*x + I) + a^2*b^2*d)
Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (d x^{2} + i\right ) + a\right )}^{3}} \,d x } \]
-1/8*((a*d^(11/2) + 2*b*d^(11/2))*x^10 - 2*(-3*I*a*d^(9/2) - 7*I*b*d^(9/2) )*x^8 - (11*a*d^(7/2) + 36*b*d^(7/2))*x^6 - 2*(I*a*d^(5/2) + 20*I*b*d^(5/2 ))*x^4 - 4*(3*a*d^(3/2) - 4*b*d^(3/2))*x^2 + ((a*d^4 + 2*b*d^4)*x^7 - (-3* I*a*d^3 - 8*I*b*d^3)*x^5 - 2*(2*a*d^2 + 5*b*d^2)*x^3 - 4*(I*a*d + I*b*d)*x )*(d*x^2 + 2*I)^(3/2) + (3*(a*d^(9/2) + 2*b*d^(9/2))*x^8 - 6*(-2*I*a*d^(7/ 2) - 5*I*b*d^(7/2))*x^6 - 2*(8*a*d^(5/2) + 25*b*d^(5/2))*x^4 - 10*(I*a*d^( 3/2) + 3*I*b*d^(3/2))*x^2 + 4*a*sqrt(d) + 4*b*sqrt(d))*(d*x^2 + 2*I) + (b* d^(11/2)*x^10 + 6*I*b*d^(9/2)*x^8 - 11*b*d^(7/2)*x^6 - 2*I*b*d^(5/2)*x^4 - 12*b*d^(3/2)*x^2 + (b*d^4*x^7 + 3*I*b*d^3*x^5 - 4*b*d^2*x^3 - 4*I*b*d*x)* (d*x^2 + 2*I)^(3/2) + (3*b*d^(9/2)*x^8 + 12*I*b*d^(7/2)*x^6 - 16*b*d^(5/2) *x^4 - 10*I*b*d^(3/2)*x^2 + 4*b*sqrt(d))*(d*x^2 + 2*I) + (3*b*d^5*x^9 + 15 *I*b*d^4*x^7 - 23*b*d^3*x^5 - 7*I*b*d^2*x^3 - 6*b*d*x)*sqrt(d*x^2 + 2*I) - 8*I*b*sqrt(d))*log(d*x^2 + sqrt(d*x^2 + 2*I)*sqrt(d)*x + I) + (3*(a*d^5 + 2*b*d^5)*x^9 - 3*(-5*I*a*d^4 - 12*I*b*d^4)*x^7 - (23*a*d^3 + 76*b*d^3)*x^ 5 - (7*I*a*d^2 + 64*I*b*d^2)*x^3 - 2*(3*a*d - 8*b*d)*x)*sqrt(d*x^2 + 2*I) - 8*I*a*sqrt(d))/(a^2*b^2*d^(11/2)*x^9 + 6*I*a^2*b^2*d^(9/2)*x^7 - 12*a^2* b^2*d^(7/2)*x^5 - 8*I*a^2*b^2*d^(5/2)*x^3 + (b^4*d^(11/2)*x^9 + 6*I*b^4*d^ (9/2)*x^7 - 12*b^4*d^(7/2)*x^5 - 8*I*b^4*d^(5/2)*x^3 + (b^4*d^4*x^6 + 3*I* b^4*d^3*x^4 - 3*b^4*d^2*x^2 - I*b^4*d)*(d*x^2 + 2*I)^(3/2) + 3*(b^4*d^(9/2 )*x^7 + 4*I*b^4*d^(7/2)*x^5 - 5*b^4*d^(5/2)*x^3 - 2*I*b^4*d^(3/2)*x)*(d...
Exception generated. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{\left (a+i b \arcsin \left (1-i d x^2\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^3} \,d x \]