Integrand size = 20, antiderivative size = 50 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\frac {5 c^2 \operatorname {CosIntegral}(\arcsin (a x))}{8 a}+\frac {5 c^2 \operatorname {CosIntegral}(3 \arcsin (a x))}{16 a}+\frac {c^2 \operatorname {CosIntegral}(5 \arcsin (a x))}{16 a} \] Output:
5/8*c^2*Ci(arcsin(a*x))/a+5/16*c^2*Ci(3*arcsin(a*x))/a+1/16*c^2*Ci(5*arcsi n(a*x))/a
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\frac {c^2 (10 \operatorname {CosIntegral}(\arcsin (a x))+5 \operatorname {CosIntegral}(3 \arcsin (a x))+\operatorname {CosIntegral}(5 \arcsin (a x)))}{16 a} \] Input:
Integrate[(c - a^2*c*x^2)^2/ArcSin[a*x],x]
Output:
(c^2*(10*CosIntegral[ArcSin[a*x]] + 5*CosIntegral[3*ArcSin[a*x]] + CosInte gral[5*ArcSin[a*x]]))/(16*a)
Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5168, 3042, 3793, 2009}
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.
\(\displaystyle \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx\) |
\(\Big \downarrow \) 5168 |
\(\displaystyle \frac {c^2 \int \frac {\left (1-a^2 x^2\right )^{5/2}}{\arcsin (a x)}d\arcsin (a x)}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c^2 \int \frac {\sin \left (\arcsin (a x)+\frac {\pi }{2}\right )^5}{\arcsin (a x)}d\arcsin (a x)}{a}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {c^2 \int \left (\frac {5 \cos (3 \arcsin (a x))}{16 \arcsin (a x)}+\frac {\cos (5 \arcsin (a x))}{16 \arcsin (a x)}+\frac {5 \sqrt {1-a^2 x^2}}{8 \arcsin (a x)}\right )d\arcsin (a x)}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c^2 \left (\frac {5}{8} \operatorname {CosIntegral}(\arcsin (a x))+\frac {5}{16} \operatorname {CosIntegral}(3 \arcsin (a x))+\frac {1}{16} \operatorname {CosIntegral}(5 \arcsin (a x))\right )}{a}\) |
Input:
Int[(c - a^2*c*x^2)^2/ArcSin[a*x],x]
Output:
(c^2*((5*CosIntegral[ArcSin[a*x]])/8 + (5*CosIntegral[3*ArcSin[a*x]])/16 + CosIntegral[5*ArcSin[a*x]]/16))/a
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[(1/(b*c))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Subst[Int[ x^n*Cos[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b , c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p, 0]
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {c^{2} \left (10 \,\operatorname {Ci}\left (\arcsin \left (a x \right )\right )+5 \,\operatorname {Ci}\left (3 \arcsin \left (a x \right )\right )+\operatorname {Ci}\left (5 \arcsin \left (a x \right )\right )\right )}{16 a}\) | \(33\) |
default | \(\frac {c^{2} \left (10 \,\operatorname {Ci}\left (\arcsin \left (a x \right )\right )+5 \,\operatorname {Ci}\left (3 \arcsin \left (a x \right )\right )+\operatorname {Ci}\left (5 \arcsin \left (a x \right )\right )\right )}{16 a}\) | \(33\) |
Input:
int((-a^2*c*x^2+c)^2/arcsin(a*x),x,method=_RETURNVERBOSE)
Output:
1/16/a*c^2*(10*Ci(arcsin(a*x))+5*Ci(3*arcsin(a*x))+Ci(5*arcsin(a*x)))
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\int { \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\arcsin \left (a x\right )} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^2/arcsin(a*x),x, algorithm="fricas")
Output:
integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arcsin(a*x), x)
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {asin}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asin}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asin}{\left (a x \right )}}\, dx\right ) \] Input:
integrate((-a**2*c*x**2+c)**2/asin(a*x),x)
Output:
c**2*(Integral(-2*a**2*x**2/asin(a*x), x) + Integral(a**4*x**4/asin(a*x), x) + Integral(1/asin(a*x), x))
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\int { \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\arcsin \left (a x\right )} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^2/arcsin(a*x),x, algorithm="maxima")
Output:
integrate((a^2*c*x^2 - c)^2/arcsin(a*x), x)
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\frac {c^{2} \operatorname {Ci}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a} + \frac {5 \, c^{2} \operatorname {Ci}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a} + \frac {5 \, c^{2} \operatorname {Ci}\left (\arcsin \left (a x\right )\right )}{8 \, a} \] Input:
integrate((-a^2*c*x^2+c)^2/arcsin(a*x),x, algorithm="giac")
Output:
1/16*c^2*cos_integral(5*arcsin(a*x))/a + 5/16*c^2*cos_integral(3*arcsin(a* x))/a + 5/8*c^2*cos_integral(arcsin(a*x))/a
Timed out. \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{\mathrm {asin}\left (a\,x\right )} \,d x \] Input:
int((c - a^2*c*x^2)^2/asin(a*x),x)
Output:
int((c - a^2*c*x^2)^2/asin(a*x), x)
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)} \, dx=c^{2} \left (\left (\int \frac {x^{4}}{\mathit {asin} \left (a x \right )}d x \right ) a^{4}-2 \left (\int \frac {x^{2}}{\mathit {asin} \left (a x \right )}d x \right ) a^{2}+\int \frac {1}{\mathit {asin} \left (a x \right )}d x \right ) \] Input:
int((-a^2*c*x^2+c)^2/asin(a*x),x)
Output:
c**2*(int(x**4/asin(a*x),x)*a**4 - 2*int(x**2/asin(a*x),x)*a**2 + int(1/as in(a*x),x))