Integrand size = 25, antiderivative size = 206 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\frac {15 \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c}+\frac {3 \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 b c}+\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c}+\frac {5 \log (a+b \arcsin (c x))}{16 b c}+\frac {15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c}+\frac {3 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 b c}+\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c} \] Output:
15/32*cos(2*a/b)*Ci(2*(a+b*arcsin(c*x))/b)/b/c+3/16*cos(4*a/b)*Ci(4*(a+b*a rcsin(c*x))/b)/b/c+1/32*cos(6*a/b)*Ci(6*(a+b*arcsin(c*x))/b)/b/c+5/16*ln(a +b*arcsin(c*x))/b/c+15/32*sin(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b/c+3/16*si n(4*a/b)*Si(4*(a+b*arcsin(c*x))/b)/b/c+1/32*sin(6*a/b)*Si(6*(a+b*arcsin(c* x))/b)/b/c
Time = 0.55 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.80 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\frac {15 \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+6 \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+18 \log (a+b \arcsin (c x))-8 \log (8 (a+b \arcsin (c x)))+15 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+6 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{32 b c} \] Input:
Integrate[(1 - c^2*x^2)^(5/2)/(a + b*ArcSin[c*x]),x]
Output:
(15*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + 6*Cos[(4*a)/b]*CosIn tegral[4*(a/b + ArcSin[c*x])] + Cos[(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c *x])] + 18*Log[a + b*ArcSin[c*x]] - 8*Log[8*(a + b*ArcSin[c*x])] + 15*Sin[ (2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 6*Sin[(4*a)/b]*SinIntegral[4 *(a/b + ArcSin[c*x])] + Sin[(6*a)/b]*SinIntegral[6*(a/b + ArcSin[c*x])])/( 32*b*c)
Time = 0.49 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.
| \(\displaystyle \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx\) |
| \(\Big \downarrow \) 5168 |
| \(\displaystyle \frac {\int \frac {\cos ^6\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}+\frac {\pi }{2}\right )^6}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle \frac {\int \left (\frac {\cos \left (\frac {6 a}{b}-\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {3 \cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {15 \cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {5}{16 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {15}{32} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {3}{16} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )+\frac {15}{32} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {3}{16} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )+\frac {5}{16} \log (a+b \arcsin (c x))}{b c}\) |
Input:
Int[(1 - c^2*x^2)^(5/2)/(a + b*ArcSin[c*x]),x]
Output:
((15*Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/32 + (3*Cos[(4*a )/b]*CosIntegral[(4*(a + b*ArcSin[c*x]))/b])/16 + (Cos[(6*a)/b]*CosIntegra l[(6*(a + b*ArcSin[c*x]))/b])/32 + (5*Log[a + b*ArcSin[c*x]])/16 + (15*Sin [(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/32 + (3*Sin[(4*a)/b]*Sin Integral[(4*(a + b*ArcSin[c*x]))/b])/16 + (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[c*x]))/b])/32)/(b*c)
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 = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {\operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )+\operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )+6 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+6 \,\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )+15 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+15 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+10 \ln \left (a +b \arcsin \left (c x \right )\right )}{32 c b}\) | \(157\) |
Input:
int((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
Output:
1/32/c*(Si(6*arcsin(c*x)+6*a/b)*sin(6*a/b)+Ci(6*arcsin(c*x)+6*a/b)*cos(6*a /b)+6*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+6*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a /b)+15*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)+15*Ci(2*arcsin(c*x)+2*a/b)*cos(2 *a/b)+10*ln(a+b*arcsin(c*x)))/b
\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \arcsin \left (c x\right ) + a} \,d x } \] Input:
integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")
Output:
integral((c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-c^2*x^2 + 1)/(b*arcsin(c*x) + a), x)
\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \] Input:
integrate((-c**2*x**2+1)**(5/2)/(a+b*asin(c*x)),x)
Output:
Integral((-(c*x - 1)*(c*x + 1))**(5/2)/(a + b*asin(c*x)), x)
\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{b \arcsin \left (c x\right ) + a} \,d x } \] Input:
integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")
Output:
integrate((-c^2*x^2 + 1)^(5/2)/(b*arcsin(c*x) + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (192) = 384\).
Time = 0.16 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.29 \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx =\text {Too large to display} \] Input:
integrate((-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="giac")
Output:
cos(a/b)^6*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) + cos(a/b)^5*sin(a/b) *sin_integral(6*a/b + 6*arcsin(c*x))/(b*c) - 3/2*cos(a/b)^4*cos_integral(6 *a/b + 6*arcsin(c*x))/(b*c) + 3/2*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin (c*x))/(b*c) - cos(a/b)^3*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b* c) + 3/2*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c) + 9 /16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) - 3/2*cos(a/b)^2* cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) + 15/16*cos(a/b)^2*cos_integral( 2*a/b + 2*arcsin(c*x))/(b*c) + 3/16*cos(a/b)*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c) - 3/4*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsi n(c*x))/(b*c) + 15/16*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x) )/(b*c) - 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c) + 3/16*cos_integr al(4*a/b + 4*arcsin(c*x))/(b*c) - 15/32*cos_integral(2*a/b + 2*arcsin(c*x) )/(b*c) + 5/16*log(b*arcsin(c*x) + a)/(b*c)
Timed out. \[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{5/2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \] Input:
int((1 - c^2*x^2)^(5/2)/(a + b*asin(c*x)),x)
Output:
int((1 - c^2*x^2)^(5/2)/(a + b*asin(c*x)), x)
\[ \int \frac {\left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {\sqrt {-c^{2} x^{2}+1}}{\mathit {asin} \left (c x \right ) b +a}d x +\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{4}}{\mathit {asin} \left (c x \right ) b +a}d x \right ) c^{4}-2 \left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {asin} \left (c x \right ) b +a}d x \right ) c^{2} \] Input:
int((-c^2*x^2+1)^(5/2)/(a+b*asin(c*x)),x)
Output:
int(sqrt( - c**2*x**2 + 1)/(asin(c*x)*b + a),x) + int((sqrt( - c**2*x**2 + 1)*x**4)/(asin(c*x)*b + a),x)*c**4 - 2*int((sqrt( - c**2*x**2 + 1)*x**2)/ (asin(c*x)*b + a),x)*c**2