Integrand size = 28, antiderivative size = 268 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\cos \left (\frac {8 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {8 (a+b \arcsin (c x))}{b}\right )}{128 b c^3}+\frac {5 \log (a+b \arcsin (c x))}{128 b c^3}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c^3}-\frac {\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 (a+b \arcsin (c x))}{b}\right )}{128 b c^3} \]
1/32*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c^3-1/32*Ci(4*(a+b*arcsin(c*x) )/b)*cos(4*a/b)/b/c^3-1/32*Ci(6*(a+b*arcsin(c*x))/b)*cos(6*a/b)/b/c^3-1/12 8*Ci(8*(a+b*arcsin(c*x))/b)*cos(8*a/b)/b/c^3+5/128*ln(a+b*arcsin(c*x))/b/c ^3+1/32*Si(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b/c^3-1/32*Si(4*(a+b*arcsin(c *x))/b)*sin(4*a/b)/b/c^3-1/32*Si(6*(a+b*arcsin(c*x))/b)*sin(6*a/b)/b/c^3-1 /128*Si(8*(a+b*arcsin(c*x))/b)*sin(8*a/b)/b/c^3
Time = 0.87 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.78 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=-\frac {-4 \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+4 \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+4 \cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\cos \left (\frac {8 a}{b}\right ) \operatorname {CosIntegral}\left (8 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+11 \log (a+b \arcsin (c x))-16 \log (8 (a+b \arcsin (c x)))-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+4 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+4 \sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\sin \left (\frac {8 a}{b}\right ) \text {Si}\left (8 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{128 b c^3} \]
-1/128*(-4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + 4*Cos[(4*a)/b ]*CosIntegral[4*(a/b + ArcSin[c*x])] + 4*Cos[(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c*x])] + Cos[(8*a)/b]*CosIntegral[8*(a/b + ArcSin[c*x])] + 11*Log[ a + b*ArcSin[c*x]] - 16*Log[8*(a + b*ArcSin[c*x])] - 4*Sin[(2*a)/b]*SinInt egral[2*(a/b + ArcSin[c*x])] + 4*Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[ c*x])] + 4*Sin[(6*a)/b]*SinIntegral[6*(a/b + ArcSin[c*x])] + Sin[(8*a)/b]* SinIntegral[8*(a/b + ArcSin[c*x])])/(b*c^3)
Time = 0.59 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5224, 4906, 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.
| \(\displaystyle \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx\) |
| \(\Big \downarrow \) 5224 |
| \(\displaystyle \frac {\int \frac {\cos ^6\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^2\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^3}\) |
| \(\Big \downarrow \) 4906 |
| \(\displaystyle \frac {\int \left (-\frac {\cos \left (\frac {8 a}{b}-\frac {8 (a+b \arcsin (c x))}{b}\right )}{128 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {6 a}{b}-\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {\cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {5}{128 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{32} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )-\frac {1}{128} \cos \left (\frac {8 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {8 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )-\frac {1}{128} \sin \left (\frac {8 a}{b}\right ) \text {Si}\left (\frac {8 (a+b \arcsin (c x))}{b}\right )+\frac {5}{128} \log (a+b \arcsin (c x))}{b c^3}\) |
((Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/32 - (Cos[(4*a)/b]* CosIntegral[(4*(a + b*ArcSin[c*x]))/b])/32 - (Cos[(6*a)/b]*CosIntegral[(6* (a + b*ArcSin[c*x]))/b])/32 - (Cos[(8*a)/b]*CosIntegral[(8*(a + b*ArcSin[c *x]))/b])/128 + (5*Log[a + b*ArcSin[c*x]])/128 + (Sin[(2*a)/b]*SinIntegral [(2*(a + b*ArcSin[c*x]))/b])/32 - (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcS in[c*x]))/b])/32 - (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[c*x]))/b])/3 2 - (Sin[(8*a)/b]*SinIntegral[(8*(a + b*ArcSin[c*x]))/b])/128)/(b*c^3)
3.4.34.3.1 Defintions of rubi rules used
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x ^2)^p] Subst[Int[x^n*Sin[-a/b + x/b]^m*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 + 2, 0] && IGtQ[m, 0]
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(-\frac {4 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+4 \,\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-4 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {Si}\left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \sin \left (\frac {8 a}{b}\right )+\operatorname {Ci}\left (8 \arcsin \left (c x \right )+\frac {8 a}{b}\right ) \cos \left (\frac {8 a}{b}\right )-4 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+4 \,\operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )+4 \,\operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )-5 \ln \left (a +b \arcsin \left (c x \right )\right )}{128 c^{3} b}\) | \(203\) |
-1/128/c^3*(4*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+4*Ci(4*arcsin(c*x)+4*a/b) *cos(4*a/b)-4*Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)+Si(8*arcsin(c*x)+8*a/b)*s in(8*a/b)+Ci(8*arcsin(c*x)+8*a/b)*cos(8*a/b)-4*Ci(2*arcsin(c*x)+2*a/b)*cos (2*a/b)+4*Si(6*arcsin(c*x)+6*a/b)*sin(6*a/b)+4*Ci(6*arcsin(c*x)+6*a/b)*cos (6*a/b)-5*ln(a+b*arcsin(c*x)))/b
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}} x^{2}}{b \arcsin \left (c x\right ) + a} \,d x } \]
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {x^{2} \left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \]
\[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {5}{2}} x^{2}}{b \arcsin \left (c x\right ) + a} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 757 vs. \(2 (250) = 500\).
Time = 0.34 (sec) , antiderivative size = 757, normalized size of antiderivative = 2.82 \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\text {Too large to display} \]
-cos(a/b)^8*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) - cos(a/b)^7*sin(a /b)*sin_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) + 2*cos(a/b)^6*cos_integra l(8*a/b + 8*arcsin(c*x))/(b*c^3) - cos(a/b)^6*cos_integral(6*a/b + 6*arcsi n(c*x))/(b*c^3) + 3/2*cos(a/b)^5*sin(a/b)*sin_integral(8*a/b + 8*arcsin(c* x))/(b*c^3) - cos(a/b)^5*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c ^3) - 5/4*cos(a/b)^4*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) + 3/2*cos (a/b)^4*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/4*cos(a/b)^4*cos_i ntegral(4*a/b + 4*arcsin(c*x))/(b*c^3) - 5/8*cos(a/b)^3*sin(a/b)*sin_integ ral(8*a/b + 8*arcsin(c*x))/(b*c^3) + cos(a/b)^3*sin(a/b)*sin_integral(6*a/ b + 6*arcsin(c*x))/(b*c^3) - 1/4*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/4*cos(a/b)^2*cos_integral(8*a/b + 8*arcsin(c*x) )/(b*c^3) - 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) + 1/4*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b) ^2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)*sin(a/b)*si n_integral(8*a/b + 8*arcsin(c*x))/(b*c^3) - 3/16*cos(a/b)*sin(a/b)*sin_int egral(6*a/b + 6*arcsin(c*x))/(b*c^3) + 1/8*cos(a/b)*sin(a/b)*sin_integral( 4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^3) - 1/128*cos_integral(8*a/b + 8*arcsin(c*x))/(b*c ^3) + 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/32*cos_integral (4*a/b + 4*arcsin(c*x))/(b*c^3) - 1/32*cos_integral(2*a/b + 2*arcsin(c*...
Timed out. \[ \int \frac {x^2 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {x^2\,{\left (1-c^2\,x^2\right )}^{5/2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \]