Integrand size = 28, antiderivative size = 144 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=-\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 b c^5}+\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b c^5}+\frac {3 \log (a+b \arcsin (c x))}{8 b c^5}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 b c^5}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b c^5} \]
-1/2*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c^5+1/8*Ci(4*(a+b*arcsin(c*x)) /b)*cos(4*a/b)/b/c^5+3/8*ln(a+b*arcsin(c*x))/b/c^5-1/2*Si(2*(a+b*arcsin(c* x))/b)*sin(2*a/b)/b/c^5+1/8*Si(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b/c^5
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.75 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\frac {-4 \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+3 \log (a+b \arcsin (c x))-4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{8 b c^5} \]
(-4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + Cos[(4*a)/b]*CosInte gral[4*(a/b + ArcSin[c*x])] + 3*Log[a + b*ArcSin[c*x]] - 4*Sin[(2*a)/b]*Si nIntegral[2*(a/b + ArcSin[c*x])] + Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSi n[c*x])])/(8*b*c^5)
Time = 0.47 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5224, 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.
| \(\displaystyle \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx\) |
| \(\Big \downarrow \) 5224 |
| \(\displaystyle \frac {\int \frac {\sin ^4\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^5}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )^4}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^5}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle \frac {\int \left (\frac {\cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 (a+b \arcsin (c x))}+\frac {3}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^5}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{2} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )-\frac {1}{2} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {3}{8} \log (a+b \arcsin (c x))}{b c^5}\) |
(-1/2*(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b]) + (Cos[(4*a)/b ]*CosIntegral[(4*(a + b*ArcSin[c*x]))/b])/8 + (3*Log[a + b*ArcSin[c*x]])/8 - (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/2 + (Sin[(4*a)/b] *SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/8)/(b*c^5)
3.4.50.3.1 Defintions of rubi rules used
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_.)*(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 = 0.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77
| method | result | size |
| default | \(\frac {\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-4 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )-4 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+3 \ln \left (a +b \arcsin \left (c x \right )\right )}{8 c^{5} b}\) | \(111\) |
1/8/c^5*(Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+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)-4*Ci(2*arcsin(c*x)+2*a/b)*cos(2* a/b)+3*ln(a+b*arcsin(c*x)))/b
\[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int { \frac {x^{4}}{\sqrt {-c^{2} x^{2} + 1} {\left (b \arcsin \left (c x\right ) + a\right )}} \,d x } \]
\[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int \frac {x^{4}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \]
\[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int { \frac {x^{4}}{\sqrt {-c^{2} x^{2} + 1} {\left (b \arcsin \left (c x\right ) + a\right )}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.76 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} + \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{5}} - \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c^{5}} - \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{5}} + \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{5}} + \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{5}} + \frac {3 \, \log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c^{5}} \]
cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + cos(a/b)^3*sin(a/ b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)^2*cos_integral(4 *a/b + 4*arcsin(c*x))/(b*c^5) - cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(c *x))/(b*c^5) - 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/( b*c^5) - cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 1 /8*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + 1/2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 3/8*log(b*arcsin(c*x) + a)/(b*c^5)
Timed out. \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int \frac {x^4}{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {1-c^2\,x^2}} \,d x \]