Integrand size = 21, antiderivative size = 104 \[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=-\frac {e (c+d x) \sqrt {1-(c+d x)^2}}{b d (a+b \arcsin (c+d x))}+\frac {e \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{b^2 d}+\frac {e \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{b^2 d} \]
e*Ci(2*(a+b*arcsin(d*x+c))/b)*cos(2*a/b)/b^2/d+e*Si(2*(a+b*arcsin(d*x+c))/ b)*sin(2*a/b)/b^2/d-e*(d*x+c)*(1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))
Time = 0.52 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.95 \[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=\frac {e \left (-\frac {b (c+d x) \sqrt {1-c^2-2 c d x-d^2 x^2}}{a+b \arcsin (c+d x)}+\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right )+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c+d x)\right )\right )\right )}{b^2 d} \]
(e*(-((b*(c + d*x)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(a + b*ArcSin[c + d* x])) + Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c + d*x])] + Sin[(2*a)/b]* SinIntegral[2*(a/b + ArcSin[c + d*x])]))/(b^2*d)
Time = 0.48 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5304, 27, 5142, 3042, 3784, 25, 3042, 3780, 3783}
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.
| \(\displaystyle \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 5304 |
| \(\displaystyle \frac {\int \frac {e (c+d x)}{(a+b \arcsin (c+d x))^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \int \frac {c+d x}{(a+b \arcsin (c+d x))^2}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 5142 |
| \(\displaystyle \frac {e \left (\frac {\int \frac {\cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e \left (\frac {\int \frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c+d x))}{b}+\frac {\pi }{2}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 3784 |
| \(\displaystyle \frac {e \left (\frac {\cos \left (\frac {2 a}{b}\right ) \int \frac {\cos \left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))-\sin \left (\frac {2 a}{b}\right ) \int -\frac {\sin \left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e \left (\frac {\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))+\cos \left (\frac {2 a}{b}\right ) \int \frac {\cos \left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e \left (\frac {\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))+\cos \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arcsin (c+d x))}{b}+\frac {\pi }{2}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 3780 |
| \(\displaystyle \frac {e \left (\frac {\cos \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arcsin (c+d x))}{b}+\frac {\pi }{2}\right )}{a+b \arcsin (c+d x)}d(a+b \arcsin (c+d x))+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
| \(\Big \downarrow \) 3783 |
| \(\displaystyle \frac {e \left (\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c+d x))}{b}\right )}{b^2}-\frac {(c+d x) \sqrt {1-(c+d x)^2}}{b (a+b \arcsin (c+d x))}\right )}{d}\) |
(e*(-(((c + d*x)*Sqrt[1 - (c + d*x)^2])/(b*(a + b*ArcSin[c + d*x]))) + (Co s[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c + d*x]))/b] + Sin[(2*a)/b]*SinIn tegral[(2*(a + b*ArcSin[c + d*x]))/b])/b^2))/d
3.3.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* e - c*f)/d] Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* f)/d] Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] && NeQ[d*e - c*f, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x ^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp [1/(b^2*c^(m + 1)*(n + 1)) Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[-a/b + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c* x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.20 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.45
| method | result | size |
| derivativedivides | \(\frac {e \left (2 \arcsin \left (d x +c \right ) \operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b +2 \arcsin \left (d x +c \right ) \sin \left (\frac {2 a}{b}\right ) \operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) b +2 \,\operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +2 \sin \left (\frac {2 a}{b}\right ) \operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) a -\sin \left (2 \arcsin \left (d x +c \right )\right ) b \right )}{2 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}}\) | \(151\) |
| default | \(\frac {e \left (2 \arcsin \left (d x +c \right ) \operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b +2 \arcsin \left (d x +c \right ) \sin \left (\frac {2 a}{b}\right ) \operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) b +2 \,\operatorname {Ci}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +2 \sin \left (\frac {2 a}{b}\right ) \operatorname {Si}\left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) a -\sin \left (2 \arcsin \left (d x +c \right )\right ) b \right )}{2 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}}\) | \(151\) |
1/2/d*e*(2*arcsin(d*x+c)*Ci(2*arcsin(d*x+c)+2*a/b)*cos(2*a/b)*b+2*arcsin(d *x+c)*sin(2*a/b)*Si(2*arcsin(d*x+c)+2*a/b)*b+2*Ci(2*arcsin(d*x+c)+2*a/b)*c os(2*a/b)*a+2*sin(2*a/b)*Si(2*arcsin(d*x+c)+2*a/b)*a-sin(2*arcsin(d*x+c))* b)/(a+b*arcsin(d*x+c))/b^2
\[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=\int { \frac {d e x + c e}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=e \left (\int \frac {c}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac {d x}{a^{2} + 2 a b \operatorname {asin}{\left (c + d x \right )} + b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}\, dx\right ) \]
e*(Integral(c/(a**2 + 2*a*b*asin(c + d*x) + b**2*asin(c + d*x)**2), x) + I ntegral(d*x/(a**2 + 2*a*b*asin(c + d*x) + b**2*asin(c + d*x)**2), x))
\[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=\int { \frac {d e x + c e}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
-((d*e*x + c*e)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1) - (b^2*d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a*b*d)*integrate((2*d^2*e*x^2 + 4*c*d*e*x + (2*c^2 - 1)*e)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)/(a*b*d^ 2*x^2 + 2*a*b*c*d*x + a*b*c^2 - a*b + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - b^2)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))), x))/(b^2* d*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + a*b*d)
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (102) = 204\).
Time = 0.37 (sec) , antiderivative size = 341, normalized size of antiderivative = 3.28 \[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=\frac {2 \, b e \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {2 \, b e \arcsin \left (d x + c\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {2 \, a e \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} + \frac {2 \, a e \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {b e \arcsin \left (d x + c\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b e}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} - \frac {a e \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{b^{3} d \arcsin \left (d x + c\right ) + a b^{2} d} \]
2*b*e*arcsin(d*x + c)*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(d*x + c))/( b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*b*e*arcsin(d*x + c)*cos(a/b)*sin(a/b) *sin_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*a*e*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin( d*x + c) + a*b^2*d) + 2*a*e*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsi n(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - b*e*arcsin(d*x + c)*cos_in tegral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - sqrt (-(d*x + c)^2 + 1)*(d*x + c)*b*e/(b^3*d*arcsin(d*x + c) + a*b^2*d) - a*e*c os_integral(2*a/b + 2*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d)
Timed out. \[ \int \frac {c e+d e x}{(a+b \arcsin (c+d x))^2} \, dx=\int \frac {c\,e+d\,e\,x}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2} \,d x \]