Integrand size = 19, antiderivative size = 491 \[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=-\frac {i (a+b \arcsin (c x))^2}{2 b e}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}+\sqrt {c^2 d+e}}\right )}{2 e} \]
-1/2*I*(a+b*arcsin(c*x))^2/b/e+1/2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2 +1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/e+1/2*(a+b*arcsin(c*x ))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2) ))/e+1/2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(- d)^(1/2)+(c^2*d+e)^(1/2)))/e+1/2*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1 )^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/e-1/2*I*b*polylog(2,-(I *c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^(1/2)))/e-1/2*I *b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^(1/2)-(c^2*d+e)^ (1/2)))/e-1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))*e^(1/2)/(I*c*(-d)^ (1/2)+(c^2*d+e)^(1/2)))/e-1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))*e^( 1/2)/(I*c*(-d)^(1/2)+(c^2*d+e)^(1/2)))/e
Time = 0.13 (sec) , antiderivative size = 399, normalized size of antiderivative = 0.81 \[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=-\frac {i \left (b \arcsin (c x)^2+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i b \arcsin (c x) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+i a \log \left (d+e x^2\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}-\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{-c \sqrt {d}+\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )+b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{c \sqrt {d}+\sqrt {c^2 d+e}}\right )\right )}{2 e} \]
((-1/2*I)*(b*ArcSin[c*x]^2 + I*b*ArcSin[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[ c*x]))/(c*Sqrt[d] - Sqrt[c^2*d + e])] + I*b*ArcSin[c*x]*Log[1 + (Sqrt[e]*E ^(I*ArcSin[c*x]))/(-(c*Sqrt[d]) + Sqrt[c^2*d + e])] + I*b*ArcSin[c*x]*Log[ 1 - (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])] + I*b*ArcSi n[c*x]*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] + Sqrt[c^2*d + e])] + I*a*Log[d + e*x^2] + b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sqrt[d] - Sqrt[c^2*d + e])] + b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(-(c*Sqrt[ d]) + Sqrt[c^2*d + e])] + b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(c*Sq rt[d] + Sqrt[c^2*d + e]))] + b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(c*S qrt[d] + Sqrt[c^2*d + e])]))/e
Time = 1.09 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5232, 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.
\(\displaystyle \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx\) |
\(\Big \downarrow \) 5232 |
\(\displaystyle \int \left (\frac {a+b \arcsin (c x)}{2 \sqrt {e} \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {a+b \arcsin (c x)}{2 \sqrt {e} \left (\sqrt {-d}-\sqrt {e} x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{-\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1-\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e}+\frac {(a+b \arcsin (c x)) \log \left (1+\frac {\sqrt {e} e^{i \arcsin (c x)}}{\sqrt {c^2 d+e}+i c \sqrt {-d}}\right )}{2 e}-\frac {i (a+b \arcsin (c x))^2}{2 b e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i c \sqrt {-d}-\sqrt {d c^2+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e}-\frac {i b \operatorname {PolyLog}\left (2,\frac {\sqrt {e} e^{i \arcsin (c x)}}{i \sqrt {-d} c+\sqrt {d c^2+e}}\right )}{2 e}\) |
((-1/2*I)*(a + b*ArcSin[c*x])^2)/(b*e) + ((a + b*ArcSin[c*x])*Log[1 - (Sqr t[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/(2*e) + ((a + b *ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2 *d + e])])/(2*e) + ((a + b*ArcSin[c*x])*Log[1 - (Sqrt[e]*E^(I*ArcSin[c*x]) )/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e) + ((a + b*ArcSin[c*x])*Log[1 + (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/(2*e) - ((I /2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e]))])/e - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^(I*ArcSin[c*x]))/(I*c*Sqrt[-d] - Sqrt[c^2*d + e])])/e - ((I/2)*b*PolyLog[2, -((Sqrt[e]*E^(I*ArcSin[c*x]) )/(I*c*Sqrt[-d] + Sqrt[c^2*d + e]))])/e - ((I/2)*b*PolyLog[2, (Sqrt[e]*E^( I*ArcSin[c*x]))/(I*c*Sqrt[-d] + Sqrt[c^2*d + e])])/e
3.7.27.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n, ( f*x)^m*(d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[p] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.59 (sec) , antiderivative size = 1965, normalized size of antiderivative = 4.00
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1965\) |
default | \(\text {Expression too large to display}\) | \(1965\) |
parts | \(\text {Expression too large to display}\) | \(1966\) |
1/c^2*(1/2*a*c^2/e*ln(c^2*e*x^2+c^2*d)+b*c^2*(-1/2*I*(2*c^2*d-2*(d*c^2*(c^ 2*d+e))^(1/2)+e)*arcsin(c*x)^2/e^2-(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2*d^2 *c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^(1/2)*e)/e^3*d*c^2/(c^2*d+e)*ln(1-e*(I*c* x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e))*arcsin(c*x) +(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e)/e^3*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2 ))^2/(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e))*c^2*d*arcsin(c*x)+1/4*I*(d*c^2 *(c^2*d+e))^(1/2)/e/(c^2*d+e)*polylog(2,e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2* c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))+I*(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2* d^2*c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^(1/2)*e)*arcsin(c*x)^2/e^2/(c^2*d+e)-1 /2*(d*c^2*(c^2*d+e))^(1/2)/e/(c^2*d+e)*arcsin(c*x)*ln(1-e*(I*c*x+(-c^2*x^2 +1)^(1/2))^2/(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))-1/4*(d*c^2*(c^2*d+e))^ (1/2)/d/c^2/(c^2*d+e)*arcsin(c*x)*ln(1-e*(I*c*x+(-c^2*x^2+1)^(1/2))^2/(2*c ^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))-1/4*(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c^2+2 *d^2*c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^(1/2)*e)/e/d/c^2/(c^2*d+e)*ln(1-e*(I* c*x+(-c^2*x^2+1)^(1/2))^2/(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e))*arcsin(c* x)+1/8*I*(d*c^2*(c^2*d+e))^(1/2)/d/c^2/(c^2*d+e)*polylog(2,e*(I*c*x+(-c^2* x^2+1)^(1/2))^2/(2*c^2*d-2*(d*c^2*(c^2*d+e))^(1/2)+e))+1/2*I*(d*c^2*(c^2*d +e))^(1/2)/e/(c^2*d+e)*arcsin(c*x)^2+1/2*I*(-2*(d*c^2*(c^2*d+e))^(1/2)*d*c ^2+2*d^2*c^4+2*c^2*e*d-(d*c^2*(c^2*d+e))^(1/2)*e)*polylog(2,e*(I*c*x+(-c^2 *x^2+1)^(1/2))^2/(2*c^2*d+2*(d*c^2*(c^2*d+e))^(1/2)+e))/e^2/(c^2*d+e)-(...
\[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x}{e x^{2} + d} \,d x } \]
\[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=\int \frac {x \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{d + e x^{2}}\, dx \]
\[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x}{e x^{2} + d} \,d x } \]
b*integrate(x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e*x^2 + d), x) + 1/2*a*log(e*x^2 + d)/e
\[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x}{e x^{2} + d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arcsin (c x))}{d+e x^2} \, dx=\int \frac {x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{e\,x^2+d} \,d x \]