Integrand size = 31, antiderivative size = 270 \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {2 b f g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}}+\frac {b g^2 x^2 \sqrt {1-c^2 x^2}}{4 c \sqrt {d-c^2 d x^2}}-\frac {2 f g \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c^2 \sqrt {d-c^2 d x^2}}-\frac {g^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{2 c^2 \sqrt {d-c^2 d x^2}}+\frac {f^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d-c^2 d x^2}}+\frac {g^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c^3 \sqrt {d-c^2 d x^2}} \]
-2*f*g*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c^2/(-c^2*d*x^2+d)^(1/2)-1/2*g^2*x*( -c^2*x^2+1)*(a+b*arcsin(c*x))/c^2/(-c^2*d*x^2+d)^(1/2)+2*b*f*g*x*(-c^2*x^2 +1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)+1/4*b*g^2*x^2*(-c^2*x^2+1)^(1/2)/c/(-c^2* d*x^2+d)^(1/2)+1/2*f^2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(-c^2*d* x^2+d)^(1/2)+1/4*g^2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c^3/(-c^2*d* x^2+d)^(1/2)
Time = 0.66 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.99 \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {-2 b \sqrt {d} \left (2 c^2 f^2+g^2\right ) \left (-1+c^2 x^2\right ) \arcsin (c x)^2-4 a \left (2 c^2 f^2+g^2\right ) \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2} \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+\sqrt {d} g \left (-1+c^2 x^2\right ) \left (4 c \left (-4 b c f x+a (4 f+g x) \sqrt {1-c^2 x^2}\right )+b g \cos (2 \arcsin (c x))\right )+2 b \sqrt {d} g \left (-1+c^2 x^2\right ) \arcsin (c x) \left (8 c f \sqrt {1-c^2 x^2}+g \sin (2 \arcsin (c x))\right )}{8 c^3 \sqrt {d} \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}} \]
(-2*b*Sqrt[d]*(2*c^2*f^2 + g^2)*(-1 + c^2*x^2)*ArcSin[c*x]^2 - 4*a*(2*c^2* f^2 + g^2)*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2* d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + Sqrt[d]*g*(-1 + c^2*x^2)*(4*c*(-4*b*c* f*x + a*(4*f + g*x)*Sqrt[1 - c^2*x^2]) + b*g*Cos[2*ArcSin[c*x]]) + 2*b*Sqr t[d]*g*(-1 + c^2*x^2)*ArcSin[c*x]*(8*c*f*Sqrt[1 - c^2*x^2] + g*Sin[2*ArcSi n[c*x]]))/(8*c^3*Sqrt[d]*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2])
Time = 0.62 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.60, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5276, 5262, 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.
| \(\displaystyle \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx\) |
| \(\Big \downarrow \) 5276 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \left (\frac {(a+b \arcsin (c x)) f^2}{\sqrt {1-c^2 x^2}}+\frac {2 g x (a+b \arcsin (c x)) f}{\sqrt {1-c^2 x^2}}+\frac {g^2 x^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right )dx}{\sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {g^2 (a+b \arcsin (c x))^2}{4 b c^3}-\frac {2 f g \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c^2}-\frac {g^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 c^2}+\frac {f^2 (a+b \arcsin (c x))^2}{2 b c}+\frac {2 b f g x}{c}+\frac {b g^2 x^2}{4 c}\right )}{\sqrt {d-c^2 d x^2}}\) |
(Sqrt[1 - c^2*x^2]*((2*b*f*g*x)/c + (b*g^2*x^2)/(4*c) - (2*f*g*Sqrt[1 - c^ 2*x^2]*(a + b*ArcSin[c*x]))/c^2 - (g^2*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c *x]))/(2*c^2) + (f^2*(a + b*ArcSin[c*x])^2)/(2*b*c) + (g^2*(a + b*ArcSin[c *x])^2)/(4*b*c^3)))/Sqrt[d - c^2*d*x^2]
3.1.45.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 - c^2*x^2)^ p] Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ [{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && Intege rQ[p - 1/2] && !GtQ[d, 0]
Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.87
| method | result | size |
| default | \(a \left (\frac {f^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}+g^{2} \left (-\frac {x \sqrt {-c^{2} d \,x^{2}+d}}{2 c^{2} d}+\frac {\arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{2} \sqrt {c^{2} d}}\right )-\frac {2 f g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}\right )+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} \left (2 c^{2} f^{2}+g^{2}\right )}{4 c^{3} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i c x \sqrt {-c^{2} x^{2}+1}-1\right ) f g \left (\arcsin \left (c x \right )+i\right )}{c^{2} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c x \sqrt {-c^{2} x^{2}+1}+c^{2} x^{2}-1\right ) f g \left (\arcsin \left (c x \right )-i\right )}{c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, g^{2}}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \arcsin \left (c x \right ) x}{8 c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{3} d \left (c^{2} x^{2}-1\right )}\right )\) | \(505\) |
| parts | \(a \left (\frac {f^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}+g^{2} \left (-\frac {x \sqrt {-c^{2} d \,x^{2}+d}}{2 c^{2} d}+\frac {\arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{2} \sqrt {c^{2} d}}\right )-\frac {2 f g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}\right )+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} \left (2 c^{2} f^{2}+g^{2}\right )}{4 c^{3} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i c x \sqrt {-c^{2} x^{2}+1}-1\right ) f g \left (\arcsin \left (c x \right )+i\right )}{c^{2} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c x \sqrt {-c^{2} x^{2}+1}+c^{2} x^{2}-1\right ) f g \left (\arcsin \left (c x \right )-i\right )}{c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, g^{2}}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \arcsin \left (c x \right ) x}{8 c^{2} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \cos \left (3 \arcsin \left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, g^{2} \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{8 c^{3} d \left (c^{2} x^{2}-1\right )}\right )\) | \(505\) |
a*(f^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+g^2*(-1/ 2*x/c^2/d*(-c^2*d*x^2+d)^(1/2)+1/2/c^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)* x/(-c^2*d*x^2+d)^(1/2)))-2*f*g/c^2/d*(-c^2*d*x^2+d)^(1/2))+b*(-1/4*(-d*(c^ 2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d/(c^2*x^2-1)*arcsin(c*x)^2*(2*c^2* f^2+g^2)-(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*f*g*( arcsin(c*x)+I)/c^2/d/(c^2*x^2-1)-(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1 /2)*x*c+c^2*x^2-1)*f*g*(arcsin(c*x)-I)/c^2/d/(c^2*x^2-1)+1/16*(-d*(c^2*x^2 -1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d/(c^2*x^2-1)*g^2+1/8*(-d*(c^2*x^2-1))^( 1/2)/c^2/d/(c^2*x^2-1)*g^2*arcsin(c*x)*x+1/16*(-d*(c^2*x^2-1))^(1/2)/c^3/d /(c^2*x^2-1)*g^2*cos(3*arcsin(c*x))+1/8*(-d*(c^2*x^2-1))^(1/2)/c^3/d/(c^2* x^2-1)*g^2*arcsin(c*x)*sin(3*arcsin(c*x)))
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
integral(-sqrt(-c^2*d*x^2 + d)*(a*g^2*x^2 + 2*a*f*g*x + a*f^2 + (b*g^2*x^2 + 2*b*f*g*x + b*f^2)*arcsin(c*x))/(c^2*d*x^2 - d), x)
Exception generated. \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
-1/2*a*g^2*(sqrt(-c^2*d*x^2 + d)*x/(c^2*d) - arcsin(c*x)/(c^3*sqrt(d))) + 1/2*b*f^2*arcsin(c*x)^2/(c*sqrt(d)) + b*g^2*integrate(x^2*arctan2(c*x, sqr t(c*x + 1)*sqrt(-c*x + 1))/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d) + 2* b*f*g*x/(c*sqrt(d)) + a*f^2*arcsin(c*x)/(c*sqrt(d)) - 2*sqrt(-c^2*d*x^2 + d)*b*f*g*arcsin(c*x)/(c^2*d) - 2*sqrt(-c^2*d*x^2 + d)*a*f*g/(c^2*d)
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\int \frac {{\left (f+g\,x\right )}^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \]