Integrand size = 31, antiderivative size = 669 \[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {b f^2 g x \sqrt {d-c^2 d x^2}}{c \sqrt {1-c^2 x^2}}+\frac {2 b g^3 x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {3 b f g^2 x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}+\frac {b g^3 x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {b c g^3 x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {1}{2} f^3 x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))-\frac {3 f g^2 x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))-\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{c^2}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^4}+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{5 c^4}+\frac {f^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{4 b c \sqrt {1-c^2 x^2}}+\frac {3 f g^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{16 b c^3 \sqrt {1-c^2 x^2}} \]
1/2*f^3*x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-3/8*f*g^2*x*(a+b*arcsin(c *x))*(-c^2*d*x^2+d)^(1/2)/c^2+3/4*f*g^2*x^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+ d)^(1/2)-f^2*g*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2-1/3 *g^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4+1/5*g^3*(-c^2 *x^2+1)^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4+b*f^2*g*x*(-c^2*d*x^2 +d)^(1/2)/c/(-c^2*x^2+1)^(1/2)+2/15*b*g^3*x*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2 *x^2+1)^(1/2)-1/4*b*c*f^3*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+3/16 *b*f*g^2*x^2*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/3*b*c*f^2*g*x^3*( -c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/45*b*g^3*x^3*(-c^2*d*x^2+d)^(1/2) /c/(-c^2*x^2+1)^(1/2)-3/16*b*c*f*g^2*x^4*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1) ^(1/2)-1/25*b*c*g^3*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/4*f^3*(a +b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)+3/16*f*g^2*( a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/c^3/(-c^2*x^2+1)^(1/2)
Time = 0.33 (sec) , antiderivative size = 356, normalized size of antiderivative = 0.53 \[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {\sqrt {d-c^2 d x^2} \left (225 a^2 \left (4 c^3 f^3+3 c f g^2\right )+30 a b \sqrt {1-c^2 x^2} \left (-16 g^3-c^2 g \left (120 f^2+45 f g x+8 g^2 x^2\right )+6 c^4 x \left (10 f^3+20 f^2 g x+15 f g^2 x^2+4 g^3 x^3\right )\right )+b^2 c x \left (480 g^3+5 c^2 g \left (720 f^2+135 f g x+16 g^2 x^2\right )-3 c^4 x \left (300 f^3+400 f^2 g x+225 f g^2 x^2+48 g^3 x^3\right )\right )+30 b \left (15 a \left (4 c^3 f^3+3 c f g^2\right )+b \sqrt {1-c^2 x^2} \left (-16 g^3-c^2 g \left (120 f^2+45 f g x+8 g^2 x^2\right )+6 c^4 x \left (10 f^3+20 f^2 g x+15 f g^2 x^2+4 g^3 x^3\right )\right )\right ) \arcsin (c x)+225 b^2 c f \left (4 c^2 f^2+3 g^2\right ) \arcsin (c x)^2\right )}{3600 b c^4 \sqrt {1-c^2 x^2}} \]
(Sqrt[d - c^2*d*x^2]*(225*a^2*(4*c^3*f^3 + 3*c*f*g^2) + 30*a*b*Sqrt[1 - c^ 2*x^2]*(-16*g^3 - c^2*g*(120*f^2 + 45*f*g*x + 8*g^2*x^2) + 6*c^4*x*(10*f^3 + 20*f^2*g*x + 15*f*g^2*x^2 + 4*g^3*x^3)) + b^2*c*x*(480*g^3 + 5*c^2*g*(7 20*f^2 + 135*f*g*x + 16*g^2*x^2) - 3*c^4*x*(300*f^3 + 400*f^2*g*x + 225*f* g^2*x^2 + 48*g^3*x^3)) + 30*b*(15*a*(4*c^3*f^3 + 3*c*f*g^2) + b*Sqrt[1 - c ^2*x^2]*(-16*g^3 - c^2*g*(120*f^2 + 45*f*g*x + 8*g^2*x^2) + 6*c^4*x*(10*f^ 3 + 20*f^2*g*x + 15*f*g^2*x^2 + 4*g^3*x^3)))*ArcSin[c*x] + 225*b^2*c*f*(4* c^2*f^2 + 3*g^2)*ArcSin[c*x]^2))/(3600*b*c^4*Sqrt[1 - c^2*x^2])
Time = 0.96 (sec) , antiderivative size = 371, normalized size of antiderivative = 0.55, 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.
| \(\displaystyle \int \sqrt {d-c^2 d x^2} (f+g x)^3 (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5276 |
| \(\displaystyle \frac {\sqrt {d-c^2 d x^2} \int (f+g x)^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {\sqrt {d-c^2 d x^2} \int \left (\sqrt {1-c^2 x^2} (a+b \arcsin (c x)) f^3+3 g x \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) f^2+3 g^2 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) f+g^3 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {d-c^2 d x^2} \left (\frac {3 f g^2 (a+b \arcsin (c x))^2}{16 b c^3}+\frac {1}{2} f^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {f^2 g \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{c^2}-\frac {3 f g^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {g^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4}-\frac {g^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4}+\frac {f^3 (a+b \arcsin (c x))^2}{4 b c}+\frac {2 b g^3 x}{15 c^3}-\frac {1}{4} b c f^3 x^2-\frac {1}{3} b c f^2 g x^3+\frac {b f^2 g x}{c}-\frac {3}{16} b c f g^2 x^4+\frac {3 b f g^2 x^2}{16 c}-\frac {1}{25} b c g^3 x^5+\frac {b g^3 x^3}{45 c}\right )}{\sqrt {1-c^2 x^2}}\) |
(Sqrt[d - c^2*d*x^2]*((b*f^2*g*x)/c + (2*b*g^3*x)/(15*c^3) - (b*c*f^3*x^2) /4 + (3*b*f*g^2*x^2)/(16*c) - (b*c*f^2*g*x^3)/3 + (b*g^3*x^3)/(45*c) - (3* b*c*f*g^2*x^4)/16 - (b*c*g^3*x^5)/25 + (f^3*x*Sqrt[1 - c^2*x^2]*(a + b*Arc Sin[c*x]))/2 - (3*f*g^2*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(8*c^2) + (3*f*g^2*x^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/4 - (f^2*g*(1 - c^2*x ^2)^(3/2)*(a + b*ArcSin[c*x]))/c^2 - (g^3*(1 - c^2*x^2)^(3/2)*(a + b*ArcSi n[c*x]))/(3*c^4) + (g^3*(1 - c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4) + (f^3*(a + b*ArcSin[c*x])^2)/(4*b*c) + (3*f*g^2*(a + b*ArcSin[c*x])^2)/(16 *b*c^3)))/Sqrt[1 - c^2*x^2]
3.1.31.3.1 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 = 0.62 (sec) , antiderivative size = 1396, normalized size of antiderivative = 2.09
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1396\) |
| parts | \(\text {Expression too large to display}\) | \(1396\) |
a*(f^3*(1/2*x*(-c^2*d*x^2+d)^(1/2)+1/2*d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2 )*x/(-c^2*d*x^2+d)^(1/2)))+g^3*(-1/5*x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d-2/15/d /c^4*(-c^2*d*x^2+d)^(3/2))+3*f*g^2*(-1/4*x*(-c^2*d*x^2+d)^(3/2)/c^2/d+1/4/ c^2*(1/2*x*(-c^2*d*x^2+d)^(1/2)+1/2*d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x /(-c^2*d*x^2+d)^(1/2))))-f^2*g*(-c^2*d*x^2+d)^(3/2)/c^2/d)+b*(-1/16*(-d*(c ^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/(c^2*x^2-1)*arcsin(c*x)^2*f*(4*c^2 *f^2+3*g^2)+1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^6*x^6-28*c^4*x^4-16*I*(-c^2 *x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2 *x^2+1)^(1/2)*x*c-1)*g^3*(I+5*arcsin(c*x))/c^4/(c^2*x^2-1)+3/256*(-d*(c^2* x^2-1))^(1/2)*(-8*I*(-c^2*x^2+1)^(1/2)*x^4*c^4+8*c^5*x^5+8*I*(-c^2*x^2+1)^ (1/2)*x^2*c^2-12*c^3*x^3-I*(-c^2*x^2+1)^(1/2)+4*c*x)*f*g^2*(4*arcsin(c*x)+ I)/c^3/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4-5*c^2*x^2-4*I*c ^3*x^3*(-c^2*x^2+1)^(1/2)+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*g*(12*I*f^2*c^2+36 *arcsin(c*x)*c^2*f^2+I*g^2+3*arcsin(c*x)*g^2)/c^4/(c^2*x^2-1)+1/16*(-d*(c^ 2*x^2-1))^(1/2)*(-2*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+2*c^3*x^3+I*(-c^2*x^2+1)^ (1/2)-2*c*x)*f^3*(I+2*arcsin(c*x))/c/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/ 2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*g*(6*I*f^2*c^2+6*arcsin(c*x)*c^2*f ^2+I*g^2+arcsin(c*x)*g^2)/c^4/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*( -c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*g*(6*arcsin(c*x)*c^2*f^2-6*I*f^2*c^2+arcs in(c*x)*g^2-I*g^2)/c^4/(c^2*x^2-1)+1/16*(-d*(c^2*x^2-1))^(1/2)*(2*I*(-c...
\[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\int { \sqrt {-c^{2} d x^{2} + d} {\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
integral((a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3 *b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d), x)
\[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\int \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )^{3}\, dx \]
\[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\int { \sqrt {-c^{2} d x^{2} + d} {\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
1/2*(sqrt(-c^2*d*x^2 + d)*x + sqrt(d)*arcsin(c*x)/c)*a*f^3 - 1/15*a*g^3*(3 *(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d)) + 3/8*a*f*g^2*(sqrt(-c^2*d*x^2 + d)*x/c^2 - 2*(-c^2*d*x^2 + d)^(3/2)*x/(c^2* d) + sqrt(d)*arcsin(c*x)/c^3) - (-c^2*d*x^2 + d)^(3/2)*a*f^2*g/(c^2*d) + s qrt(d)*integrate((b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*sqrt(c* x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x)
Exception generated. \[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve cteur & l) Error: Bad Argument Value
Timed out. \[ \int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\int {\left (f+g\,x\right )}^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2} \,d x \]