Integrand size = 29, antiderivative size = 517 \[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\frac {b d^2 g x \sqrt {d-c^2 d x^2}}{7 c \sqrt {1-c^2 x^2}}-\frac {25 b c d^2 f x^2 \sqrt {d-c^2 d x^2}}{96 \sqrt {1-c^2 x^2}}-\frac {b c d^2 g x^3 \sqrt {d-c^2 d x^2}}{7 \sqrt {1-c^2 x^2}}+\frac {5 b c^3 d^2 f x^4 \sqrt {d-c^2 d x^2}}{96 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d^2 g x^5 \sqrt {d-c^2 d x^2}}{35 \sqrt {1-c^2 x^2}}-\frac {b c^5 d^2 g x^7 \sqrt {d-c^2 d x^2}}{49 \sqrt {1-c^2 x^2}}+\frac {b d^2 f \left (1-c^2 x^2\right )^{5/2} \sqrt {d-c^2 d x^2}}{36 c}+\frac {5}{16} d^2 f x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))+\frac {5}{24} d^2 f x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))+\frac {1}{6} d^2 f x \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))-\frac {d^2 g \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{7 c^2}+\frac {5 d^2 f \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{32 b c \sqrt {1-c^2 x^2}} \]
1/36*b*d^2*f*(-c^2*x^2+1)^(5/2)*(-c^2*d*x^2+d)^(1/2)/c+5/16*d^2*f*x*(a+b*a rcsin(c*x))*(-c^2*d*x^2+d)^(1/2)+5/24*d^2*f*x*(-c^2*x^2+1)*(a+b*arcsin(c*x ))*(-c^2*d*x^2+d)^(1/2)+1/6*d^2*f*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))*(-c^2 *d*x^2+d)^(1/2)-1/7*d^2*g*(-c^2*x^2+1)^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^ (1/2)/c^2+1/7*b*d^2*g*x*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-25/96*b* c*d^2*f*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/7*b*c*d^2*g*x^3*(-c^ 2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5/96*b*c^3*d^2*f*x^4*(-c^2*d*x^2+d)^(1 /2)/(-c^2*x^2+1)^(1/2)+3/35*b*c^3*d^2*g*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2 +1)^(1/2)-1/49*b*c^5*d^2*g*x^7*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+5/3 2*d^2*f*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)
Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.49 \[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\frac {d^2 \sqrt {d-c^2 d x^2} \left (11025 a^2 c f+210 a b \sqrt {1-c^2 x^2} \left (48 g \left (-1+c^2 x^2\right )^3+7 c^2 f x \left (33-26 c^2 x^2+8 c^4 x^4\right )\right )+b^2 c x \left (-245 c^2 f x \left (99-39 c^2 x^2+8 c^4 x^4\right )-288 g \left (-35+35 c^2 x^2-21 c^4 x^4+5 c^6 x^6\right )\right )+210 b \left (105 a c f+b \sqrt {1-c^2 x^2} \left (48 g \left (-1+c^2 x^2\right )^3+7 c^2 f x \left (33-26 c^2 x^2+8 c^4 x^4\right )\right )\right ) \arcsin (c x)+11025 b^2 c f \arcsin (c x)^2\right )}{70560 b c^2 \sqrt {1-c^2 x^2}} \]
(d^2*Sqrt[d - c^2*d*x^2]*(11025*a^2*c*f + 210*a*b*Sqrt[1 - c^2*x^2]*(48*g* (-1 + c^2*x^2)^3 + 7*c^2*f*x*(33 - 26*c^2*x^2 + 8*c^4*x^4)) + b^2*c*x*(-24 5*c^2*f*x*(99 - 39*c^2*x^2 + 8*c^4*x^4) - 288*g*(-35 + 35*c^2*x^2 - 21*c^4 *x^4 + 5*c^6*x^6)) + 210*b*(105*a*c*f + b*Sqrt[1 - c^2*x^2]*(48*g*(-1 + c^ 2*x^2)^3 + 7*c^2*f*x*(33 - 26*c^2*x^2 + 8*c^4*x^4)))*ArcSin[c*x] + 11025*b ^2*c*f*ArcSin[c*x]^2))/(70560*b*c^2*Sqrt[1 - c^2*x^2])
Time = 0.63 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, 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.
| \(\displaystyle \int \left (d-c^2 d x^2\right )^{5/2} (f+g x) (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5276 |
| \(\displaystyle \frac {d^2 \sqrt {d-c^2 d x^2} \int (f+g x) \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d^2 \sqrt {d-c^2 d x^2} \int \left (f (a+b \arcsin (c x)) \left (1-c^2 x^2\right )^{5/2}+g x (a+b \arcsin (c x)) \left (1-c^2 x^2\right )^{5/2}\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \sqrt {d-c^2 d x^2} \left (\frac {1}{6} f x \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))+\frac {5}{24} f x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {5}{16} f x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {g \left (1-c^2 x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^2}+\frac {5 f (a+b \arcsin (c x))^2}{32 b c}-\frac {1}{49} b c^5 g x^7+\frac {5}{96} b c^3 f x^4+\frac {3}{35} b c^3 g x^5+\frac {b f \left (1-c^2 x^2\right )^3}{36 c}-\frac {25}{96} b c f x^2-\frac {1}{7} b c g x^3+\frac {b g x}{7 c}\right )}{\sqrt {1-c^2 x^2}}\) |
(d^2*Sqrt[d - c^2*d*x^2]*((b*g*x)/(7*c) - (25*b*c*f*x^2)/96 - (b*c*g*x^3)/ 7 + (5*b*c^3*f*x^4)/96 + (3*b*c^3*g*x^5)/35 - (b*c^5*g*x^7)/49 + (b*f*(1 - c^2*x^2)^3)/(36*c) + (5*f*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/16 + ( 5*f*x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/24 + (f*x*(1 - c^2*x^2)^(5/ 2)*(a + b*ArcSin[c*x]))/6 - (g*(1 - c^2*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(7 *c^2) + (5*f*(a + b*ArcSin[c*x])^2)/(32*b*c)))/Sqrt[1 - c^2*x^2]
3.1.42.3.1 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 = 0.63 (sec) , antiderivative size = 1423, normalized size of antiderivative = 2.75
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1423\) |
| parts | \(\text {Expression too large to display}\) | \(1423\) |
1/6*a*f*x*(-c^2*d*x^2+d)^(5/2)+5/24*a*f*d*x*(-c^2*d*x^2+d)^(3/2)+5/16*a*f* d^2*x*(-c^2*d*x^2+d)^(1/2)+5/16*a*f*d^3/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2) *x/(-c^2*d*x^2+d)^(1/2))-1/7*a*g*(-c^2*d*x^2+d)^(7/2)/c^2/d+b*(-5/32*(-d*( c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*f*d^2+1/6 272*(-d*(c^2*x^2-1))^(1/2)*(64*c^8*x^8-144*c^6*x^6-64*I*c^7*x^7*(-c^2*x^2+ 1)^(1/2)+104*c^4*x^4+112*I*(-c^2*x^2+1)^(1/2)*x^5*c^5-25*c^2*x^2-56*I*(-c^ 2*x^2+1)^(1/2)*x^3*c^3+7*I*(-c^2*x^2+1)^(1/2)*x*c+1)*g*(I+7*arcsin(c*x))*d ^2/c^2/(c^2*x^2-1)+1/2304*(-d*(c^2*x^2-1))^(1/2)*(-32*I*(-c^2*x^2+1)^(1/2) *c^6*x^6+32*c^7*x^7+48*I*(-c^2*x^2+1)^(1/2)*x^4*c^4-64*c^5*x^5-18*I*(-c^2* x^2+1)^(1/2)*x^2*c^2+38*c^3*x^3+I*(-c^2*x^2+1)^(1/2)-6*c*x)*f*(I+6*arcsin( c*x))*d^2/c/(c^2*x^2-1)-5/128*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+ 1)^(1/2)*x*c-1)*g*(arcsin(c*x)+I)*d^2/c^2/(c^2*x^2-1)-5/128*(-d*(c^2*x^2-1 ))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*g*(arcsin(c*x)-I)*d^2/c^2/(c ^2*x^2-1)+15/256*(-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*(-I+2*arcsin(c*x))*d^2/c/(c^2*x^2-1) +1/128*(-d*(c^2*x^2-1))^(1/2)*(4*I*c^3*x^3*(-c^2*x^2+1)^(1/2)+4*c^4*x^4-3* I*(-c^2*x^2+1)^(1/2)*x*c-5*c^2*x^2+1)*g*(-I+3*arcsin(c*x))*d^2/c^2/(c^2*x^ 2-1)-1/7840*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*g* (11*I+70*arcsin(c*x))*cos(6*arcsin(c*x))*d^2/c^2/(c^2*x^2-1)-3/15680*(-d*( c^2*x^2-1))^(1/2)*(I*c^2*x^2-c*x*(-c^2*x^2+1)^(1/2)-I)*g*(9*I+35*arcsin...
\[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (g x + f\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
integral((a*c^4*d^2*g*x^5 + a*c^4*d^2*f*x^4 - 2*a*c^2*d^2*g*x^3 - 2*a*c^2* d^2*f*x^2 + a*d^2*g*x + a*d^2*f + (b*c^4*d^2*g*x^5 + b*c^4*d^2*f*x^4 - 2*b *c^2*d^2*g*x^3 - 2*b*c^2*d^2*f*x^2 + b*d^2*g*x + b*d^2*f)*arcsin(c*x))*sqr t(-c^2*d*x^2 + d), x)
Timed out. \[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]
\[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (g x + f\right )} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
1/48*(8*(-c^2*d*x^2 + d)^(5/2)*x + 10*(-c^2*d*x^2 + d)^(3/2)*d*x + 15*sqrt (-c^2*d*x^2 + d)*d^2*x + 15*d^(5/2)*arcsin(c*x)/c)*a*f - 1/7*(-c^2*d*x^2 + d)^(7/2)*a*g/(c^2*d) + sqrt(d)*integrate((b*c^4*d^2*g*x^5 + b*c^4*d^2*f*x ^4 - 2*b*c^2*d^2*g*x^3 - 2*b*c^2*d^2*f*x^2 + b*d^2*g*x + b*d^2*f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x)
Exception generated. \[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\text {Exception raised: RuntimeError} \]
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. \[ \int (f+g x) \left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x)) \, dx=\int \left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2} \,d x \]