Integrand size = 27, antiderivative size = 375 \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\frac {16 b d x \sqrt {d-c^2 d x^2}}{1155 c^7 \sqrt {1-c^2 x^2}}+\frac {8 b d x^3 \sqrt {d-c^2 d x^2}}{3465 c^5 \sqrt {1-c^2 x^2}}+\frac {2 b d x^5 \sqrt {d-c^2 d x^2}}{1925 c^3 \sqrt {1-c^2 x^2}}+\frac {b d x^7 \sqrt {d-c^2 d x^2}}{1617 c \sqrt {1-c^2 x^2}}-\frac {4 b c d x^9 \sqrt {d-c^2 d x^2}}{297 \sqrt {1-c^2 x^2}}+\frac {b c^3 d x^{11} \sqrt {d-c^2 d x^2}}{121 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^8 d}+\frac {3 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^8 d^2}-\frac {\left (d-c^2 d x^2\right )^{9/2} (a+b \arcsin (c x))}{3 c^8 d^3}+\frac {\left (d-c^2 d x^2\right )^{11/2} (a+b \arcsin (c x))}{11 c^8 d^4} \]
-1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))/c^8/d+3/7*(-c^2*d*x^2+d)^(7/2) *(a+b*arcsin(c*x))/c^8/d^2-1/3*(-c^2*d*x^2+d)^(9/2)*(a+b*arcsin(c*x))/c^8/ d^3+1/11*(-c^2*d*x^2+d)^(11/2)*(a+b*arcsin(c*x))/c^8/d^4+16/1155*b*d*x*(-c ^2*d*x^2+d)^(1/2)/c^7/(-c^2*x^2+1)^(1/2)+8/3465*b*d*x^3*(-c^2*d*x^2+d)^(1/ 2)/c^5/(-c^2*x^2+1)^(1/2)+2/1925*b*d*x^5*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*x^ 2+1)^(1/2)+1/1617*b*d*x^7*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-4/297* b*c*d*x^9*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/121*b*c^3*d*x^11*(-c^2 *d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)
Time = 0.12 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.46 \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\frac {d \sqrt {d-c^2 d x^2} \left (-3465 a \left (1-c^2 x^2\right )^{5/2} \left (16+40 c^2 x^2+70 c^4 x^4+105 c^6 x^6\right )+b c x \left (55440+9240 c^2 x^2+4158 c^4 x^4+2475 c^6 x^6-53900 c^8 x^8+33075 c^{10} x^{10}\right )-3465 b \left (1-c^2 x^2\right )^{5/2} \left (16+40 c^2 x^2+70 c^4 x^4+105 c^6 x^6\right ) \arcsin (c x)\right )}{4002075 c^8 \sqrt {1-c^2 x^2}} \]
(d*Sqrt[d - c^2*d*x^2]*(-3465*a*(1 - c^2*x^2)^(5/2)*(16 + 40*c^2*x^2 + 70* c^4*x^4 + 105*c^6*x^6) + b*c*x*(55440 + 9240*c^2*x^2 + 4158*c^4*x^4 + 2475 *c^6*x^6 - 53900*c^8*x^8 + 33075*c^10*x^10) - 3465*b*(1 - c^2*x^2)^(5/2)*( 16 + 40*c^2*x^2 + 70*c^4*x^4 + 105*c^6*x^6)*ArcSin[c*x]))/(4002075*c^8*Sqr t[1 - c^2*x^2])
Time = 0.58 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.60, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5194, 27, 2341, 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.
| \(\displaystyle \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {d \left (1-c^2 x^2\right )^2 \left (105 c^6 x^6+70 c^4 x^4+40 c^2 x^2+16\right )}{1155 c^8}dx}{\sqrt {1-c^2 x^2}}+\frac {\left (d-c^2 d x^2\right )^{11/2} (a+b \arcsin (c x))}{11 c^8 d^4}-\frac {\left (d-c^2 d x^2\right )^{9/2} (a+b \arcsin (c x))}{3 c^8 d^3}+\frac {3 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^8 d^2}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^8 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b d \sqrt {d-c^2 d x^2} \int \left (1-c^2 x^2\right )^2 \left (105 c^6 x^6+70 c^4 x^4+40 c^2 x^2+16\right )dx}{1155 c^7 \sqrt {1-c^2 x^2}}+\frac {\left (d-c^2 d x^2\right )^{11/2} (a+b \arcsin (c x))}{11 c^8 d^4}-\frac {\left (d-c^2 d x^2\right )^{9/2} (a+b \arcsin (c x))}{3 c^8 d^3}+\frac {3 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^8 d^2}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^8 d}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {b d \sqrt {d-c^2 d x^2} \int \left (105 c^{10} x^{10}-140 c^8 x^8+5 c^6 x^6+6 c^4 x^4+8 c^2 x^2+16\right )dx}{1155 c^7 \sqrt {1-c^2 x^2}}+\frac {\left (d-c^2 d x^2\right )^{11/2} (a+b \arcsin (c x))}{11 c^8 d^4}-\frac {\left (d-c^2 d x^2\right )^{9/2} (a+b \arcsin (c x))}{3 c^8 d^3}+\frac {3 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^8 d^2}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^8 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (d-c^2 d x^2\right )^{11/2} (a+b \arcsin (c x))}{11 c^8 d^4}-\frac {\left (d-c^2 d x^2\right )^{9/2} (a+b \arcsin (c x))}{3 c^8 d^3}+\frac {3 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{7 c^8 d^2}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^8 d}+\frac {b d \left (\frac {105 c^{10} x^{11}}{11}-\frac {140 c^8 x^9}{9}+\frac {5 c^6 x^7}{7}+\frac {6 c^4 x^5}{5}+\frac {8 c^2 x^3}{3}+16 x\right ) \sqrt {d-c^2 d x^2}}{1155 c^7 \sqrt {1-c^2 x^2}}\) |
(b*d*Sqrt[d - c^2*d*x^2]*(16*x + (8*c^2*x^3)/3 + (6*c^4*x^5)/5 + (5*c^6*x^ 7)/7 - (140*c^8*x^9)/9 + (105*c^10*x^11)/11))/(1155*c^7*Sqrt[1 - c^2*x^2]) - ((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^8*d) + (3*(d - c^2*d*x ^2)^(7/2)*(a + b*ArcSin[c*x]))/(7*c^8*d^2) - ((d - c^2*d*x^2)^(9/2)*(a + b *ArcSin[c*x]))/(3*c^8*d^3) + ((d - c^2*d*x^2)^(11/2)*(a + b*ArcSin[c*x]))/ (11*c^8*d^4)
3.1.77.3.1 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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_) , x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin [c*x]) u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[Sim plifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 1781, normalized size of antiderivative = 4.75
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1781\) |
| parts | \(\text {Expression too large to display}\) | \(1781\) |
a*(-1/11*x^6*(-c^2*d*x^2+d)^(5/2)/c^2/d+6/11/c^2*(-1/9*x^4*(-c^2*d*x^2+d)^ (5/2)/c^2/d+4/9/c^2*(-1/7*x^2*(-c^2*d*x^2+d)^(5/2)/c^2/d-2/35/d/c^4*(-c^2* d*x^2+d)^(5/2))))+b*(-1/247808*(-d*(c^2*x^2-1))^(1/2)*(1+11*I*(-c^2*x^2+1) ^(1/2)*x*c+1024*c^12*x^12+2816*I*(-c^2*x^2+1)^(1/2)*x^9*c^9-61*c^2*x^2-235 2*c^6*x^6+620*c^4*x^4-2816*I*(-c^2*x^2+1)^(1/2)*x^7*c^7-3328*c^10*x^10+409 6*c^8*x^8+1232*I*(-c^2*x^2+1)^(1/2)*x^5*c^5-1024*I*(-c^2*x^2+1)^(1/2)*x^11 *c^11-220*I*(-c^2*x^2+1)^(1/2)*x^3*c^3)*(I+11*arcsin(c*x))*d/c^8/(c^2*x^2- 1)-1/55296*(-d*(c^2*x^2-1))^(1/2)*(256*c^10*x^10-704*c^8*x^8-256*I*(-c^2*x ^2+1)^(1/2)*x^9*c^9+688*c^6*x^6+576*I*(-c^2*x^2+1)^(1/2)*x^7*c^7-280*c^4*x ^4-432*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+41*c^2*x^2+120*I*(-c^2*x^2+1)^(1/2)*x^ 3*c^3-9*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(I+9*arcsin(c*x))*d/c^8/(c^2*x^2-1)+1/ 100352*(-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)*(I+7*arcsin(c*x))* d/c^8/(c^2*x^2-1)+11/51200*(-d*(c^2*x^2-1))^(1/2)*(16*c^6*x^6-28*c^4*x^4-1 6*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)*(I+5*arcsin(c*x))*d/c^8/(c^2*x^2-1)+1/3072*( -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)*(I+3*arcsin(c*x))*d/c^8/(c^2*x^2-1)-7/1024*( -d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)...
Time = 0.26 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.66 \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=-\frac {{\left (33075 \, b c^{11} d x^{11} - 53900 \, b c^{9} d x^{9} + 2475 \, b c^{7} d x^{7} + 4158 \, b c^{5} d x^{5} + 9240 \, b c^{3} d x^{3} + 55440 \, b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 3465 \, {\left (105 \, a c^{12} d x^{12} - 245 \, a c^{10} d x^{10} + 145 \, a c^{8} d x^{8} + a c^{6} d x^{6} + 2 \, a c^{4} d x^{4} + 8 \, a c^{2} d x^{2} - 16 \, a d + {\left (105 \, b c^{12} d x^{12} - 245 \, b c^{10} d x^{10} + 145 \, b c^{8} d x^{8} + b c^{6} d x^{6} + 2 \, b c^{4} d x^{4} + 8 \, b c^{2} d x^{2} - 16 \, b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{4002075 \, {\left (c^{10} x^{2} - c^{8}\right )}} \]
-1/4002075*((33075*b*c^11*d*x^11 - 53900*b*c^9*d*x^9 + 2475*b*c^7*d*x^7 + 4158*b*c^5*d*x^5 + 9240*b*c^3*d*x^3 + 55440*b*c*d*x)*sqrt(-c^2*d*x^2 + d)* sqrt(-c^2*x^2 + 1) + 3465*(105*a*c^12*d*x^12 - 245*a*c^10*d*x^10 + 145*a*c ^8*d*x^8 + a*c^6*d*x^6 + 2*a*c^4*d*x^4 + 8*a*c^2*d*x^2 - 16*a*d + (105*b*c ^12*d*x^12 - 245*b*c^10*d*x^10 + 145*b*c^8*d*x^8 + b*c^6*d*x^6 + 2*b*c^4*d *x^4 + 8*b*c^2*d*x^2 - 16*b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^10*x^ 2 - c^8)
Timed out. \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.71 \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=-\frac {1}{1155} \, {\left (\frac {105 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{6}}{c^{2} d} + \frac {70 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}}{c^{4} d} + \frac {40 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{6} d} + \frac {16 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{8} d}\right )} b \arcsin \left (c x\right ) - \frac {1}{1155} \, {\left (\frac {105 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{6}}{c^{2} d} + \frac {70 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}}{c^{4} d} + \frac {40 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{6} d} + \frac {16 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{8} d}\right )} a + \frac {{\left (33075 \, c^{10} d^{\frac {3}{2}} x^{11} - 53900 \, c^{8} d^{\frac {3}{2}} x^{9} + 2475 \, c^{6} d^{\frac {3}{2}} x^{7} + 4158 \, c^{4} d^{\frac {3}{2}} x^{5} + 9240 \, c^{2} d^{\frac {3}{2}} x^{3} + 55440 \, d^{\frac {3}{2}} x\right )} b}{4002075 \, c^{7}} \]
-1/1155*(105*(-c^2*d*x^2 + d)^(5/2)*x^6/(c^2*d) + 70*(-c^2*d*x^2 + d)^(5/2 )*x^4/(c^4*d) + 40*(-c^2*d*x^2 + d)^(5/2)*x^2/(c^6*d) + 16*(-c^2*d*x^2 + d )^(5/2)/(c^8*d))*b*arcsin(c*x) - 1/1155*(105*(-c^2*d*x^2 + d)^(5/2)*x^6/(c ^2*d) + 70*(-c^2*d*x^2 + d)^(5/2)*x^4/(c^4*d) + 40*(-c^2*d*x^2 + d)^(5/2)* x^2/(c^6*d) + 16*(-c^2*d*x^2 + d)^(5/2)/(c^8*d))*a + 1/4002075*(33075*c^10 *d^(3/2)*x^11 - 53900*c^8*d^(3/2)*x^9 + 2475*c^6*d^(3/2)*x^7 + 4158*c^4*d^ (3/2)*x^5 + 9240*c^2*d^(3/2)*x^3 + 55440*d^(3/2)*x)*b/c^7
Exception generated. \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int x^7 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x)) \, dx=\int x^7\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]