Integrand size = 27, antiderivative size = 263 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{42 x^6 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{140 x^4 \sqrt {1-c^2 x^2}}+\frac {2 b c^5 \sqrt {d-c^2 d x^2}}{105 x^2 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{105 d x^3}-\frac {8 b c^7 \sqrt {d-c^2 d x^2} \log (x)}{105 \sqrt {1-c^2 x^2}} \]
-1/7*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/d/x^7-4/35*c^2*(-c^2*d*x^2+d)^ (3/2)*(a+b*arcsin(c*x))/d/x^5-8/105*c^4*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c *x))/d/x^3-1/42*b*c*(-c^2*d*x^2+d)^(1/2)/x^6/(-c^2*x^2+1)^(1/2)+1/140*b*c^ 3*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)+2/105*b*c^5*(-c^2*d*x^2+d)^( 1/2)/x^2/(-c^2*x^2+1)^(1/2)-8/105*b*c^7*ln(x)*(-c^2*d*x^2+d)^(1/2)/(-c^2*x ^2+1)^(1/2)
Time = 0.25 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (-50 b c x+15 b c^3 x^3+40 b c^5 x^5+392 b c^7 x^7-300 a \sqrt {1-c^2 x^2}+60 a c^2 x^2 \sqrt {1-c^2 x^2}+80 a c^4 x^4 \sqrt {1-c^2 x^2}+160 a c^6 x^6 \sqrt {1-c^2 x^2}+20 b \sqrt {1-c^2 x^2} \left (-15+3 c^2 x^2+4 c^4 x^4+8 c^6 x^6\right ) \arcsin (c x)-160 b c^7 x^7 \log (x)\right )}{2100 x^7 \sqrt {1-c^2 x^2}} \]
(Sqrt[d - c^2*d*x^2]*(-50*b*c*x + 15*b*c^3*x^3 + 40*b*c^5*x^5 + 392*b*c^7* x^7 - 300*a*Sqrt[1 - c^2*x^2] + 60*a*c^2*x^2*Sqrt[1 - c^2*x^2] + 80*a*c^4* x^4*Sqrt[1 - c^2*x^2] + 160*a*c^6*x^6*Sqrt[1 - c^2*x^2] + 20*b*Sqrt[1 - c^ 2*x^2]*(-15 + 3*c^2*x^2 + 4*c^4*x^4 + 8*c^6*x^6)*ArcSin[c*x] - 160*b*c^7*x ^7*Log[x]))/(2100*x^7*Sqrt[1 - c^2*x^2])
Time = 0.43 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.66, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5194, 27, 2010, 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.
| \(\displaystyle \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {-8 c^6 x^6-4 c^4 x^4-3 c^2 x^2+15}{105 x^7}dx}{\sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{105 d x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {-8 c^6 x^6-4 c^4 x^4-3 c^2 x^2+15}{x^7}dx}{105 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{105 d x^3}\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (-\frac {8 c^6}{x}-\frac {4 c^4}{x^3}-\frac {3 c^2}{x^5}+\frac {15}{x^7}\right )dx}{105 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{105 d x^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{105 d x^3}+\frac {b c \sqrt {d-c^2 d x^2} \left (-8 c^6 \log (x)+\frac {2 c^4}{x^2}+\frac {3 c^2}{4 x^4}-\frac {5}{2 x^6}\right )}{105 \sqrt {1-c^2 x^2}}\) |
-1/7*((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(d*x^7) - (4*c^2*(d - c^2 *d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(35*d*x^5) - (8*c^4*(d - c^2*d*x^2)^(3/ 2)*(a + b*ArcSin[c*x]))/(105*d*x^3) + (b*c*Sqrt[d - c^2*d*x^2]*(-5/(2*x^6) + (3*c^2)/(4*x^4) + (2*c^4)/x^2 - 8*c^6*Log[x]))/(105*Sqrt[1 - c^2*x^2])
3.1.61.3.1 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[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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 = 0.25 (sec) , antiderivative size = 2751, normalized size of antiderivative = 10.46
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(2751\) |
| parts | \(\text {Expression too large to display}\) | \(2751\) |
225/7*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2 *x^2+225)/x^7/(c^2*x^2-1)*arcsin(c*x)+73/20*b*(-d*(c^2*x^2-1))^(1/2)/(280* c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)/(c^2*x^2-1)*c^7*(-c^2*x^2+ 1)^(1/2)+8/105*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*ln( (I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*c^7-8*I*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8* x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^6/(c^2*x^2-1)*(-c^2*x^2+1)^( 1/2)*arcsin(c*x)*c^13-8/5*I*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6* x^6-21*c^4*x^4-315*c^2*x^2+225)*x^4/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin( c*x)*c^11-24*I*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^ 4-315*c^2*x^2+225)*x^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^9+64/3 *I*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^ 2+225)*x^8/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^15-302/105*I*b*(-d *(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x ^5/(c^2*x^2-1)*(-c^2*x^2+1)*c^12+a*(-1/7/d/x^7*(-c^2*d*x^2+d)^(3/2)+4/7*c^ 2*(-1/5/d/x^5*(-c^2*d*x^2+d)^(3/2)-2/15*c^2/d/x^3*(-c^2*d*x^2+d)^(3/2)))+6 4/3*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x ^2+225)*x^9/(c^2*x^2-1)*arcsin(c*x)*c^16-56/3*b*(-d*(c^2*x^2-1))^(1/2)/(28 0*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^7/(c^2*x^2-1)*arcsin(c *x)*c^14-16/3*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4 -315*c^2*x^2+225)*x^6/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*c^13-4/15*b*(-d*(c...
Time = 0.31 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.16 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=\left [\frac {16 \, {\left (b c^{9} x^{9} - b c^{7} x^{7}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} + \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} - d}{c^{2} x^{4} - x^{2}}\right ) - {\left (8 \, b c^{5} x^{5} - {\left (8 \, b c^{5} + 3 \, b c^{3} - 10 \, b c\right )} x^{7} + 3 \, b c^{3} x^{3} - 10 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 4 \, {\left (8 \, a c^{8} x^{8} - 4 \, a c^{6} x^{6} - a c^{4} x^{4} - 18 \, a c^{2} x^{2} + {\left (8 \, b c^{8} x^{8} - 4 \, b c^{6} x^{6} - b c^{4} x^{4} - 18 \, b c^{2} x^{2} + 15 \, b\right )} \arcsin \left (c x\right ) + 15 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{420 \, {\left (c^{2} x^{9} - x^{7}\right )}}, -\frac {32 \, {\left (b c^{9} x^{9} - b c^{7} x^{7}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} {\left (x^{2} + 1\right )} \sqrt {-d}}{c^{2} d x^{4} - {\left (c^{2} + 1\right )} d x^{2} + d}\right ) + {\left (8 \, b c^{5} x^{5} - {\left (8 \, b c^{5} + 3 \, b c^{3} - 10 \, b c\right )} x^{7} + 3 \, b c^{3} x^{3} - 10 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 4 \, {\left (8 \, a c^{8} x^{8} - 4 \, a c^{6} x^{6} - a c^{4} x^{4} - 18 \, a c^{2} x^{2} + {\left (8 \, b c^{8} x^{8} - 4 \, b c^{6} x^{6} - b c^{4} x^{4} - 18 \, b c^{2} x^{2} + 15 \, b\right )} \arcsin \left (c x\right ) + 15 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{420 \, {\left (c^{2} x^{9} - x^{7}\right )}}\right ] \]
[1/420*(16*(b*c^9*x^9 - b*c^7*x^7)*sqrt(d)*log((c^2*d*x^6 + c^2*d*x^2 - d* x^4 + sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^4 - 1)*sqrt(d) - d)/(c^2* x^4 - x^2)) - (8*b*c^5*x^5 - (8*b*c^5 + 3*b*c^3 - 10*b*c)*x^7 + 3*b*c^3*x^ 3 - 10*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 4*(8*a*c^8*x^8 - 4 *a*c^6*x^6 - a*c^4*x^4 - 18*a*c^2*x^2 + (8*b*c^8*x^8 - 4*b*c^6*x^6 - b*c^4 *x^4 - 18*b*c^2*x^2 + 15*b)*arcsin(c*x) + 15*a)*sqrt(-c^2*d*x^2 + d))/(c^2 *x^9 - x^7), -1/420*(32*(b*c^9*x^9 - b*c^7*x^7)*sqrt(-d)*arctan(sqrt(-c^2* d*x^2 + d)*sqrt(-c^2*x^2 + 1)*(x^2 + 1)*sqrt(-d)/(c^2*d*x^4 - (c^2 + 1)*d* x^2 + d)) + (8*b*c^5*x^5 - (8*b*c^5 + 3*b*c^3 - 10*b*c)*x^7 + 3*b*c^3*x^3 - 10*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 4*(8*a*c^8*x^8 - 4*a *c^6*x^6 - a*c^4*x^4 - 18*a*c^2*x^2 + (8*b*c^8*x^8 - 4*b*c^6*x^6 - b*c^4*x ^4 - 18*b*c^2*x^2 + 15*b)*arcsin(c*x) + 15*a)*sqrt(-c^2*d*x^2 + d))/(c^2*x ^9 - x^7)]
\[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=\int \frac {\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{x^{8}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=-\frac {1}{420} \, {\left (32 \, c^{6} \sqrt {d} \log \left (x\right ) - \frac {8 \, c^{4} \sqrt {d} x^{4} + 3 \, c^{2} \sqrt {d} x^{2} - 10 \, \sqrt {d}}{x^{6}}\right )} b c - \frac {1}{105} \, {\left (\frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4}}{d x^{3}} + \frac {12 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{5}} + \frac {15 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{7}}\right )} b \arcsin \left (c x\right ) - \frac {1}{105} \, {\left (\frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4}}{d x^{3}} + \frac {12 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{5}} + \frac {15 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{7}}\right )} a \]
-1/420*(32*c^6*sqrt(d)*log(x) - (8*c^4*sqrt(d)*x^4 + 3*c^2*sqrt(d)*x^2 - 1 0*sqrt(d))/x^6)*b*c - 1/105*(8*(-c^2*d*x^2 + d)^(3/2)*c^4/(d*x^3) + 12*(-c ^2*d*x^2 + d)^(3/2)*c^2/(d*x^5) + 15*(-c^2*d*x^2 + d)^(3/2)/(d*x^7))*b*arc sin(c*x) - 1/105*(8*(-c^2*d*x^2 + d)^(3/2)*c^4/(d*x^3) + 12*(-c^2*d*x^2 + d)^(3/2)*c^2/(d*x^5) + 15*(-c^2*d*x^2 + d)^(3/2)/(d*x^7))*a
Exception generated. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{x^8} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2}}{x^8} \,d x \]