Integrand size = 27, antiderivative size = 282 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=-\frac {b c^3 d^2 \sqrt {d-c^2 d x^2}}{189 x^6 \sqrt {1-c^2 x^2}}+\frac {b c^5 d^2 \sqrt {d-c^2 d x^2}}{42 x^4 \sqrt {1-c^2 x^2}}-\frac {b c^7 d^2 \sqrt {d-c^2 d x^2}}{21 x^2 \sqrt {1-c^2 x^2}}-\frac {b c d^2 \left (1-c^2 x^2\right )^{7/2} \sqrt {d-c^2 d x^2}}{72 x^8}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}-\frac {2 b c^9 d^2 \sqrt {d-c^2 d x^2} \log (x)}{63 \sqrt {1-c^2 x^2}} \]
-1/9*(-c^2*d*x^2+d)^(7/2)*(a+b*arcsin(c*x))/d/x^9-2/63*c^2*(-c^2*d*x^2+d)^ (7/2)*(a+b*arcsin(c*x))/d/x^7-1/72*b*c*d^2*(-c^2*x^2+1)^(7/2)*(-c^2*d*x^2+ d)^(1/2)/x^8-1/189*b*c^3*d^2*(-c^2*d*x^2+d)^(1/2)/x^6/(-c^2*x^2+1)^(1/2)+1 /42*b*c^5*d^2*(-c^2*d*x^2+d)^(1/2)/x^4/(-c^2*x^2+1)^(1/2)-1/21*b*c^7*d^2*( -c^2*d*x^2+d)^(1/2)/x^2/(-c^2*x^2+1)^(1/2)-2/63*b*c^9*d^2*ln(x)*(-c^2*d*x^ 2+d)^(1/2)/(-c^2*x^2+1)^(1/2)
Time = 0.31 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.82 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=\frac {d^2 \sqrt {d-c^2 d x^2} \left (-735 b c x+2660 b c^3 x^3-3150 b c^5 x^5+420 b c^7 x^7+4566 b c^9 x^9-5880 a \sqrt {1-c^2 x^2}+15960 a c^2 x^2 \sqrt {1-c^2 x^2}-12600 a c^4 x^4 \sqrt {1-c^2 x^2}+840 a c^6 x^6 \sqrt {1-c^2 x^2}+1680 a c^8 x^8 \sqrt {1-c^2 x^2}-840 b \left (1-c^2 x^2\right )^{7/2} \left (7+2 c^2 x^2\right ) \arcsin (c x)-1680 b c^9 x^9 \log (x)\right )}{52920 x^9 \sqrt {1-c^2 x^2}} \]
(d^2*Sqrt[d - c^2*d*x^2]*(-735*b*c*x + 2660*b*c^3*x^3 - 3150*b*c^5*x^5 + 4 20*b*c^7*x^7 + 4566*b*c^9*x^9 - 5880*a*Sqrt[1 - c^2*x^2] + 15960*a*c^2*x^2 *Sqrt[1 - c^2*x^2] - 12600*a*c^4*x^4*Sqrt[1 - c^2*x^2] + 840*a*c^6*x^6*Sqr t[1 - c^2*x^2] + 1680*a*c^8*x^8*Sqrt[1 - c^2*x^2] - 840*b*(1 - c^2*x^2)^(7 /2)*(7 + 2*c^2*x^2)*ArcSin[c*x] - 1680*b*c^9*x^9*Log[x]))/(52920*x^9*Sqrt[ 1 - c^2*x^2])
Time = 0.40 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.60, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5194, 27, 354, 87, 49, 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.
\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx\) |
\(\Big \downarrow \) 5194 |
\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {d^2 \left (1-c^2 x^2\right )^3 \left (2 c^2 x^2+7\right )}{63 x^9}dx}{\sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b c d^2 \sqrt {d-c^2 d x^2} \int \frac {\left (1-c^2 x^2\right )^3 \left (2 c^2 x^2+7\right )}{x^9}dx}{63 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {b c d^2 \sqrt {d-c^2 d x^2} \int \frac {\left (1-c^2 x^2\right )^3 \left (2 c^2 x^2+7\right )}{x^{10}}dx^2}{126 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {b c d^2 \sqrt {d-c^2 d x^2} \left (2 c^2 \int \frac {\left (1-c^2 x^2\right )^3}{x^8}dx^2-\frac {7 \left (1-c^2 x^2\right )^4}{4 x^8}\right )}{126 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {b c d^2 \sqrt {d-c^2 d x^2} \left (2 c^2 \int \left (-\frac {c^6}{x^2}+\frac {3 c^4}{x^4}-\frac {3 c^2}{x^6}+\frac {1}{x^8}\right )dx^2-\frac {7 \left (1-c^2 x^2\right )^4}{4 x^8}\right )}{126 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{9 d x^9}-\frac {2 c^2 \left (d-c^2 d x^2\right )^{7/2} (a+b \arcsin (c x))}{63 d x^7}+\frac {b c d^2 \sqrt {d-c^2 d x^2} \left (2 c^2 \left (c^6 \left (-\log \left (x^2\right )\right )-\frac {3 c^4}{x^2}+\frac {3 c^2}{2 x^4}-\frac {1}{3 x^6}\right )-\frac {7 \left (1-c^2 x^2\right )^4}{4 x^8}\right )}{126 \sqrt {1-c^2 x^2}}\) |
-1/9*((d - c^2*d*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(d*x^9) - (2*c^2*(d - c^2 *d*x^2)^(7/2)*(a + b*ArcSin[c*x]))/(63*d*x^7) + (b*c*d^2*Sqrt[d - c^2*d*x^ 2]*((-7*(1 - c^2*x^2)^4)/(4*x^8) + 2*c^2*(-1/3*1/x^6 + (3*c^2)/(2*x^4) - ( 3*c^4)/x^2 - c^6*Log[x^2])))/(126*Sqrt[1 - c^2*x^2])
3.1.91.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/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 = 0.32 (sec) , antiderivative size = 5324, normalized size of antiderivative = 18.88
method | result | size |
default | \(\text {Expression too large to display}\) | \(5324\) |
parts | \(\text {Expression too large to display}\) | \(5324\) |
Time = 0.34 (sec) , antiderivative size = 747, normalized size of antiderivative = 2.65 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=\left [\frac {24 \, {\left (b c^{11} d^{2} x^{11} - b c^{9} d^{2} x^{9}\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 (12 \, b c^{7} d^{2} x^{7} - 90 \, b c^{5} d^{2} x^{5} - {\left (12 \, b c^{7} - 90 \, b c^{5} + 76 \, b c^{3} - 21 \, b c\right )} d^{2} x^{9} + 76 \, b c^{3} d^{2} x^{3} - 21 \, b c d^{2} x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 24 \, {\left (2 \, a c^{10} d^{2} x^{10} - a c^{8} d^{2} x^{8} - 16 \, a c^{6} d^{2} x^{6} + 34 \, a c^{4} d^{2} x^{4} - 26 \, a c^{2} d^{2} x^{2} + 7 \, a d^{2} + {\left (2 \, b c^{10} d^{2} x^{10} - b c^{8} d^{2} x^{8} - 16 \, b c^{6} d^{2} x^{6} + 34 \, b c^{4} d^{2} x^{4} - 26 \, b c^{2} d^{2} x^{2} + 7 \, b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{1512 \, {\left (c^{2} x^{11} - x^{9}\right )}}, -\frac {48 \, {\left (b c^{11} d^{2} x^{11} - b c^{9} d^{2} x^{9}\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 (12 \, b c^{7} d^{2} x^{7} - 90 \, b c^{5} d^{2} x^{5} - {\left (12 \, b c^{7} - 90 \, b c^{5} + 76 \, b c^{3} - 21 \, b c\right )} d^{2} x^{9} + 76 \, b c^{3} d^{2} x^{3} - 21 \, b c d^{2} x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} - 24 \, {\left (2 \, a c^{10} d^{2} x^{10} - a c^{8} d^{2} x^{8} - 16 \, a c^{6} d^{2} x^{6} + 34 \, a c^{4} d^{2} x^{4} - 26 \, a c^{2} d^{2} x^{2} + 7 \, a d^{2} + {\left (2 \, b c^{10} d^{2} x^{10} - b c^{8} d^{2} x^{8} - 16 \, b c^{6} d^{2} x^{6} + 34 \, b c^{4} d^{2} x^{4} - 26 \, b c^{2} d^{2} x^{2} + 7 \, b d^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{1512 \, {\left (c^{2} x^{11} - x^{9}\right )}}\right ] \]
[1/1512*(24*(b*c^11*d^2*x^11 - b*c^9*d^2*x^9)*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)) - (12*b*c^7*d^2*x^7 - 90*b*c^5*d^2*x^5 - (12*b*c^7 - 90*b*c^5 + 76*b*c^3 - 21*b*c)*d^2*x^9 + 76*b*c^3*d^2*x^3 - 21*b*c*d^2*x) *sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 24*(2*a*c^10*d^2*x^10 - a*c^8*d ^2*x^8 - 16*a*c^6*d^2*x^6 + 34*a*c^4*d^2*x^4 - 26*a*c^2*d^2*x^2 + 7*a*d^2 + (2*b*c^10*d^2*x^10 - b*c^8*d^2*x^8 - 16*b*c^6*d^2*x^6 + 34*b*c^4*d^2*x^4 - 26*b*c^2*d^2*x^2 + 7*b*d^2)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^1 1 - x^9), -1/1512*(48*(b*c^11*d^2*x^11 - b*c^9*d^2*x^9)*sqrt(-d)*arctan(sq rt(-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)) + (12*b*c^7*d^2*x^7 - 90*b*c^5*d^2*x^5 - (12*b*c^7 - 90* b*c^5 + 76*b*c^3 - 21*b*c)*d^2*x^9 + 76*b*c^3*d^2*x^3 - 21*b*c*d^2*x)*sqrt (-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) - 24*(2*a*c^10*d^2*x^10 - a*c^8*d^2*x^ 8 - 16*a*c^6*d^2*x^6 + 34*a*c^4*d^2*x^4 - 26*a*c^2*d^2*x^2 + 7*a*d^2 + (2* b*c^10*d^2*x^10 - b*c^8*d^2*x^8 - 16*b*c^6*d^2*x^6 + 34*b*c^4*d^2*x^4 - 26 *b*c^2*d^2*x^2 + 7*b*d^2)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d))/(c^2*x^11 - x ^9)]
Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.57 \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=-\frac {1}{1512} \, {\left (48 \, c^{8} d^{\frac {5}{2}} \log \left (x\right ) - \frac {12 \, c^{6} d^{\frac {5}{2}} x^{6} - 90 \, c^{4} d^{\frac {5}{2}} x^{4} + 76 \, c^{2} d^{\frac {5}{2}} x^{2} - 21 \, d^{\frac {5}{2}}}{x^{8}}\right )} b c - \frac {1}{63} \, b {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} c^{2}}{d x^{7}} + \frac {7 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{9}}\right )} \arcsin \left (c x\right ) - \frac {1}{63} \, a {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}} c^{2}}{d x^{7}} + \frac {7 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {7}{2}}}{d x^{9}}\right )} \]
-1/1512*(48*c^8*d^(5/2)*log(x) - (12*c^6*d^(5/2)*x^6 - 90*c^4*d^(5/2)*x^4 + 76*c^2*d^(5/2)*x^2 - 21*d^(5/2))/x^8)*b*c - 1/63*b*(2*(-c^2*d*x^2 + d)^( 7/2)*c^2/(d*x^7) + 7*(-c^2*d*x^2 + d)^(7/2)/(d*x^9))*arcsin(c*x) - 1/63*a* (2*(-c^2*d*x^2 + d)^(7/2)*c^2/(d*x^7) + 7*(-c^2*d*x^2 + d)^(7/2)/(d*x^9))
Exception generated. \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{x^{10}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^{5/2}}{x^{10}} \,d x \]