Integrand size = 27, antiderivative size = 212 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b}{6 c^5 d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {x (a+b \arcsin (c x))}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {2 b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \]
1/3*x^3*(a+b*arcsin(c*x))/c^2/d/(-c^2*d*x^2+d)^(3/2)-x*(a+b*arcsin(c*x))/c ^4/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b/c^5/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d )^(1/2)+1/2*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c^5/d^2/(-c^2*d*x^2+d )^(1/2)-2/3*b*ln(-c^2*x^2+1)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(-c^2*d*x^2+d)^(1/ 2)
Time = 0.40 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {-3 b \sqrt {d} \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)^2-6 a \left (-1+c^2 x^2\right ) \sqrt {d-c^2 d x^2} \arctan \left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+\sqrt {d} \left (6 a c x-8 a c^3 x^3+b \sqrt {1-c^2 x^2}+4 b \left (1-c^2 x^2\right )^{3/2} \log \left (1-c^2 x^2\right )\right )+2 b \sqrt {d} \arcsin (c x) \sin (3 \arcsin (c x))}{6 c^5 d^{5/2} \left (-1+c^2 x^2\right ) \sqrt {d-c^2 d x^2}} \]
(-3*b*Sqrt[d]*(1 - c^2*x^2)^(3/2)*ArcSin[c*x]^2 - 6*a*(-1 + c^2*x^2)*Sqrt[ d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + Sqrt[d]*(6*a*c*x - 8*a*c^3*x^3 + b*Sqrt[1 - c^2*x^2] + 4*b*(1 - c^2*x^2) ^(3/2)*Log[1 - c^2*x^2]) + 2*b*Sqrt[d]*ArcSin[c*x]*Sin[3*ArcSin[c*x]])/(6* c^5*d^(5/2)*(-1 + c^2*x^2)*Sqrt[d - c^2*d*x^2])
Time = 0.72 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5206, 243, 49, 2009, 5206, 240, 5152}
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.
| \(\displaystyle \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5206 |
| \(\displaystyle -\frac {\int \frac {x^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \frac {x^3}{\left (1-c^2 x^2\right )^2}dx}{3 c d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle -\frac {\int \frac {x^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \frac {x^2}{\left (1-c^2 x^2\right )^2}dx^2}{6 c d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle -\frac {\int \frac {x^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \left (\frac {1}{c^2 \left (c^2 x^2-1\right )}+\frac {1}{c^2 \left (c^2 x^2-1\right )^2}\right )dx^2}{6 c d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\int \frac {x^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}}dx}{c^2 d}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b \sqrt {1-c^2 x^2} \left (\frac {1}{c^4 \left (1-c^2 x^2\right )}+\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5206 |
| \(\displaystyle -\frac {-\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {d-c^2 d x^2}}dx}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \int \frac {x}{1-c^2 x^2}dx}{c d \sqrt {d-c^2 d x^2}}+\frac {x (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}}{c^2 d}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b \sqrt {1-c^2 x^2} \left (\frac {1}{c^4 \left (1-c^2 x^2\right )}+\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle -\frac {-\frac {\int \frac {a+b \arcsin (c x)}{\sqrt {d-c^2 d x^2}}dx}{c^2 d}+\frac {x (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}}}{c^2 d}+\frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b \sqrt {1-c^2 x^2} \left (\frac {1}{c^4 \left (1-c^2 x^2\right )}+\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5152 |
| \(\displaystyle \frac {x^3 (a+b \arcsin (c x))}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {\frac {x (a+b \arcsin (c x))}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}}}{c^2 d}-\frac {b \sqrt {1-c^2 x^2} \left (\frac {1}{c^4 \left (1-c^2 x^2\right )}+\frac {\log \left (1-c^2 x^2\right )}{c^4}\right )}{6 c d^2 \sqrt {d-c^2 d x^2}}\) |
(x^3*(a + b*ArcSin[c*x]))/(3*c^2*d*(d - c^2*d*x^2)^(3/2)) - (b*Sqrt[1 - c^ 2*x^2]*(1/(c^4*(1 - c^2*x^2)) + Log[1 - c^2*x^2]/c^4))/(6*c*d^2*Sqrt[d - c ^2*d*x^2]) - ((x*(a + b*ArcSin[c*x]))/(c^2*d*Sqrt[d - c^2*d*x^2]) - (Sqrt[ 1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c^3*d*Sqrt[d - c^2*d*x^2]) + (b*S qrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c^3*d*Sqrt[d - c^2*d*x^2]))/(c^2*d)
3.2.31.3.1 Defintions of rubi rules used
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[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + (-Simp[f^2*((m - 1)/(2*e*(p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Simp [b*f*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{ a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IG tQ[m, 1]
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.69
| method | result | size |
| default | \(\frac {a \,x^{3}}{3 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {a x}{c^{4} d^{2} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{4} d^{2} \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (3 \arcsin \left (c x \right )^{2} x^{4} c^{4}+8 i \arcsin \left (c x \right ) x^{4} c^{4}-8 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+8 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3} x^{3}-6 \arcsin \left (c x \right )^{2} x^{2} c^{2}-16 i \arcsin \left (c x \right ) x^{2} c^{2}+16 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-6 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x c +c^{2} x^{2}+3 \arcsin \left (c x \right )^{2}+8 i \arcsin \left (c x \right )-8 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )-1\right )}{6 d^{3} \left (c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1\right ) c^{5}}\) | \(359\) |
| parts | \(\frac {a \,x^{3}}{3 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {a x}{c^{4} d^{2} \sqrt {-c^{2} d \,x^{2}+d}}+\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{4} d^{2} \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (3 \arcsin \left (c x \right )^{2} x^{4} c^{4}+8 i \arcsin \left (c x \right ) x^{4} c^{4}-8 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+8 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) c^{3} x^{3}-6 \arcsin \left (c x \right )^{2} x^{2} c^{2}-16 i \arcsin \left (c x \right ) x^{2} c^{2}+16 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-6 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x c +c^{2} x^{2}+3 \arcsin \left (c x \right )^{2}+8 i \arcsin \left (c x \right )-8 \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )-1\right )}{6 d^{3} \left (c^{6} x^{6}-3 c^{4} x^{4}+3 c^{2} x^{2}-1\right ) c^{5}}\) | \(359\) |
1/3*a*x^3/c^2/d/(-c^2*d*x^2+d)^(3/2)-a/c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)+a/c^ 4/d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/6*b*(-d *(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*(3*arcsin(c*x)^2*x^4*c^4+8*I*arcsin (c*x)*x^4*c^4-8*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*x^4*c^4+8*(-c^2*x^2+1)^ (1/2)*arcsin(c*x)*c^3*x^3-6*arcsin(c*x)^2*x^2*c^2-16*I*arcsin(c*x)*x^2*c^2 +16*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*x^2*c^2-6*(-c^2*x^2+1)^(1/2)*arcsin (c*x)*x*c+c^2*x^2+3*arcsin(c*x)^2+8*I*arcsin(c*x)-8*ln(1+(I*c*x+(-c^2*x^2+ 1)^(1/2))^2)-1)/d^3/(c^6*x^6-3*c^4*x^4+3*c^2*x^2-1)/c^5
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(b*x^4*arcsin(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{4} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
1/3*(x*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c ^4*d)) - x/(sqrt(-c^2*d*x^2 + d)*c^4*d^2) + 3*arcsin(c*x)/(c^5*d^(5/2)))*a + b*integrate(x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^ 4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d)
Exception generated. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {x^4 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]