Integrand size = 27, antiderivative size = 310 \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b c^3 \sqrt {d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^3 x^2 \sqrt {1-c^2 x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}-\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 b c^3 \sqrt {d-c^2 d x^2} \log (x)}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {4 b c^3 \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{3 d^3 \sqrt {1-c^2 x^2}} \]
1/3*(-a-b*arcsin(c*x))/d/x^3/(-c^2*d*x^2+d)^(3/2)-2*c^2*(a+b*arcsin(c*x))/ d/x/(-c^2*d*x^2+d)^(3/2)+8/3*c^4*x*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(3/2 )+16/3*c^4*x*(a+b*arcsin(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)-1/6*b*c^3*(-c^2*d* x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(3/2)-1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^3/x^2/( -c^2*x^2+1)^(1/2)+8/3*b*c^3*ln(x)*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1 /2)+4/3*b*c^3*ln(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(1/2)
Time = 0.36 (sec) , antiderivative size = 277, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d-c^2 d x^2} \left (b c x-b c^3 x^3+2 a \sqrt {1-c^2 x^2}+12 a c^2 x^2 \sqrt {1-c^2 x^2}-48 a c^4 x^4 \sqrt {1-c^2 x^2}+32 a c^6 x^6 \sqrt {1-c^2 x^2}+2 b \sqrt {1-c^2 x^2} \left (1+6 c^2 x^2-24 c^4 x^4+16 c^6 x^6\right ) \arcsin (c x)+8 b c^3 x^3 \left (-1+c^2 x^2\right )^2 \log \left (1-\frac {1}{c^2 x^2}\right )-16 b c^3 x^3 \log \left (1-c^2 x^2\right )+32 b c^5 x^5 \log \left (1-c^2 x^2\right )-16 b c^7 x^7 \log \left (1-c^2 x^2\right )\right )}{6 d^3 x^3 \left (1-c^2 x^2\right )^{5/2}} \]
-1/6*(Sqrt[d - c^2*d*x^2]*(b*c*x - b*c^3*x^3 + 2*a*Sqrt[1 - c^2*x^2] + 12* a*c^2*x^2*Sqrt[1 - c^2*x^2] - 48*a*c^4*x^4*Sqrt[1 - c^2*x^2] + 32*a*c^6*x^ 6*Sqrt[1 - c^2*x^2] + 2*b*Sqrt[1 - c^2*x^2]*(1 + 6*c^2*x^2 - 24*c^4*x^4 + 16*c^6*x^6)*ArcSin[c*x] + 8*b*c^3*x^3*(-1 + c^2*x^2)^2*Log[1 - 1/(c^2*x^2) ] - 16*b*c^3*x^3*Log[1 - c^2*x^2] + 32*b*c^5*x^5*Log[1 - c^2*x^2] - 16*b*c ^7*x^7*Log[1 - c^2*x^2]))/(d^3*x^3*(1 - c^2*x^2)^(5/2))
Time = 0.61 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5194, 27, 2331, 2123, 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.
| \(\displaystyle \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {16 c^6 x^6-24 c^4 x^4+6 c^2 x^2+1}{3 d^3 x^3 \left (1-c^2 x^2\right )^2}dx}{\sqrt {1-c^2 x^2}}-\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {16 c^6 x^6-24 c^4 x^4+6 c^2 x^2+1}{x^3 \left (1-c^2 x^2\right )^2}dx}{3 d^3 \sqrt {1-c^2 x^2}}-\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2331 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {16 c^6 x^6-24 c^4 x^4+6 c^2 x^2+1}{x^4 \left (1-c^2 x^2\right )^2}dx^2}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (\frac {8 c^4}{c^2 x^2-1}-\frac {c^4}{\left (c^2 x^2-1\right )^2}+\frac {8 c^2}{x^2}+\frac {1}{x^4}\right )dx^2}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 c^2 (a+b \arcsin (c x))}{d x \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \left (d-c^2 d x^2\right )^{3/2}}+\frac {16 c^4 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b c \sqrt {d-c^2 d x^2} \left (-\frac {c^2}{1-c^2 x^2}+8 c^2 \log \left (x^2\right )+8 c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )}{6 d^3 \sqrt {1-c^2 x^2}}\) |
-1/3*(a + b*ArcSin[c*x])/(d*x^3*(d - c^2*d*x^2)^(3/2)) - (2*c^2*(a + b*Arc Sin[c*x]))/(d*x*(d - c^2*d*x^2)^(3/2)) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3* d*(d - c^2*d*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSin[c*x]))/(3*d^2*Sqrt[d - c^2*d*x^2]) + (b*c*Sqrt[d - c^2*d*x^2]*(-x^(-2) - c^2/(1 - c^2*x^2) + 8*c^ 2*Log[x^2] + 8*c^2*Log[1 - c^2*x^2]))/(6*d^3*Sqrt[1 - c^2*x^2])
3.2.39.3.1 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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && 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 = 0.21 (sec) , antiderivative size = 1877, normalized size of antiderivative = 6.05
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1877\) |
| parts | \(\text {Expression too large to display}\) | \(1877\) |
-64*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^ 2-1)/d^3*x^6*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^9+128*I*b*(-d*(c^2*x^2-1))^( 1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^4*arcsin(c*x)*( -c^2*x^2+1)^(1/2)*c^7-176/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6* x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*c^5+2* b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d ^3*c^3*(-c^2*x^2+1)^(1/2)+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6* x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x^3*arcsin(c*x)+a*(-1/3/d/x^3/(-c^2*d*x^2 +d)^(3/2)+2*c^2*(-1/d/x/(-c^2*d*x^2+d)^(3/2)+4*c^2*(1/3/d*x/(-c^2*d*x^2+d) ^(3/2)+2/3/d^2*x/(-c^2*d*x^2+d)^(1/2))))-8/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^ 2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^4-1)*c^3-64*b *(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^ 3*x^7*arcsin(c*x)*c^10+160*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6 +35*c^4*x^4-10*c^2*x^2-1)/d^3*x^5*arcsin(c*x)*c^8-344/3*b*(-d*(c^2*x^2-1)) ^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x^3*arcsin(c*x) *c^6-2*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x ^2-1)/d^3*x^2*c^5*(-c^2*x^2+1)^(1/2)+12*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x ^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3*x*arcsin(c*x)*c^4+6*b*(-d*(c^2* x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^4*x^4-10*c^2*x^2-1)/d^3/x*arcsin (c*x)*c^2-280/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(12*c^8*x^8-36*c^6*x^6+35*c^...
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}} \,d x } \]
integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^10 - 3*c^4*d ^3*x^8 + 3*c^2*d^3*x^6 - d^3*x^4), x)
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{4} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {8 \, c^{2} \log \left (c x + 1\right )}{d^{\frac {5}{2}}} + \frac {8 \, c^{2} \log \left (c x - 1\right )}{d^{\frac {5}{2}}} + \frac {16 \, c^{2} \log \left (x\right )}{d^{\frac {5}{2}}} + \frac {1}{c^{2} d^{\frac {5}{2}} x^{4} - d^{\frac {5}{2}} x^{2}}\right )} + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {6 \, c^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} b \arcsin \left (c x\right ) + \frac {1}{3} \, {\left (\frac {16 \, c^{4} x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {8 \, c^{4} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d} - \frac {6 \, c^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x} - \frac {1}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d x^{3}}\right )} a \]
1/6*b*c*(8*c^2*log(c*x + 1)/d^(5/2) + 8*c^2*log(c*x - 1)/d^(5/2) + 16*c^2* log(x)/d^(5/2) + 1/(c^2*d^(5/2)*x^4 - d^(5/2)*x^2)) + 1/3*(16*c^4*x/(sqrt( -c^2*d*x^2 + d)*d^2) + 8*c^4*x/((-c^2*d*x^2 + d)^(3/2)*d) - 6*c^2/((-c^2*d *x^2 + d)^(3/2)*d*x) - 1/((-c^2*d*x^2 + d)^(3/2)*d*x^3))*b*arcsin(c*x) + 1 /3*(16*c^4*x/(sqrt(-c^2*d*x^2 + d)*d^2) + 8*c^4*x/((-c^2*d*x^2 + d)^(3/2)* d) - 6*c^2/((-c^2*d*x^2 + d)^(3/2)*d*x) - 1/((-c^2*d*x^2 + d)^(3/2)*d*x^3) )*a
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]