Integrand size = 27, antiderivative size = 238 \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^2 x^2 \sqrt {1-c^2 x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}-\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}+\frac {5 b c^3 \sqrt {d-c^2 d x^2} \log (x)}{3 d^2 \sqrt {1-c^2 x^2}}+\frac {b c^3 \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{2 d^2 \sqrt {1-c^2 x^2}} \]
1/3*(-a-b*arcsin(c*x))/d/x^3/(-c^2*d*x^2+d)^(1/2)-4/3*c^2*(a+b*arcsin(c*x) )/d/x/(-c^2*d*x^2+d)^(1/2)+8/3*c^4*x*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(1 /2)-1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^2/x^2/(-c^2*x^2+1)^(1/2)+5/3*b*c^3*ln(x )*(-c^2*d*x^2+d)^(1/2)/d^2/(-c^2*x^2+1)^(1/2)+1/2*b*c^3*ln(-c^2*x^2+1)*(-c ^2*d*x^2+d)^(1/2)/d^2/(-c^2*x^2+1)^(1/2)
Time = 0.59 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d-c^2 d x^2} \left (b c x-b c^3 x^3+2 a \sqrt {1-c^2 x^2}+8 a c^2 x^2 \sqrt {1-c^2 x^2}-16 a c^4 x^4 \sqrt {1-c^2 x^2}-2 b \sqrt {1-c^2 x^2} \left (-1-4 c^2 x^2+8 c^4 x^4\right ) \arcsin (c x)-5 b c^3 x^3 \left (-1+c^2 x^2\right ) \log \left (1-\frac {1}{c^2 x^2}\right )-8 b c^3 x^3 \log \left (1-c^2 x^2\right )+8 b c^5 x^5 \log \left (1-c^2 x^2\right )\right )}{6 d^2 x^3 \left (1-c^2 x^2\right )^{3/2}} \]
-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] + 8*a *c^2*x^2*Sqrt[1 - c^2*x^2] - 16*a*c^4*x^4*Sqrt[1 - c^2*x^2] - 2*b*Sqrt[1 - c^2*x^2]*(-1 - 4*c^2*x^2 + 8*c^4*x^4)*ArcSin[c*x] - 5*b*c^3*x^3*(-1 + c^2 *x^2)*Log[1 - 1/(c^2*x^2)] - 8*b*c^3*x^3*Log[1 - c^2*x^2] + 8*b*c^5*x^5*Lo g[1 - c^2*x^2]))/(d^2*x^3*(1 - c^2*x^2)^(3/2))
Time = 0.46 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.73, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5194, 27, 1578, 1195, 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.
| \(\displaystyle \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5194 |
| \(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {-8 c^4 x^4+4 c^2 x^2+1}{3 d^2 x^3 \left (1-c^2 x^2\right )}dx}{\sqrt {1-c^2 x^2}}-\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {-8 c^4 x^4+4 c^2 x^2+1}{x^3 \left (1-c^2 x^2\right )}dx}{3 d^2 \sqrt {1-c^2 x^2}}-\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {-8 c^4 x^4+4 c^2 x^2+1}{x^4 \left (1-c^2 x^2\right )}dx^2}{6 d^2 \sqrt {1-c^2 x^2}}-\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 1195 |
| \(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (\frac {3 c^4}{c^2 x^2-1}+\frac {5 c^2}{x^2}+\frac {1}{x^4}\right )dx^2}{6 d^2 \sqrt {1-c^2 x^2}}-\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 c^2 (a+b \arcsin (c x))}{3 d x \sqrt {d-c^2 d x^2}}-\frac {a+b \arcsin (c x)}{3 d x^3 \sqrt {d-c^2 d x^2}}+\frac {8 c^4 x (a+b \arcsin (c x))}{3 d \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {d-c^2 d x^2} \left (5 c^2 \log \left (x^2\right )+3 c^2 \log \left (1-c^2 x^2\right )-\frac {1}{x^2}\right )}{6 d^2 \sqrt {1-c^2 x^2}}\) |
-1/3*(a + b*ArcSin[c*x])/(d*x^3*Sqrt[d - c^2*d*x^2]) - (4*c^2*(a + b*ArcSi n[c*x]))/(3*d*x*Sqrt[d - c^2*d*x^2]) + (8*c^4*x*(a + b*ArcSin[c*x]))/(3*d* Sqrt[d - c^2*d*x^2]) + (b*c*Sqrt[d - c^2*d*x^2]*(-x^(-2) + 5*c^2*Log[x^2] + 3*c^2*Log[1 - c^2*x^2]))/(6*d^2*Sqrt[1 - c^2*x^2])
3.2.28.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x ] && IGtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(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 = 1048, normalized size of antiderivative = 4.40
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1048\) |
| parts | \(\text {Expression too large to display}\) | \(1048\) |
a*(-1/3/d/x^3/(-c^2*d*x^2+d)^(1/2)+4/3*c^2*(-1/d/x/(-c^2*d*x^2+d)^(1/2)+2* c^2/d*x/(-c^2*d*x^2+d)^(1/2)))+4*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c ^2*x^2-1)/d^2*x^3*c^6+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2 -1)/d^2*x^5*(-c^2*x^2+1)*c^8-64/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7* c^2*x^2-1)/d^2*x^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^5-16/3*I*b*(-d*(c^2*x^ 2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^3*(-c^2*x^2+1)*c^6-4/3*I*b*(-d*( c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*(-c^2*x^2+1)*c^4-16*I*b*(- d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^5*c^8-64/3*b*(-d*(c^2*x ^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^3*arcsin(c*x)*c^6-8/3*I*b*(-d*( c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*(-c^2*x^2+1)^(1/2)*arcsin(c* x)*c^3+32/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x^7*c^1 0+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*c^4+8*b*(-d *(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*x*arcsin(c*x)*c^4+16/3*I*b *(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*c^3 +4/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2*c^3*(-c^2*x^2+1) ^(1/2)+4*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2/x*arcsin(c*x )*c^2+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2/x^2*(-c^2*x ^2+1)^(1/2)*c+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^4*x^4-7*c^2*x^2-1)/d^2/x^3 *arcsin(c*x)-5/3*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/(c^2*x^2- 1)*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2-1)*c^3-b*(-d*(c^2*x^2-1))^(1/2)*(-c^...
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^4*d^2*x^8 - 2*c^2*d^2 *x^6 + d^2*x^4), x)
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{4} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
1/3*(8*c^4*x/(sqrt(-c^2*d*x^2 + d)*d) - 4*c^2/(sqrt(-c^2*d*x^2 + d)*d*x) - 1/(sqrt(-c^2*d*x^2 + d)*d*x^3))*a - b*integrate(arctan2(c*x, sqrt(c*x + 1 )*sqrt(-c*x + 1))/((c^2*d*x^6 - d*x^4)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/s qrt(d)
\[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x^4 \left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4\,{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]