Integrand size = 27, antiderivative size = 224 \[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b c \sqrt {d-c^2 d x^2} \log (x)}{d^3 \sqrt {1-c^2 x^2}}+\frac {5 b c \sqrt {d-c^2 d x^2} \log \left (1-c^2 x^2\right )}{6 d^3 \sqrt {1-c^2 x^2}} \]
(-a-b*arcsin(c*x))/d/x/(-c^2*d*x^2+d)^(3/2)+4/3*c^2*x*(a+b*arcsin(c*x))/d/ (-c^2*d*x^2+d)^(3/2)+8/3*c^2*x*(a+b*arcsin(c*x))/d^2/(-c^2*d*x^2+d)^(1/2)- 1/6*b*c*(-c^2*d*x^2+d)^(1/2)/d^3/(-c^2*x^2+1)^(3/2)+b*c*ln(x)*(-c^2*d*x^2+ d)^(1/2)/d^3/(-c^2*x^2+1)^(1/2)+5/6*b*c*ln(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2 )/d^3/(-c^2*x^2+1)^(1/2)
Time = 0.51 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d-c^2 d x^2} \left (b c x-b c^3 x^3+6 a \sqrt {1-c^2 x^2}-24 a c^2 x^2 \sqrt {1-c^2 x^2}+16 a c^4 x^4 \sqrt {1-c^2 x^2}+2 b \sqrt {1-c^2 x^2} \left (3-12 c^2 x^2+8 c^4 x^4\right ) \arcsin (c x)+3 b c x \left (-1+c^2 x^2\right )^2 \log \left (1-\frac {1}{c^2 x^2}\right )-8 b c x \log \left (1-c^2 x^2\right )+16 b c^3 x^3 \log \left (1-c^2 x^2\right )-8 b c^5 x^5 \log \left (1-c^2 x^2\right )\right )}{6 d^3 x \left (1-c^2 x^2\right )^{5/2}} \]
-1/6*(Sqrt[d - c^2*d*x^2]*(b*c*x - b*c^3*x^3 + 6*a*Sqrt[1 - c^2*x^2] - 24* 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]*(3 - 12*c^2*x^2 + 8*c^4*x^4)*ArcSin[c*x] + 3*b*c*x*(-1 + c^2*x^ 2)^2*Log[1 - 1/(c^2*x^2)] - 8*b*c*x*Log[1 - c^2*x^2] + 16*b*c^3*x^3*Log[1 - c^2*x^2] - 8*b*c^5*x^5*Log[1 - c^2*x^2]))/(d^3*x*(1 - c^2*x^2)^(5/2))
Time = 0.46 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.75, 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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5194 |
\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {8 c^4 x^4-12 c^2 x^2+3}{3 d^3 x \left (1-c^2 x^2\right )^2}dx}{\sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {8 c^4 x^4-12 c^2 x^2+3}{x \left (1-c^2 x^2\right )^2}dx}{3 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {8 c^4 x^4-12 c^2 x^2+3}{x^2 \left (1-c^2 x^2\right )^2}dx^2}{6 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1195 |
\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (\frac {5 c^2}{c^2 x^2-1}-\frac {c^2}{\left (c^2 x^2-1\right )^2}+\frac {3}{x^2}\right )dx^2}{6 d^3 \sqrt {1-c^2 x^2}}+\frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8 c^2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {4 c^2 x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{d x \left (d-c^2 d x^2\right )^{3/2}}+\frac {b c \sqrt {d-c^2 d x^2} \left (\frac {1}{c^2 x^2-1}+5 \log \left (1-c^2 x^2\right )+3 \log \left (x^2\right )\right )}{6 d^3 \sqrt {1-c^2 x^2}}\) |
-((a + b*ArcSin[c*x])/(d*x*(d - c^2*d*x^2)^(3/2))) + (4*c^2*x*(a + b*ArcSi n[c*x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (8*c^2*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]*((-1 + c^2*x^2)^(-1) + 3 *Log[x^2] + 5*Log[1 - c^2*x^2]))/(6*d^3*Sqrt[1 - c^2*x^2])
3.2.37.3.1 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 = 1346, normalized size of antiderivative = 6.01
method | result | size |
default | \(\text {Expression too large to display}\) | \(1346\) |
parts | \(\text {Expression too large to display}\) | \(1346\) |
a*(-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)))+9*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4 +26*c^2*x^2-9)/d^3/x*arcsin(c*x)+3/2*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-2 5*c^4*x^4+26*c^2*x^2-9)/d^3*(-c^2*x^2+1)^(1/2)*c-4/3*b*(-d*(c^2*x^2-1))^(1 /2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^2*(-c^2*x^2+1)^(1/2)*c^3-5/3 *b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/(c^2*x^2-1)*ln(1+(I*c*x+( -c^2*x^2+1)^(1/2))^2)*c-64/3*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^ 4+26*c^2*x^2-9)/d^3*x^5*arcsin(c*x)*c^6+56*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6 *x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^3*arcsin(c*x)*c^4-44*b*(-d*(c^2*x^2-1) )^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*arcsin(c*x)*c^2-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))^2-1)*c+20*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2* x^2-9)/d^3*x^3*(-c^2*x^2+1)*c^4+4*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25 *c^4*x^4+26*c^2*x^2-9)/d^3*x*c^2-24*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6- 25*c^4*x^4+26*c^2*x^2-9)/d^3*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c-112/3*I*b*(- d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^7*c^8-4*I*b *(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x*(-c^2*x^ 2+1)*c^2+136/3*I*b*(-d*(c^2*x^2-1))^(1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2 -9)/d^3*x^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*c^3-80/3*I*b*(-d*(c^2*x^2-1))^( 1/2)/(8*c^6*x^6-25*c^4*x^4+26*c^2*x^2-9)/d^3*x^5*(-c^2*x^2+1)*c^6-24*I*...
\[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \]
integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^8 - 3*c^4*d^ 3*x^6 + 3*c^2*d^3*x^4 - d^3*x^2), x)
\[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x^{2} \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \]
1/3*a*(8*c^2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + 4*c^2*x/((-c^2*d*x^2 + d)^(3/2 )*d) - 3/((-c^2*d*x^2 + d)^(3/2)*d*x)) + b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2*x^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(d)
\[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x^2 \left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]