Integrand size = 24, antiderivative size = 154 \[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {2 x (a+b \arcsin (c x))}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{3 c d^2 \sqrt {d-c^2 d x^2}} \]
1/3*x*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(3/2)+2/3*x*(a+b*arcsin(c*x))/d^2 /(-c^2*d*x^2+d)^(1/2)-1/6*b/c/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+ 1/3*b*ln(-c^2*x^2+1)*(-c^2*x^2+1)^(1/2)/c/d^2/(-c^2*d*x^2+d)^(1/2)
Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73 \[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d-c^2 d x^2} \left (-6 a c x+4 a c^3 x^3+b \sqrt {1-c^2 x^2}+2 b c x \left (-3+2 c^2 x^2\right ) \arcsin (c x)-2 b \left (1-c^2 x^2\right )^{3/2} \log \left (-1+c^2 x^2\right )\right )}{6 c d^3 \left (-1+c^2 x^2\right )^2} \]
-1/6*(Sqrt[d - c^2*d*x^2]*(-6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 - c^2*x^2] + 2*b*c*x*(-3 + 2*c^2*x^2)*ArcSin[c*x] - 2*b*(1 - c^2*x^2)^(3/2)*Log[-1 + c^ 2*x^2]))/(c*d^3*(-1 + c^2*x^2)^2)
Time = 0.36 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5162, 241, 5160, 240}
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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5162 |
\(\displaystyle \frac {2 \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{3/2}}dx}{3 d}-\frac {b c \sqrt {1-c^2 x^2} \int \frac {x}{\left (1-c^2 x^2\right )^2}dx}{3 d^2 \sqrt {d-c^2 d x^2}}+\frac {x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {2 \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{3/2}}dx}{3 d}+\frac {x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 5160 |
\(\displaystyle \frac {2 \left (\frac {x (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} \int \frac {x}{1-c^2 x^2}dx}{d \sqrt {d-c^2 d x^2}}\right )}{3 d}+\frac {x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {x (a+b \arcsin (c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {2 \left (\frac {x (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )}{2 c d \sqrt {d-c^2 d x^2}}\right )}{3 d}-\frac {b}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}\) |
-1/6*b/(c*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d - c^2*d*x^2]) + (x*(a + b*ArcSin[c* x]))/(3*d*(d - c^2*d*x^2)^(3/2)) + (2*((x*(a + b*ArcSin[c*x]))/(d*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[1 - c^2*x^2])/(2*c*d*Sqrt[d - c^2* d*x^2])))/(3*d)
3.2.35.3.1 Defintions of rubi rules used
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_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x _Symbol] :> Simp[x*((a + b*ArcSin[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp[b *c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcSin[c*x ])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*d*(p + 1 ))), x] + (Simp[(2*p + 3)/(2*d*(p + 1)) Int[(d + e*x^2)^(p + 1)*(a + b*Ar cSin[c*x])^n, x], x] + Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2 *x^2)^p] Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x ]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 1072, normalized size of antiderivative = 6.96
method | result | size |
default | \(\text {Expression too large to display}\) | \(1072\) |
parts | \(\text {Expression too large to display}\) | \(1072\) |
a*(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))+2/3*I*b*(- d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*(-c^2*x^2 +1)*x^5+I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4) *(-c^2*x^2+1)*x-I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^ 2*x^2-4)*x-2*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x ^2-4)*c^3*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x^4-2*b*(-d*(c^2*x^2-1))^(1/2)/d^ 3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^4*arcsin(c*x)*x^5+8/3*I*b*(-d*(c^2 *x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*x^3-5/3*I*b*(-d *(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*(-c^2*x^2+ 1)*x^3+4/3*I*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d^3/(c^2*x^2-1) *arcsin(c*x)-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2 *x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^2+17/3*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6* x^6-10*c^4*x^4+11*c^2*x^2-4)*c^2*arcsin(c*x)*x^3-8/3*I*b*(-d*(c^2*x^2-1))^ (1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*arcsin(c*x)*(-c^2*x^2+1)^( 1/2)-7/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4 )*c^4*x^5+14/3*I*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2 *x^2-4)*c*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x^2+2/3*b*(-d*(c^2*x^2-1))^(1/2)/ d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)/c*(-c^2*x^2+1)^(1/2)-4*b*(-d*(c^2* x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*x+2/3*I* b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*c^6*x^...
\[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^6*d^3*x^6 - 3*c^4*d^ 3*x^4 + 3*c^2*d^3*x^2 - d^3), x)
\[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} x^{2} - c^{2} d^{\frac {5}{2}}} + \frac {2 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}}} + \frac {2 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {2 \, x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, x}{\sqrt {-c^{2} d x^{2} + d} d^{2}} + \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} d}\right )} \]
1/6*b*c*(1/(c^4*d^(5/2)*x^2 - c^2*d^(5/2)) + 2*log(c*x + 1)/(c^2*d^(5/2)) + 2*log(c*x - 1)/(c^2*d^(5/2))) + 1/3*b*(2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d))*arcsin(c*x) + 1/3*a*(2*x/(sqrt(-c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d))
Exception generated. \[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {a+b \arcsin (c x)}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]