Integrand size = 31, antiderivative size = 271 \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=-\frac {b x \left (2 f g+\left (c^2 f^2+g^2\right ) x\right )}{6 c d^2 \sqrt {1-c^2 x^2} \sqrt {d-c^2 d x^2}}+\frac {2 f \left (g+c^2 f x\right ) (a+b \arcsin (c x))}{3 c^2 d^2 \sqrt {d-c^2 d x^2}}+\frac {x (f+g x)^2 (a+b \arcsin (c x))}{3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b (2 c f-g) (c f+g) \sqrt {1-c^2 x^2} \log (1-c x)}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b (c f-g) (2 c f+g) \sqrt {1-c^2 x^2} \log (1+c x)}{6 c^3 d^2 \sqrt {d-c^2 d x^2}} \]
2/3*f*(c^2*f*x+g)*(a+b*arcsin(c*x))/c^2/d^2/(-c^2*d*x^2+d)^(1/2)+1/3*x*(g* x+f)^2*(a+b*arcsin(c*x))/d^2/(-c^2*x^2+1)/(-c^2*d*x^2+d)^(1/2)-1/6*b*x*(2* f*g+(c^2*f^2+g^2)*x)/c/d^2/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+1/6*b*( 2*c*f-g)*(c*f+g)*ln(-c*x+1)*(-c^2*x^2+1)^(1/2)/c^3/d^2/(-c^2*d*x^2+d)^(1/2 )+1/6*b*(c*f-g)*(2*c*f+g)*ln(c*x+1)*(-c^2*x^2+1)^(1/2)/c^3/d^2/(-c^2*d*x^2 +d)^(1/2)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.75 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.05 \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {c \sqrt {d-c^2 d x^2} \left (2 i b c^2 f g \left (1-c^2 x^2\right )^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-c^2} x\right ),1\right )-\sqrt {-c^2} \left (-4 a c f g-6 a c^3 f^2 x+4 a c^5 f^2 x^3-2 a c^3 g^2 x^3+b c^2 f^2 \sqrt {1-c^2 x^2}+b g^2 \sqrt {1-c^2 x^2}+2 b c^2 f g x \sqrt {1-c^2 x^2}+2 b c \left (-2 f g-c^2 g^2 x^3+c^2 f^2 x \left (-3+2 c^2 x^2\right )\right ) \arcsin (c x)-b \left (2 c^2 f^2-g^2\right ) \left (1-c^2 x^2\right )^{3/2} \log \left (-1+c^2 x^2\right )\right )\right )}{6 \left (-c^2\right )^{5/2} d^3 \left (-1+c^2 x^2\right )^2} \]
(c*Sqrt[d - c^2*d*x^2]*((2*I)*b*c^2*f*g*(1 - c^2*x^2)^(3/2)*EllipticF[I*Ar cSinh[Sqrt[-c^2]*x], 1] - Sqrt[-c^2]*(-4*a*c*f*g - 6*a*c^3*f^2*x + 4*a*c^5 *f^2*x^3 - 2*a*c^3*g^2*x^3 + b*c^2*f^2*Sqrt[1 - c^2*x^2] + b*g^2*Sqrt[1 - c^2*x^2] + 2*b*c^2*f*g*x*Sqrt[1 - c^2*x^2] + 2*b*c*(-2*f*g - c^2*g^2*x^3 + c^2*f^2*x*(-3 + 2*c^2*x^2))*ArcSin[c*x] - b*(2*c^2*f^2 - g^2)*(1 - c^2*x^ 2)^(3/2)*Log[-1 + c^2*x^2])))/(6*(-c^2)^(5/2)*d^3*(-1 + c^2*x^2)^2)
Time = 0.54 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {5276, 5260, 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.
\(\displaystyle \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5276 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{d^2 \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 5260 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-b c \int \left (\frac {x (f+g x)^2}{3 \left (1-c^2 x^2\right )^2}+\frac {2 f \left (f x c^2+g\right )}{3 c^2 \left (1-c^2 x^2\right )}\right )dx+\frac {x (f+g x)^2 (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 f \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \sqrt {1-c^2 x^2}}\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {x (f+g x)^2 (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 f \left (c^2 f x+g\right ) (a+b \arcsin (c x))}{3 c^2 \sqrt {1-c^2 x^2}}-b c \left (\frac {f g \text {arctanh}(c x)}{3 c^3}-\frac {f^2 \log \left (1-c^2 x^2\right )}{3 c^2}+\frac {(f+g x)^2}{6 c^2 \left (1-c^2 x^2\right )}+\frac {g^2 \log \left (1-c^2 x^2\right )}{6 c^4}\right )\right )}{d^2 \sqrt {d-c^2 d x^2}}\) |
(Sqrt[1 - c^2*x^2]*((x*(f + g*x)^2*(a + b*ArcSin[c*x]))/(3*(1 - c^2*x^2)^( 3/2)) + (2*f*(g + c^2*f*x)*(a + b*ArcSin[c*x]))/(3*c^2*Sqrt[1 - c^2*x^2]) - b*c*((f + g*x)^2/(6*c^2*(1 - c^2*x^2)) + (f*g*ArcTanh[c*x])/(3*c^3) - (f ^2*Log[1 - c^2*x^2])/(3*c^2) + (g^2*Log[1 - c^2*x^2])/(6*c^4))))/(d^2*Sqrt [d - c^2*d*x^2])
3.1.55.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 - c^2*x^2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IG tQ[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3] )
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 - c^2*x^2)^ p] Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ [{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && Intege rQ[p - 1/2] && !GtQ[d, 0]
Result contains complex when optimal does not.
Time = 0.88 (sec) , antiderivative size = 3783, normalized size of antiderivative = 13.96
method | result | size |
default | \(\text {Expression too large to display}\) | \(3783\) |
parts | \(\text {Expression too large to display}\) | \(3783\) |
-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 ^4*x^6*f*g+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^2*(-c^2*x^2+1)*x^5*g^2+16/3*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^4*f*g-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*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*f ^2+4/3*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)*x*f*g-8/3*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^4*arcsin(c*x)*x^6*f*g+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)*(-c^2*x^2+1)*arcsin(c*x)*x^3*g^2+7*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^5*g^2-2*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^7*g^2-2*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*f^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^2*x^2+1)*x*f^2-6*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*f*g+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*ar csin(c*x)*x^3*f^2-1/2*b*(-d*(c^2*x^2-1))^(1/2)/d^3/(3*c^6*x^6-10*c^4*x^4+1 1*c^2*x^2-4)*c*(-c^2*x^2+1)^(1/2)*x^2*f^2-1/2*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*g^2+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^2*arcsin(c...
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(-c^2*d*x^2 + d)*(a*g^2*x^2 + 2*a*f*g*x + a*f^2 + (b*g^2*x^2 + 2*b*f*g*x + b*f^2)*arcsin(c*x))/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^ 3*x^2 - d^3), x)
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )^{2}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
1/6*b*c*f^2*(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*f^2*(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*f^2*(2*x/(sqrt(- c^2*d*x^2 + d)*d^2) + x/((-c^2*d*x^2 + d)^(3/2)*d)) - 1/3*a*g^2*(x/(sqrt(- c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d)) - sqrt(d)*inte grate((b*g^2*x^2 + 2*b*f*g*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sq rt(c*x + 1)*sqrt(-c*x + 1))/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x) + 2/3*a*f*g/((-c^2*d*x^2 + d)^(3/2)*c^2*d)
Exception generated. \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(f+g x)^2 (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {{\left (f+g\,x\right )}^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]