Integrand size = 33, antiderivative size = 738 \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=-\frac {2 a b g^3 x \sqrt {1-c^2 x^2}}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 g^3 \left (1-c^2 x^2\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {2 b^2 g^3 x \sqrt {1-c^2 x^2} \arcsin (c x)}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {g \left (3 c^2 f^2+g^2\right ) (a+b \arcsin (c x))^2}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {f \left (f^2+\frac {3 g^2}{c^2}\right ) x (a+b \arcsin (c x))^2}{d \sqrt {d-c^2 d x^2}}-\frac {i f \left (c^2 f^2+3 g^2\right ) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {g^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {f g^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{b c^3 d \sqrt {d-c^2 d x^2}}+\frac {4 i b g \left (3 c^2 f^2+g^2\right ) \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 b f \left (c^2 f^2+3 g^2\right ) \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {2 i b^2 g \left (3 c^2 f^2+g^2\right ) \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}+\frac {2 i b^2 g \left (3 c^2 f^2+g^2\right ) \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^4 d \sqrt {d-c^2 d x^2}}-\frac {i b^2 f \left (c^2 f^2+3 g^2\right ) \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}} \]
-2*b^2*g^3*(-c^2*x^2+1)/c^4/d/(-c^2*d*x^2+d)^(1/2)+g*(3*c^2*f^2+g^2)*(a+b* arcsin(c*x))^2/c^4/d/(-c^2*d*x^2+d)^(1/2)+f*(f^2+3*g^2/c^2)*x*(a+b*arcsin( c*x))^2/d/(-c^2*d*x^2+d)^(1/2)+g^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c^4/d/ (-c^2*d*x^2+d)^(1/2)-2*a*b*g^3*x*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^( 1/2)-2*b^2*g^3*x*arcsin(c*x)*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2) -I*f*(c^2*f^2+3*g^2)*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d* x^2+d)^(1/2)-f*g^2*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c^3/d/(-c^2*d* x^2+d)^(1/2)+4*I*b*g*(3*c^2*f^2+g^2)*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2* x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/c^4/d/(-c^2*d*x^2+d)^(1/2)+2*b*f*(c^2*f^2 +3*g^2)*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^ (1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)-2*I*b^2*g*(3*c^2*f^2+g^2)*polylog(2,-I*(I *c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)/c^4/d/(-c^2*d*x^2+d)^(1/2)+2* I*b^2*g*(3*c^2*f^2+g^2)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+ 1)^(1/2)/c^4/d/(-c^2*d*x^2+d)^(1/2)-I*b^2*f*(c^2*f^2+3*g^2)*polylog(2,-(I* c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)
Time = 2.52 (sec) , antiderivative size = 325, normalized size of antiderivative = 0.44 \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {1-c^2 x^2} \left (2 g^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {2 c f g^2 (a+b \arcsin (c x))^3}{b}-4 b g^3 \left (a c x+b \sqrt {1-c^2 x^2}+b c x \arcsin (c x)\right )+(c f-g)^3 \left (-(a+b \arcsin (c x))^2 \cot \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )+i \left ((a+b \arcsin (c x)) \left (a+b \arcsin (c x)-4 i b \log \left (1+i e^{-i \arcsin (c x)}\right )\right )+4 b^2 \operatorname {PolyLog}\left (2,-i e^{-i \arcsin (c x)}\right )\right )\right )-(c f+g)^3 \left (i \left ((a+b \arcsin (c x)) \left (a+b \arcsin (c x)+4 i b \log \left (1+i e^{i \arcsin (c x)}\right )\right )+4 b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )\right )-(a+b \arcsin (c x))^2 \tan \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )\right )}{2 c^4 d \sqrt {d-c^2 d x^2}} \]
(Sqrt[1 - c^2*x^2]*(2*g^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2 - (2*c*f *g^2*(a + b*ArcSin[c*x])^3)/b - 4*b*g^3*(a*c*x + b*Sqrt[1 - c^2*x^2] + b*c *x*ArcSin[c*x]) + (c*f - g)^3*(-((a + b*ArcSin[c*x])^2*Cot[(Pi + 2*ArcSin[ c*x])/4]) + I*((a + b*ArcSin[c*x])*(a + b*ArcSin[c*x] - (4*I)*b*Log[1 + I/ E^(I*ArcSin[c*x])]) + 4*b^2*PolyLog[2, (-I)/E^(I*ArcSin[c*x])])) - (c*f + g)^3*(I*((a + b*ArcSin[c*x])*(a + b*ArcSin[c*x] + (4*I)*b*Log[1 + I*E^(I*A rcSin[c*x])]) + 4*b^2*PolyLog[2, (-I)*E^(I*ArcSin[c*x])]) - (a + b*ArcSin[ c*x])^2*Tan[(Pi + 2*ArcSin[c*x])/4])))/(2*c^4*d*Sqrt[d - c^2*d*x^2])
Time = 1.27 (sec) , antiderivative size = 447, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5276, 5274, 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.
| \(\displaystyle \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5276 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \left (-\frac {x (a+b \arcsin (c x))^2 g^3}{c^2 \sqrt {1-c^2 x^2}}-\frac {3 f (a+b \arcsin (c x))^2 g^2}{c^2 \sqrt {1-c^2 x^2}}+\frac {\left (c^2 f^3+3 g^2 f+g \left (3 c^2 f^2+g^2\right ) x\right ) (a+b \arcsin (c x))^2}{c^2 \left (1-c^2 x^2\right )^{3/2}}\right )dx}{d \sqrt {d-c^2 d x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {4 i b g \left (3 c^2 f^2+g^2\right ) \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c^4}-\frac {f g^2 (a+b \arcsin (c x))^3}{b c^3}+\frac {f x \left (\frac {3 g^2}{c^2}+f^2\right ) (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {g \left (3 c^2 f^2+g^2\right ) (a+b \arcsin (c x))^2}{c^4 \sqrt {1-c^2 x^2}}+\frac {g^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c^4}-\frac {i f \left (c^2 f^2+3 g^2\right ) (a+b \arcsin (c x))^2}{c^3}+\frac {2 b f \left (c^2 f^2+3 g^2\right ) \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c^3}-\frac {2 a b g^3 x}{c^3}-\frac {2 b^2 g^3 x \arcsin (c x)}{c^3}-\frac {2 i b^2 g \left (3 c^2 f^2+g^2\right ) \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c^4}+\frac {2 i b^2 g \left (3 c^2 f^2+g^2\right ) \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^4}-\frac {i b^2 f \left (c^2 f^2+3 g^2\right ) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c^3}-\frac {2 b^2 g^3 \sqrt {1-c^2 x^2}}{c^4}\right )}{d \sqrt {d-c^2 d x^2}}\) |
(Sqrt[1 - c^2*x^2]*((-2*a*b*g^3*x)/c^3 - (2*b^2*g^3*Sqrt[1 - c^2*x^2])/c^4 - (2*b^2*g^3*x*ArcSin[c*x])/c^3 - (I*f*(c^2*f^2 + 3*g^2)*(a + b*ArcSin[c* x])^2)/c^3 + (g*(3*c^2*f^2 + g^2)*(a + b*ArcSin[c*x])^2)/(c^4*Sqrt[1 - c^2 *x^2]) + (f*(f^2 + (3*g^2)/c^2)*x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] + (g^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/c^4 - (f*g^2*(a + b*ArcSi n[c*x])^3)/(b*c^3) + ((4*I)*b*g*(3*c^2*f^2 + g^2)*(a + b*ArcSin[c*x])*ArcT an[E^(I*ArcSin[c*x])])/c^4 + (2*b*f*(c^2*f^2 + 3*g^2)*(a + b*ArcSin[c*x])* Log[1 + E^((2*I)*ArcSin[c*x])])/c^3 - ((2*I)*b^2*g*(3*c^2*f^2 + g^2)*PolyL og[2, (-I)*E^(I*ArcSin[c*x])])/c^4 + ((2*I)*b^2*g*(3*c^2*f^2 + g^2)*PolyLo g[2, I*E^(I*ArcSin[c*x])])/c^4 - (I*b^2*f*(c^2*f^2 + 3*g^2)*PolyLog[2, -E^ ((2*I)*ArcSin[c*x])])/c^3))/(d*Sqrt[d - c^2*d*x^2])
3.1.75.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSin[c*x] )^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1527 vs. \(2 (731 ) = 1462\).
Time = 1.19 (sec) , antiderivative size = 1528, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1528\) |
| parts | \(\text {Expression too large to display}\) | \(1528\) |
a^2*(f^3/d*x/(-c^2*d*x^2+d)^(1/2)+g^3*(-x^2/c^2/d/(-c^2*d*x^2+d)^(1/2)+2/d /c^4/(-c^2*d*x^2+d)^(1/2))+3*f*g^2*(x/c^2/d/(-c^2*d*x^2+d)^(1/2)-1/c^2/d/( c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2)))+3*f^2*g/c^2/d/( -c^2*d*x^2+d)^(1/2))+b^2*((-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^ 3/(c^2*x^2-1)*arcsin(c*x)^3*f*g^2+1/2*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(- c^2*x^2+1)^(1/2)*x*c-1)*g^3*(arcsin(c*x)^2-2+2*I*arcsin(c*x))/d^2/c^4/(c^2 *x^2-1)+1/2*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*g^ 3*(arcsin(c*x)^2-2-2*I*arcsin(c*x))/d^2/c^4/(c^2*x^2-1)-(-d*(c^2*x^2-1))^( 1/2)/d^2/c^4/(c^2*x^2-1)*arcsin(c*x)^2*(I*(-c^2*x^2+1)^(1/2)*c^3*f^3+c^4*f ^3*x+3*I*(-c^2*x^2+1)^(1/2)*c*f*g^2+3*c^2*f*g^2*x+3*f^2*g*c^2+g^3)+(-d*(c^ 2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)*(2*I*arcsin(c*x)^2*c^3*f^3+I*polylog(2, -(I*c*x+(-c^2*x^2+1)^(1/2))^2)*c^3*f^3-2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2 )*arcsin(c*x)*c^3*f^3+6*I*arcsin(c*x)^2*c*f*g^2-6*arcsin(c*x)*ln(1+I*(I*c* x+(-c^2*x^2+1)^(1/2)))*c^2*f^2*g+6*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^ (1/2)))*c^2*f^2*g+6*I*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))*c^2*f^2*g-6*I* dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*c^2*f^2*g+3*I*polylog(2,-(I*c*x+(-c^ 2*x^2+1)^(1/2))^2)*c*f*g^2-6*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*arcsin(c*x )*c*f*g^2-2*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))*g^3+2*arcsin(c* x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*g^3+2*I*dilog(1+I*(I*c*x+(-c^2*x^2+1 )^(1/2)))*g^3-2*I*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*g^3)/d^2/c^4/(c...
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
integral((a^2*g^3*x^3 + 3*a^2*f*g^2*x^2 + 3*a^2*f^2*g*x + a^2*f^3 + (b^2*g ^3*x^3 + 3*b^2*f*g^2*x^2 + 3*b^2*f^2*g*x + b^2*f^3)*arcsin(c*x)^2 + 2*(a*b *g^3*x^3 + 3*a*b*f*g^2*x^2 + 3*a*b*f^2*g*x + a*b*f^3)*arcsin(c*x))*sqrt(-c ^2*d*x^2 + d)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2} \left (f + g x\right )^{3}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (g x + f\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
-a^2*g^3*(x^2/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 2/(sqrt(-c^2*d*x^2 + d)*c^4*d )) + 3*a^2*f*g^2*(x/(sqrt(-c^2*d*x^2 + d)*c^2*d) - arcsin(c*x)/(c^3*d^(3/2 ))) + 2*a*b*f^3*x*arcsin(c*x)/(sqrt(-c^2*d*x^2 + d)*d) + a^2*f^3*x/(sqrt(- c^2*d*x^2 + d)*d) - a*b*f^3*log(x^2 - 1/c^2)/(c*d^(3/2)) + 3*a^2*f^2*g/(sq rt(-c^2*d*x^2 + d)*c^2*d) - sqrt(d)*integrate(((b^2*g^3*x^3 + 3*b^2*f*g^2* x^2 + 3*b^2*f^2*g*x + b^2*f^3)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^ 2 + 2*(a*b*g^3*x^3 + 3*a*b*f*g^2*x^2 + 3*a*b*f^2*g*x)*arctan2(c*x, sqrt(c* x + 1)*sqrt(-c*x + 1)))/((c^2*d^2*x^2 - d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)) , x)
Exception generated. \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(f+g x)^3 (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {{\left (f+g\,x\right )}^3\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]