Integrand size = 32, antiderivative size = 896 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=-\frac {2 b^2 e^2 \left (1-c^2 x^2\right )^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b^2 e^2 x \left (1-c^2 x^2\right )^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b^2 e^2 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b e^2 x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b c e^2 x^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 e^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {e^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {c^2 e^2 x^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 e^2 x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {i e^2 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 i b e^2 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 b e^2 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 i b^2 e^2 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i b^2 e^2 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {i b^2 e^2 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
-2/3*b^2*e^2*(-c^2*x^2+1)^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2/3*b^2*e^2 *x*(-c^2*x^2+1)^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*b^2*e^2*(-c^2*x^2+1 )^(5/2)*arcsin(c*x)/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*b*e^2*(-c^2*x^2 +1)^(3/2)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2/3*b*e^2*x *(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3 *b*c*e^2*x^2*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*e*x+ e)^(5/2)-2/3*e^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e* x+e)^(5/2)+1/3*e^2*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c* e*x+e)^(5/2)+1/3*c^2*e^2*x^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5 /2)/(-c*e*x+e)^(5/2)+2/3*e^2*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))^2/(c*d*x+d )^(5/2)/(-c*e*x+e)^(5/2)-1/3*I*b^2*e^2*(-c^2*x^2+1)^(5/2)*polylog(2,-(I*c* x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-2/3*I*b^2*e^2* (-c^2*x^2+1)^(5/2)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(5/ 2)/(-c*e*x+e)^(5/2)+2/3*b*e^2*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*ln(1+(I *c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+2/3*I*b^2*e ^2*(-c^2*x^2+1)^(5/2)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d) ^(5/2)/(-c*e*x+e)^(5/2)-1/3*I*e^2*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2/c /(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-4/3*I*b*e^2*(-c^2*x^2+1)^(5/2)*(a+b*arcs in(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/ 2)
Time = 8.37 (sec) , antiderivative size = 369, normalized size of antiderivative = 0.41 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=-\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (\frac {2 a^2 (2+c x)}{(1+c x)^2}+\frac {b^2 \left (\cot \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right ) \left (4+\arcsin (c x)^2 \left (2+\csc ^2\left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )\right )+2 \arcsin (c x) \left (-i \arcsin (c x)+\csc ^2\left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )-4 \log \left (1+i e^{-i \arcsin (c x)}\right )\right )-8 i \operatorname {PolyLog}\left (2,-i e^{-i \arcsin (c x)}\right )\right )}{\sqrt {1-c^2 x^2}}+\frac {2 a b \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left (2+3 \arcsin (c x)-6 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+\cos \left (\frac {3}{2} \arcsin (c x)\right ) \left (\arcsin (c x)+2 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+2 \left (1+\left (-1+\sqrt {1-c^2 x^2}\right ) \arcsin (c x)-2 \left (2+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^3}\right )}{6 c d^3 e} \]
-1/6*(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((2*a^2*(2 + c*x))/(1 + c*x)^2 + (b^ 2*(Cot[(Pi + 2*ArcSin[c*x])/4]*(4 + ArcSin[c*x]^2*(2 + Csc[(Pi + 2*ArcSin[ c*x])/4]^2)) + 2*ArcSin[c*x]*((-I)*ArcSin[c*x] + Csc[(Pi + 2*ArcSin[c*x])/ 4]^2 - 4*Log[1 + I/E^(I*ArcSin[c*x])]) - (8*I)*PolyLog[2, (-I)/E^(I*ArcSin [c*x])]))/Sqrt[1 - c^2*x^2] + (2*a*b*(Cos[ArcSin[c*x]/2]*(2 + 3*ArcSin[c*x ] - 6*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + Cos[(3*ArcSin[c*x])/ 2]*(ArcSin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + 2*(1 + (-1 + Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 2*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(Sqrt[1 - c^2*x ^2]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3)))/(c*d^3*e)
Time = 1.37 (sec) , antiderivative size = 465, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5178, 27, 5262, 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.
\(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{(c d x+d)^{5/2} \sqrt {e-c e x}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {e^2 (1-c x)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
\(\Big \downarrow \) 5262 |
\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {c^2 x^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}-\frac {2 c x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}+\frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {4 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {b c x^2 (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )}+\frac {2 x (a+b \arcsin (c x))^2}{3 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}+\frac {2 b x (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )}-\frac {2 (a+b \arcsin (c x))^2}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {b (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )}+\frac {c^2 x^3 (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {i (a+b \arcsin (c x))^2}{3 c}+\frac {2 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {2 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-\frac {i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c}-\frac {b^2 \arcsin (c x)}{3 c}+\frac {2 b^2 x}{3 \sqrt {1-c^2 x^2}}-\frac {2 b^2}{3 c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(e^2*(1 - c^2*x^2)^(5/2)*((-2*b^2)/(3*c*Sqrt[1 - c^2*x^2]) + (2*b^2*x)/(3* Sqrt[1 - c^2*x^2]) - (b^2*ArcSin[c*x])/(3*c) - (b*(a + b*ArcSin[c*x]))/(3* c*(1 - c^2*x^2)) + (2*b*x*(a + b*ArcSin[c*x]))/(3*(1 - c^2*x^2)) - (b*c*x^ 2*(a + b*ArcSin[c*x]))/(3*(1 - c^2*x^2)) - ((I/3)*(a + b*ArcSin[c*x])^2)/c - (2*(a + b*ArcSin[c*x])^2)/(3*c*(1 - c^2*x^2)^(3/2)) + (x*(a + b*ArcSin[ c*x])^2)/(3*(1 - c^2*x^2)^(3/2)) + (c^2*x^3*(a + b*ArcSin[c*x])^2)/(3*(1 - c^2*x^2)^(3/2)) + (2*x*(a + b*ArcSin[c*x])^2)/(3*Sqrt[1 - c^2*x^2]) - ((( 4*I)/3)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (2*b*(a + b*A rcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(3*c) + (((2*I)/3)*b^2*PolyLog [2, (-I)*E^(I*ArcSin[c*x])])/c - (((2*I)/3)*b^2*PolyLog[2, I*E^(I*ArcSin[c *x])])/c - ((I/3)*b^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/c))/((d + c*d*x) ^(5/2)*(e - c*e*x)^(5/2))
3.6.63.3.1 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[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 2)^q) Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {5}{2}} \sqrt {-c e x +e}}d x\]
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c e x + e}} \,d x } \]
integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sq rt(-c*e*x + e)/(c^4*d^3*e*x^4 + 2*c^3*d^3*e*x^3 - 2*c*d^3*e*x - d^3*e), x)
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {5}{2}} \sqrt {- e \left (c x - 1\right )}}\, dx \]
Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c e x + e}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} \sqrt {e-c e x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{5/2}\,\sqrt {e-c\,e\,x}} \,d x \]