Integrand size = 32, antiderivative size = 709 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\frac {b^2 d \left (1-c^2 x^2\right )^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {b^2 d x \left (1-c^2 x^2\right )^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b d \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {b d x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {d x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 d x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{3 (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i d \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {2 i b d \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 b d \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {i b^2 d \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
1/3*b^2*d*(-c^2*x^2+1)^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*b^2*d*x*(- c^2*x^2+1)^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*b*d*(-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)-1/3*b*d*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*d*(-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*d*x*(-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*d*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)-2/3*I*d*(-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)+2/3*I* b*d*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(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)+4/3*b*d*(-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 )-1/3*I*b^2*d*(-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*b^2*d*(-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*I*b ^2*d*(-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)
Time = 9.90 (sec) , antiderivative size = 764, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\frac {\sqrt {-e (-1+c x)} \sqrt {d (1+c x)} \left (\frac {a^2}{6 d^2 e^3 (-1+c x)^2}-\frac {5 a^2}{12 d^2 e^3 (-1+c x)}-\frac {a^2}{4 d^2 e^3 (1+c x)}\right )}{c}-\frac {a b \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2} \left (2 \arcsin (c x) (2 c x+\cos (2 \arcsin (c x)))+\sqrt {1-c^2 x^2} \left (-1+5 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+3 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-c x \left (5 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+3 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )\right )\right )}{3 c d e^2 \sqrt {(-d-c d x) (e-c e x)} \sqrt {-d e \left (1-c^2 x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^3 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}-\frac {b^2 \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2} \left (9 i \pi \arcsin (c x)-\frac {(-2+\arcsin (c x)) \arcsin (c x)}{-1+c x}+(1-4 i) \arcsin (c x)^2+16 \pi \log \left (1+e^{-i \arcsin (c x)}\right )+3 (\pi +2 \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )-5 (\pi -2 \arcsin (c x)) \log \left (1+i e^{i \arcsin (c x)}\right )-16 \pi \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )+5 \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-3 \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-10 i \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-6 i \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )+\frac {2 \arcsin (c x)^2 \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^3}+\frac {\left (4+5 \arcsin (c x)^2\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )}+\frac {3 \arcsin (c x)^2 \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )}\right )}{6 c d e^2 \sqrt {(-d-c d x) (e-c e x)} \sqrt {-d e \left (1-c^2 x^2\right )}} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*(a^2/(6*d^2*e^3*(-1 + c*x)^2) - ( 5*a^2)/(12*d^2*e^3*(-1 + c*x)) - a^2/(4*d^2*e^3*(1 + c*x))))/c - (a*b*Sqrt [d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2]*(2*ArcSin[c*x]*(2*c*x + Cos[ 2*ArcSin[c*x]]) + Sqrt[1 - c^2*x^2]*(-1 + 5*Log[Cos[ArcSin[c*x]/2] - Sin[A rcSin[c*x]/2]] + 3*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] - c*x*(5*L og[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 3*Log[Cos[ArcSin[c*x]/2] + S in[ArcSin[c*x]/2]]))))/(3*c*d*e^2*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[-(d* e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^3*(Cos[ArcSin[ c*x]/2] + Sin[ArcSin[c*x]/2])) - (b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt [1 - c^2*x^2]*((9*I)*Pi*ArcSin[c*x] - ((-2 + ArcSin[c*x])*ArcSin[c*x])/(-1 + c*x) + (1 - 4*I)*ArcSin[c*x]^2 + 16*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 3*(Pi + 2*ArcSin[c*x])*Log[1 - I*E^(I*ArcSin[c*x])] - 5*(Pi - 2*ArcSin[c*x ])*Log[1 + I*E^(I*ArcSin[c*x])] - 16*Pi*Log[Cos[ArcSin[c*x]/2]] + 5*Pi*Log [-Cos[(Pi + 2*ArcSin[c*x])/4]] - 3*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - ( 10*I)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (6*I)*PolyLog[2, I*E^(I*ArcSin[ c*x])] + (2*ArcSin[c*x]^2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[Ar cSin[c*x]/2])^3 + ((4 + 5*ArcSin[c*x]^2)*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c *x]/2] - Sin[ArcSin[c*x]/2]) + (3*ArcSin[c*x]^2*Sin[ArcSin[c*x]/2])/(Cos[A rcSin[c*x]/2] + Sin[ArcSin[c*x]/2])))/(6*c*d*e^2*Sqrt[(-d - c*d*x)*(e - c* e*x)]*Sqrt[-(d*e*(1 - c^2*x^2))])
Time = 1.00 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.54, 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.
| \(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{(c d x+d)^{3/2} (e-c e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d (c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {c x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}+\frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \left (1-c^2 x^2\right )^{5/2} \left (\frac {2 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {b x (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )}-\frac {b (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )}+\frac {2 x (a+b \arcsin (c x))^2}{3 \sqrt {1-c^2 x^2}}+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}+\frac {(a+b \arcsin (c x))^2}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {2 i (a+b \arcsin (c x))^2}{3 c}+\frac {4 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c}+\frac {i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-\frac {2 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c}+\frac {b^2 x}{3 \sqrt {1-c^2 x^2}}+\frac {b^2}{3 c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(d*(1 - c^2*x^2)^(5/2)*(b^2/(3*c*Sqrt[1 - c^2*x^2]) + (b^2*x)/(3*Sqrt[1 - c^2*x^2]) - (b*(a + b*ArcSin[c*x]))/(3*c*(1 - c^2*x^2)) - (b*x*(a + b*ArcS in[c*x]))/(3*(1 - c^2*x^2)) - (((2*I)/3)*(a + b*ArcSin[c*x])^2)/c + (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)) + (2*x*(a + b*ArcSin[c*x])^2)/(3*Sqrt[1 - c^2*x^2]) + (((2*I)/3)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (4*b*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(3*c) - ((I/3)*b^2*PolyLog[ 2, (-I)*E^(I*ArcSin[c*x])])/c + ((I/3)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])] )/c - (((2*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))
3.6.74.3.1 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))
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
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^5*d^2*e^3*x^5 - c^4*d^2*e^3*x^4 - 2*c^3*d^2*e^3*x^3 + 2* c^2*d^2*e^3*x^2 + c*d^2*e^3*x - d^2*e^3), x)
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
1/6*a*b*c*(2*sqrt(d)*sqrt(e)/(c^3*d^2*e^3*x - c^2*d^2*e^3) + 3*log(c*x + 1 )/(c^2*d^(3/2)*e^(5/2)) + 5*log(c*x - 1)/(c^2*d^(3/2)*e^(5/2))) - 2/3*a*b* (1/(sqrt(-c^2*d*e*x^2 + d*e)*c^2*d*e^2*x - sqrt(-c^2*d*e*x^2 + d*e)*c*d*e^ 2) - 2*x/(sqrt(-c^2*d*e*x^2 + d*e)*d*e^2))*arcsin(c*x) - 1/3*a^2*(1/(sqrt( -c^2*d*e*x^2 + d*e)*c^2*d*e^2*x - sqrt(-c^2*d*e*x^2 + d*e)*c*d*e^2) - 2*x/ (sqrt(-c^2*d*e*x^2 + d*e)*d*e^2)) + b^2*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2/((c^3*d*e^2*x^3 - c^2*d*e^2*x^2 - c*d*e^2*x + d*e^2)* sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/(sqrt(d)*sqrt(e))
\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{3/2}\,{\left (e-c\,e\,x\right )}^{5/2}} \,d x \]