Integrand size = 32, antiderivative size = 559 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=-\frac {68 b^2 e^3 \left (1-c^2 x^2\right )}{9 c \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {3 b^2 e^3 x \left (1-c^2 x^2\right )}{4 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 e^3 \left (1-c^2 x^2\right )^2}{27 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {3 b^2 e^3 \sqrt {1-c^2 x^2} \arcsin (c x)}{4 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {22 b e^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {3 b c e^3 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b c^2 e^3 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{9 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {11 e^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {3 e^3 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {c e^3 x^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {5 e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{6 b c \sqrt {d+c d x} \sqrt {e-c e x}} \]
-68/9*b^2*e^3*(-c^2*x^2+1)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+3/4*b^2*e^3* x*(-c^2*x^2+1)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2/27*b^2*e^3*(-c^2*x^2+1)^ 2/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+11/3*e^3*(-c^2*x^2+1)*(a+b*arcsin(c*x ))^2/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-3/2*e^3*x*(-c^2*x^2+1)*(a+b*arcsin (c*x))^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/3*c*e^3*x^2*(-c^2*x^2+1)*(a+b* arcsin(c*x))^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-3/4*b^2*e^3*arcsin(c*x)*(- c^2*x^2+1)^(1/2)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-22/3*b*e^3*x*(a+b*arcs in(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+3/2*b*c*e^3*x ^2*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2 /9*b*c^2*e^3*x^3*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c* e*x+e)^(1/2)+5/6*e^3*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^ (1/2)/(-c*e*x+e)^(1/2)
Time = 13.34 (sec) , antiderivative size = 473, normalized size of antiderivative = 0.85 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\frac {180 b^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-540 a^2 \sqrt {d} e^{5/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )-6 b e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (264 b c x+8 b c^3 x^3-270 a \sqrt {1-c^2 x^2}+108 a c x \sqrt {1-c^2 x^2}+27 b \cos (2 \arcsin (c x))+6 a \cos (3 \arcsin (c x))\right )+18 b e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 \left (30 a+45 b \sqrt {1-c^2 x^2}-b \cos (3 \arcsin (c x))-9 b \sin (2 \arcsin (c x))\right )+e^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (-1620 a b c x+792 a^2 \sqrt {1-c^2 x^2}-1620 b^2 \sqrt {1-c^2 x^2}-324 a^2 c x \sqrt {1-c^2 x^2}+72 a^2 c^2 x^2 \sqrt {1-c^2 x^2}-162 a b \cos (2 \arcsin (c x))+4 b^2 \cos (3 \arcsin (c x))+81 b^2 \sin (2 \arcsin (c x))+12 a b \sin (3 \arcsin (c x))\right )}{216 c d \sqrt {1-c^2 x^2}} \]
(180*b^2*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 540*a^2*Sqrt[ d]*e^(5/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/ (Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] - 6*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x ]*ArcSin[c*x]*(264*b*c*x + 8*b*c^3*x^3 - 270*a*Sqrt[1 - c^2*x^2] + 108*a*c *x*Sqrt[1 - c^2*x^2] + 27*b*Cos[2*ArcSin[c*x]] + 6*a*Cos[3*ArcSin[c*x]]) + 18*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^2*(30*a + 45*b*Sqrt[ 1 - c^2*x^2] - b*Cos[3*ArcSin[c*x]] - 9*b*Sin[2*ArcSin[c*x]]) + e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-1620*a*b*c*x + 792*a^2*Sqrt[1 - c^2*x^2] - 162 0*b^2*Sqrt[1 - c^2*x^2] - 324*a^2*c*x*Sqrt[1 - c^2*x^2] + 72*a^2*c^2*x^2*S qrt[1 - c^2*x^2] - 162*a*b*Cos[2*ArcSin[c*x]] + 4*b^2*Cos[3*ArcSin[c*x]] + 81*b^2*Sin[2*ArcSin[c*x]] + 12*a*b*Sin[3*ArcSin[c*x]]))/(216*c*d*Sqrt[1 - c^2*x^2])
Time = 0.84 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.54, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5178, 27, 5272, 3042, 3798, 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.
| \(\displaystyle \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {c d x+d}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {e^3 (1-c x)^3 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^3 \sqrt {1-c^2 x^2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5272 |
| \(\displaystyle \frac {e^3 \sqrt {1-c^2 x^2} \int \left (c-c^2 x\right )^3 (a+b \arcsin (c x))^2d\arcsin (c x)}{c^4 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^3 \sqrt {1-c^2 x^2} \int (a+b \arcsin (c x))^2 (c-c \sin (\arcsin (c x)))^3d\arcsin (c x)}{c^4 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3798 |
| \(\displaystyle \frac {e^3 \sqrt {1-c^2 x^2} \int \left (-x^3 (a+b \arcsin (c x))^2 c^6+3 x^2 (a+b \arcsin (c x))^2 c^5-3 x (a+b \arcsin (c x))^2 c^4+(a+b \arcsin (c x))^2 c^3\right )d\arcsin (c x)}{c^4 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^3 \sqrt {1-c^2 x^2} \left (-\frac {2}{9} b c^6 x^3 (a+b \arcsin (c x))+\frac {3}{2} b c^5 x^2 (a+b \arcsin (c x))-\frac {22}{3} b c^4 x (a+b \arcsin (c x))+\frac {5 c^3 (a+b \arcsin (c x))^3}{6 b}+\frac {1}{3} c^5 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {3}{2} c^4 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {11}{3} c^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {3}{4} b^2 c^3 \arcsin (c x)+\frac {3}{4} b^2 c^4 x \sqrt {1-c^2 x^2}+\frac {2}{27} b^2 c^3 \left (1-c^2 x^2\right )^{3/2}-\frac {68}{9} b^2 c^3 \sqrt {1-c^2 x^2}\right )}{c^4 \sqrt {c d x+d} \sqrt {e-c e x}}\) |
(e^3*Sqrt[1 - c^2*x^2]*((-68*b^2*c^3*Sqrt[1 - c^2*x^2])/9 + (3*b^2*c^4*x*S qrt[1 - c^2*x^2])/4 + (2*b^2*c^3*(1 - c^2*x^2)^(3/2))/27 - (3*b^2*c^3*ArcS in[c*x])/4 - (22*b*c^4*x*(a + b*ArcSin[c*x]))/3 + (3*b*c^5*x^2*(a + b*ArcS in[c*x]))/2 - (2*b*c^6*x^3*(a + b*ArcSin[c*x]))/9 + (11*c^3*Sqrt[1 - c^2*x ^2]*(a + b*ArcSin[c*x])^2)/3 - (3*c^4*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c* x])^2)/2 + (c^5*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/3 + (5*c^3*(a + b*ArcSin[c*x])^3)/(6*b)))/(c^4*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
3.6.55.3.1 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[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ m, 0] || NeQ[a^2 - b^2, 0])
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_.))/Sq rt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[1/(c^(m + 1)*Sqrt[d]) Subst[In t[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c , d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (G tQ[m, 0] || IGtQ[n, 0])
\[\int \frac {\left (-c e x +e \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}}d x\]
\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d}} \,d x } \]
integral((a^2*c^2*e^2*x^2 - 2*a^2*c*e^2*x + a^2*e^2 + (b^2*c^2*e^2*x^2 - 2 *b^2*c*e^2*x + b^2*e^2)*arcsin(c*x)^2 + 2*(a*b*c^2*e^2*x^2 - 2*a*b*c*e^2*x + a*b*e^2)*arcsin(c*x))*sqrt(-c*e*x + e)/sqrt(c*d*x + d), x)
Timed out. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d}} \,d x } \]
Timed out. \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{\sqrt {d+c d x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (e-c\,e\,x\right )}^{5/2}}{\sqrt {d+c\,d\,x}} \,d x \]