Integrand size = 40, antiderivative size = 275 \[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 i b \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 b^2 \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}-\frac {3 i b^3 \operatorname {PolyLog}\left (4,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c} \] Output:
1/4*I*(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^4/b/c-(a+b*arcsin((-c*x+1 )^(1/2)/(c*x+1)^(1/2)))^3*ln(1-(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1) /(c*x+1))^(1/2))^2)/c+3/2*I*b*(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2 *polylog(2,(I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/ c-3/2*b^2*(a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))*polylog(3,(I*(-c*x+1) ^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c-3/4*I*b^3*polylog(4, (I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))^2)/c
\[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx \] Input:
Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]
Output:
Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2), x]
Time = 0.85 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {7232, 5136, 3042, 25, 4200, 25, 2620, 3011, 7163, 2720, 7143}
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.
\(\displaystyle \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{1-c^2 x^2} \, dx\) |
\(\Big \downarrow \) 7232 |
\(\displaystyle -\frac {\int \frac {\sqrt {c x+1} \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{\sqrt {1-c x}}d\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}{c}\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle -\frac {\int \frac {\sqrt {c x+1} \sqrt {1-\frac {1-c x}{c x+1}} \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{\sqrt {1-c x}}d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3 \tan \left (\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3 \tan \left (\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )+\frac {\pi }{2}\right )d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle -\frac {2 i \int -\frac {e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )} \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}}d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {-2 i \int \frac {e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )} \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}}d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-\frac {3}{2} i b \int \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2 \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-\frac {3}{2} i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \int \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right ) \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-\frac {3}{2} i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \left (\frac {1}{2} i b \int \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )d\arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )\right )\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-\frac {3}{2} i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \left (\frac {1}{4} b \int \frac {\sqrt {c x+1} \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{\sqrt {1-c x}}de^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}-\frac {1}{2} i \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )\right )\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3-\frac {3}{2} i b \left (\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2-i b \left (\frac {1}{4} b \operatorname {PolyLog}\left (4,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (3,e^{2 i \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )\right )\right )\right )-\frac {i \left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b}}{c}\) |
Input:
Int[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]
Output:
-((((-1/4*I)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^4)/b - (2*I)*((I/ 2)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*Log[1 - E^((2*I)*ArcSin[S qrt[1 - c*x]/Sqrt[1 + c*x]])] - ((3*I)/2)*b*((I/2)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*PolyLog[2, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c *x]])] - I*b*((-1/2*I)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog [3, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])] + (b*PolyLog[4, E^((2*I )*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/4))))/c)
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol ] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^ m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] , x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.) *(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^2), x_Symbol] :> Simp[2*e*(g/(C*(e*f - d *g))) Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1170 vs. \(2 (300 ) = 600\).
Time = 0.40 (sec) , antiderivative size = 1171, normalized size of antiderivative = 4.26
method | result | size |
default | \(\text {Expression too large to display}\) | \(1171\) |
parts | \(\text {Expression too large to display}\) | \(1171\) |
Input:
int((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x,method=_RE TURNVERBOSE)
Output:
-1/2*a^3/c*ln(c*x-1)+1/2*a^3/c*ln(c*x+1)-b^3*(-1/4*I/c*arcsin((-c*x+1)^(1/ 2)/(c*x+1)^(1/2))^4+1/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1+I*(-c* x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-3*I/c*arcsin((-c*x+1) ^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x +1)/(c*x+1))^(1/2))+6/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,-I* (-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+6*I/c*polylog(4,- I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+1/c*arcsin((-c* x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1 )/(c*x+1))^(1/2))-3*I/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,I *(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+6/c*arcsin((-c*x +1)^(1/2)/(c*x+1)^(1/2))*polylog(3,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x +1)/(c*x+1))^(1/2))+6*I/c*polylog(4,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c* x+1)/(c*x+1))^(1/2)))-3*a*b^2*(-1/3*I/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2 ))^3+1/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1-I*(-c*x+1)^(1/2)/(c*x +1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-2*I/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^ (1/2))*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2) )+2/c*polylog(3,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2)) +1/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+I*(-c*x+1)^(1/2)/(c*x+1)^ (1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-2*I/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2 ))*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2)...
\[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg orithm="fricas")
Output:
integral(-(b^3*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1))^3 + 3*a*b^2*arcsin(sqr t(-c*x + 1)/sqrt(c*x + 1))^2 + 3*a^2*b*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1) ) + a^3)/(c^2*x^2 - 1), x)
Timed out. \[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*asin((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**3/(-c**2*x**2+1),x)
Output:
Timed out
\[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg orithm="maxima")
Output:
1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) - integrate((b^3*arctan2(sqrt(-c *x + 1), sqrt(2)*sqrt(c)*sqrt(x))^3 + 3*a*b^2*arctan2(sqrt(-c*x + 1), sqrt (2)*sqrt(c)*sqrt(x))^2 + 3*a^2*b*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)*s qrt(x)))/(c^2*x^2 - 1), x)
\[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int { -\frac {{\left (b \arcsin \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{3}}{c^{2} x^{2} - 1} \,d x } \] Input:
integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, alg orithm="giac")
Output:
integrate(-(b*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^3/(c^2*x^2 - 1), x )
Timed out. \[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\int -\frac {{\left (a+b\,\mathrm {asin}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \] Input:
int(-(a + b*asin((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1),x)
Output:
int(-(a + b*asin((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1), x)
\[ \int \frac {\left (a+b \arcsin \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx=\frac {-6 \left (\int \frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )}{c^{2} x^{2}-1}d x \right ) a^{2} b c -2 \left (\int \frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3}}{c^{2} x^{2}-1}d x \right ) b^{3} c -6 \left (\int \frac {\mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{c^{2} x^{2}-1}d x \right ) a \,b^{2} c -\mathrm {log}\left (c^{2} x -c \right ) a^{3}+\mathrm {log}\left (c^{2} x +c \right ) a^{3}}{2 c} \] Input:
int((a+b*asin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x)
Output:
( - 6*int(asin(sqrt( - c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1),x)*a**2*b*c - 2*int(asin(sqrt( - c*x + 1)/sqrt(c*x + 1))**3/(c**2*x**2 - 1),x)*b**3*c - 6*int(asin(sqrt( - c*x + 1)/sqrt(c*x + 1))**2/(c**2*x**2 - 1),x)*a*b**2 *c - log(c**2*x - c)*a**3 + log(c**2*x + c)*a**3)/(2*c)