Integrand size = 25, antiderivative size = 128 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=b c^3 d^2 \sqrt {1-c^2 x^2}-\frac {b c d^2 \sqrt {1-c^2 x^2}}{6 x^2}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}+c^4 d^2 x (a+b \arcsin (c x))+\frac {11}{6} b c^3 d^2 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \]
-1/3*d^2*(a+b*arcsin(c*x))/x^3+2*c^2*d^2*(a+b*arcsin(c*x))/x+c^4*d^2*x*(a+ b*arcsin(c*x))+11/6*b*c^3*d^2*arctanh((-c^2*x^2+1)^(1/2))+b*c^3*d^2*(-c^2* x^2+1)^(1/2)-1/6*b*c*d^2*(-c^2*x^2+1)^(1/2)/x^2
Time = 0.07 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=\frac {d^2 \left (-2 a+12 a c^2 x^2+6 a c^4 x^4-b c x \sqrt {1-c^2 x^2}+6 b c^3 x^3 \sqrt {1-c^2 x^2}+2 b \left (-1+6 c^2 x^2+3 c^4 x^4\right ) \arcsin (c x)-11 b c^3 x^3 \log (x)+11 b c^3 x^3 \log \left (1+\sqrt {1-c^2 x^2}\right )\right )}{6 x^3} \]
(d^2*(-2*a + 12*a*c^2*x^2 + 6*a*c^4*x^4 - b*c*x*Sqrt[1 - c^2*x^2] + 6*b*c^ 3*x^3*Sqrt[1 - c^2*x^2] + 2*b*(-1 + 6*c^2*x^2 + 3*c^4*x^4)*ArcSin[c*x] - 1 1*b*c^3*x^3*Log[x] + 11*b*c^3*x^3*Log[1 + Sqrt[1 - c^2*x^2]]))/(6*x^3)
Time = 0.38 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5192, 27, 1578, 1192, 25, 1471, 25, 27, 299, 219}
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.
\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 5192 |
\(\displaystyle -b c \int -\frac {d^2 \left (-3 c^4 x^4-6 c^2 x^2+1\right )}{3 x^3 \sqrt {1-c^2 x^2}}dx+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} b c d^2 \int \frac {-3 c^4 x^4-6 c^2 x^2+1}{x^3 \sqrt {1-c^2 x^2}}dx+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{6} b c d^2 \int \frac {-3 c^4 x^4-6 c^2 x^2+1}{x^4 \sqrt {1-c^2 x^2}}dx^2+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 1192 |
\(\displaystyle -\frac {b d^2 \int -\frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {1-c^2 x^2}}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b d^2 \int \frac {3 c^4 x^8-12 c^4 x^4+8 c^4}{\left (1-x^4\right )^2}d\sqrt {1-c^2 x^2}}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle -\frac {b d^2 \left (\frac {1}{2} \int -\frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {1-c^2 x^2}+\frac {c^4 \sqrt {1-c^2 x^2}}{2 \left (1-x^4\right )}\right )}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {b d^2 \left (\frac {c^4 \sqrt {1-c^2 x^2}}{2 \left (1-x^4\right )}-\frac {1}{2} \int \frac {c^4 \left (17-6 x^4\right )}{1-x^4}d\sqrt {1-c^2 x^2}\right )}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b d^2 \left (\frac {c^4 \sqrt {1-c^2 x^2}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \int \frac {17-6 x^4}{1-x^4}d\sqrt {1-c^2 x^2}\right )}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle -\frac {b d^2 \left (\frac {c^4 \sqrt {1-c^2 x^2}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \int \frac {1}{1-x^4}d\sqrt {1-c^2 x^2}+6 \sqrt {1-c^2 x^2}\right )\right )}{3 c}+c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle c^4 d^2 x (a+b \arcsin (c x))+\frac {2 c^2 d^2 (a+b \arcsin (c x))}{x}-\frac {d^2 (a+b \arcsin (c x))}{3 x^3}-\frac {b d^2 \left (\frac {c^4 \sqrt {1-c^2 x^2}}{2 \left (1-x^4\right )}-\frac {1}{2} c^4 \left (11 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )+6 \sqrt {1-c^2 x^2}\right )\right )}{3 c}\) |
-1/3*(d^2*(a + b*ArcSin[c*x]))/x^3 + (2*c^2*d^2*(a + b*ArcSin[c*x]))/x + c ^4*d^2*x*(a + b*ArcSin[c*x]) - (b*d^2*((c^4*Sqrt[1 - c^2*x^2])/(2*(1 - x^4 )) - (c^4*(6*Sqrt[1 - c^2*x^2] + 11*ArcTanh[Sqrt[1 - c^2*x^2]]))/2))/(3*c)
3.1.18.3.1 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_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^ 2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp[ (a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c ^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0 ] && IGtQ[p, 0]
Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88
method | result | size |
parts | \(d^{2} a \left (c^{4} x +\frac {2 c^{2}}{x}-\frac {1}{3 x^{3}}\right )+d^{2} b \,c^{3} \left (c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}+\frac {2 \arcsin \left (c x \right )}{c x}+\sqrt {-c^{2} x^{2}+1}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\) | \(113\) |
derivativedivides | \(c^{3} \left (d^{2} a \left (c x -\frac {1}{3 c^{3} x^{3}}+\frac {2}{c x}\right )+d^{2} b \left (c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}+\frac {2 \arcsin \left (c x \right )}{c x}+\sqrt {-c^{2} x^{2}+1}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(115\) |
default | \(c^{3} \left (d^{2} a \left (c x -\frac {1}{3 c^{3} x^{3}}+\frac {2}{c x}\right )+d^{2} b \left (c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}+\frac {2 \arcsin \left (c x \right )}{c x}+\sqrt {-c^{2} x^{2}+1}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(115\) |
d^2*a*(c^4*x+2*c^2/x-1/3/x^3)+d^2*b*c^3*(c*x*arcsin(c*x)-1/3/c^3/x^3*arcsi n(c*x)+2/c/x*arcsin(c*x)+(-c^2*x^2+1)^(1/2)-1/6/c^2/x^2*(-c^2*x^2+1)^(1/2) +11/6*arctanh(1/(-c^2*x^2+1)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.27 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=\frac {12 \, a c^{4} d^{2} x^{4} + 11 \, b c^{3} d^{2} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - 11 \, b c^{3} d^{2} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 24 \, a c^{2} d^{2} x^{2} - 4 \, a d^{2} + 4 \, {\left (3 \, b c^{4} d^{2} x^{4} + 6 \, b c^{2} d^{2} x^{2} - b d^{2}\right )} \arcsin \left (c x\right ) + 2 \, {\left (6 \, b c^{3} d^{2} x^{3} - b c d^{2} x\right )} \sqrt {-c^{2} x^{2} + 1}}{12 \, x^{3}} \]
1/12*(12*a*c^4*d^2*x^4 + 11*b*c^3*d^2*x^3*log(sqrt(-c^2*x^2 + 1) + 1) - 11 *b*c^3*d^2*x^3*log(sqrt(-c^2*x^2 + 1) - 1) + 24*a*c^2*d^2*x^2 - 4*a*d^2 + 4*(3*b*c^4*d^2*x^4 + 6*b*c^2*d^2*x^2 - b*d^2)*arcsin(c*x) + 2*(6*b*c^3*d^2 *x^3 - b*c*d^2*x)*sqrt(-c^2*x^2 + 1))/x^3
Time = 3.22 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.82 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=a c^{4} d^{2} x + \frac {2 a c^{2} d^{2}}{x} - \frac {a d^{2}}{3 x^{3}} + b c^{4} d^{2} \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) - 2 b c^{3} d^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) + \frac {2 b c^{2} d^{2} \operatorname {asin}{\left (c x \right )}}{x} + \frac {b c d^{2} \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} - \frac {b d^{2} \operatorname {asin}{\left (c x \right )}}{3 x^{3}} \]
a*c**4*d**2*x + 2*a*c**2*d**2/x - a*d**2/(3*x**3) + b*c**4*d**2*Piecewise( (0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) - 2*b*c**3*d* *2*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), Tr ue)) + 2*b*c**2*d**2*asin(c*x)/x + b*c*d**2*Piecewise((-c**2*acosh(1/(c*x) )/2 + c/(2*x*sqrt(-1 + 1/(c**2*x**2))) - 1/(2*c*x**3*sqrt(-1 + 1/(c**2*x** 2))), 1/Abs(c**2*x**2) > 1), (I*c**2*asin(1/(c*x))/2 - I*c*sqrt(1 - 1/(c** 2*x**2))/(2*x), True))/3 - b*d**2*asin(c*x)/(3*x**3)
Time = 0.30 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.33 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=a c^{4} d^{2} x + {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b c^{3} d^{2} + 2 \, {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b c^{2} d^{2} - \frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d^{2} + \frac {2 \, a c^{2} d^{2}}{x} - \frac {a d^{2}}{3 \, x^{3}} \]
a*c^4*d^2*x + (c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*b*c^3*d^2 + 2*(c*log( 2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*b*c^2*d^2 - 1/6*( (c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2) *c + 2*arcsin(c*x)/x^3)*b*d^2 + 2*a*c^2*d^2/x - 1/3*a*d^2/x^3
Leaf count of result is larger than twice the leaf count of optimal. 1409 vs. \(2 (116) = 232\).
Time = 59.12 (sec) , antiderivative size = 1409, normalized size of antiderivative = 11.01 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=\text {Too large to display} \]
-1/24*b*c^11*d^2*x^8*arcsin(c*x)/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^ 3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^8) - 1/24*a*c^1 1*d^2*x^8/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^8) + 1/24*b*c^10*d^2*x^7/((c^5*x^5/(sq rt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x ^2 + 1) + 1)^7) + 5/6*b*c^9*d^2*x^6*arcsin(c*x)/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^ 6) + 5/6*a*c^9*d^2*x^6/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqr t(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^6) - 11/6*b*c^8*d^2*x^5*l og(abs(c)*abs(x))/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^ 2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^5) + 11/6*b*c^8*d^2*x^5*log(sq rt(-c^2*x^2 + 1) + 1)/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt (-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^5) - 23/24*b*c^8*d^2*x^5/( (c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)* (sqrt(-c^2*x^2 + 1) + 1)^5) + 15/4*b*c^7*d^2*x^4*arcsin(c*x)/((c^5*x^5/(sq rt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x ^2 + 1) + 1)^4) + 15/4*a*c^7*d^2*x^4/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^4) - 11/6*b *c^6*d^2*x^3*log(abs(c)*abs(x))/((c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^3 *x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^3) + 11/6*b*c...
Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2}{x^4} \,d x \]