\(\int \frac {(d-c^2 d x^2)^2 (a+b \arcsin (c x))}{x^2} \, dx\) [16]

Optimal result

Integrand size = 25, antiderivative size = 123 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=-\frac {5}{3} b c d^2 \sqrt {1-c^2 x^2}-\frac {1}{9} b c d^2 \left (1-c^2 x^2\right )^{3/2}-\frac {d^2 (a+b \arcsin (c x))}{x}-2 c^2 d^2 x (a+b \arcsin (c x))+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-b c d^2 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:

-5/3*b*c*d^2*(-c^2*x^2+1)^(1/2)-1/9*b*c*d^2*(-c^2*x^2+1)^(3/2)-d^2*(a+b*ar 
csin(c*x))/x-2*c^2*d^2*x*(a+b*arcsin(c*x))+1/3*c^4*d^2*x^3*(a+b*arcsin(c*x 
))-b*c*d^2*arctanh((-c^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {d^2 \left (-9 a-18 a c^2 x^2+3 a c^4 x^4-16 b c x \sqrt {1-c^2 x^2}+b c^3 x^3 \sqrt {1-c^2 x^2}+3 b \left (-3-6 c^2 x^2+c^4 x^4\right ) \arcsin (c x)+9 b c x \log (x)-9 b c x \log \left (1+\sqrt {1-c^2 x^2}\right )\right )}{9 x} \] Input:

Integrate[((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]))/x^2,x]
 

Output:

(d^2*(-9*a - 18*a*c^2*x^2 + 3*a*c^4*x^4 - 16*b*c*x*Sqrt[1 - c^2*x^2] + b*c 
^3*x^3*Sqrt[1 - c^2*x^2] + 3*b*(-3 - 6*c^2*x^2 + c^4*x^4)*ArcSin[c*x] + 9* 
b*c*x*Log[x] - 9*b*c*x*Log[1 + Sqrt[1 - c^2*x^2]]))/(9*x)
 

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5192, 27, 1578, 1192, 25, 1467, 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.

\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx\)

\(\Big \downarrow \) 5192

\(\displaystyle -b c \int -\frac {d^2 \left (-c^4 x^4+6 c^2 x^2+3\right )}{3 x \sqrt {1-c^2 x^2}}dx+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} b c d^2 \int \frac {-c^4 x^4+6 c^2 x^2+3}{x \sqrt {1-c^2 x^2}}dx+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {1}{6} b c d^2 \int \frac {-c^4 x^4+6 c^2 x^2+3}{x^2 \sqrt {1-c^2 x^2}}dx^2+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 1192

\(\displaystyle \frac {b d^2 \int -\frac {-c^4 x^8-4 c^4 x^4+8 c^4}{1-x^4}d\sqrt {1-c^2 x^2}}{3 c^3}+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b d^2 \int \frac {-c^4 x^8-4 c^4 x^4+8 c^4}{1-x^4}d\sqrt {1-c^2 x^2}}{3 c^3}+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 1467

\(\displaystyle -\frac {b d^2 \int \left (x^4 c^4+\frac {3 c^4}{1-x^4}+5 c^4\right )d\sqrt {1-c^2 x^2}}{3 c^3}+\frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} c^4 d^2 x^3 (a+b \arcsin (c x))-2 c^2 d^2 x (a+b \arcsin (c x))-\frac {d^2 (a+b \arcsin (c x))}{x}+\frac {b d^2 \left (-3 c^4 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )-\frac {1}{3} c^4 x^6-5 c^4 \sqrt {1-c^2 x^2}\right )}{3 c^3}\)

Input:

Int[((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]))/x^2,x]
 

Output:

-((d^2*(a + b*ArcSin[c*x]))/x) - 2*c^2*d^2*x*(a + b*ArcSin[c*x]) + (c^4*d^ 
2*x^3*(a + b*ArcSin[c*x]))/3 + (b*d^2*(-1/3*(c^4*x^6) - 5*c^4*Sqrt[1 - c^2 
*x^2] - 3*c^4*ArcTanh[Sqrt[1 - c^2*x^2]]))/(3*c^3)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1192

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]
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, 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] && IGtQ[q, -2]
 

rule 1578

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5192

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]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93

method	result	size
parts	\(a \,d^{2} \left (\frac {c^{4} x^{3}}{3}-2 c^{2} x -\frac {1}{x}\right )+d^{2} b c \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-2 c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {-c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\)	\(115\)
derivativedivides	\(c \left (a \,d^{2} \left (\frac {c^{3} x^{3}}{3}-2 c x -\frac {1}{c x}\right )+d^{2} b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-2 c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {-c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\)	\(117\)
default	\(c \left (a \,d^{2} \left (\frac {c^{3} x^{3}}{3}-2 c x -\frac {1}{c x}\right )+d^{2} b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-2 c x \arcsin \left (c x \right )-\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {16 \sqrt {-c^{2} x^{2}+1}}{9}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\)	\(117\)

Input:

int((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^2,x,method=_RETURNVERBOSE)
 

Output:

a*d^2*(1/3*c^4*x^3-2*c^2*x-1/x)+d^2*b*c*(1/3*c^3*x^3*arcsin(c*x)-2*c*x*arc 
sin(c*x)-arcsin(c*x)/c/x+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)-16/9*(-c^2*x^2+1)^ 
(1/2)-arctanh(1/(-c^2*x^2+1)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {6 \, a c^{4} d^{2} x^{4} - 36 \, a c^{2} d^{2} x^{2} - 9 \, b c d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + 9 \, b c d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) - 18 \, a d^{2} + 6 \, {\left (b c^{4} d^{2} x^{4} - 6 \, b c^{2} d^{2} x^{2} - 3 \, b d^{2}\right )} \arcsin \left (c x\right ) + 2 \, {\left (b c^{3} d^{2} x^{3} - 16 \, b c d^{2} x\right )} \sqrt {-c^{2} x^{2} + 1}}{18 \, x} \] Input:

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^2,x, algorithm="fricas")
 

Output:

1/18*(6*a*c^4*d^2*x^4 - 36*a*c^2*d^2*x^2 - 9*b*c*d^2*x*log(sqrt(-c^2*x^2 + 
 1) + 1) + 9*b*c*d^2*x*log(sqrt(-c^2*x^2 + 1) - 1) - 18*a*d^2 + 6*(b*c^4*d 
^2*x^4 - 6*b*c^2*d^2*x^2 - 3*b*d^2)*arcsin(c*x) + 2*(b*c^3*d^2*x^3 - 16*b* 
c*d^2*x)*sqrt(-c^2*x^2 + 1))/x
 

Sympy [A] (verification not implemented)

Time = 2.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.50 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {a c^{4} d^{2} x^{3}}{3} - 2 a c^{2} d^{2} x - \frac {a d^{2}}{x} - \frac {b c^{5} d^{2} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{3} + \frac {b c^{4} d^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} - 2 b c^{2} d^{2} \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) + b c d^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d^{2} \operatorname {asin}{\left (c x \right )}}{x} \] Input:

integrate((-c**2*d*x**2+d)**2*(a+b*asin(c*x))/x**2,x)
 

Output:

a*c**4*d**2*x**3/3 - 2*a*c**2*d**2*x - a*d**2/x - b*c**5*d**2*Piecewise((- 
x**2*sqrt(-c**2*x**2 + 1)/(3*c**2) - 2*sqrt(-c**2*x**2 + 1)/(3*c**4), Ne(c 
**2, 0)), (x**4/4, True))/3 + b*c**4*d**2*x**3*asin(c*x)/3 - 2*b*c**2*d**2 
*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) + 
b*c*d**2*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x 
)), True)) - b*d**2*asin(c*x)/x
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.30 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {1}{3} \, a c^{4} d^{2} x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{2} - 2 \, a c^{2} d^{2} x - 2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b c d^{2} - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d^{2} - \frac {a d^{2}}{x} \] Input:

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^2,x, algorithm="maxima")
 

Output:

1/3*a*c^4*d^2*x^3 + 1/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 
 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*c^4*d^2 - 2*a*c^2*d^2*x - 2*(c*x*arcsin(c* 
x) + sqrt(-c^2*x^2 + 1))*b*c*d^2 - (c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/ 
abs(x)) + arcsin(c*x)/x)*b*d^2 - a*d^2/x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2717 vs. \(2 (111) = 222\).

Time = 6.09 (sec) , antiderivative size = 2717, normalized size of antiderivative = 22.09 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\text {Too large to display} \] Input:

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^2,x, algorithm="giac")
 

Output:

-1/2*b*c^9*d^2*x^8*arcsin(c*x)/((c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^ 
5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + 
c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^8) - 1/2*a*c^9*d^2* 
x^8/((c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 
 1)^5 + 3*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1 
))*(sqrt(-c^2*x^2 + 1) + 1)^8) + b*c^8*d^2*x^7*log(abs(c)*abs(x))/((c^7*x^ 
7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^ 
3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^ 
2*x^2 + 1) + 1)^7) - b*c^8*d^2*x^7*log(sqrt(-c^2*x^2 + 1) + 1)/((c^7*x^7/( 
sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^3*x 
^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x 
^2 + 1) + 1)^7) + 16/9*b*c^8*d^2*x^7/((c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 
+ 3*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1 
)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^7) - 6*b*c^7* 
d^2*x^6*arcsin(c*x)/((c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^5*x^5/(sqrt 
(-c^2*x^2 + 1) + 1)^5 + 3*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqrt(- 
c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^6) - 6*a*c^7*d^2*x^6/((c^7*x^7 
/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^3 
*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2 
*x^2 + 1) + 1)^6) + 3*b*c^6*d^2*x^5*log(abs(c)*abs(x))/((c^7*x^7/(sqrt(...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\left \{\begin {array}{cl} b\,c^4\,d^2\,\left (\frac {\sqrt {\frac {1}{c^2}-x^2}\,\left (\frac {2}{c^2}+x^2\right )}{9}+\frac {x^3\,\mathrm {asin}\left (c\,x\right )}{3}\right )-\frac {a\,d^2\,\left (-c^4\,x^4+6\,c^2\,x^2+3\right )}{3\,x}-2\,b\,c\,d^2\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )-b\,c\,d^2\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-c^2\,x^2}}\right )-\frac {b\,d^2\,\mathrm {asin}\left (c\,x\right )}{x} & \text {\ if\ \ }0<c\\ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2}{x^2} \,d x & \text {\ if\ \ }\neg 0<c \end {array}\right . \] Input:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^2)/x^2,x)
                                                                                    
                                                                                    
 

Output:

piecewise(0 < c, - b*c*d^2*atanh(1/(- c^2*x^2 + 1)^(1/2)) - (a*d^2*(6*c^2* 
x^2 - c^4*x^4 + 3))/(3*x) - 2*b*c*d^2*((- c^2*x^2 + 1)^(1/2) + c*x*asin(c* 
x)) + b*c^4*d^2*(((1/c^2 - x^2)^(1/2)*(2/c^2 + x^2))/9 + (x^3*asin(c*x))/3 
) - (b*d^2*asin(c*x))/x, ~0 < c, int(((a + b*asin(c*x))*(d - c^2*d*x^2)^2) 
/x^2, x))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {d^{2} \left (3 \mathit {asin} \left (c x \right ) b \,c^{4} x^{4}-18 \mathit {asin} \left (c x \right ) b \,c^{2} x^{2}-9 \mathit {asin} \left (c x \right ) b +\sqrt {-c^{2} x^{2}+1}\, b \,c^{3} x^{3}-16 \sqrt {-c^{2} x^{2}+1}\, b c x +9 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b c x +3 a \,c^{4} x^{4}-18 a \,c^{2} x^{2}-9 a \right )}{9 x} \] Input:

int((-c^2*d*x^2+d)^2*(a+b*asin(c*x))/x^2,x)
 

Output:

(d**2*(3*asin(c*x)*b*c**4*x**4 - 18*asin(c*x)*b*c**2*x**2 - 9*asin(c*x)*b 
+ sqrt( - c**2*x**2 + 1)*b*c**3*x**3 - 16*sqrt( - c**2*x**2 + 1)*b*c*x + 9 
*log(tan(asin(c*x)/2))*b*c*x + 3*a*c**4*x**4 - 18*a*c**2*x**2 - 9*a))/(9*x 
)