\(\int \frac {(d+e x^2)^2 (a+b \arcsin (c x))}{x^2} \, dx\) [438]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b e^2 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac {d^2 (a+b \arcsin (c x))}{x}+2 d e x (a+b \arcsin (c x))+\frac {1}{3} e^2 x^3 (a+b \arcsin (c x))-b c d^2 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:

1/3*b*e*(6*c^2*d+e)*(-c^2*x^2+1)^(1/2)/c^3-1/9*b*e^2*(-c^2*x^2+1)^(3/2)/c^ 
3-d^2*(a+b*arcsin(c*x))/x+2*d*e*x*(a+b*arcsin(c*x))+1/3*e^2*x^3*(a+b*arcsi 
n(c*x))-b*c*d^2*arctanh((-c^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {1}{9} \left (-\frac {9 a d^2}{x}+18 a d e x+3 a e^2 x^3+\frac {b e \sqrt {1-c^2 x^2} \left (2 e+c^2 \left (18 d+e x^2\right )\right )}{c^3}+\frac {3 b \left (-3 d^2+6 d e x^2+e^2 x^4\right ) \arcsin (c x)}{x}+9 b c d^2 \log (x)-9 b c d^2 \log \left (1+\sqrt {1-c^2 x^2}\right )\right ) \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5230, 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.

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

\(\Big \downarrow \) 5230

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1192

\(\displaystyle \frac {b \int -\frac {-e^2 x^8+2 e \left (3 d c^2+e\right ) x^4+3 c^4 d^2-e^2-6 c^2 d e}{1-x^4}d\sqrt {1-c^2 x^2}}{3 c^3}-\frac {d^2 (a+b \arcsin (c x))}{x}+2 d e x (a+b \arcsin (c x))+\frac {1}{3} e^2 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {b \int \frac {-e^2 x^8+2 e \left (3 d c^2+e\right ) x^4+3 c^4 d^2-e^2-6 c^2 d e}{1-x^4}d\sqrt {1-c^2 x^2}}{3 c^3}-\frac {d^2 (a+b \arcsin (c x))}{x}+2 d e x (a+b \arcsin (c x))+\frac {1}{3} e^2 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 1467

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((d^2*(a + b*ArcSin[c*x]))/x) + 2*d*e*x*(a + b*ArcSin[c*x]) + (e^2*x^3*(a 
 + b*ArcSin[c*x]))/3 + (b*(-1/3*(e^2*x^6) + e*(6*c^2*d + e)*Sqrt[1 - c^2*x 
^2] - 3*c^4*d^2*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 5230

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] && NeQ[c^2*d + e, 
0] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.25

method	result	size
parts	\(a \left (\frac {e^{2} x^{3}}{3}+2 d e x -\frac {d^{2}}{x}\right )+b c \left (\frac {\arcsin \left (c x \right ) x^{3} e^{2}}{3 c}+\frac {2 \arcsin \left (c x \right ) x d e}{c}-\frac {\arcsin \left (c x \right ) d^{2}}{c x}-\frac {e^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )+3 c^{4} d^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )-6 c^{2} d e \sqrt {-c^{2} x^{2}+1}}{3 c^{4}}\right )\)	\(157\)
derivativedivides	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arcsin \left (c x \right ) c^{3} d e x +\frac {\arcsin \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{2}}{x}-\frac {e^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}-c^{4} d^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+2 c^{2} d e \sqrt {-c^{2} x^{2}+1}\right )}{c^{4}}\right )\)	\(168\)
default	\(c \left (\frac {a \left (2 c^{3} d e x +\frac {e^{2} c^{3} x^{3}}{3}-\frac {c^{3} d^{2}}{x}\right )}{c^{4}}+\frac {b \left (2 \arcsin \left (c x \right ) c^{3} d e x +\frac {\arcsin \left (c x \right ) e^{2} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{2}}{x}-\frac {e^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}-c^{4} d^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+2 c^{2} d e \sqrt {-c^{2} x^{2}+1}\right )}{c^{4}}\right )\)	\(168\)

Input:

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

Output:

a*(1/3*e^2*x^3+2*d*e*x-d^2/x)+b*c*(1/3/c*arcsin(c*x)*x^3*e^2+2/c*arcsin(c* 
x)*x*d*e-arcsin(c*x)*d^2/c/x-1/3/c^4*(e^2*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2) 
-2/3*(-c^2*x^2+1)^(1/2))+3*c^4*d^2*arctanh(1/(-c^2*x^2+1)^(1/2))-6*c^2*d*e 
*(-c^2*x^2+1)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.37 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {6 \, a c^{3} e^{2} x^{4} - 9 \, b c^{4} d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + 9 \, b c^{4} d^{2} x \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 36 \, a c^{3} d e x^{2} - 18 \, a c^{3} d^{2} + 6 \, {\left (b c^{3} e^{2} x^{4} + 6 \, b c^{3} d e x^{2} - 3 \, b c^{3} d^{2}\right )} \arcsin \left (c x\right ) + 2 \, {\left (b c^{2} e^{2} x^{3} + 2 \, {\left (9 \, b c^{2} d e + b e^{2}\right )} x\right )} \sqrt {-c^{2} x^{2} + 1}}{18 \, c^{3} x} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 2.02 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.33 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=- \frac {a d^{2}}{x} + 2 a d e x + \frac {a e^{2} x^{3}}{3} + b c d^{2} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b c e^{2} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{3} - \frac {b d^{2} \operatorname {asin}{\left (c x \right )}}{x} + 2 b d e \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) + \frac {b e^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} \] Input:

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

Output:

-a*d**2/x + 2*a*d*e*x + a*e**2*x**3/3 + b*c*d**2*Piecewise((-acosh(1/(c*x) 
), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), True)) - b*c*e**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*d**2*asin(c*x)/x + 2*b*d*e*Piecewise((0, E 
q(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) + b*e**2*x**3*asin 
(c*x)/3
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {1}{3} \, a e^{2} x^{3} - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d^{2} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e^{2} + 2 \, a d e x + \frac {2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d e}{c} - \frac {a d^{2}}{x} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4243 vs. \(2 (114) = 228\).

Time = 1.79 (sec) , antiderivative size = 4243, normalized size of antiderivative = 33.67 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*b*c^12*d^2*x^8*arcsin(c*x)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3* 
c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 
+ c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^8) - 1/2*a*c^12 
*d^2*x^8/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 
+ 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 
+ 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^8) + b*c^11*d^2*x^7*log(abs(c)*abs(x)) 
/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1 
)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1 
))*(sqrt(-c^2*x^2 + 1) + 1)^7) - b*c^11*d^2*x^7*log(sqrt(-c^2*x^2 + 1) + 1 
)/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 
1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 
1))*(sqrt(-c^2*x^2 + 1) + 1)^7) - 2*b*c^10*d^2*x^6*arcsin(c*x)/((c^10*x^7/ 
(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6* 
x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^ 
2*x^2 + 1) + 1)^6) - 2*a*c^10*d^2*x^6/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^ 
7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 
 1)^3 + c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^6) + 3*b* 
c^9*d^2*x^5*log(abs(c)*abs(x))/((c^10*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 3*c 
^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 3*c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + 
 c^4*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^5) - 3*b*c^9*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.29 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arcsin (c x))}{x^2} \, dx=\frac {-9 \mathit {asin} \left (c x \right ) b \,c^{3} d^{2}+18 \mathit {asin} \left (c x \right ) b \,c^{3} d e \,x^{2}+3 \mathit {asin} \left (c x \right ) b \,c^{3} e^{2} x^{4}+18 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} d e x +\sqrt {-c^{2} x^{2}+1}\, b \,c^{2} e^{2} x^{3}+2 \sqrt {-c^{2} x^{2}+1}\, b \,e^{2} x +9 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{4} d^{2} x -9 a \,c^{3} d^{2}+18 a \,c^{3} d e \,x^{2}+3 a \,c^{3} e^{2} x^{4}}{9 c^{3} x} \] Input:

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

Output:

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