\(\int \frac {(d+e x^2)^3 (a+b \arcsin (c x))}{x^4} \, dx\) [449]

Optimal result

Integrand size = 21, antiderivative size = 186 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {b e^2 \left (9 c^2 d+e\right ) \sqrt {1-c^2 x^2}}{3 c^3}-\frac {b c d^3 \sqrt {1-c^2 x^2}}{6 x^2}-\frac {b e^3 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))-\frac {1}{6} b c d^2 \left (c^2 d+18 e\right ) \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {1}{6} \left (-\frac {2 a d^3}{x^3}-\frac {18 a d^2 e}{x}+18 a d e^2 x+2 a e^3 x^3+\frac {b \sqrt {1-c^2 x^2} \left (-3 c^4 d^3+4 e^3 x^2+2 c^2 e^2 x^2 \left (27 d+e x^2\right )\right )}{3 c^3 x^2}+\frac {2 b \left (-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6\right ) \arcsin (c x)}{x^3}+b c d^2 \left (c^2 d+18 e\right ) \log (x)-b c d^2 \left (c^2 d+18 e\right ) \log \left (1+\sqrt {1-c^2 x^2}\right )\right ) \] Input:

Integrate[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]
 

Output:

((-2*a*d^3)/x^3 - (18*a*d^2*e)/x + 18*a*d*e^2*x + 2*a*e^3*x^3 + (b*Sqrt[1 
- c^2*x^2]*(-3*c^4*d^3 + 4*e^3*x^2 + 2*c^2*e^2*x^2*(27*d + e*x^2)))/(3*c^3 
*x^2) + (2*b*(-d^3 - 9*d^2*e*x^2 + 9*d*e^2*x^4 + e^3*x^6)*ArcSin[c*x])/x^3 
 + b*c*d^2*(c^2*d + 18*e)*Log[x] - b*c*d^2*(c^2*d + 18*e)*Log[1 + Sqrt[1 - 
 c^2*x^2]])/6
 

Rubi [A] (warning: unable to verify)

Time = 0.69 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5230, 27, 2331, 2124, 27, 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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx\)

\(\Big \downarrow \) 5230

\(\displaystyle -b c \int -\frac {-e^3 x^6-9 d e^2 x^4+9 d^2 e x^2+d^3}{3 x^3 \sqrt {1-c^2 x^2}}dx-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} b c \int \frac {-e^3 x^6-9 d e^2 x^4+9 d^2 e x^2+d^3}{x^3 \sqrt {1-c^2 x^2}}dx-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 2331

\(\displaystyle \frac {1}{6} b c \int \frac {-e^3 x^6-9 d e^2 x^4+9 d^2 e x^2+d^3}{x^4 \sqrt {1-c^2 x^2}}dx^2-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 2124

\(\displaystyle \frac {1}{6} b c \left (-\int -\frac {-2 e^3 x^4-18 d e^2 x^2+d^2 \left (d c^2+18 e\right )}{2 x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} b c \left (\frac {1}{2} \int \frac {-2 e^3 x^4-18 d e^2 x^2+d^2 \left (d c^2+18 e\right )}{x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 1192

\(\displaystyle \frac {1}{6} b c \left (\frac {\int -\frac {-2 e^3 x^8+2 e^2 \left (9 d c^2+2 e\right ) x^4+c^6 d^3-2 e^3-18 c^2 d e^2+18 c^4 d^2 e}{1-x^4}d\sqrt {1-c^2 x^2}}{c^4}-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} b c \left (-\frac {\int \frac {-2 e^3 x^8+2 e^2 \left (9 d c^2+2 e\right ) x^4+c^6 d^3-2 e^3-18 c^2 d e^2+18 c^4 d^2 e}{1-x^4}d\sqrt {1-c^2 x^2}}{c^4}-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 1467

\(\displaystyle \frac {1}{6} b c \left (-\frac {\int \left (2 e^3 x^4-2 e^2 \left (9 d c^2+e\right )+\frac {d^3 c^6+18 d^2 e c^4}{1-x^4}\right )d\sqrt {1-c^2 x^2}}{c^4}-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d^3 (a+b \arcsin (c x))}{3 x^3}-\frac {3 d^2 e (a+b \arcsin (c x))}{x}+3 d e^2 x (a+b \arcsin (c x))+\frac {1}{3} e^3 x^3 (a+b \arcsin (c x))+\frac {1}{6} b c \left (\frac {-c^4 d^2 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \left (c^2 d+18 e\right )+2 e^2 \sqrt {1-c^2 x^2} \left (9 c^2 d+e\right )-\frac {2}{3} e^3 x^6}{c^4}-\frac {d^3 \sqrt {1-c^2 x^2}}{x^2}\right )\)

Input:

Int[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]
 

Output:

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

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 2009

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

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

rule 2331

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2   S 
ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; 
 FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
 

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 = 0.24 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.26

method	result	size
parts	\(a \left (\frac {e^{3} x^{3}}{3}+3 d \,e^{2} x -\frac {d^{3}}{3 x^{3}}-\frac {3 d^{2} e}{x}\right )+b \,c^{3} \left (\frac {\arcsin \left (c x \right ) e^{3} x^{3}}{3 c^{3}}+\frac {3 \arcsin \left (c x \right ) x d \,e^{2}}{c^{3}}-\frac {\arcsin \left (c x \right ) d^{3}}{3 c^{3} x^{3}}-\frac {3 \arcsin \left (c x \right ) d^{2} e}{c^{3} x}-\frac {e^{3} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )-c^{6} d^{3} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )-9 c^{2} d \,e^{2} \sqrt {-c^{2} x^{2}+1}+9 c^{4} d^{2} e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{3 c^{6}}\right )\)	\(235\)
derivativedivides	\(c^{3} \left (\frac {a \left (3 c^{3} d \,e^{2} x +\frac {e^{3} c^{3} x^{3}}{3}-\frac {c^{3} d^{3}}{3 x^{3}}-\frac {3 c^{3} d^{2} e}{x}\right )}{c^{6}}+\frac {b \left (3 \arcsin \left (c x \right ) c^{3} d \,e^{2} x +\frac {\arcsin \left (c x \right ) e^{3} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{3}}{3 x^{3}}-\frac {3 \arcsin \left (c x \right ) c^{3} d^{2} e}{x}+\frac {c^{6} d^{3} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}-\frac {e^{3} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+3 c^{2} d \,e^{2} \sqrt {-c^{2} x^{2}+1}-3 c^{4} d^{2} e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{6}}\right )\)	\(249\)
default	\(c^{3} \left (\frac {a \left (3 c^{3} d \,e^{2} x +\frac {e^{3} c^{3} x^{3}}{3}-\frac {c^{3} d^{3}}{3 x^{3}}-\frac {3 c^{3} d^{2} e}{x}\right )}{c^{6}}+\frac {b \left (3 \arcsin \left (c x \right ) c^{3} d \,e^{2} x +\frac {\arcsin \left (c x \right ) e^{3} c^{3} x^{3}}{3}-\frac {\arcsin \left (c x \right ) c^{3} d^{3}}{3 x^{3}}-\frac {3 \arcsin \left (c x \right ) c^{3} d^{2} e}{x}+\frac {c^{6} d^{3} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}-\frac {e^{3} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+3 c^{2} d \,e^{2} \sqrt {-c^{2} x^{2}+1}-3 c^{4} d^{2} e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{6}}\right )\)	\(249\)

Input:

int((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x,method=_RETURNVERBOSE)
 

Output:

a*(1/3*e^3*x^3+3*d*e^2*x-1/3*d^3/x^3-3*d^2*e/x)+b*c^3*(1/3/c^3*arcsin(c*x) 
*e^3*x^3+3/c^3*arcsin(c*x)*x*d*e^2-1/3*arcsin(c*x)*d^3/c^3/x^3-3/c^3*arcsi 
n(c*x)*d^2*e/x-1/3/c^6*(e^3*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2 
+1)^(1/2))-c^6*d^3*(-1/2/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/2*arctanh(1/(-c^2*x^ 
2+1)^(1/2)))-9*c^2*d*e^2*(-c^2*x^2+1)^(1/2)+9*c^4*d^2*e*arctanh(1/(-c^2*x^ 
2+1)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.32 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {12 \, a c^{3} e^{3} x^{6} + 108 \, a c^{3} d e^{2} x^{4} - 108 \, a c^{3} d^{2} e x^{2} - 12 \, a c^{3} d^{3} - 3 \, {\left (b c^{6} d^{3} + 18 \, b c^{4} d^{2} e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + 3 \, {\left (b c^{6} d^{3} + 18 \, b c^{4} d^{2} e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 12 \, {\left (b c^{3} e^{3} x^{6} + 9 \, b c^{3} d e^{2} x^{4} - 9 \, b c^{3} d^{2} e x^{2} - b c^{3} d^{3}\right )} \arcsin \left (c x\right ) + 2 \, {\left (2 \, b c^{2} e^{3} x^{5} - 3 \, b c^{4} d^{3} x + 2 \, {\left (27 \, b c^{2} d e^{2} + 2 \, b e^{3}\right )} x^{3}\right )} \sqrt {-c^{2} x^{2} + 1}}{36 \, c^{3} x^{3}} \] Input:

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="fricas")
 

Output:

1/36*(12*a*c^3*e^3*x^6 + 108*a*c^3*d*e^2*x^4 - 108*a*c^3*d^2*e*x^2 - 12*a* 
c^3*d^3 - 3*(b*c^6*d^3 + 18*b*c^4*d^2*e)*x^3*log(sqrt(-c^2*x^2 + 1) + 1) + 
 3*(b*c^6*d^3 + 18*b*c^4*d^2*e)*x^3*log(sqrt(-c^2*x^2 + 1) - 1) + 12*(b*c^ 
3*e^3*x^6 + 9*b*c^3*d*e^2*x^4 - 9*b*c^3*d^2*e*x^2 - b*c^3*d^3)*arcsin(c*x) 
 + 2*(2*b*c^2*e^3*x^5 - 3*b*c^4*d^3*x + 2*(27*b*c^2*d*e^2 + 2*b*e^3)*x^3)* 
sqrt(-c^2*x^2 + 1))/(c^3*x^3)
 

Sympy [A] (verification not implemented)

Time = 3.55 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.67 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=- \frac {a d^{3}}{3 x^{3}} - \frac {3 a d^{2} e}{x} + 3 a d e^{2} x + \frac {a e^{3} x^{3}}{3} + \frac {b c d^{3} \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} + 3 b c d^{2} e \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b c e^{3} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{3} - \frac {b d^{3} \operatorname {asin}{\left (c x \right )}}{3 x^{3}} - \frac {3 b d^{2} e \operatorname {asin}{\left (c x \right )}}{x} + 3 b d e^{2} \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) + \frac {b e^{3} x^{3} \operatorname {asin}{\left (c x \right )}}{3} \] Input:

integrate((e*x**2+d)**3*(a+b*asin(c*x))/x**4,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {1}{3} \, a e^{3} x^{3} - \frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d^{3} - 3 \, {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d^{2} e + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e^{3} + 3 \, a d e^{2} x + \frac {3 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d e^{2}}{c} - \frac {3 \, a d^{2} e}{x} - \frac {a d^{3}}{3 \, x^{3}} \] Input:

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="maxima")
 

Output:

1/3*a*e^3*x^3 - 1/6*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sq 
rt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*b*d^3 - 3*(c*log(2*sqrt(-c^2* 
x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*b*d^2*e + 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*e^3 + 3 
*a*d*e^2*x + 3*(c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*b*d*e^2/c - 3*a*d^2* 
e/x - 1/3*a*d^3/x^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7971 vs. \(2 (166) = 332\).

Time = 8.18 (sec) , antiderivative size = 7971, normalized size of antiderivative = 42.85 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\text {Too large to display} \] Input:

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="giac")
 

Output:

-1/24*b*c^18*d^3*x^12*arcsin(c*x)/((c^12*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 
3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^12) - 1/ 
24*a*c^18*d^3*x^12/((c^12*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 3*c^10*x^7/(sqr 
t(-c^2*x^2 + 1) + 1)^7 + 3*c^8*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + c^6*x^3/(s 
qrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^12) + 1/24*b*c^17*d^3*x 
^11/((c^12*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 
1) + 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^11) - 1/4*b*c^16*d^3*x^10*arcsin(c*x)/ 
((c^12*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 1) + 
 1)^3)*(sqrt(-c^2*x^2 + 1) + 1)^10) - 1/4*a*c^16*d^3*x^10/((c^12*x^9/(sqrt 
(-c^2*x^2 + 1) + 1)^9 + 3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(-c^ 
2*x^2 + 1) + 1)^10) + 1/6*b*c^15*d^3*x^9*log(abs(c)*abs(x))/((c^12*x^9/(sq 
rt(-c^2*x^2 + 1) + 1)^9 + 3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3)*(sqrt(- 
c^2*x^2 + 1) + 1)^9) - 1/6*b*c^15*d^3*x^9*log(sqrt(-c^2*x^2 + 1) + 1)/((c^ 
12*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 3*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 + c^6*x^3/(sqrt(-c^2*x^2 + 1) + ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^3}{x^4} \,d x \] Input:

int(((a + b*asin(c*x))*(d + e*x^2)^3)/x^4,x)
 

Output:

int(((a + b*asin(c*x))*(d + e*x^2)^3)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.33 \[ \int \frac {\left (d+e x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {-6 \mathit {asin} \left (c x \right ) b \,c^{3} d^{3}-54 \mathit {asin} \left (c x \right ) b \,c^{3} d^{2} e \,x^{2}+54 \mathit {asin} \left (c x \right ) b \,c^{3} d \,e^{2} x^{4}+6 \mathit {asin} \left (c x \right ) b \,c^{3} e^{3} x^{6}-3 \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} d^{3} x +54 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} d \,e^{2} x^{3}+2 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} e^{3} x^{5}+4 \sqrt {-c^{2} x^{2}+1}\, b \,e^{3} x^{3}+3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{6} d^{3} x^{3}+54 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{4} d^{2} e \,x^{3}-6 a \,c^{3} d^{3}-54 a \,c^{3} d^{2} e \,x^{2}+54 a \,c^{3} d \,e^{2} x^{4}+6 a \,c^{3} e^{3} x^{6}}{18 c^{3} x^{3}} \] Input:

int((e*x^2+d)^3*(a+b*asin(c*x))/x^4,x)
 

Output:

( - 6*asin(c*x)*b*c**3*d**3 - 54*asin(c*x)*b*c**3*d**2*e*x**2 + 54*asin(c* 
x)*b*c**3*d*e**2*x**4 + 6*asin(c*x)*b*c**3*e**3*x**6 - 3*sqrt( - c**2*x**2 
 + 1)*b*c**4*d**3*x + 54*sqrt( - c**2*x**2 + 1)*b*c**2*d*e**2*x**3 + 2*sqr 
t( - c**2*x**2 + 1)*b*c**2*e**3*x**5 + 4*sqrt( - c**2*x**2 + 1)*b*e**3*x** 
3 + 3*log(tan(asin(c*x)/2))*b*c**6*d**3*x**3 + 54*log(tan(asin(c*x)/2))*b* 
c**4*d**2*e*x**3 - 6*a*c**3*d**3 - 54*a*c**3*d**2*e*x**2 + 54*a*c**3*d*e** 
2*x**4 + 6*a*c**3*e**3*x**6)/(18*c**3*x**3)