\(\int \frac {(d+e x^2) (a+b \arcsin (c x))}{x^4} \, dx\) [431]

Optimal result

Integrand size = 19, antiderivative size = 85 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2}-\frac {d (a+b \arcsin (c x))}{3 x^3}-\frac {e (a+b \arcsin (c x))}{x}-\frac {1}{6} b c \left (c^2 d+6 e\right ) \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.28 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=-\frac {a d}{3 x^3}-\frac {a e}{x}-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2}-\frac {b d \arcsin (c x)}{3 x^3}-\frac {b e \arcsin (c x)}{x}-\frac {1}{6} b c^3 d \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )-b c e \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5230, 27, 354, 87, 73, 221}

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.

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

\(\Big \downarrow \) 5230

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && 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 = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33

method	result	size
parts	\(a \left (-\frac {d}{3 x^{3}}-\frac {e}{x}\right )+b \,c^{3} \left (-\frac {\arcsin \left (c x \right ) d}{3 c^{3} x^{3}}-\frac {\arcsin \left (c x \right ) e}{c^{3} x}-\frac {-d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )+3 e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{3 c^{2}}\right )\)	\(113\)
derivativedivides	\(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arcsin \left (c x \right ) e}{c x}-\frac {\arcsin \left (c x \right ) d}{3 c \,x^{3}}+\frac {d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}-e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{2}}\right )\)	\(120\)
default	\(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arcsin \left (c x \right ) e}{c x}-\frac {\arcsin \left (c x \right ) d}{3 c \,x^{3}}+\frac {d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}-e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{2}}\right )\)	\(120\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.35 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=-\frac {{\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 2 \, \sqrt {-c^{2} x^{2} + 1} b c d x + 12 \, a e x^{2} + 4 \, a d + 4 \, {\left (3 \, b e x^{2} + b d\right )} \arcsin \left (c x\right )}{12 \, x^{3}} \] Input:

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

Output:

-1/12*((b*c^3*d + 6*b*c*e)*x^3*log(sqrt(-c^2*x^2 + 1) + 1) - (b*c^3*d + 6* 
b*c*e)*x^3*log(sqrt(-c^2*x^2 + 1) - 1) + 2*sqrt(-c^2*x^2 + 1)*b*c*d*x + 12 
*a*e*x^2 + 4*a*d + 4*(3*b*e*x^2 + b*d)*arcsin(c*x))/x^3
 

Sympy [A] (verification not implemented)

Time = 2.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.98 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=- \frac {a d}{3 x^{3}} - \frac {a e}{x} + \frac {b c d \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} + b c e \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d \operatorname {asin}{\left (c x \right )}}{3 x^{3}} - \frac {b e \operatorname {asin}{\left (c x \right )}}{x} \] Input:

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

Output:

-a*d/(3*x**3) - a*e/x + b*c*d*Piecewise((-c**2*acosh(1/(c*x))/2 + c/(2*x*s 
qrt(-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*c*e*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*a 
sin(1/(c*x)), True)) - b*d*asin(c*x)/(3*x**3) - b*e*asin(c*x)/x
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (75) = 150\).

Time = 120.87 (sec) , antiderivative size = 424, normalized size of antiderivative = 4.99 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=-\frac {b c^{6} d x^{3} \arcsin \left (c x\right )}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} - \frac {a c^{6} d x^{3}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} + \frac {b c^{5} d x^{2}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} - \frac {b c^{4} d x \arcsin \left (c x\right )}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{4} d x}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} + \frac {1}{6} \, b c^{3} d \log \left ({\left | c \right |} {\left | x \right |}\right ) - \frac {1}{6} \, b c^{3} d \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c^{2} e x \arcsin \left (c x\right )}{2 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {b c^{2} d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right )}{8 \, x} - \frac {a c^{2} e x}{2 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{2} d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{8 \, x} + b c e \log \left ({\left | c \right |} {\left | x \right |}\right ) - b c e \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}}{24 \, x^{2}} - \frac {b e {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right )}{2 \, x} - \frac {b d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3} \arcsin \left (c x\right )}{24 \, x^{3}} - \frac {a e {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{2 \, x} - \frac {a d {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}}{24 \, x^{3}} \] Input:

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

Output:

-1/24*b*c^6*d*x^3*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^3 - 1/24*a*c^6*d*x^ 
3/(sqrt(-c^2*x^2 + 1) + 1)^3 + 1/24*b*c^5*d*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 
 - 1/8*b*c^4*d*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1) - 1/8*a*c^4*d*x/(sqr 
t(-c^2*x^2 + 1) + 1) + 1/6*b*c^3*d*log(abs(c)*abs(x)) - 1/6*b*c^3*d*log(sq 
rt(-c^2*x^2 + 1) + 1) - 1/2*b*c^2*e*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1) 
 - 1/8*b*c^2*d*(sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)/x - 1/2*a*c^2*e*x/(sqr 
t(-c^2*x^2 + 1) + 1) - 1/8*a*c^2*d*(sqrt(-c^2*x^2 + 1) + 1)/x + b*c*e*log( 
abs(c)*abs(x)) - b*c*e*log(sqrt(-c^2*x^2 + 1) + 1) - 1/24*b*c*d*(sqrt(-c^2 
*x^2 + 1) + 1)^2/x^2 - 1/2*b*e*(sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)/x - 1/ 
24*b*d*(sqrt(-c^2*x^2 + 1) + 1)^3*arcsin(c*x)/x^3 - 1/2*a*e*(sqrt(-c^2*x^2 
 + 1) + 1)/x - 1/24*a*d*(sqrt(-c^2*x^2 + 1) + 1)^3/x^3
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arcsin (c x))}{x^4} \, dx=\frac {-2 \mathit {asin} \left (c x \right ) b d -6 \mathit {asin} \left (c x \right ) b e \,x^{2}-\sqrt {-c^{2} x^{2}+1}\, b c d x +\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{3} d \,x^{3}+6 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b c e \,x^{3}-2 a d -6 a e \,x^{2}}{6 x^{3}} \] Input:

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

Output:

( - 2*asin(c*x)*b*d - 6*asin(c*x)*b*e*x**2 - sqrt( - c**2*x**2 + 1)*b*c*d* 
x + log(tan(asin(c*x)/2))*b*c**3*d*x**3 + 6*log(tan(asin(c*x)/2))*b*c*e*x* 
*3 - 2*a*d - 6*a*e*x**2)/(6*x**3)