3.1.6 \(\int \frac {(d-c^2 d x^2) (a+b \arcsin (c x))}{x} \, dx\) [6]

3.1.6.1 Optimal result
3.1.6.2 Mathematica [A] (verified)
3.1.6.3 Rubi [A] (verified)
3.1.6.4 Maple [A] (verified)
3.1.6.5 Fricas [F]
3.1.6.6 Sympy [F]
3.1.6.7 Maxima [F]
3.1.6.8 Giac [F]
3.1.6.9 Mupad [F(-1)]

3.1.6.1 Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {\left (d-c^2 d x^2\right ) (a+b \arcsin (c x))}{x} \, dx=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \arcsin (c x)+\frac {1}{2} d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))-\frac {i d (a+b \arcsin (c x))^2}{2 b}+d (a+b \arcsin (c x)) \log \left (1-e^{2 i \arcsin (c x)}\right )-\frac {1}{2} i b d \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right ) \]

output
-1/4*b*d*arcsin(c*x)+1/2*d*(-c^2*x^2+1)*(a+b*arcsin(c*x))-1/2*I*d*(a+b*arc 
sin(c*x))^2/b+d*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/2*I 
*b*d*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/4*b*c*d*x*(-c^2*x^2+1)^(1/2 
)
 
3.1.6.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d-c^2 d x^2\right ) (a+b \arcsin (c x))}{x} \, dx=-\frac {1}{2} a c^2 d x^2-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}+\frac {1}{4} b d \arcsin (c x)-\frac {1}{2} b c^2 d x^2 \arcsin (c x)+b d \arcsin (c x) \log \left (1-e^{2 i \arcsin (c x)}\right )+a d \log (x)-\frac {1}{2} i b d \left (\arcsin (c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right ) \]

input
Integrate[((d - c^2*d*x^2)*(a + b*ArcSin[c*x]))/x,x]
 
output
-1/2*(a*c^2*d*x^2) - (b*c*d*x*Sqrt[1 - c^2*x^2])/4 + (b*d*ArcSin[c*x])/4 - 
 (b*c^2*d*x^2*ArcSin[c*x])/2 + b*d*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x 
])] + a*d*Log[x] - (I/2)*b*d*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c 
*x])])
 
3.1.6.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {5188, 211, 223, 5136, 3042, 25, 4200, 25, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 5188

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 5136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4200

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle \frac {1}{2} d \left (1-c^2 x^2\right ) (a+b \arcsin (c x))+d \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {1}{2} b c d \left (\frac {\arcsin (c x)}{2 c}+\frac {1}{2} x \sqrt {1-c^2 x^2}\right )\)

input
Int[((d - c^2*d*x^2)*(a + b*ArcSin[c*x]))/x,x]
 
output
(d*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/2 - (b*c*d*((x*Sqrt[1 - c^2*x^2])/2 
+ ArcSin[c*x]/(2*c)))/2 + d*(((-1/2*I)*(a + b*ArcSin[c*x])^2)/b - (2*I)*(( 
I/2)*(a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])] + (b*PolyLog[2, E^ 
((2*I)*ArcSin[c*x])])/4))
 

3.1.6.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 211
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 2620
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4200
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol 
] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^ 
m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] 
, x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
 

rule 5136
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( 
a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
 

rule 5188
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.))/(x_), 
x_Symbol] :> Simp[(d + e*x^2)^p*((a + b*ArcSin[c*x])/(2*p)), x] + (Simp[d 
 Int[(d + e*x^2)^(p - 1)*((a + b*ArcSin[c*x])/x), x], x] - Simp[b*c*(d^p/(2 
*p))   Int[(1 - c^2*x^2)^(p - 1/2), x], x]) /; FreeQ[{a, b, c, d, e}, x] && 
 EqQ[c^2*d + e, 0] && IGtQ[p, 0]
 
3.1.6.4 Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.30

method result size
parts \(-d a \left (\frac {c^{2} x^{2}}{2}-\ln \left (x \right )\right )-d b \left (\frac {i \arcsin \left (c x \right )^{2}}{2}-\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {\arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right )}{4}+\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{8}\right )\) \(157\)
derivativedivides \(-d a \left (\frac {c^{2} x^{2}}{2}-\ln \left (c x \right )\right )-d b \left (\frac {i \arcsin \left (c x \right )^{2}}{2}-\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {\arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right )}{4}+\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{8}\right )\) \(159\)
default \(-d a \left (\frac {c^{2} x^{2}}{2}-\ln \left (c x \right )\right )-d b \left (\frac {i \arcsin \left (c x \right )^{2}}{2}-\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {\arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right )}{4}+\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{8}\right )\) \(159\)

input
int((-c^2*d*x^2+d)*(a+b*arcsin(c*x))/x,x,method=_RETURNVERBOSE)
 
output
-d*a*(1/2*c^2*x^2-ln(x))-d*b*(1/2*I*arcsin(c*x)^2-arcsin(c*x)*ln(1+I*c*x+( 
-c^2*x^2+1)^(1/2))+I*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-arcsin(c*x)*ln(1 
-I*c*x-(-c^2*x^2+1)^(1/2))+I*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-1/4*arcsi 
n(c*x)*cos(2*arcsin(c*x))+1/8*sin(2*arcsin(c*x)))
 
3.1.6.5 Fricas [F]

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

input
integrate((-c^2*d*x^2+d)*(a+b*arcsin(c*x))/x,x, algorithm="fricas")
 
output
integral(-(a*c^2*d*x^2 - a*d + (b*c^2*d*x^2 - b*d)*arcsin(c*x))/x, x)
 
3.1.6.6 Sympy [F]

\[ \int \frac {\left (d-c^2 d x^2\right ) (a+b \arcsin (c x))}{x} \, dx=- d \left (\int \left (- \frac {a}{x}\right )\, dx + \int a c^{2} x\, dx + \int \left (- \frac {b \operatorname {asin}{\left (c x \right )}}{x}\right )\, dx + \int b c^{2} x \operatorname {asin}{\left (c x \right )}\, dx\right ) \]

input
integrate((-c**2*d*x**2+d)*(a+b*asin(c*x))/x,x)
 
output
-d*(Integral(-a/x, x) + Integral(a*c**2*x, x) + Integral(-b*asin(c*x)/x, x 
) + Integral(b*c**2*x*asin(c*x), x))
 
3.1.6.7 Maxima [F]

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

input
integrate((-c^2*d*x^2+d)*(a+b*arcsin(c*x))/x,x, algorithm="maxima")
 
output
-1/2*a*c^2*d*x^2 + a*d*log(x) - integrate((b*c^2*d*x^2 - b*d)*arctan2(c*x, 
 sqrt(c*x + 1)*sqrt(-c*x + 1))/x, x)
 
3.1.6.8 Giac [F]

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

input
integrate((-c^2*d*x^2+d)*(a+b*arcsin(c*x))/x,x, algorithm="giac")
 
output
integrate(-(c^2*d*x^2 - d)*(b*arcsin(c*x) + a)/x, x)
 
3.1.6.9 Mupad [F(-1)]

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

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