3.1.30 \(\int \frac {x^2 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx\) [30]

3.1.30.1 Optimal result

Integrand size = 25, antiderivative size = 124 \[ \int \frac {x^2 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}-\frac {x (a+b \arcsin (c x))}{c^2 d}-\frac {2 i (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c^3 d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c^3 d}-\frac {i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^3 d} \]

output

-x*(a+b*arcsin(c*x))/c^2/d-2*I*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1) 
^(1/2))/c^3/d+I*b*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d-I*b*polyl 
og(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d-b*(-c^2*x^2+1)^(1/2)/c^3/d
 

3.1.30.2 Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.92 \[ \int \frac {x^2 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=-\frac {2 a c x+2 b \sqrt {1-c^2 x^2}+i b \pi \arcsin (c x)+2 b c x \arcsin (c x)-b \pi \log \left (1-i e^{i \arcsin (c x)}\right )-2 b \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-b \pi \log \left (1+i e^{i \arcsin (c x)}\right )+2 b \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )+a \log (1-c x)-a \log (1+c x)+b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-2 i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+2 i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{2 c^3 d} \]

input

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

output

-1/2*(2*a*c*x + 2*b*Sqrt[1 - c^2*x^2] + I*b*Pi*ArcSin[c*x] + 2*b*c*x*ArcSi 
n[c*x] - b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 2*b*ArcSin[c*x]*Log[1 - I*E^( 
I*ArcSin[c*x])] - b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 2*b*ArcSin[c*x]*Log[ 
1 + I*E^(I*ArcSin[c*x])] + a*Log[1 - c*x] - a*Log[1 + c*x] + b*Pi*Log[-Cos 
[(Pi + 2*ArcSin[c*x])/4]] + b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (2*I)* 
b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (2*I)*b*PolyLog[2, I*E^(I*ArcSin[c* 
x])])/(c^3*d)
 

3.1.30.3 Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {5210, 27, 241, 5164, 3042, 4669, 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.

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

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 5164

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle \frac {-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))+i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^3 d}-\frac {x (a+b \arcsin (c x))}{c^2 d}-\frac {b \sqrt {1-c^2 x^2}}{c^3 d}\)

input

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

output

-((b*Sqrt[1 - c^2*x^2])/(c^3*d)) - (x*(a + b*ArcSin[c*x]))/(c^2*d) + ((-2* 
I)*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])] + I*b*PolyLog[2, (-I)*E^( 
I*ArcSin[c*x])] - I*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^3*d)
 

3.1.30.3.1 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 241

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

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 4669

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

rule 5164

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> Simp[1/(c*d)   Subst[Int[(a + b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] 
/; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]
 

rule 5210

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. 
)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + 
 b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Simp[f^2*((m - 1)/(c^2*(m + 2*p 
 + 1)))   Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + S 
imp[b*f*(n/(c*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Int[(f* 
x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; 
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m 
, 1] && NeQ[m + 2*p + 1, 0]
 

3.1.30.4 Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.54

method	result	size
derivativedivides	\(\frac {-\frac {a \left (c x +\frac {\ln \left (c x -1\right )}{2}-\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \sqrt {-c^{2} x^{2}+1}}{d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b \arcsin \left (c x \right ) c x}{d}-\frac {i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}}{c^{3}}\)	\(191\)
default	\(\frac {-\frac {a \left (c x +\frac {\ln \left (c x -1\right )}{2}-\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \sqrt {-c^{2} x^{2}+1}}{d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b \arcsin \left (c x \right ) c x}{d}-\frac {i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}}{c^{3}}\)	\(191\)
parts	\(-\frac {a \left (\frac {x}{c^{2}}+\frac {\ln \left (c x -1\right )}{2 c^{3}}-\frac {\ln \left (c x +1\right )}{2 c^{3}}\right )}{d}-\frac {b \sqrt {-c^{2} x^{2}+1}}{c^{3} d}-\frac {b \arcsin \left (c x \right ) x}{d \,c^{2}}+\frac {i b \operatorname {dilog}\left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \,c^{3}}-\frac {i b \operatorname {dilog}\left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \,c^{3}}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \,c^{3}}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \,c^{3}}\)	\(212\)

input

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

output

1/c^3*(-a/d*(c*x+1/2*ln(c*x-1)-1/2*ln(c*x+1))-b/d*(-c^2*x^2+1)^(1/2)-b/d*a 
rcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+b/d*arcsin(c*x)*ln(1-I*(I*c* 
x+(-c^2*x^2+1)^(1/2)))-b/d*arcsin(c*x)*c*x-I*b/d*dilog(1-I*(I*c*x+(-c^2*x^ 
2+1)^(1/2)))+I*b/d*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2))))
 

3.1.30.5 Fricas [F]

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

input

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

output

integral(-(b*x^2*arcsin(c*x) + a*x^2)/(c^2*d*x^2 - d), x)
 

3.1.30.6 Sympy [F]

\[ \int \frac {x^2 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=- \frac {\int \frac {a x^{2}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{2} \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

input

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

output

-(Integral(a*x**2/(c**2*x**2 - 1), x) + Integral(b*x**2*asin(c*x)/(c**2*x* 
*2 - 1), x))/d
 

3.1.30.7 Maxima [F]

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

input

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

output

-1/2*a*(2*x/(c^2*d) - log(c*x + 1)/(c^3*d) + log(c*x - 1)/(c^3*d)) + 1/2*( 
2*c^3*d*integrate(-1/2*(2*c*x - log(c*x + 1) + log(-c*x + 1))*sqrt(c*x + 1 
)*sqrt(-c*x + 1)/(c^4*d*x^2 - c^2*d), x) - 2*c*x*arctan2(c*x, sqrt(c*x + 1 
)*sqrt(-c*x + 1)) + arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1 
) - arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1))*b/(c^3*d)
 

3.1.30.8 Giac [F]

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

input

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

output

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

3.1.30.9 Mupad [F(-1)]

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

input

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

output

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