\(\int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx\) [29]

Optimal result

Integrand size = 25, antiderivative size = 144 \[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=-\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}+\frac {b \arcsin (c x)}{4 c^4 d}-\frac {x^2 (a+b \arcsin (c x))}{2 c^2 d}+\frac {i (a+b \arcsin (c x))^2}{2 b c^4 d}-\frac {(a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c^4 d}+\frac {i b \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{2 c^4 d} \] Output:

-1/4*b*x*(-c^2*x^2+1)^(1/2)/c^3/d+1/4*b*arcsin(c*x)/c^4/d-1/2*x^2*(a+b*arc 
sin(c*x))/c^2/d+1/2*I*(a+b*arcsin(c*x))^2/b/c^4/d-(a+b*arcsin(c*x))*ln(1+( 
I*c*x+(-c^2*x^2+1)^(1/2))^2)/c^4/d+1/2*I*b*polylog(2,-(I*c*x+(-c^2*x^2+1)^ 
(1/2))^2)/c^4/d
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(312\) vs. \(2(144)=288\).

Time = 0.34 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.17 \[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=-\frac {2 a c^2 x^2+b c x \sqrt {1-c^2 x^2}+4 i b \pi \arcsin (c x)+2 b c^2 x^2 \arcsin (c x)-2 i b \arcsin (c x)^2-2 b \arctan \left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )+8 b \pi \log \left (1+e^{-i \arcsin (c x)}\right )+2 b \pi \log \left (1-i e^{i \arcsin (c x)}\right )+4 b \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-2 b \pi \log \left (1+i e^{i \arcsin (c x)}\right )+4 b \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )+2 a \log \left (1-c^2 x^2\right )-8 b \pi \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-2 b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-4 i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-4 i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{4 c^4 d} \] Input:

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

Output:

-1/4*(2*a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2] + (4*I)*b*Pi*ArcSin[c*x] + 2*b 
*c^2*x^2*ArcSin[c*x] - (2*I)*b*ArcSin[c*x]^2 - 2*b*ArcTan[(c*x)/(-1 + Sqrt 
[1 - c^2*x^2])] + 8*b*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 2*b*Pi*Log[1 - I* 
E^(I*ArcSin[c*x])] + 4*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 2*b*Pi 
*Log[1 + I*E^(I*ArcSin[c*x])] + 4*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x] 
)] + 2*a*Log[1 - c^2*x^2] - 8*b*Pi*Log[Cos[ArcSin[c*x]/2]] + 2*b*Pi*Log[-C 
os[(Pi + 2*ArcSin[c*x])/4]] - 2*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (4 
*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (4*I)*b*PolyLog[2, I*E^(I*ArcSi 
n[c*x])])/(c^4*d)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5210, 27, 262, 223, 5180, 3042, 4202, 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.

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

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 5180

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )\right )}{c^4 d}-\frac {x^2 (a+b \arcsin (c x))}{2 c^2 d}+\frac {b \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )}{2 c d}\)

Input:

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

Output:

-1/2*(x^2*(a + b*ArcSin[c*x]))/(c^2*d) + (b*(-1/2*(x*Sqrt[1 - c^2*x^2])/c^ 
2 + ArcSin[c*x]/(2*c^3)))/(2*c*d) + (((I/2)*(a + b*ArcSin[c*x])^2)/b - (2* 
I)*((-1/2*I)*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])] - (b*PolyL 
og[2, -E^((2*I)*ArcSin[c*x])])/4))/(c^4*d)
 

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 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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

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 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5180

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), 
x_Symbol] :> Simp[-e^(-1)   Subst[Int[(a + b*x)^n*Tan[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]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.11

method	result	size
derivativedivides	\(\frac {-\frac {a \left (\frac {c^{2} x^{2}}{2}+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}+\frac {i b \arcsin \left (c x \right )^{2}}{2 d}-\frac {b \sqrt {-c^{2} x^{2}+1}\, c x}{4 d}-\frac {b \arcsin \left (c x \right ) c^{2} x^{2}}{2 d}+\frac {b \arcsin \left (c x \right )}{4 d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\)	\(160\)
default	\(\frac {-\frac {a \left (\frac {c^{2} x^{2}}{2}+\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}+\frac {i b \arcsin \left (c x \right )^{2}}{2 d}-\frac {b \sqrt {-c^{2} x^{2}+1}\, c x}{4 d}-\frac {b \arcsin \left (c x \right ) c^{2} x^{2}}{2 d}+\frac {b \arcsin \left (c x \right )}{4 d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b \operatorname {polylog}\left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}}{c^{4}}\)	\(160\)
parts	\(-\frac {a \,x^{2}}{2 d \,c^{2}}-\frac {a \ln \left (c^{2} x^{2}-1\right )}{2 d \,c^{4}}+\frac {i b \arcsin \left (c x \right )^{2}}{2 d \,c^{4}}-\frac {b x \sqrt {-c^{2} x^{2}+1}}{4 c^{3} d}-\frac {b \arcsin \left (c x \right ) x^{2}}{2 d \,c^{2}}+\frac {b \arcsin \left (c x \right )}{4 c^{4} d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d \,c^{4}}+\frac {i b \operatorname {polylog}\left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{4} d}\)	\(170\)

Input:

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

Output:

1/c^4*(-a/d*(1/2*c^2*x^2+1/2*ln(c*x-1)+1/2*ln(c*x+1))+1/2*I*b/d*arcsin(c*x 
)^2-1/4*b/d*(-c^2*x^2+1)^(1/2)*c*x-1/2*b/d*arcsin(c*x)*c^2*x^2+1/4*b*arcsi 
n(c*x)/d-b/d*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+1/2*I*b*polylo 
g(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d)
 

Fricas [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} - d} \,d x } \] Input:

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

Output:

integral(-(b*x^3*arcsin(c*x) + a*x^3)/(c^2*d*x^2 - d), x)
 

Sympy [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=- \frac {\int \frac {a x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \] Input:

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

Output:

-(Integral(a*x**3/(c**2*x**2 - 1), x) + Integral(b*x**3*asin(c*x)/(c**2*x* 
*2 - 1), x))/d
 

Maxima [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} - d} \,d x } \] Input:

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

Output:

-1/2*a*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) - 1/2*(2*c^4*d*integrate(1 
/2*(c^2*x^2*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)) + e^(1/2*log(c*x + 1) 
 + 1/2*log(-c*x + 1))*log(c*x + 1) + e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 
1))*log(-c*x + 1))/(c^7*d*x^4 - c^5*d*x^2 + (c^5*d*x^2 - c^3*d)*e^(log(c*x 
 + 1) + log(-c*x + 1))), x) + c^2*x^2*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^4*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 (a+b \arcsin (c x))}{d-c^2 d x^2} \, dx=\frac {-2 \mathit {asin} \left (c x \right ) b \,c^{2} x^{2}+\mathit {asin} \left (c x \right ) b -\sqrt {-c^{2} x^{2}+1}\, b c x -4 \left (\int \frac {\mathit {asin} \left (c x \right ) x}{c^{2} x^{2}-1}d x \right ) b \,c^{2}-2 \,\mathrm {log}\left (c^{2} x -c \right ) a -2 \,\mathrm {log}\left (c^{2} x +c \right ) a -2 a \,c^{2} x^{2}}{4 c^{4} d} \] Input:

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

Output:

( - 2*asin(c*x)*b*c**2*x**2 + asin(c*x)*b - sqrt( - c**2*x**2 + 1)*b*c*x - 
 4*int((asin(c*x)*x)/(c**2*x**2 - 1),x)*b*c**2 - 2*log(c**2*x - c)*a - 2*l 
og(c**2*x + c)*a - 2*a*c**2*x**2)/(4*c**4*d)