\(\int \frac {a+b \arcsin (c x)}{d-c^2 d x^2} \, dx\) [32]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {a+b \arcsin (c x)}{d-c^2 d x^2} \, dx=-\frac {2 i (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c d}-\frac {i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c d} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(207\) vs. \(2(84)=168\).

Time = 0.03 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.46 \[ \int \frac {a+b \arcsin (c x)}{d-c^2 d x^2} \, dx=\frac {-i b \pi \arcsin (c x)+b \pi \log \left (1-i e^{i \arcsin (c x)}\right )+2 b \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )+b \pi \log \left (1+i e^{i \arcsin (c x)}\right )-2 b \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )-a \log (1-c x)+a \log (1+c x)-b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+2 i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-2 i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{2 c d} \] Input:

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

Output:

((-I)*b*Pi*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])] + 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])])/(2*c*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {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.

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

\(\Big \downarrow \) 5164

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4669

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

((-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*d)
 

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.67

method	result	size
derivativedivides	\(\frac {\frac {a \,\operatorname {arctanh}\left (c x \right )}{d}-\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \arcsin \left (c x \right )+i \operatorname {arctanh}\left (c x \right ) \left (\ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )-\ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )-i \operatorname {dilog}\left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{d}}{c}\)	\(140\)
default	\(\frac {\frac {a \,\operatorname {arctanh}\left (c x \right )}{d}-\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \arcsin \left (c x \right )+i \operatorname {arctanh}\left (c x \right ) \left (\ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )-\ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )-i \operatorname {dilog}\left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{d}}{c}\)	\(140\)
parts	\(-\frac {a \ln \left (c x -1\right )}{2 d c}+\frac {a \ln \left (c x +1\right )}{2 d c}-\frac {b \left (-\operatorname {arctanh}\left (c x \right ) \arcsin \left (c x \right )+i \operatorname {arctanh}\left (c x \right ) \left (\ln \left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )-\ln \left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )-i \operatorname {dilog}\left (1+\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (c x +1\right )}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{d c}\)	\(160\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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