\(\int \frac {d+e x}{a+b \arcsin (c x)} \, dx\) [18]

Optimal result

Integrand size = 16, antiderivative size = 115 \[ \int \frac {d+e x}{a+b \arcsin (c x)} \, dx=\frac {d \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{b c}-\frac {e \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right ) \sin \left (\frac {2 a}{b}\right )}{2 b c^2}+\frac {d \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{b c}+\frac {e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b c^2} \] Output:

d*cos(a/b)*Ci(a/b+arcsin(c*x))/b/c-1/2*e*Ci(2*a/b+2*arcsin(c*x))*sin(2*a/b 
)/b/c^2+d*sin(a/b)*Si(a/b+arcsin(c*x))/b/c+1/2*e*cos(2*a/b)*Si(2*a/b+2*arc 
sin(c*x))/b/c^2
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.85 \[ \int \frac {d+e x}{a+b \arcsin (c x)} \, dx=\frac {2 c d \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )-e \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+2 c d \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{2 b c^2} \] Input:

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

Output:

(2*c*d*Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]] - e*CosIntegral[2*(a/b + Ar 
cSin[c*x])]*Sin[(2*a)/b] + 2*c*d*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]] + 
 e*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])])/(2*b*c^2)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5246, 7293, 2009}

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.

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

\(\Big \downarrow \) 5246

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {c d \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{b}-\frac {e \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b}+\frac {c d \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{b}+\frac {e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b}}{c^2}\)

Input:

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

Output:

((c*d*Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]])/b - (e*CosIntegral[(2*a)/b 
+ 2*ArcSin[c*x]]*Sin[(2*a)/b])/(2*b) + (c*d*Sin[a/b]*SinIntegral[a/b + Arc 
Sin[c*x]])/b + (e*Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b) 
)/c^2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5246

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_.) + (e_.)*(x_))^(m_.), x_S 
ymbol] :> Simp[1/c^(m + 1)   Subst[Int[(a + b*x)^n*Cos[x]*(c*d + e*Sin[x])^ 
m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[m, 0]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {\frac {d \left (\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )\right )}{b}+\frac {e \left (\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )-\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )\right )}{2 c b}}{c}\)	\(103\)
default	\(\frac {\frac {d \left (\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )+\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )\right )}{b}+\frac {e \left (\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )-\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )\right )}{2 c b}}{c}\)	\(103\)

Input:

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

Output:

1/c*(d*(Si(arcsin(c*x)+a/b)*sin(a/b)+Ci(arcsin(c*x)+a/b)*cos(a/b))/b+1/2/c 
*e/b*(Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)-Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b 
)))
 

Fricas [F]

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

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

Output:

integral((e*x + d)/(b*arcsin(c*x) + a), x)
 

Sympy [F]

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

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

Output:

Integral((d + e*x)/(a + b*asin(c*x)), x)
 

Maxima [F]

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

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

Output:

integrate((e*x + d)/(b*arcsin(c*x) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.21 \[ \int \frac {d+e x}{a+b \arcsin (c x)} \, dx=\frac {d \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac {e \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{2}} + \frac {e \cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c^{2}} + \frac {d \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b c} - \frac {e \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{2}} \] Input:

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

Output:

d*cos(a/b)*cos_integral(a/b + arcsin(c*x))/(b*c) - e*cos(a/b)*cos_integral 
(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b*c^2) + e*cos(a/b)^2*sin_integral(2*a/b 
 + 2*arcsin(c*x))/(b*c^2) + d*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b* 
c) - 1/2*e*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x/(asin(c*x)*b + a),x)*e + int(1/(asin(c*x)*b + a),x)*d