\(\int \frac {d+e x^2}{a+b \arccos (c x)} \, dx\) [671]

Optimal result

Integrand size = 18, antiderivative size = 179 \[ \int \frac {d+e x^2}{a+b \arccos (c x)} \, dx=\frac {d \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{b c}+\frac {e \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c^3}-\frac {e \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c^3}+\frac {d \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b c}+\frac {e \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c^3}-\frac {e \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c^3} \] Output:

d*cos(a/b)*Ci((a+b*arccos(c*x))/b)/b/c+1/4*e*cos(a/b)*Ci((a+b*arccos(c*x)) 
/b)/b/c^3-1/4*e*cos(3*a/b)*Ci(3*(a+b*arccos(c*x))/b)/b/c^3+d*sin(a/b)*Si(( 
a+b*arccos(c*x))/b)/b/c+1/4*e*sin(a/b)*Si((a+b*arccos(c*x))/b)/b/c^3-1/4*e 
*sin(3*a/b)*Si(3*(a+b*arccos(c*x))/b)/b/c^3
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.70 \[ \int \frac {d+e x^2}{a+b \arccos (c x)} \, dx=\frac {\left (4 c^2 d+e\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )+e \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )-4 c^2 d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )-e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )-e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{4 b c^3} \] Input:

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

Output:

((4*c^2*d + e)*CosIntegral[a/b + ArcCos[c*x]]*Sin[a/b] + e*CosIntegral[3*( 
a/b + ArcCos[c*x])]*Sin[(3*a)/b] - 4*c^2*d*Cos[a/b]*SinIntegral[a/b + ArcC 
os[c*x]] - e*Cos[a/b]*SinIntegral[a/b + ArcCos[c*x]] - e*Cos[(3*a)/b]*SinI 
ntegral[3*(a/b + ArcCos[c*x])])/(4*b*c^3)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5173, 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.

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

\(\Big \downarrow \) 5173

\(\displaystyle \int \left (\frac {d}{a+b \arccos (c x)}+\frac {e x^2}{a+b \arccos (c x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c^3}+\frac {e \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c^3}-\frac {e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c^3}-\frac {e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c^3}+\frac {d \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )}{b c}-\frac {d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{b c}\)

Input:

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

Output:

(d*CosIntegral[(a + b*ArcCos[c*x])/b]*Sin[a/b])/(b*c) + (e*CosIntegral[(a 
+ b*ArcCos[c*x])/b]*Sin[a/b])/(4*b*c^3) + (e*CosIntegral[(3*(a + b*ArcCos[ 
c*x]))/b]*Sin[(3*a)/b])/(4*b*c^3) - (d*Cos[a/b]*SinIntegral[(a + b*ArcCos[ 
c*x])/b])/(b*c) - (e*Cos[a/b]*SinIntegral[(a + b*ArcCos[c*x])/b])/(4*b*c^3 
) - (e*Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcCos[c*x]))/b])/(4*b*c^3)
 

Defintions of rubi rules used

rule 2009

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

rule 5173

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Int[ExpandIntegrand[(a + b*ArcCos[c*x])^n, (d + e*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (G 
tQ[p, 0] || IGtQ[n, 0])
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {-\frac {d \left (\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{b}-\frac {e \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )\right )}{4 c^{2} b}-\frac {e \left (\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{4 c^{2} b}}{c}\)	\(152\)
default	\(\frac {-\frac {d \left (\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{b}-\frac {e \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )\right )}{4 c^{2} b}-\frac {e \left (\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{4 c^{2} b}}{c}\)	\(152\)

Input:

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

Output:

1/c*(-d*(Si(arccos(c*x)+a/b)*cos(a/b)-Ci(arccos(c*x)+a/b)*sin(a/b))/b-1/4* 
e/c^2*(Si(3*arccos(c*x)+3*a/b)*cos(3*a/b)-Ci(3*arccos(c*x)+3*a/b)*sin(3*a/ 
b))/b-1/4*e/c^2*(Si(arccos(c*x)+a/b)*cos(a/b)-Ci(arccos(c*x)+a/b)*sin(a/b) 
)/b)
 

Fricas [F]

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

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

Output:

integral((e*x^2 + d)/(b*arccos(c*x) + a), x)
 

Sympy [F]

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

integrate((e*x**2+d)/(a+b*acos(c*x)),x)
 

Output:

Integral((d + e*x**2)/(a + b*acos(c*x)), x)
 

Maxima [F]

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

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

Output:

integrate((e*x^2 + d)/(b*arccos(c*x) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.28 \[ \int \frac {d+e x^2}{a+b \arccos (c x)} \, dx=\frac {e \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{3}} + \frac {d \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c} - \frac {e \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b c^{3}} - \frac {d \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{b c} - \frac {e \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{3}} + \frac {e \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c^{3}} + \frac {3 \, e \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, b c^{3}} - \frac {e \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, b c^{3}} \] Input:

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

Output:

e*cos(a/b)^2*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c^3) + d*cos_ 
integral(a/b + arccos(c*x))*sin(a/b)/(b*c) - e*cos(a/b)^3*sin_integral(3*a 
/b + 3*arccos(c*x))/(b*c^3) - d*cos(a/b)*sin_integral(a/b + arccos(c*x))/( 
b*c) - 1/4*e*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c^3) + 1/4*e* 
cos_integral(a/b + arccos(c*x))*sin(a/b)/(b*c^3) + 3/4*e*cos(a/b)*sin_inte 
gral(3*a/b + 3*arccos(c*x))/(b*c^3) - 1/4*e*cos(a/b)*sin_integral(a/b + ar 
ccos(c*x))/(b*c^3)
 

Mupad [F(-1)]

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

int((d + e*x^2)/(a + b*acos(c*x)),x)
 

Output:

int((d + e*x^2)/(a + b*acos(c*x)), x)
 

Reduce [F]

\[ \int \frac {d+e x^2}{a+b \arccos (c x)} \, dx=\left (\int \frac {x^{2}}{\mathit {acos} \left (c x \right ) b +a}d x \right ) e +\left (\int \frac {1}{\mathit {acos} \left (c x \right ) b +a}d x \right ) d \] Input:

int((e*x^2+d)/(a+b*acos(c*x)),x)
 

Output:

int(x**2/(acos(c*x)*b + a),x)*e + int(1/(acos(c*x)*b + a),x)*d