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

Optimal result

Integrand size = 22, antiderivative size = 125 \[ \int \frac {d-c^2 d x^2}{a+b \arccos (c x)} \, dx=\frac {3 d \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b c}-\frac {d \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b c}-\frac {3 d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )}{4 b c}+\frac {d \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )}{4 b c} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.74 \[ \int \frac {d-c^2 d x^2}{a+b \arccos (c x)} \, dx=\frac {d \left (3 \operatorname {CosIntegral}\left (\frac {a}{b}+\arccos (c x)\right ) \sin \left (\frac {a}{b}\right )-\operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )-3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arccos (c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arccos (c x)\right )\right )\right )}{4 b c} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5169, 25, 3042, 3793, 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.

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

\(\Big \downarrow \) 5169

\(\displaystyle -\frac {d \int -\frac {\sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \int \frac {\sin ^3\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {d \int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )^3}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c}\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {d \int \left (\frac {3 \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{4 (a+b \arccos (c x))}-\frac {\sin \left (\frac {3 a}{b}-\frac {3 (a+b \arccos (c x))}{b}\right )}{4 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b c}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d \left (-\frac {3}{4} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos (c x)}{b}\right )+\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arccos (c x))}{b}\right )+\frac {3}{4} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arccos (c x)}{b}\right )-\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arccos (c x))}{b}\right )\right )}{b c}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 5169

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(-(b*c)^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Subst[ 
Int[x^n*Sin[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{ 
a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p, 0]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {d \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-3 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+3 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{4 c b}\)	\(94\)
default	\(\frac {d \left (\operatorname {Si}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\operatorname {Ci}\left (3 \arccos \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-3 \,\operatorname {Si}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+3 \,\operatorname {Ci}\left (\arccos \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )\right )}{4 c b}\)	\(94\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.42 \[ \int \frac {d-c^2 d x^2}{a+b \arccos (c x)} \, dx=-\frac {d \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c} + \frac {d \cos \left (\frac {a}{b}\right )^{3} \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{b c} + \frac {d \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c} + \frac {3 \, d \operatorname {Ci}\left (\frac {a}{b} + \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{4 \, b c} - \frac {3 \, d \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arccos \left (c x\right )\right )}{4 \, b c} - \frac {3 \, d \cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arccos \left (c x\right )\right )}{4 \, b c} \] Input:

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

Output:

-d*cos(a/b)^2*cos_integral(3*a/b + 3*arccos(c*x))*sin(a/b)/(b*c) + d*cos(a 
/b)^3*sin_integral(3*a/b + 3*arccos(c*x))/(b*c) + 1/4*d*cos_integral(3*a/b 
 + 3*arccos(c*x))*sin(a/b)/(b*c) + 3/4*d*cos_integral(a/b + arccos(c*x))*s 
in(a/b)/(b*c) - 3/4*d*cos(a/b)*sin_integral(3*a/b + 3*arccos(c*x))/(b*c) - 
 3/4*d*cos(a/b)*sin_integral(a/b + arccos(c*x))/(b*c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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