Integrand size = 28, antiderivative size = 82 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{8 b c^3}+\frac {\log (a+b \arccos (c x))}{8 b c^3}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )}{8 b c^3} \] Output:
-1/8*cos(4*a/b)*Ci(4*(a+b*arccos(c*x))/b)/b/c^3+1/8*ln(a+b*arccos(c*x))/b/ c^3-1/8*sin(4*a/b)*Si(4*(a+b*arccos(c*x))/b)/b/c^3
Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )-\log (8 (a+b \arccos (c x)))+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arccos (c x)\right )\right )}{8 b c^3} \] Input:
Integrate[(x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcCos[c*x]),x]
Output:
(Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcCos[c*x])] - Log[8*(a + b*ArcCos[c*x ])] + Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcCos[c*x])])/(8*b*c^3)
Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {5225, 4906, 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.
\(\displaystyle \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx\) |
\(\Big \downarrow \) 5225 |
\(\displaystyle -\frac {\int \frac {\cos ^2\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin ^2\left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^3}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle -\frac {\int \left (\frac {1}{8 (a+b \arccos (c x))}-\frac {\cos \left (\frac {4 a}{b}-\frac {4 (a+b \arccos (c x))}{b}\right )}{8 (a+b \arccos (c x))}\right )d(a+b \arccos (c x))}{b c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arccos (c x))}{b}\right )-\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arccos (c x))}{b}\right )+\frac {1}{8} \log (a+b \arccos (c x))}{b c^3}\) |
Input:
Int[(x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcCos[c*x]),x]
Output:
-((-1/8*(Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcCos[c*x]))/b]) + Log[a + b* ArcCos[c*x]]/8 - (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcCos[c*x]))/b])/8)/ (b*c^3))
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[(-(b*c^(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c ^2*x^2)^p] Subst[Int[x^n*Cos[-a/b + x/b]^m*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 + 2, 0] && IGtQ[m, 0]
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {\operatorname {Si}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\operatorname {Ci}\left (4 \arccos \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-\ln \left (4 b \arccos \left (c x \right )+4 a \right )}{8 c^{3} b}\) | \(68\) |
Input:
int(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x)),x,method=_RETURNVERBOSE)
Output:
1/8/c^3*(Si(4*arccos(c*x)+4*a/b)*sin(4*a/b)+Ci(4*arccos(c*x)+4*a/b)*cos(4* a/b)-ln(4*b*arccos(c*x)+4*a))/b
\[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\int { \frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{b \arccos \left (c x\right ) + a} \,d x } \] Input:
integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x)),x, algorithm="fricas")
Output:
integral(sqrt(-c^2*x^2 + 1)*x^2/(b*arccos(c*x) + a), x)
\[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\int \frac {x^{2} \sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}{a + b \operatorname {acos}{\left (c x \right )}}\, dx \] Input:
integrate(x**2*(-c**2*x**2+1)**(1/2)/(a+b*acos(c*x)),x)
Output:
Integral(x**2*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*acos(c*x)), x)
\[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\int { \frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{b \arccos \left (c x\right ) + a} \,d x } \] Input:
integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x)),x, algorithm="maxima")
Output:
integrate(sqrt(-c^2*x^2 + 1)*x^2/(b*arccos(c*x) + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (76) = 152\).
Time = 0.16 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.05 \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{3}} + \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{b c^{3}} - \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{2 \, b c^{3}} + \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arccos \left (c x\right )\right )}{8 \, b c^{3}} - \frac {\log \left (b \arccos \left (c x\right ) + a\right )}{8 \, b c^{3}} \] Input:
integrate(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arccos(c*x)),x, algorithm="giac")
Output:
cos(a/b)^4*cos_integral(4*a/b + 4*arccos(c*x))/(b*c^3) + cos(a/b)^3*sin(a/ b)*sin_integral(4*a/b + 4*arccos(c*x))/(b*c^3) - cos(a/b)^2*cos_integral(4 *a/b + 4*arccos(c*x))/(b*c^3) - 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arccos(c*x))/(b*c^3) + 1/8*cos_integral(4*a/b + 4*arccos(c*x))/(b*c^3) - 1/8*log(b*arccos(c*x) + a)/(b*c^3)
Timed out. \[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\int \frac {x^2\,\sqrt {1-c^2\,x^2}}{a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \] Input:
int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*acos(c*x)),x)
Output:
int((x^2*(1 - c^2*x^2)^(1/2))/(a + b*acos(c*x)), x)
\[ \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \arccos (c x)} \, dx=\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {acos} \left (c x \right ) b +a}d x \] Input:
int(x^2*(-c^2*x^2+1)^(1/2)/(a+b*acos(c*x)),x)
Output:
int((sqrt( - c**2*x**2 + 1)*x**2)/(acos(c*x)*b + a),x)