Integrand size = 20, antiderivative size = 50 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=-\frac {5 c^2 \text {Si}(\arccos (a x))}{8 a}+\frac {5 c^2 \text {Si}(3 \arccos (a x))}{16 a}-\frac {c^2 \text {Si}(5 \arccos (a x))}{16 a} \] Output:
-5/8*c^2*Si(arccos(a*x))/a+5/16*c^2*Si(3*arccos(a*x))/a-1/16*c^2*Si(5*arcc os(a*x))/a
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=-\frac {c^2 (10 \text {Si}(\arccos (a x))-5 \text {Si}(3 \arccos (a x))+\text {Si}(5 \arccos (a x)))}{16 a} \] Input:
Integrate[(c - a^2*c*x^2)^2/ArcCos[a*x],x]
Output:
-1/16*(c^2*(10*SinIntegral[ArcCos[a*x]] - 5*SinIntegral[3*ArcCos[a*x]] + S inIntegral[5*ArcCos[a*x]]))/a
Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.80, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5169, 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.
\(\displaystyle \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx\) |
\(\Big \downarrow \) 5169 |
\(\displaystyle -\frac {c^2 \int \frac {\left (1-a^2 x^2\right )^{5/2}}{\arccos (a x)}d\arccos (a x)}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {c^2 \int \frac {\sin (\arccos (a x))^5}{\arccos (a x)}d\arccos (a x)}{a}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle -\frac {c^2 \int \left (-\frac {5 \sin (3 \arccos (a x))}{16 \arccos (a x)}+\frac {\sin (5 \arccos (a x))}{16 \arccos (a x)}+\frac {5 \sqrt {1-a^2 x^2}}{8 \arccos (a x)}\right )d\arccos (a x)}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^2 \left (\frac {5}{8} \text {Si}(\arccos (a x))-\frac {5}{16} \text {Si}(3 \arccos (a x))+\frac {1}{16} \text {Si}(5 \arccos (a x))\right )}{a}\) |
Input:
Int[(c - a^2*c*x^2)^2/ArcCos[a*x],x]
Output:
-((c^2*((5*SinIntegral[ArcCos[a*x]])/8 - (5*SinIntegral[3*ArcCos[a*x]])/16 + SinIntegral[5*ArcCos[a*x]]/16))/a)
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]))
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]
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {c^{2} \left (5 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )-\operatorname {Si}\left (5 \arccos \left (a x \right )\right )-10 \,\operatorname {Si}\left (\arccos \left (a x \right )\right )\right )}{16 a}\) | \(35\) |
default | \(\frac {c^{2} \left (5 \,\operatorname {Si}\left (3 \arccos \left (a x \right )\right )-\operatorname {Si}\left (5 \arccos \left (a x \right )\right )-10 \,\operatorname {Si}\left (\arccos \left (a x \right )\right )\right )}{16 a}\) | \(35\) |
Input:
int((-a^2*c*x^2+c)^2/arccos(a*x),x,method=_RETURNVERBOSE)
Output:
1/16/a*c^2*(5*Si(3*arccos(a*x))-Si(5*arccos(a*x))-10*Si(arccos(a*x)))
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=\int { \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\arccos \left (a x\right )} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^2/arccos(a*x),x, algorithm="fricas")
Output:
integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arccos(a*x), x)
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {acos}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {acos}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {acos}{\left (a x \right )}}\, dx\right ) \] Input:
integrate((-a**2*c*x**2+c)**2/acos(a*x),x)
Output:
c**2*(Integral(-2*a**2*x**2/acos(a*x), x) + Integral(a**4*x**4/acos(a*x), x) + Integral(1/acos(a*x), x))
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=\int { \frac {{\left (a^{2} c x^{2} - c\right )}^{2}}{\arccos \left (a x\right )} \,d x } \] Input:
integrate((-a^2*c*x^2+c)^2/arccos(a*x),x, algorithm="maxima")
Output:
integrate((a^2*c*x^2 - c)^2/arccos(a*x), x)
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=-\frac {c^{2} \operatorname {Si}\left (5 \, \arccos \left (a x\right )\right )}{16 \, a} + \frac {5 \, c^{2} \operatorname {Si}\left (3 \, \arccos \left (a x\right )\right )}{16 \, a} - \frac {5 \, c^{2} \operatorname {Si}\left (\arccos \left (a x\right )\right )}{8 \, a} \] Input:
integrate((-a^2*c*x^2+c)^2/arccos(a*x),x, algorithm="giac")
Output:
-1/16*c^2*sin_integral(5*arccos(a*x))/a + 5/16*c^2*sin_integral(3*arccos(a *x))/a - 5/8*c^2*sin_integral(arccos(a*x))/a
Timed out. \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=\int \frac {{\left (c-a^2\,c\,x^2\right )}^2}{\mathrm {acos}\left (a\,x\right )} \,d x \] Input:
int((c - a^2*c*x^2)^2/acos(a*x),x)
Output:
int((c - a^2*c*x^2)^2/acos(a*x), x)
\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arccos (a x)} \, dx=c^{2} \left (\left (\int \frac {x^{4}}{\mathit {acos} \left (a x \right )}d x \right ) a^{4}-2 \left (\int \frac {x^{2}}{\mathit {acos} \left (a x \right )}d x \right ) a^{2}+\int \frac {1}{\mathit {acos} \left (a x \right )}d x \right ) \] Input:
int((-a^2*c*x^2+c)^2/acos(a*x),x)
Output:
c**2*(int(x**4/acos(a*x),x)*a**4 - 2*int(x**2/acos(a*x),x)*a**2 + int(1/ac os(a*x),x))