Integrand size = 24, antiderivative size = 112 \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\frac {b}{6 c \pi ^{5/2} \left (1-c^2 x^2\right )}+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -c^2 \pi x^2\right )^{3/2}}+\frac {2 x (a+b \arccos (c x))}{3 \pi ^2 \sqrt {\pi -c^2 \pi x^2}}-\frac {b \log \left (1-c^2 x^2\right )}{3 c \pi ^{5/2}} \] Output:
1/6*b/c/Pi^(5/2)/(-c^2*x^2+1)+1/3*x*(a+b*arccos(c*x))/Pi/(-Pi*c^2*x^2+Pi)^ (3/2)+2/3*x*(a+b*arccos(c*x))/Pi^2/(-Pi*c^2*x^2+Pi)^(1/2)-1/3*b*ln(-c^2*x^ 2+1)/c/Pi^(5/2)
Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\frac {b-b c^2 x^2+6 a c x \sqrt {1-c^2 x^2}-4 a c^3 x^3 \sqrt {1-c^2 x^2}-2 b c x \sqrt {1-c^2 x^2} \left (-3+2 c^2 x^2\right ) \arccos (c x)-2 b \left (-1+c^2 x^2\right )^2 \log \left (1-c^2 x^2\right )}{6 c \pi ^{5/2} \left (-1+c^2 x^2\right )^2} \] Input:
Integrate[(a + b*ArcCos[c*x])/(Pi - c^2*Pi*x^2)^(5/2),x]
Output:
(b - b*c^2*x^2 + 6*a*c*x*Sqrt[1 - c^2*x^2] - 4*a*c^3*x^3*Sqrt[1 - c^2*x^2] - 2*b*c*x*Sqrt[1 - c^2*x^2]*(-3 + 2*c^2*x^2)*ArcCos[c*x] - 2*b*(-1 + c^2* x^2)^2*Log[1 - c^2*x^2])/(6*c*Pi^(5/2)*(-1 + c^2*x^2)^2)
Time = 0.37 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5163, 241, 5161, 240}
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.
\(\displaystyle \int \frac {a+b \arccos (c x)}{\left (\pi -\pi c^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 5163 |
\(\displaystyle \frac {2 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{3/2}}dx}{3 \pi }+\frac {b c \int \frac {x}{\left (1-c^2 x^2\right )^2}dx}{3 \pi ^{5/2}}+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {2 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{3/2}}dx}{3 \pi }+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 5161 |
\(\displaystyle \frac {2 \left (\frac {b c \int \frac {x}{1-c^2 x^2}dx}{\pi ^{3/2}}+\frac {x (a+b \arccos (c x))}{\pi \sqrt {\pi -\pi c^2 x^2}}\right )}{3 \pi }+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {2 \left (\frac {x (a+b \arccos (c x))}{\pi \sqrt {\pi -\pi c^2 x^2}}-\frac {b \log \left (1-c^2 x^2\right )}{2 \pi ^{3/2} c}\right )}{3 \pi }+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\) |
Input:
Int[(a + b*ArcCos[c*x])/(Pi - c^2*Pi*x^2)^(5/2),x]
Output:
b/(6*c*Pi^(5/2)*(1 - c^2*x^2)) + (x*(a + b*ArcCos[c*x]))/(3*Pi*(Pi - c^2*P i*x^2)^(3/2)) + (2*((x*(a + b*ArcCos[c*x]))/(Pi*Sqrt[Pi - c^2*Pi*x^2]) - ( b*Log[1 - c^2*x^2])/(2*c*Pi^(3/2))))/(3*Pi)
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x _Symbol] :> Simp[x*((a + b*ArcCos[c*x])^n/(d*Sqrt[d + e*x^2])), x] + Simp[b *c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcCos[c*x ])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcCos[c*x])^n/(2*d*(p + 1 ))), x] + (Simp[(2*p + 3)/(2*d*(p + 1)) Int[(d + e*x^2)^(p + 1)*(a + b*Ar cCos[c*x])^n, x], x] - Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2 *x^2)^p] Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcCos[c*x])^(n - 1), x], x ]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]
Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.49
method | result | size |
default | \(a \left (\frac {x}{3 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {2 x}{3 \pi ^{2} \sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )-\frac {b \left (2 \ln \left (-c^{2} x^{2}+1\right ) x^{4} c^{4}+4 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}-4 \ln \left (-c^{2} x^{2}+1\right ) x^{2} c^{2}-6 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c +c^{2} x^{2}+2 \ln \left (-c^{2} x^{2}+1\right )-1\right )}{6 c \,\pi ^{\frac {5}{2}} \left (c^{2} x^{2}-1\right )^{2}}\) | \(167\) |
parts | \(a \left (\frac {x}{3 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {2 x}{3 \pi ^{2} \sqrt {-\pi \,c^{2} x^{2}+\pi }}\right )-\frac {b \left (2 \ln \left (-c^{2} x^{2}+1\right ) x^{4} c^{4}+4 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}-4 \ln \left (-c^{2} x^{2}+1\right ) x^{2} c^{2}-6 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c +c^{2} x^{2}+2 \ln \left (-c^{2} x^{2}+1\right )-1\right )}{6 c \,\pi ^{\frac {5}{2}} \left (c^{2} x^{2}-1\right )^{2}}\) | \(167\) |
Input:
int((a+b*arccos(c*x))/(-Pi*c^2*x^2+Pi)^(5/2),x,method=_RETURNVERBOSE)
Output:
a*(1/3/Pi*x/(-Pi*c^2*x^2+Pi)^(3/2)+2/3/Pi^2*x/(-Pi*c^2*x^2+Pi)^(1/2))-1/6* b/c/Pi^(5/2)*(2*ln(-c^2*x^2+1)*x^4*c^4+4*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x^ 3*c^3-4*ln(-c^2*x^2+1)*x^2*c^2-6*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x*c+c^2*x^ 2+2*ln(-c^2*x^2+1)-1)/(c^2*x^2-1)^2
\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\int { \frac {b \arccos \left (c x\right ) + a}{{\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(5/2),x, algorithm="fricas")
Output:
integral(-sqrt(pi - pi*c^2*x^2)*(b*arccos(c*x) + a)/(pi^3*c^6*x^6 - 3*pi^3 *c^4*x^4 + 3*pi^3*c^2*x^2 - pi^3), x)
\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\frac {\int \frac {a}{c^{4} x^{4} \sqrt {- c^{2} x^{2} + 1} - 2 c^{2} x^{2} \sqrt {- c^{2} x^{2} + 1} + \sqrt {- c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {acos}{\left (c x \right )}}{c^{4} x^{4} \sqrt {- c^{2} x^{2} + 1} - 2 c^{2} x^{2} \sqrt {- c^{2} x^{2} + 1} + \sqrt {- c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \] Input:
integrate((a+b*acos(c*x))/(-pi*c**2*x**2+pi)**(5/2),x)
Output:
(Integral(a/(c**4*x**4*sqrt(-c**2*x**2 + 1) - 2*c**2*x**2*sqrt(-c**2*x**2 + 1) + sqrt(-c**2*x**2 + 1)), x) + Integral(b*acos(c*x)/(c**4*x**4*sqrt(-c **2*x**2 + 1) - 2*c**2*x**2*sqrt(-c**2*x**2 + 1) + sqrt(-c**2*x**2 + 1)), x))/pi**(5/2)
Time = 0.14 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.26 \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {1}{\pi ^{\frac {5}{2}} c^{4} x^{2} - \pi ^{\frac {5}{2}} c^{2}} + \frac {2 \, \log \left (c x + 1\right )}{\pi ^{\frac {5}{2}} c^{2}} + \frac {2 \, \log \left (c x - 1\right )}{\pi ^{\frac {5}{2}} c^{2}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{\pi {\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x}{\pi ^{2} \sqrt {\pi - \pi c^{2} x^{2}}}\right )} \arccos \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{\pi {\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, x}{\pi ^{2} \sqrt {\pi - \pi c^{2} x^{2}}}\right )} \] Input:
integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(5/2),x, algorithm="maxima")
Output:
-1/6*b*c*(1/(pi^(5/2)*c^4*x^2 - pi^(5/2)*c^2) + 2*log(c*x + 1)/(pi^(5/2)*c ^2) + 2*log(c*x - 1)/(pi^(5/2)*c^2)) + 1/3*b*(x/(pi*(pi - pi*c^2*x^2)^(3/2 )) + 2*x/(pi^2*sqrt(pi - pi*c^2*x^2)))*arccos(c*x) + 1/3*a*(x/(pi*(pi - pi *c^2*x^2)^(3/2)) + 2*x/(pi^2*sqrt(pi - pi*c^2*x^2)))
Exception generated. \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {acos}\left (c\,x\right )}{{\left (\Pi -\Pi \,c^2\,x^2\right )}^{5/2}} \,d x \] Input:
int((a + b*acos(c*x))/(Pi - Pi*c^2*x^2)^(5/2),x)
Output:
int((a + b*acos(c*x))/(Pi - Pi*c^2*x^2)^(5/2), x)
\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{2} x^{2}-3 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}-2 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}}d x \right ) b +2 a \,c^{2} x^{3}-3 a x}{3 \sqrt {\pi }\, \sqrt {-c^{2} x^{2}+1}\, \pi ^{2} \left (c^{2} x^{2}-1\right )} \] Input:
int((a+b*acos(c*x))/(-Pi*c^2*x^2+Pi)^(5/2),x)
Output:
(3*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 + 1)*c**4*x**4 - 2*sqrt( - c**2*x**2 + 1)*c**2*x**2 + sqrt( - c**2*x**2 + 1)),x)*b*c**2*x **2 - 3*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 + 1)*c**4* x**4 - 2*sqrt( - c**2*x**2 + 1)*c**2*x**2 + sqrt( - c**2*x**2 + 1)),x)*b + 2*a*c**2*x**3 - 3*a*x)/(3*sqrt(pi)*sqrt( - c**2*x**2 + 1)*pi**2*(c**2*x** 2 - 1))