\(\int \frac {a+b \arccos (c x)}{(\pi -c^2 \pi x^2)^{7/2}} \, dx\) [43]

Optimal result

Integrand size = 24, antiderivative size = 168 \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\frac {b}{20 c \pi ^{7/2} \left (1-c^2 x^2\right )^2}+\frac {2 b}{15 c \pi ^{7/2} \left (1-c^2 x^2\right )}+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -c^2 \pi x^2\right )^{5/2}}+\frac {4 x (a+b \arccos (c x))}{15 \pi ^2 \left (\pi -c^2 \pi x^2\right )^{3/2}}+\frac {8 x (a+b \arccos (c x))}{15 \pi ^3 \sqrt {\pi -c^2 \pi x^2}}-\frac {4 b \log \left (1-c^2 x^2\right )}{15 c \pi ^{7/2}} \] Output:

1/20*b/c/Pi^(7/2)/(-c^2*x^2+1)^2+2/15*b/c/Pi^(7/2)/(-c^2*x^2+1)+1/5*x*(a+b 
*arccos(c*x))/Pi/(-Pi*c^2*x^2+Pi)^(5/2)+4/15*x*(a+b*arccos(c*x))/Pi^2/(-Pi 
*c^2*x^2+Pi)^(3/2)+8/15*x*(a+b*arccos(c*x))/Pi^3/(-Pi*c^2*x^2+Pi)^(1/2)-4/ 
15*b*ln(-c^2*x^2+1)/c/Pi^(7/2)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\frac {-11 b+19 b c^2 x^2-8 b c^4 x^4-60 a c x \sqrt {1-c^2 x^2}+80 a c^3 x^3 \sqrt {1-c^2 x^2}-32 a c^5 x^5 \sqrt {1-c^2 x^2}-4 b c x \sqrt {1-c^2 x^2} \left (15-20 c^2 x^2+8 c^4 x^4\right ) \arccos (c x)-16 b \left (-1+c^2 x^2\right )^3 \log \left (1-c^2 x^2\right )}{60 c \pi ^{7/2} \left (-1+c^2 x^2\right )^3} \] Input:

Integrate[(a + b*ArcCos[c*x])/(Pi - c^2*Pi*x^2)^(7/2),x]
 

Output:

(-11*b + 19*b*c^2*x^2 - 8*b*c^4*x^4 - 60*a*c*x*Sqrt[1 - c^2*x^2] + 80*a*c^ 
3*x^3*Sqrt[1 - c^2*x^2] - 32*a*c^5*x^5*Sqrt[1 - c^2*x^2] - 4*b*c*x*Sqrt[1 
- c^2*x^2]*(15 - 20*c^2*x^2 + 8*c^4*x^4)*ArcCos[c*x] - 16*b*(-1 + c^2*x^2) 
^3*Log[1 - c^2*x^2])/(60*c*Pi^(7/2)*(-1 + c^2*x^2)^3)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5163, 241, 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.

\(\displaystyle \int \frac {a+b \arccos (c x)}{\left (\pi -\pi c^2 x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 5163

\(\displaystyle \frac {4 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}}dx}{5 \pi }+\frac {b c \int \frac {x}{\left (1-c^2 x^2\right )^3}dx}{5 \pi ^{7/2}}+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {4 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{5/2}}dx}{5 \pi }+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}+\frac {b}{20 \pi ^{7/2} c \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 5163

\(\displaystyle \frac {4 \left (\frac {2 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{3/2}}dx}{3 \pi }+\frac {b c \int \frac {x}{\left (1-c^2 x^2\right )^2}dx}{3 \pi ^{5/2}}+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}\right )}{5 \pi }+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}+\frac {b}{20 \pi ^{7/2} c \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {4 \left (\frac {2 \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{3/2}}dx}{3 \pi }+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\right )}{5 \pi }+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}+\frac {b}{20 \pi ^{7/2} c \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 5161

\(\displaystyle \frac {4 \left (\frac {2 \left (\frac {b c \int \frac {x}{1-c^2 x^2}dx}{\pi ^{3/2}}+\frac {x (a+b \arccos (c x))}{\pi \sqrt {\pi -\pi c^2 x^2}}\right )}{3 \pi }+\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\right )}{5 \pi }+\frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}+\frac {b}{20 \pi ^{7/2} c \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 240

\(\displaystyle \frac {x (a+b \arccos (c x))}{5 \pi \left (\pi -\pi c^2 x^2\right )^{5/2}}+\frac {4 \left (\frac {x (a+b \arccos (c x))}{3 \pi \left (\pi -\pi c^2 x^2\right )^{3/2}}+\frac {2 \left (\frac {x (a+b \arccos (c x))}{\pi \sqrt {\pi -\pi c^2 x^2}}-\frac {b \log \left (1-c^2 x^2\right )}{2 \pi ^{3/2} c}\right )}{3 \pi }+\frac {b}{6 \pi ^{5/2} c \left (1-c^2 x^2\right )}\right )}{5 \pi }+\frac {b}{20 \pi ^{7/2} c \left (1-c^2 x^2\right )^2}\)

Input:

Int[(a + b*ArcCos[c*x])/(Pi - c^2*Pi*x^2)^(7/2),x]
 

Output:

b/(20*c*Pi^(7/2)*(1 - c^2*x^2)^2) + (x*(a + b*ArcCos[c*x]))/(5*Pi*(Pi - c^ 
2*Pi*x^2)^(5/2)) + (4*(b/(6*c*Pi^(5/2)*(1 - c^2*x^2)) + (x*(a + b*ArcCos[c 
*x]))/(3*Pi*(Pi - c^2*Pi*x^2)^(3/2)) + (2*((x*(a + b*ArcCos[c*x]))/(Pi*Sqr 
t[Pi - c^2*Pi*x^2]) - (b*Log[1 - c^2*x^2])/(2*c*Pi^(3/2))))/(3*Pi)))/(5*Pi 
)
 

Defintions of rubi rules used

rule 240

Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x 
^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
 

rule 241

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]
 

rule 5161

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]
 

rule 5163

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]
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.45

method	result	size
default	\(a \left (\frac {x}{5 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {8 x}{15 \pi ^{2} \sqrt {-\pi \,c^{2} x^{2}+\pi }}}{\pi }\right )-\frac {b \left (16 \ln \left (-c^{2} x^{2}+1\right ) x^{6} c^{6}+32 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{5} c^{5}-48 \ln \left (-c^{2} x^{2}+1\right ) x^{4} c^{4}-80 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}+8 c^{4} x^{4}+48 \ln \left (-c^{2} x^{2}+1\right ) x^{2} c^{2}+60 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c -19 c^{2} x^{2}-16 \ln \left (-c^{2} x^{2}+1\right )+11\right )}{60 c \,\pi ^{\frac {7}{2}} \left (c^{2} x^{2}-1\right )^{3}}\)	\(244\)
parts	\(a \left (\frac {x}{5 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 \pi \left (-\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {8 x}{15 \pi ^{2} \sqrt {-\pi \,c^{2} x^{2}+\pi }}}{\pi }\right )-\frac {b \left (16 \ln \left (-c^{2} x^{2}+1\right ) x^{6} c^{6}+32 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{5} c^{5}-48 \ln \left (-c^{2} x^{2}+1\right ) x^{4} c^{4}-80 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x^{3} c^{3}+8 c^{4} x^{4}+48 \ln \left (-c^{2} x^{2}+1\right ) x^{2} c^{2}+60 \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right ) x c -19 c^{2} x^{2}-16 \ln \left (-c^{2} x^{2}+1\right )+11\right )}{60 c \,\pi ^{\frac {7}{2}} \left (c^{2} x^{2}-1\right )^{3}}\)	\(244\)

Input:

int((a+b*arccos(c*x))/(-Pi*c^2*x^2+Pi)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

a*(1/5/Pi*x/(-Pi*c^2*x^2+Pi)^(5/2)+4/5/Pi*(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)))-1/60*b/c/Pi^(7/2)*(16*ln(-c^2*x^2+1)* 
x^6*c^6+32*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x^5*c^5-48*ln(-c^2*x^2+1)*x^4*c^ 
4-80*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x^3*c^3+8*c^4*x^4+48*ln(-c^2*x^2+1)*x^ 
2*c^2+60*(-c^2*x^2+1)^(1/2)*arccos(c*x)*x*c-19*c^2*x^2-16*ln(-c^2*x^2+1)+1 
1)/(c^2*x^2-1)^3
 

Fricas [F]

\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\int { \frac {b \arccos \left (c x\right ) + a}{{\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(7/2),x, algorithm="fricas")
 

Output:

integral(sqrt(pi - pi*c^2*x^2)*(b*arccos(c*x) + a)/(pi^4*c^8*x^8 - 4*pi^4* 
c^6*x^6 + 6*pi^4*c^4*x^4 - 4*pi^4*c^2*x^2 + pi^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*acos(c*x))/(-pi*c**2*x**2+pi)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\int { \frac {b \arccos \left (c x\right ) + a}{{\left (\pi - \pi c^{2} x^{2}\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(7/2),x, algorithm="maxima")
 

Output:

1/15*a*(3*x/(pi*(pi - pi*c^2*x^2)^(5/2)) + 4*x/(pi^2*(pi - pi*c^2*x^2)^(3/ 
2)) + 8*x/(pi^3*sqrt(pi - pi*c^2*x^2))) - b*integrate(arctan2(sqrt(c*x + 1 
)*sqrt(-c*x + 1), c*x)/((pi^3*c^6*x^6 - 3*pi^3*c^4*x^4 + 3*pi^3*c^2*x^2 - 
pi^3)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x)/sqrt(pi)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*arccos(c*x))/(-pi*c^2*x^2+pi)^(7/2),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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\int \frac {a+b\,\mathrm {acos}\left (c\,x\right )}{{\left (\Pi -\Pi \,c^2\,x^2\right )}^{7/2}} \,d x \] Input:

int((a + b*acos(c*x))/(Pi - Pi*c^2*x^2)^(7/2),x)
 

Output:

int((a + b*acos(c*x))/(Pi - Pi*c^2*x^2)^(7/2), x)
 

Reduce [F]

\[ \int \frac {a+b \arccos (c x)}{\left (\pi -c^2 \pi x^2\right )^{7/2}} \, dx=\frac {-15 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{6} x^{6}-3 \sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{4} x^{4}+30 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{6} x^{6}-3 \sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) b \,c^{2} x^{2}-15 \sqrt {-c^{2} x^{2}+1}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{6} x^{6}-3 \sqrt {-c^{2} x^{2}+1}\, c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) b +8 a \,c^{4} x^{5}-20 a \,c^{2} x^{3}+15 a x}{15 \sqrt {\pi }\, \sqrt {-c^{2} x^{2}+1}\, \pi ^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )} \] Input:

int((a+b*acos(c*x))/(-Pi*c^2*x^2+Pi)^(7/2),x)
 

Output:

( - 15*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 + 1)*c**6*x 
**6 - 3*sqrt( - c**2*x**2 + 1)*c**4*x**4 + 3*sqrt( - c**2*x**2 + 1)*c**2*x 
**2 - sqrt( - c**2*x**2 + 1)),x)*b*c**4*x**4 + 30*sqrt( - c**2*x**2 + 1)*i 
nt(acos(c*x)/(sqrt( - c**2*x**2 + 1)*c**6*x**6 - 3*sqrt( - c**2*x**2 + 1)* 
c**4*x**4 + 3*sqrt( - c**2*x**2 + 1)*c**2*x**2 - sqrt( - c**2*x**2 + 1)),x 
)*b*c**2*x**2 - 15*sqrt( - c**2*x**2 + 1)*int(acos(c*x)/(sqrt( - c**2*x**2 
 + 1)*c**6*x**6 - 3*sqrt( - c**2*x**2 + 1)*c**4*x**4 + 3*sqrt( - c**2*x**2 
 + 1)*c**2*x**2 - sqrt( - c**2*x**2 + 1)),x)*b + 8*a*c**4*x**5 - 20*a*c**2 
*x**3 + 15*a*x)/(15*sqrt(pi)*sqrt( - c**2*x**2 + 1)*pi**3*(c**4*x**4 - 2*c 
**2*x**2 + 1))