\(\int \frac {x}{\arccos (a x)^{7/2}} \, dx\) [116]

Optimal result

Integrand size = 10, antiderivative size = 119 \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}+\frac {8 x^2}{15 \arccos (a x)^{3/2}}-\frac {32 x \sqrt {1-a^2 x^2}}{15 a \sqrt {\arccos (a x)}}+\frac {32 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{15 a^2} \] Output:

2/5*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(5/2)-4/15/a^2/arccos(a*x)^(3/2)+8/ 
15*x^2/arccos(a*x)^(3/2)-32/15*x*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(1/2)+32 
/15*Pi^(1/2)*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))/a^2
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63 \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\frac {\frac {4 \cos (2 \arccos (a x))}{\arccos (a x)^{3/2}}+32 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )-\frac {\left (-3+16 \arccos (a x)^2\right ) \sin (2 \arccos (a x))}{\arccos (a x)^{5/2}}}{15 a^2} \] Input:

Integrate[x/ArcCos[a*x]^(7/2),x]
 

Output:

((4*Cos[2*ArcCos[a*x]])/ArcCos[a*x]^(3/2) + 32*Sqrt[Pi]*FresnelC[(2*Sqrt[A 
rcCos[a*x]])/Sqrt[Pi]] - ((-3 + 16*ArcCos[a*x]^2)*Sin[2*ArcCos[a*x]])/ArcC 
os[a*x]^(5/2))/(15*a^2)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5145, 5153, 5223, 5143, 25, 3042, 3785, 3833}

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.

\(\displaystyle \int \frac {x}{\arccos (a x)^{7/2}} \, dx\)

\(\Big \downarrow \) 5145

\(\displaystyle -\frac {2 \int \frac {1}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx}{5 a}+\frac {4}{5} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}\)

\(\Big \downarrow \) 5153

\(\displaystyle \frac {4}{5} a \int \frac {x^2}{\sqrt {1-a^2 x^2} \arccos (a x)^{5/2}}dx+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 5223

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \int \frac {x}{\arccos (a x)^{3/2}}dx}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 5143

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 \int -\frac {\cos (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}+\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \int \frac {\cos (2 \arccos (a x))}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \int \frac {\sin \left (2 \arccos (a x)+\frac {\pi }{2}\right )}{\sqrt {\arccos (a x)}}d\arccos (a x)}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 3785

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {4 \int \cos (2 \arccos (a x))d\sqrt {\arccos (a x)}}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {4}{5} a \left (\frac {2 x^2}{3 a \arccos (a x)^{3/2}}-\frac {4 \left (\frac {2 x \sqrt {1-a^2 x^2}}{a \sqrt {\arccos (a x)}}-\frac {2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos (a x)}}{\sqrt {\pi }}\right )}{a^2}\right )}{3 a}\right )+\frac {2 x \sqrt {1-a^2 x^2}}{5 a \arccos (a x)^{5/2}}-\frac {4}{15 a^2 \arccos (a x)^{3/2}}\)

Input:

Int[x/ArcCos[a*x]^(7/2),x]
 

Output:

(2*x*Sqrt[1 - a^2*x^2])/(5*a*ArcCos[a*x]^(5/2)) - 4/(15*a^2*ArcCos[a*x]^(3 
/2)) + (4*a*((2*x^2)/(3*a*ArcCos[a*x]^(3/2)) - (4*((2*x*Sqrt[1 - a^2*x^2]) 
/(a*Sqrt[ArcCos[a*x]]) - (2*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[P 
i]])/a^2))/(3*a)))/5
 

Defintions of rubi rules used

rule 25

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

rule 3042

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

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 5143

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[( 
-x^m)*Sqrt[1 - c^2*x^2]*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] - S 
imp[1/(b^2*c^(m + 1)*(n + 1))   Subst[Int[ExpandTrigReduce[x^(n + 1), Cos[- 
a/b + x/b]^(m - 1)*(m - (m + 1)*Cos[-a/b + x/b]^2), x], x], x, a + b*ArcCos 
[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
 

rule 5145

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[( 
-x^m)*Sqrt[1 - c^2*x^2]*((a + b*ArcCos[c*x])^(n + 1)/(b*c*(n + 1))), x] + ( 
-Simp[c*((m + 1)/(b*(n + 1)))   Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n + 1)/ 
Sqrt[1 - c^2*x^2]), x], x] + Simp[m/(b*c*(n + 1))   Int[x^(m - 1)*((a + b*A 
rcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x]) /; FreeQ[{a, b, c}, x] && I 
GtQ[m, 0] && LtQ[n, -2]
 

rule 5153

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S 
ymbol] :> Simp[(-(b*c*(n + 1))^(-1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2] 
]*(a + b*ArcCos[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^ 
2*d + e, 0] && NeQ[n, -1]
 

rule 5223

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) 
+ (e_.)*(x_)^2], x_Symbol] :> Simp[(-(f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c 
^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] + Simp[f*(m/(b*c*( 
n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]   Int[(f*x)^(m - 1)*(a + b 
*ArcCos[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2 
*d + e, 0] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {-32 \sqrt {\pi }\, \operatorname {FresnelC}\left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {5}{2}}+16 \sin \left (2 \arccos \left (a x \right )\right ) \arccos \left (a x \right )^{2}-4 \cos \left (2 \arccos \left (a x \right )\right ) \arccos \left (a x \right )-3 \sin \left (2 \arccos \left (a x \right )\right )}{15 a^{2} \arccos \left (a x \right )^{\frac {5}{2}}}\)	\(73\)

Input:

int(x/arccos(a*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15/a^2*(-32*Pi^(1/2)*FresnelC(2*arccos(a*x)^(1/2)/Pi^(1/2))*arccos(a*x) 
^(5/2)+16*sin(2*arccos(a*x))*arccos(a*x)^2-4*cos(2*arccos(a*x))*arccos(a*x 
)-3*sin(2*arccos(a*x)))/arccos(a*x)^(5/2)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x/arccos(a*x)^(7/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\int \frac {x}{\operatorname {acos}^{\frac {7}{2}}{\left (a x \right )}}\, dx \] Input:

integrate(x/acos(a*x)**(7/2),x)
 

Output:

Integral(x/acos(a*x)**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\arccos (a x)^{7/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x/arccos(a*x)^(7/2),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F]

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

integrate(x/arccos(a*x)^(7/2),x, algorithm="giac")
 

Output:

integrate(x/arccos(a*x)^(7/2), x)
 

Mupad [F(-1)]

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

int(x/acos(a*x)^(7/2),x)
                                                                                    
                                                                                    
 

Output:

int(x/acos(a*x)^(7/2), x)
 

Reduce [F]

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

int(x/acos(a*x)^(7/2),x)
 

Output:

(2*( - 6*acos(a*x)**3*int((sqrt( - a**2*x**2 + 1)*sqrt(acos(a*x))*x**2)/(a 
cos(a*x)**3*a**2*x**2 - acos(a*x)**3),x)*a**3 - 2*sqrt(acos(a*x))*acos(a*x 
) + 3*sqrt( - a**2*x**2 + 1)*sqrt(acos(a*x))*a*x))/(15*acos(a*x)**3*a**2)