3.1.37 \(\int \arccos (a+b x)^{5/2} \, dx\) [37]

3.1.37.1 Optimal result

Integrand size = 10, antiderivative size = 111 \[ \int \arccos (a+b x)^{5/2} \, dx=-\frac {15 (a+b x) \sqrt {\arccos (a+b x)}}{4 b}-\frac {5 \sqrt {1-(a+b x)^2} \arccos (a+b x)^{3/2}}{2 b}+\frac {(a+b x) \arccos (a+b x)^{5/2}}{b}+\frac {15 \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a+b x)}\right )}{4 b} \]

output

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

3.1.37.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.71 \[ \int \arccos (a+b x)^{5/2} \, dx=-\frac {i \left (\sqrt {-i \arccos (a+b x)} \Gamma \left (\frac {7}{2},-i \arccos (a+b x)\right )-\sqrt {i \arccos (a+b x)} \Gamma \left (\frac {7}{2},i \arccos (a+b x)\right )\right )}{2 b \sqrt {\arccos (a+b x)}} \]

input

Integrate[ArcCos[a + b*x]^(5/2),x]
 

output

((-1/2*I)*(Sqrt[(-I)*ArcCos[a + b*x]]*Gamma[7/2, (-I)*ArcCos[a + b*x]] - S 
qrt[I*ArcCos[a + b*x]]*Gamma[7/2, I*ArcCos[a + b*x]]))/(b*Sqrt[ArcCos[a + 
b*x]])
 

3.1.37.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5303, 5131, 5183, 5131, 5225, 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.

\(\displaystyle \int \arccos (a+b x)^{5/2} \, dx\)

\(\Big \downarrow \) 5303

\(\displaystyle \frac {\int \arccos (a+b x)^{5/2}d(a+b x)}{b}\)

\(\Big \downarrow \) 5131

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

\(\Big \downarrow \) 5183

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

\(\Big \downarrow \) 5131

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

\(\Big \downarrow \) 5225

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3833

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

input

Int[ArcCos[a + b*x]^(5/2),x]
 

output

((a + b*x)*ArcCos[a + b*x]^(5/2) + (5*(-(Sqrt[1 - (a + b*x)^2]*ArcCos[a + 
b*x]^(3/2)) - (3*((a + b*x)*Sqrt[ArcCos[a + b*x]] - Sqrt[Pi/2]*FresnelC[Sq 
rt[2/Pi]*Sqrt[ArcCos[a + b*x]]]))/2))/2)/b
 

3.1.37.3.1 Defintions of rubi rules used

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 5131

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

rule 5183

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

rule 5225

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]
 

rule 5303

Int[((a_.) + ArcCos[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d 
  Subst[Int[(a + b*ArcCos[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, 
n}, x]
 

3.1.37.4 Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, b x -4 \arccos \left (b x +a \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, a +10 \arccos \left (b x +a \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+15 \sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, b x +15 \sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}\, \sqrt {\pi }\, a -15 \pi \,\operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (b x +a \right )}}{\sqrt {\pi }}\right )\right )}{8 b \sqrt {\pi }}\)	\(140\)

input

int(arccos(b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

output

-1/8/b*2^(1/2)*(-4*arccos(b*x+a)^(5/2)*2^(1/2)*Pi^(1/2)*b*x-4*arccos(b*x+a 
)^(5/2)*2^(1/2)*Pi^(1/2)*a+10*arccos(b*x+a)^(3/2)*2^(1/2)*Pi^(1/2)*(-b^2*x 
^2-2*a*b*x-a^2+1)^(1/2)+15*2^(1/2)*arccos(b*x+a)^(1/2)*Pi^(1/2)*b*x+15*2^( 
1/2)*arccos(b*x+a)^(1/2)*Pi^(1/2)*a-15*Pi*FresnelC(2^(1/2)/Pi^(1/2)*arccos 
(b*x+a)^(1/2)))/Pi^(1/2)
 

3.1.37.5 Fricas [F(-2)]

Exception generated. \[ \int \arccos (a+b x)^{5/2} \, dx=\text {Exception raised: TypeError} \]

input

integrate(arccos(b*x+a)^(5/2),x, algorithm="fricas")
 

output

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

3.1.37.6 Sympy [F]

\[ \int \arccos (a+b x)^{5/2} \, dx=\int \operatorname {acos}^{\frac {5}{2}}{\left (a + b x \right )}\, dx \]

input

integrate(acos(b*x+a)**(5/2),x)
 

output

Integral(acos(a + b*x)**(5/2), x)
 

3.1.37.7 Maxima [F(-2)]

Exception generated. \[ \int \arccos (a+b x)^{5/2} \, dx=\text {Exception raised: RuntimeError} \]

input

integrate(arccos(b*x+a)^(5/2),x, algorithm="maxima")
 

output

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

3.1.37.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.65 \[ \int \arccos (a+b x)^{5/2} \, dx=\frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {\arccos \left (b x + a\right )^{\frac {5}{2}} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{2 \, b} + \frac {5 i \, \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{4 \, b} - \frac {5 i \, \arccos \left (b x + a\right )^{\frac {3}{2}} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{4 \, b} - \frac {\left (15 i + 15\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (b x + a\right )}\right )}{32 \, b} + \frac {\left (15 i - 15\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\arccos \left (b x + a\right )}\right )}{32 \, b} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (i \, \arccos \left (b x + a\right )\right )}}{8 \, b} - \frac {15 \, \sqrt {\arccos \left (b x + a\right )} e^{\left (-i \, \arccos \left (b x + a\right )\right )}}{8 \, b} \]

input

integrate(arccos(b*x+a)^(5/2),x, algorithm="giac")
 

output

1/2*arccos(b*x + a)^(5/2)*e^(I*arccos(b*x + a))/b + 1/2*arccos(b*x + a)^(5 
/2)*e^(-I*arccos(b*x + a))/b + 5/4*I*arccos(b*x + a)^(3/2)*e^(I*arccos(b*x 
 + a))/b - 5/4*I*arccos(b*x + a)^(3/2)*e^(-I*arccos(b*x + a))/b - (15/32*I 
 + 15/32)*sqrt(2)*sqrt(pi)*erf((1/2*I - 1/2)*sqrt(2)*sqrt(arccos(b*x + a)) 
)/b + (15/32*I - 15/32)*sqrt(2)*sqrt(pi)*erf(-(1/2*I + 1/2)*sqrt(2)*sqrt(a 
rccos(b*x + a)))/b - 15/8*sqrt(arccos(b*x + a))*e^(I*arccos(b*x + a))/b - 
15/8*sqrt(arccos(b*x + a))*e^(-I*arccos(b*x + a))/b
 

3.1.37.9 Mupad [F(-1)]

Timed out. \[ \int \arccos (a+b x)^{5/2} \, dx=\int {\mathrm {acos}\left (a+b\,x\right )}^{5/2} \,d x \]

input

int(acos(a + b*x)^(5/2),x)
 

output

int(acos(a + b*x)^(5/2), x)